Integrand size = 103, antiderivative size = 20 \[ \int \frac {e^{-2 x \log \left (\frac {x}{3+x}\right )+x^2 \log (-3+x) \log \left (\frac {x}{3+x}\right )} \left (-9+18 x-5 x^2+\left (18 x+x^3+x^4\right ) \log \left (\frac {x}{3+x}\right )+\log (-3+x) \left (-9 x^2+3 x^3+\left (-18 x^2+2 x^4\right ) \log \left (\frac {x}{3+x}\right )\right )\right )}{-9+x^2} \, dx=x \left (\frac {x}{3+x}\right )^{x (-2+x \log (-3+x))} \]
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-2 x \log \left (\frac {x}{3+x}\right )+x^2 \log (-3+x) \log \left (\frac {x}{3+x}\right )} \left (-9+18 x-5 x^2+\left (18 x+x^3+x^4\right ) \log \left (\frac {x}{3+x}\right )+\log (-3+x) \left (-9 x^2+3 x^3+\left (-18 x^2+2 x^4\right ) \log \left (\frac {x}{3+x}\right )\right )\right )}{-9+x^2} \, dx=(-3+x)^{x^2 \log \left (\frac {x}{3+x}\right )} x \left (\frac {x}{3+x}\right )^{-2 x} \]
Integrate[(E^(-2*x*Log[x/(3 + x)] + x^2*Log[-3 + x]*Log[x/(3 + x)])*(-9 + 18*x - 5*x^2 + (18*x + x^3 + x^4)*Log[x/(3 + x)] + Log[-3 + x]*(-9*x^2 + 3 *x^3 + (-18*x^2 + 2*x^4)*Log[x/(3 + x)])))/(-9 + x^2),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{x^2 \log (x-3) \log \left (\frac {x}{x+3}\right )-2 x \log \left (\frac {x}{x+3}\right )} \left (-5 x^2+\left (x^4+x^3+18 x\right ) \log \left (\frac {x}{x+3}\right )+\log (x-3) \left (3 x^3-9 x^2+\left (2 x^4-18 x^2\right ) \log \left (\frac {x}{x+3}\right )\right )+18 x-9\right )}{x^2-9} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (-\frac {9 x^2 \log (x-3) \left (\frac {x}{x+3}\right )^{x (x \log (x-3)-2)}}{x^2-9}+\frac {x \left (x^2+2 x^2 \log (x-3)-2 x-6 x \log (x-3)+6\right ) \log \left (\frac {x}{x+3}\right ) \left (\frac {x}{x+3}\right )^{x (x \log (x-3)-2)}}{x-3}-\frac {5 x^2 \left (\frac {x}{x+3}\right )^{x (x \log (x-3)-2)}}{x^2-9}+\frac {18 x \left (\frac {x}{x+3}\right )^{x (x \log (x-3)-2)}}{x^2-9}-\frac {9 \left (\frac {x}{x+3}\right )^{x (x \log (x-3)-2)}}{x^2-9}+\frac {3 x^3 \log (x-3) \left (\frac {x}{x+3}\right )^{x (x \log (x-3)-2)}}{x^2-9}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (\frac {x}{x+3}\right )^{x^2 \log (x-3)-2 x} \left (5 x^2-\left ((x-3) x^2 \log (x-3) \left (2 (x+3) \log \left (\frac {x}{x+3}\right )+3\right )\right )-\left (x^3+x^2+18\right ) x \log \left (\frac {x}{x+3}\right )-18 x+9\right )}{9-x^2}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {\left (3 x^2 \log (x-3)-5 x+3\right ) \left (\frac {x}{x+3}\right )^{x^2 \log (x-3)-2 x}}{x+3}+\frac {x \left (x^2+2 x^2 \log (x-3)-2 x-6 x \log (x-3)+6\right ) \log \left (\frac {x}{x+3}\right ) \left (\frac {x}{x+3}\right )^{x^2 \log (x-3)-2 x}}{x-3}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (3 x^2 \log (x-3)-5 x+3\right ) \left (\frac {x}{x+3}\right )^{x^2 \log (x-3)-2 x}}{x+3}+\frac {x \left (-x^2-2 