Integrand size = 195, antiderivative size = 31 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (x \left (\frac {e^{\frac {5}{-2+\log (5)}} x}{4+x}-\log (-9+2 x)\right )\right ) \] Output:
ln(x*(exp(5/(ln(5)-2))*x/(4+x)-ln(2*x-9)))*x
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (x \left (\frac {e^{\frac {5}{-2+\log (5)}} x}{4+x}-\log (-9+2 x)\right )\right ) \] Input:
Integrate[(32*x + 16*x^2 + 2*x^3 + E^(5/(-2 + Log[5]))*(72*x - 7*x^2 - 2*x ^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x] + (E^(5/(-2 + Log[5]))*( 36*x + x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x])*Log[(E^ (5/(-2 + Log[5]))*x^2 + (-4*x - x^2)*Log[-9 + 2*x])/(4 + x)])/(E^(5/(-2 + Log[5]))*(36*x + x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x ]),x]
Output:
x*Log[x*((E^(5/(-2 + Log[5]))*x)/(4 + x) - Log[-9 + 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.
| \(\displaystyle \int \frac {2 x^3+16 x^2+\left (-2 x^3-7 x^2+72 x\right ) e^{\frac {5}{\log (5)-2}}+\left (2 x^3+7 x^2-40 x-144\right ) \log (2 x-9)+\left (\left (-2 x^3+x^2+36 x\right ) e^{\frac {5}{\log (5)-2}}+\left (2 x^3+7 x^2-40 x-144\right ) \log (2 x-9)\right ) \log \left (\frac {x^2 e^{\frac {5}{\log (5)-2}}+\left (-x^2-4 x\right ) \log (2 x-9)}{x+4}\right )+32 x}{\left (-2 x^3+x^2+36 x\right ) e^{\frac {5}{\log (5)-2}}+\left (2 x^3+7 x^2-40 x-144\right ) \log (2 x-9)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 x^3+16 x^2+\left (-2 x^3-7 x^2+72 x\right ) e^{\frac {5}{\log (5)-2}}+\left (2 x^3+7 x^2-40 x-144\right ) \log (2 x-9)+\left (\left (-2 x^3+x^2+36 x\right ) e^{\frac {5}{\log (5)-2}}+\left (2 x^3+7 x^2-40 x-144\right ) \log (2 x-9)\right ) \log \left (\frac {x^2 e^{\frac {5}{\log (5)-2}}+\left (-x^2-4 x\right ) \log (2 x-9)}{x+4}\right )+32 x}{\left (-2 x^2+x+36\right ) \left (x (-\log (2 x-9))+x e^{\frac {5}{\log (5)-2}}-4 \log (2 x-9)\right )}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (-\frac {2 x^3}{(x+4) (2 x-9) \left (x (-\log (2 x-9))+x e^{\frac {5}{\log (5)-2}}-4 \log (2 x-9)\right )}-\frac {16 x^2}{(x+4) (2 x-9) \left (x (-\log (2 x-9))+x e^{\frac {5}{\log (5)-2}}-4 \log (2 x-9)\right )}+\log \left (\frac {x^2 e^{\frac {5}{\log (5)-2}}}{x+4}-x \log (2 x-9)\right )+\frac {(x+8) x e^{\frac {5}{\log (5)-2}}}{(x+4) \left (x (-\log (2 x-9))+x e^{\frac {5}{\log (5)-2}}-4 \log (2 x-9)\right )}-\frac {32 x}{(x+4) (2 x-9) \left (x (-\log (2 x-9))+x e^{\frac {5}{\log (5)-2}}-4 \log (2 x-9)\right )}+\frac {(x+4) \log (2 x-9)}{x \log (2 x-9)-x e^{\frac {5}{\log (5)-2}}+4 \log (2 x-9)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2}{x+4}-x \log (2 x-9)\right )dx+4 e^{-\frac {5}{2-\log (5)}} \int \frac {1}{-\log (2 x-9) x+e^{\frac {5}{-2+\log (5)}} x-4 \log (2 x-9)}dx-\frac {17}{2} \int \frac {1}{-\log (2 x-9) x+e^{\frac {5}{-2+\log (5)}} x-4 \log (2 x-9)}dx-\int \frac {x}{-\log (2 x-9) x+e^{\frac {5}{-2+\log (5)}} x-4 \log (2 x-9)}dx-16 e^{-\frac {5}{2-\log (5)}} \int \frac {1}{(x+4) \left (-\log (2 x-9) x+e^{\frac {5}{-2+\log (5)}} x-4 \log (2 x-9)\right )}dx-\frac {153}{2} \int \frac {1}{(2 x-9) \left (-\log (2 x-9) x+e^{\frac {5}{-2+\log (5)}} x-4 \log (2 x-9)\right )}dx+x\) |
Input:
Int[(32*x + 16*x^2 + 2*x^3 + E^(5/(-2 + Log[5]))*(72*x - 7*x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x] + (E^(5/(-2 + Log[5]))*(36*x + x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x])*Log[(E^(5/(-2 + Log[5]))*x^2 + (-4*x - x^2)*Log[-9 + 2*x])/(4 + x)])/(E^(5/(-2 + Log[5] ))*(36*x + x^2 - 2*x^3) + (-144 - 40*x + 7*x^2 + 2*x^3)*Log[-9 + 2*x]),x]
Output:
$Aborted
Time = 7.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29
| method | result | size |
| parallelrisch | \(\ln \left (\frac {\left (-x^{2}-4 x \right ) \ln \left (2 x -9\right )+x^{2} {\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}}{4+x}\right ) x\) | \(40\) |
