\(\int \frac {(16 x+16 x^2+e^{2 x} (-x-x^2)) \log (-\frac {1}{x+x^2})+(9+18 x) \log ^{18}(-\frac {1}{x+x^2})+\log ^9(-\frac {1}{x+x^2}) (e^x (-9-18 x)+e^x (x+x^2) \log (-\frac {1}{x+x^2}))}{(8 x+8 x^2) \log (-\frac {1}{x+x^2})} \, dx\) [498]

Optimal result

Integrand size = 120, antiderivative size = 28 \[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=2 x-\frac {1}{16} \left (-e^x+\log ^9\left (-\frac {1}{x+x^2}\right )\right )^2 \] Output:

2*x-1/4*(ln(-1/(x^2+x))^9-exp(x))*(1/4*ln(-1/(x^2+x))^9-1/4*exp(x))
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=\frac {1}{8} e^x \log ^9\left (-\frac {1}{x+x^2}\right )+\frac {1}{8} \left (-\frac {e^{2 x}}{2}+16 x-\frac {1}{2} \log ^{18}\left (-\frac {1}{x+x^2}\right )\right ) \] Input:

Integrate[((16*x + 16*x^2 + E^(2*x)*(-x - x^2))*Log[-(x + x^2)^(-1)] + (9 
+ 18*x)*Log[-(x + x^2)^(-1)]^18 + Log[-(x + x^2)^(-1)]^9*(E^x*(-9 - 18*x) 
+ E^x*(x + x^2)*Log[-(x + x^2)^(-1)]))/((8*x + 8*x^2)*Log[-(x + x^2)^(-1)] 
),x]
 

Output:

(E^x*Log[-(x + x^2)^(-1)]^9)/8 + (-1/2*E^(2*x) + 16*x - Log[-(x + x^2)^(-1 
)]^18/2)/8
 

Rubi [F]

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.

\(\displaystyle \int \frac {(18 x+9) \log ^{18}\left (-\frac {1}{x^2+x}\right )+\left (e^x \left (x^2+x\right ) \log \left (-\frac {1}{x^2+x}\right )+e^x (-18 x-9)\right ) \log ^9\left (-\frac {1}{x^2+x}\right )+\left (16 x^2+e^{2 x} \left (-x^2-x\right )+16 x\right ) \log \left (-\frac {1}{x^2+x}\right )}{\left (8 x^2+8 x\right ) \log \left (-\frac {1}{x^2+x}\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {(18 x+9) \log ^{18}\left (-\frac {1}{x^2+x}\right )+\left (e^x \left (x^2+x\right ) \log \left (-\frac {1}{x^2+x}\right )+e^x (-18 x-9)\right ) \log ^9\left (-\frac {1}{x^2+x}\right )+\left (16 x^2+e^{2 x} \left (-x^2-x\right )+16 x\right ) \log \left (-\frac {1}{x^2+x}\right )}{x (8 x+8) \log \left (-\frac {1}{x^2+x}\right )}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {(18 x+9) \log ^{18}\left (-\frac {1}{x^2+x}\right )+\left (e^x \left (x^2+x\right ) \log \left (-\frac {1}{x^2+x}\right )+e^x (-18 x-9)\right ) \log ^9\left (-\frac {1}{x^2+x}\right )+\left (16 x^2+e^{2 x} \left (-x^2-x\right )+16 x\right ) \log \left (-\frac {1}{x^2+x}\right )}{x (8 x+8) \log \left (-\frac {1}{x (x+1)}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {16 x^2+18 x \log ^{17}\left (-\frac {1}{x^2+x}\right )+9 \log ^{17}\left (-\frac {1}{x^2+x}\right )+16 x}{8 x (x+1)}+\frac {e^x \left (x^2 \log \left (-\frac {1}{x^2+x}\right )+x \log \left (-\frac {1}{x^2+x}\right )-18 x-9\right ) \log ^8\left (-\frac {1}{x (x+1)}\right )}{8 x (x+1)}-\frac {e^{2 x}}{8}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {9}{8} \int \frac {\log ^{17}\left (-\frac {1}{x (x+1)}\right )}{x}dx+\frac {9}{8} \int \frac {\log ^{17}\left (-\frac {1}{x (x+1)}\right )}{x+1}dx+\frac {1}{8} \int e^x \log ^9\left (-\frac {1}{x (x+1)}\right )dx+\frac {9}{8} \int \frac {e^x \log ^8\left (-\frac {1}{x (x+1)}\right )}{-x-1}dx-\frac {9}{8} \int \frac {e^x \log ^8\left (-\frac {1}{x (x+1)}\right )}{x}dx+2 x-\frac {e^{2 x}}{16}\)