x^2 \log (x-3)+2 x+6 x \log (x-3)-6\right ) \log \left (\frac {x}{x+3}\right ) \left (\frac {x}{x+3}\right )^{x^2 \log (x-3)-2 x}}{3-x}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {\left (3 x^2 \log (x-3)-5 x+3\right ) \left (\frac {x}{x+3}\right )^{x^2 \log (x-3)-2 x}}{x+3}+\frac {x \left (-x^2-2 x^2 \log (x-3)+2 x+6 x \log (x-3)-6\right ) \log \left (\frac {x}{x+3}\right ) \left (\frac {x}{x+3}\right )^{x^2 \log (x-3)-2 x}}{3-x}\right )dx\) |
Int[(E^(-2*x*Log[x/(3 + x)] + x^2*Log[-3 + x]*Log[x/(3 + x)])*(-9 + 18*x - 5*x^2 + (18*x + x^3 + x^4)*Log[x/(3 + x)] + Log[-3 + x]*(-9*x^2 + 3*x^3 + (-18*x^2 + 2*x^4)*Log[x/(3 + x)])))/(-9 + x^2),x]
3.25.38.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 5.96 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \({\mathrm e}^{x \left (\ln \left (-3+x \right ) x -2\right ) \ln \left (\frac {x}{3+x}\right )} x\) | \(22\) |
risch | \(x \left (-3+x \right )^{\frac {x^{2} \left (-i \pi \,\operatorname {csgn}\left (\frac {i x}{3+x}\right )+i \pi \,\operatorname {csgn}\left (i x \right )+i \pi \,\operatorname {csgn}\left (\frac {i}{3+x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i x}{3+x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{3+x}\right )+2 \ln \left (x \right )-2 \ln \left (3+x \right )\right )}{2}} x^{-2 x} \left (3+x \right )^{2 x} {\mathrm e}^{-i \pi \,\operatorname {csgn}\left (\frac {i x}{3+x}\right ) x \left (\operatorname {csgn}\left (\frac {i x}{3+x}\right )-\operatorname {csgn}\left (\frac {i}{3+x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i x}{3+x}\right )+\operatorname {csgn}\left (i x \right )\right )}\) | \(155\) |
int((((2*x^4-18*x^2)*ln(x/(3+x))+3*x^3-9*x^2)*ln(-3+x)+(x^4+x^3+18*x)*ln(x /(3+x))-5*x^2+18*x-9)*exp(x^2*ln(x/(3+x))*ln(-3+x)-2*x*ln(x/(3+x)))/(x^2-9 ),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {e^{-2 x \log \left (\frac {x}{3+x}\right )+x^2 \log (-3+x) \log \left (\frac {x}{3+x}\right )} \left (-9+18 x-5 x^2+\left (18 x+x^3+x^4\right ) \log \left (\frac {x}{3+x}\right )+\log (-3+x) \left (-9 x^2+3 x^3+\left (-18 x^2+2 x^4\right ) \log \left (\frac {x}{3+x}\right )\right )\right )}{-9+x^2} \, dx=x e^{\left (x^{2} \log \left (x - 3\right ) \log \left (\frac {x}{x + 3}\right ) - 2 \, x \log \left (\frac {x}{x + 3}\right )\right )} \]
integrate((((2*x^4-18*x^2)*log(x/(3+x))+3*x^3-9*x^2)*log(-3+x)+(x^4+x^3+18 *x)*log(x/(3+x))-5*x^2+18*x-9)*exp(x^2*log(x/(3+x))*log(-3+x)-2*x*log(x/(3 +x)))/(x^2-9),x, algorithm=\
Time = 7.38 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-2 x \log \left (\frac {x}{3+x}\right )+x^2 \log (-3+x) \log \left (\frac {x}{3+x}\right )} \left (-9+18 x-5 x^2+\left (18 x+x^3+x^4\right ) \log \left (\frac {x}{3+x}\right )+\log (-3+x) \left (-9 x^2+3 x^3+\left (-18 x^2+2 x^4\right ) \log \left (\frac {x}{3+x}\right )\right )\right )}{-9+x^2} \, dx=x e^{x^{2} \log {\left (\frac {x}{x + 3} \right )} \log {\left (x - 3 \right )} - 2 x \log {\left (\frac {x}{x + 3} \right )}} \]