| risch | \(x \ln \left (x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}-\ln \left (2 x -9\right ) x -4 \ln \left (2 x -9\right )\right )-x \ln \left (4+x \right )+x \ln \left (x \right )-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) \operatorname {csgn}\left (i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{3}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right ) {\operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right ) \operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right ) \operatorname {csgn}\left (i x \right )}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{3}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i x \left (\ln \left (2 x -9\right ) x -x \,{\mathrm e}^{\frac {5}{\ln \left (5\right )-2}}+4 \ln \left (2 x -9\right )\right )}{4+x}\right )}^{2} \operatorname {csgn}\left (i x \right )}{2}\) | \(575\) |
Input:
int((((2*x^3+7*x^2-40*x-144)*ln(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(ln(5)-2))) *ln(((-x^2-4*x)*ln(2*x-9)+x^2*exp(5/(ln(5)-2)))/(4+x))+(2*x^3+7*x^2-40*x-1 44)*ln(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(ln(5)-2))+2*x^3+16*x^2+32*x)/((2* x^3+7*x^2-40*x-144)*ln(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(ln(5)-2))),x,method =_RETURNVERBOSE)
Output:
ln(((-x^2-4*x)*ln(2*x-9)+x^2*exp(5/(ln(5)-2)))/(4+x))*x
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (\frac {x^{2} e^{\left (\frac {5}{\log \left (5\right ) - 2}\right )} - {\left (x^{2} + 4 \, x\right )} \log \left (2 \, x - 9\right )}{x + 4}\right ) \] Input:
integrate((((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log (5)-2)))*log(((-x^2-4*x)*log(2*x-9)+x^2*exp(5/(log(5)-2)))/(4+x))+(2*x^3+7 *x^2-40*x-144)*log(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(log(5)-2))+2*x^3+16*x ^2+32*x)/((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5 )-2))),x, algorithm="fricas")
Output:
x*log((x^2*e^(5/(log(5) - 2)) - (x^2 + 4*x)*log(2*x - 9))/(x + 4))
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (26) = 52\).
Time = 1.37 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.42 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=\left (x - \frac {1}{24}\right ) \log {\left (\frac {\frac {x^{2}}{e^{- \frac {5}{-2 + \log {\left (5 \right )}}}} + \left (- x^{2} - 4 x\right ) \log {\left (2 x - 9 \right )}}{x + 4} \right )} + \frac {\log {\left (x \right )}}{24} + \frac {\log {\left (- \frac {x}{x e^{\frac {5}{2 - \log {\left (5 \right )}}} + 4 e^{\frac {5}{2 - \log {\left (5 \right )}}}} + \log {\left (2 x - 9 \right )} \right )}}{24} \] Input:
integrate((((2*x**3+7*x**2-40*x-144)*ln(2*x-9)+(-2*x**3+x**2+36*x)*exp(5/( ln(5)-2)))*ln(((-x**2-4*x)*ln(2*x-9)+x**2*exp(5/(ln(5)-2)))/(4+x))+(2*x**3 +7*x**2-40*x-144)*ln(2*x-9)+(-2*x**3-7*x**2+72*x)*exp(5/(ln(5)-2))+2*x**3+ 16*x**2+32*x)/((2*x**3+7*x**2-40*x-144)*ln(2*x-9)+(-2*x**3+x**2+36*x)*exp( 5/(ln(5)-2))),x)
Output:
(x - 1/24)*log((x**2*exp(5/(-2 + log(5))) + (-x**2 - 4*x)*log(2*x - 9))/(x + 4)) + log(x)/24 + log(-x/(x*exp(5/(2 - log(5))) + 4*exp(5/(2 - log(5))) ) + log(2*x - 9))/24
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (x {\left (e^{\left (\frac {5}{\log \left (5\right ) - 2}\right )} - \log \left (2 \, x - 9\right )\right )} - 4 \, \log \left (2 \, x - 9\right )\right ) - x \log \left (x + 4\right ) + x \log \left (x\right ) \] Input:
integrate((((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log (5)-2)))*log(((-x^2-4*x)*log(2*x-9)+x^2*exp(5/(log(5)-2)))/(4+x))+(2*x^3+7 *x^2-40*x-144)*log(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(log(5)-2))+2*x^3+16*x ^2+32*x)/((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5 )-2))),x, algorithm="maxima")
Output:
x*log(x*(e^(5/(log(5) - 2)) - log(2*x - 9)) - 4*log(2*x - 9)) - x*log(x + 4) + x*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (30) = 60\).