Input:

Int[((16*x + 16*x^2 + E^(2*x)*(-x - x^2))*Log[-(x + x^2)^(-1)] + (9 + 18*x 
)*Log[-(x + x^2)^(-1)]^18 + Log[-(x + x^2)^(-1)]^9*(E^x*(-9 - 18*x) + E^x* 
(x + x^2)*Log[-(x + x^2)^(-1)]))/((8*x + 8*x^2)*Log[-(x + x^2)^(-1)]),x]
 

Output:

$Aborted
 

Maple [F(-1)]

Timed out.

\[\int \frac {\left (18 x +9\right ) \ln \left (-\frac {1}{x^{2}+x}\right )^{18}+\left (\left (x^{2}+x \right ) {\mathrm e}^{x} \ln \left (-\frac {1}{x^{2}+x}\right )+\left (-18 x -9\right ) {\mathrm e}^{x}\right ) \ln \left (-\frac {1}{x^{2}+x}\right )^{9}+\left (\left (-x^{2}-x \right ) {\mathrm e}^{2 x}+16 x^{2}+16 x \right ) \ln \left (-\frac {1}{x^{2}+x}\right )}{\left (8 x^{2}+8 x \right ) \ln \left (-\frac {1}{x^{2}+x}\right )}d x\]

Input:

int(((18*x+9)*ln(-1/(x^2+x))^18+((x^2+x)*exp(x)*ln(-1/(x^2+x))+(-18*x-9)*e 
xp(x))*ln(-1/(x^2+x))^9+((-x^2-x)*exp(x)^2+16*x^2+16*x)*ln(-1/(x^2+x)))/(8 
*x^2+8*x)/ln(-1/(x^2+x)),x)
 

Output:

int(((18*x+9)*ln(-1/(x^2+x))^18+((x^2+x)*exp(x)*ln(-1/(x^2+x))+(-18*x-9)*e 
xp(x))*ln(-1/(x^2+x))^9+((-x^2-x)*exp(x)^2+16*x^2+16*x)*ln(-1/(x^2+x)))/(8 
*x^2+8*x)/ln(-1/(x^2+x)),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=-\frac {1}{16} \, \log \left (-\frac {1}{x^{2} + x}\right )^{18} + \frac {1}{8} \, e^{x} \log \left (-\frac {1}{x^{2} + x}\right )^{9} + 2 \, x - \frac {1}{16} \, e^{\left (2 \, x\right )} \] Input:

integrate(((18*x+9)*log(-1/(x^2+x))^18+((x^2+x)*exp(x)*log(-1/(x^2+x))+(-1 
8*x-9)*exp(x))*log(-1/(x^2+x))^9+((-x^2-x)*exp(x)^2+16*x^2+16*x)*log(-1/(x 
^2+x)))/(8*x^2+8*x)/log(-1/(x^2+x)),x, algorithm="fricas")
 

Output:

-1/16*log(-1/(x^2 + x))^18 + 1/8*e^x*log(-1/(x^2 + x))^9 + 2*x - 1/16*e^(2 
*x)
 

Sympy [A] (verification not implemented)

