I want to know how to integrate this function I have tried many things many substitutions but none works.
I even try to expand the numerator by $\sin3x$ and $\cos3x$ properties and tried to convert higher powers to multiple angles
Integrate $ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ $
indefinite-integralstrigonometric-integrals
Related Solutions
The preliminary substitution $x=u^2$ yields the integral
$$2\int u\sqrt{1+u^3}\mathrm du$$
This is indeed in the form of an elliptic integral, where the cubic under the square root factorizes as $(u+1)(u^2-u+1)$.
For handling integrals like these, one first performs a preliminary Möbius substitution. In this case, we let
$$u=-\frac{-1+\sqrt{(-1)^2-(-1)+1}+(-1-\sqrt{(-1)^2-(-1)+1})v}{1+v}=\frac{2\sqrt{3}}{1+v}-(1+\sqrt{3})$$
to give
$$12\int\frac{(1-\sqrt{3}+(1+\sqrt{3})v)}{(1+v)^5}\sqrt{(1-v^2)\left(2\sqrt{3}-3+(2\sqrt{3}+3\right) v^2)}\mathrm dv$$
We can now use the Jacobian elliptic functions. Letting $v=\mathrm{cn}(w|m)$, $\mathrm dv=-\mathrm{sn}(w|m)\mathrm{dn}(w|m)\mathrm dw$ and using the Pythagorean formula $\mathrm{sn}^2(w|m)+\mathrm{cn}^2(w|m)=1$ yields
$$-12\int\frac{(1-\sqrt{3}+(1+\sqrt{3})\mathrm{cn}(w|m))}{(1+\mathrm{cn}(w|m))^5}\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(w|m)}\mathrm{sn}^2(w|m)\mathrm{dn}(w|m)\mathrm dw$$
We perform a further application of the Pythagorean formula, and factor out a constant:
$$-24\sqrt[4]{3}\int\frac{(1-\sqrt{3}+(1+\sqrt{3})\mathrm{cn}(w|m))}{(1+\mathrm{cn}(w|m))^5}\sqrt{1-\frac{2+\sqrt{3}}{4}\mathrm{sn}^2(w|m)}\mathrm{sn}^2(w|m)\mathrm{dn}(w|m)\mathrm dw$$
From that, if we let $m=\dfrac{2+\sqrt{3}}{4}$, we can then exploit the identity $\mathrm{dn}^2(w|m)+m\,\mathrm{sn}^2(w|m)=1$:
$$-24\sqrt[4]{3}\int\frac{(1-\sqrt{3}+(1+\sqrt{3})\mathrm{cn}(w|m))}{(1+\mathrm{cn}(w|m))^5}\mathrm{sn}^2(w|m)\mathrm{dn}^2(w|m)\;\mathrm dw$$
We can express everything in terms of $\mathrm{cn}$; using the Pythagorean relations and splitting into partial fractions yields
$$2\sqrt[4]{3}\left(3(5+3\sqrt{3})w-6(13+8\sqrt{3})\varrho_1+48(3+2\sqrt{3})\varrho_2-96(1+\sqrt{3})\varrho_3+48\sqrt{3}\varrho_4\right)$$
where for brevity
$$\varrho_k=\int\frac{\mathrm dw}{(1+\mathrm{cn}(w\mid m))^k}$$
Evaluating $\varrho_k$ is algebraically a rather complicated affair; for brevity, I will instead refer you to formulae 341.52-55 in Byrd and Friedman, where a recursion relation is listed. The required members are:
$$\begin{align*}\varrho_1&=w-\varepsilon\left(w \mid \frac{2+\sqrt{3}}{4}\right)+\frac{\mathrm{sn}\left(w \mid \frac{2+\sqrt{3}}{4}\right)\mathrm{dn}\left(w \mid \frac{2+\sqrt{3}}{4}\right)}{1+\mathrm{cn}\left(w \mid \frac{2+\sqrt{3}}{4}\right)}\\\varrho_2&=\frac13\left((3+\sqrt{3})\varrho_1-\frac12(2+\sqrt{3})w+\frac{\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^2}\right)\\\varrho_3&=\frac15\left(2\left(3+\sqrt{3}\right)\varrho_2-\frac32(2+\sqrt{3})\varrho_1+\frac{2+\sqrt{3}}{4}w+\frac{\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right) \mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^3}\right)\\\varrho_4&=\frac17\left(3\left(3+\sqrt{3}\right)\varrho_3-\frac52\left(2+\sqrt{3}\right)\varrho_2+\frac{2+\sqrt{3}}{2}\varrho_1+\frac{\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right) \mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^4}\right)\end{align*}$$