Time = 0.50 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.26 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x \log \left (-5 \cdot 5^{\frac {1}{4}} x^{2} \log \left (2 \, x - 9\right ) + x^{2} e^{\left (\frac {5 \, {\left (\log \left (5\right )^{2} - 2 \, \log \left (5\right ) + 4\right )}}{4 \, {\left (\log \left (5\right ) - 2\right )}}\right )} - 20 \cdot 5^{\frac {1}{4}} x \log \left (2 \, x - 9\right )\right ) - x \log \left (5 \cdot 5^{\frac {1}{4}} x + 20 \cdot 5^{\frac {1}{4}}\right ) \] Input:
integrate((((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log (5)-2)))*log(((-x^2-4*x)*log(2*x-9)+x^2*exp(5/(log(5)-2)))/(4+x))+(2*x^3+7 *x^2-40*x-144)*log(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(log(5)-2))+2*x^3+16*x ^2+32*x)/((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5 )-2))),x, algorithm="giac")
Output:
x*log(-5*5^(1/4)*x^2*log(2*x - 9) + x^2*e^(5/4*(log(5)^2 - 2*log(5) + 4)/( log(5) - 2)) - 20*5^(1/4)*x*log(2*x - 9)) - x*log(5*5^(1/4)*x + 20*5^(1/4) )
Time = 3.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=x\,\ln \left (-\frac {\ln \left (2\,x-9\right )\,\left (x^2+4\,x\right )-x^2\,{\mathrm {e}}^{\frac {5}{\ln \left (5\right )-2}}}{x+4}\right ) \] Input:
int((32*x + log(-(log(2*x - 9)*(4*x + x^2) - x^2*exp(5/(log(5) - 2)))/(x + 4))*(exp(5/(log(5) - 2))*(36*x + x^2 - 2*x^3) - log(2*x - 9)*(40*x - 7*x^ 2 - 2*x^3 + 144)) - log(2*x - 9)*(40*x - 7*x^2 - 2*x^3 + 144) + 16*x^2 + 2 *x^3 - exp(5/(log(5) - 2))*(7*x^2 - 72*x + 2*x^3))/(exp(5/(log(5) - 2))*(3 6*x + x^2 - 2*x^3) - log(2*x - 9)*(40*x - 7*x^2 - 2*x^3 + 144)),x)
Output:
x*log(-(log(2*x - 9)*(4*x + x^2) - x^2*exp(5/(log(5) - 2)))/(x + 4))
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {32 x+16 x^2+2 x^3+e^{\frac {5}{-2+\log (5)}} \left (72 x-7 x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)+\left (e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)\right ) \log \left (\frac {e^{\frac {5}{-2+\log (5)}} x^2+\left (-4 x-x^2\right ) \log (-9+2 x)}{4+x}\right )}{e^{\frac {5}{-2+\log (5)}} \left (36 x+x^2-2 x^3\right )+\left (-144-40 x+7 x^2+2 x^3\right ) \log (-9+2 x)} \, dx=\mathrm {log}\left (\frac {e^{\frac {5}{\mathrm {log}\left (5\right )-2}} x^{2}-\mathrm {log}\left (2 x -9\right ) x^{2}-4 \,\mathrm {log}\left (2 x -9\right ) x}{x +4}\right ) x \] Input:
int((((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2) ))*log(((-x^2-4*x)*log(2*x-9)+x^2*exp(5/(log(5)-2)))/(4+x))+(2*x^3+7*x^2-4 0*x-144)*log(2*x-9)+(-2*x^3-7*x^2+72*x)*exp(5/(log(5)-2))+2*x^3+16*x^2+32* x)/((2*x^3+7*x^2-40*x-144)*log(2*x-9)+(-2*x^3+x^2+36*x)*exp(5/(log(5)-2))) ,x)
Output:
log((e**(5/(log(5) - 2))*x**2 - log(2*x - 9)*x**2 - 4*log(2*x - 9)*x)/(x + 4))*x