integrate((((2*x**4-18*x**2)*ln(x/(3+x))+3*x**3-9*x**2)*ln(-3+x)+(x**4+x** 3+18*x)*ln(x/(3+x))-5*x**2+18*x-9)*exp(x**2*ln(x/(3+x))*ln(-3+x)-2*x*ln(x/ (3+x)))/(x**2-9),x)
Time = 0.35 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95 \[ \int \frac {e^{-2 x \log \left (\frac {x}{3+x}\right )+x^2 \log (-3+x) \log \left (\frac {x}{3+x}\right )} \left (-9+18 x-5 x^2+\left (18 x+x^3+x^4\right ) \log \left (\frac {x}{3+x}\right )+\log (-3+x) \left (-9 x^2+3 x^3+\left (-18 x^2+2 x^4\right ) \log \left (\frac {x}{3+x}\right )\right )\right )}{-9+x^2} \, dx=x e^{\left (-x^{2} \log \left (x + 3\right ) \log \left (x - 3\right ) + x^{2} \log \left (x - 3\right ) \log \left (x\right ) + 2 \, x \log \left (x + 3\right ) - 2 \, x \log \left (x\right )\right )} \]
integrate((((2*x^4-18*x^2)*log(x/(3+x))+3*x^3-9*x^2)*log(-3+x)+(x^4+x^3+18 *x)*log(x/(3+x))-5*x^2+18*x-9)*exp(x^2*log(x/(3+x))*log(-3+x)-2*x*log(x/(3 +x)))/(x^2-9),x, algorithm=\
\[ \int \frac {e^{-2 x \log \left (\frac {x}{3+x}\right )+x^2 \log (-3+x) \log \left (\frac {x}{3+x}\right )} \left (-9+18 x-5 x^2+\left (18 x+x^3+x^4\right ) \log \left (\frac {x}{3+x}\right )+\log (-3+x) \left (-9 x^2+3 x^3+\left (-18 x^2+2 x^4\right ) \log \left (\frac {x}{3+x}\right )\right )\right )}{-9+x^2} \, dx=\int { -\frac {{\left (5 \, x^{2} - {\left (3 \, x^{3} - 9 \, x^{2} + 2 \, {\left (x^{4} - 9 \, x^{2}\right )} \log \left (\frac {x}{x + 3}\right )\right )} \log \left (x - 3\right ) - {\left (x^{4} + x^{3} + 18 \, x\right )} \log \left (\frac {x}{x + 3}\right ) - 18 \, x + 9\right )} e^{\left (x^{2} \log \left (x - 3\right ) \log \left (\frac {x}{x + 3}\right ) - 2 \, x \log \left (\frac {x}{x + 3}\right )\right )}}{x^{2} - 9} \,d x } \]
integrate((((2*x^4-18*x^2)*log(x/(3+x))+3*x^3-9*x^2)*log(-3+x)+(x^4+x^3+18 *x)*log(x/(3+x))-5*x^2+18*x-9)*exp(x^2*log(x/(3+x))*log(-3+x)-2*x*log(x/(3 +x)))/(x^2-9),x, algorithm=\
integrate(-(5*x^2 - (3*x^3 - 9*x^2 + 2*(x^4 - 9*x^2)*log(x/(x + 3)))*log(x - 3) - (x^4 + x^3 + 18*x)*log(x/(x + 3)) - 18*x + 9)*e^(x^2*log(x - 3)*lo g(x/(x + 3)) - 2*x*log(x/(x + 3)))/(x^2 - 9), x)
Time = 13.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {e^{-2 x \log \left (\frac {x}{3+x}\right )+x^2 \log (-3+x) \log \left (\frac {x}{3+x}\right )} \left (-9+18 x-5 x^2+\left (18 x+x^3+x^4\right ) \log \left (\frac {x}{3+x}\right )+\log (-3+x) \left (-9 x^2+3 x^3+\left (-18 x^2+2 x^4\right ) \log \left (\frac {x}{3+x}\right )\right )\right )}{-9+x^2} \, dx=\frac {x\,{\left (x-3\right )}^{x^2\,\ln \left (\frac {x}{x+3}\right )}}{{\left (\frac {x}{x+3}\right )}^{2\,x}} \]