where $\varepsilon(w|m)=E(\mathrm{am}(w|m)|m)$ is the Jacobi epsilon function. After much algebra and tears, we end up with
$$\begin{split}\frac{\sqrt[4]{3}}{7} \left((3-\sqrt{3})w-6\,\varepsilon\left(w \mid \frac{2+\sqrt{3}}{4}\right)+\frac{2\,\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right) \mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^4}\left(3\mathrm{cn}^3\left(w\mid \frac{2+\sqrt{3}}{4}\right)+\right.\right. \\ \left.\left.(21+8\sqrt{3})\mathrm{cn}^2\left(w\mid \frac{2+\sqrt{3}}{4}\right)+(9-8\sqrt{3})\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)+8 \sqrt{3}-9\right)\right)\end{split}$$
Undoing the substitutions yields
$$\begin{split}\frac2{7\sqrt[4]{3}}\left(\frac{2\sqrt[4]{3}\sqrt{1+x^{3/2}}(3+x(1+\sqrt{3}+\sqrt{x}))}{1+\sqrt{3}+\sqrt{x}}-6\sqrt{3} E\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\sqrt{x}}-1\right)\mid \frac{2+\sqrt{3}}{4}\right)\right. \\ \left.+3(\sqrt{3}-1)F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\sqrt{x}}-1\right)\mid \frac{2+\sqrt{3}}{4}\right)\right)\end{split}$$
You can check yourself that the derivative of this expression is $\sqrt{1+x^\frac32}$.
Let $f_n(x) = \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^n$ and evaluate
$$\frac{df_n(x)}{dx} = -\frac{n(\cos x - x \sin x)^{n-1}(x^2+2)}{(x\cos x + \sin x)^{n+1}}$$
For $n=1$ and $3$, respectively \begin{align} &\frac{df_1(x)}{dx} = -\frac{x^2+2}{(x\cos x + \sin x)^{2}}\\ &\frac{df_3(x)}{dx}=- \frac{3(\cos x - x \sin x)^2(x^2+2)}{(x\cos x + \sin x)^{4}} \end{align}
which leads to
$$\frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4} = -\frac{d}{dx}\left(\frac13f_3(x)+f_1(x)\right)$$
and
$$\begin{align} & \int \frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4}dx \\ =& -\frac13f_3(x)-f_1(x) \\ =& -\frac13 \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^3 - \frac {\cos x - x \sin x}{x\cos x + \sin x} \\ =& -\frac13 \left( \frac {1- x \tan x}{x + \tan x} \right)^3 - \frac {1- x \tan x}{x + \tan x} \\ \end{align}$$
Best Answer
$$ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ =\int \frac{3\sin x -4\sin^3x+4\cos^3x-3\cos x}{ \sin^3 x + \cos^3 x } dx=\int \frac{(\cos x -\sin x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)(1- \sin x\cos x) } dx=\int \frac{(\cos^2 x -\sin^2 x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)^2(1- \sin x\cos x) } dx=\int \frac{\cos 2x (1+ 2\sin 2x)}{ (1+ \sin 2x)(1- 0.5\sin 2x) } dx=0.5\int \frac{1+ 2\sin 2x}{ (1+ \sin 2x)(1- 0.5\sin 2x) } d(\sin 2x)$$ Hope you can finish from here.