Time = 4.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=2 x - \frac {e^{2 x}}{16} + \frac {e^{x} \log {\left (- \frac {1}{x^{2} + x} \right )}^{9}}{8} - \frac {\log {\left (- \frac {1}{x^{2} + x} \right )}^{18}}{16} \] Input:

integrate(((18*x+9)*ln(-1/(x**2+x))**18+((x**2+x)*exp(x)*ln(-1/(x**2+x))+( 
-18*x-9)*exp(x))*ln(-1/(x**2+x))**9+((-x**2-x)*exp(x)**2+16*x**2+16*x)*ln( 
-1/(x**2+x)))/(8*x**2+8*x)/ln(-1/(x**2+x)),x)
 

Output:

2*x - exp(2*x)/16 + exp(x)*log(-1/(x**2 + x))**9/8 - log(-1/(x**2 + x))**1 
8/16
 

Maxima [F]

\[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=\int { \frac {9 \, {\left (2 \, x + 1\right )} \log \left (-\frac {1}{x^{2} + x}\right )^{18} + {\left ({\left (x^{2} + x\right )} e^{x} \log \left (-\frac {1}{x^{2} + x}\right ) - 9 \, {\left (2 \, x + 1\right )} e^{x}\right )} \log \left (-\frac {1}{x^{2} + x}\right )^{9} + {\left (16 \, x^{2} - {\left (x^{2} + x\right )} e^{\left (2 \, x\right )} + 16 \, x\right )} \log \left (-\frac {1}{x^{2} + x}\right )}{8 \, {\left (x^{2} + x\right )} \log \left (-\frac {1}{x^{2} + x}\right )} \,d x } \] Input:

integrate(((18*x+9)*log(-1/(x^2+x))^18+((x^2+x)*exp(x)*log(-1/(x^2+x))+(-1 
8*x-9)*exp(x))*log(-1/(x^2+x))^9+((-x^2-x)*exp(x)^2+16*x^2+16*x)*log(-1/(x 
^2+x)))/(8*x^2+8*x)/log(-1/(x^2+x)),x, algorithm="maxima")
 

Output:

-1/16*log(x)^18 - 21879/8*log(x)^8*log(-x - 1)^10 - 1989*log(x)^7*log(-x - 
 1)^11 - 4641/4*log(x)^6*log(-x - 1)^12 - 1071/2*log(x)^5*log(-x - 1)^13 - 
 765/4*log(x)^4*log(-x - 1)^14 - 51*log(x)^3*log(-x - 1)^15 - 153/16*log(x 
)^2*log(-x - 1)^16 - 9/8*log(x)*log(-x - 1)^17 - 1/16*log(-x - 1)^18 - 1/8 
*e^x*log(x)^9 - 1/8*(24310*log(x)^9 + e^x)*log(-x - 1)^9 - 9/8*(2431*log(x 
)^10 + e^x*log(x))*log(-x - 1)^8 - 9/2*(442*log(x)^11 + e^x*log(x)^2)*log( 
-x - 1)^7 - 21/4*(221*log(x)^12 + 2*e^x*log(x)^3)*log(-x - 1)^6 - 63/4*(34 
*log(x)^13 + e^x*log(x)^4)*log(-x - 1)^5 - 9/4*(85*log(x)^14 + 7*e^x*log(x 
)^5)*log(-x - 1)^4 - 3/2*(34*log(x)^15 + 7*e^x*log(x)^6)*log(-x - 1)^3 - 9 
/16*(17*log(x)^16 + 8*e^x*log(x)^7)*log(-x - 1)^2 + 1/8*e^(-2)*exp_integra 
l_e(1, -2*x - 2) - 9/8*(log(x)^17 + e^x*log(x)^8)*log(-x - 1) + 2*x - 1/8* 
integrate(x*e^(2*x)/(x + 1), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=\text {Timed out} \] Input:

integrate(((18*x+9)*log(-1/(x^2+x))^18+((x^2+x)*exp(x)*log(-1/(x^2+x))+(-1 
8*x-9)*exp(x))*log(-1/(x^2+x))^9+((-x^2-x)*exp(x)^2+16*x^2+16*x)*log(-1/(x 
^2+x)))/(8*x^2+8*x)/log(-1/(x^2+x)),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 4.84 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=-\frac {{\ln \left (-\frac {1}{x^2+x}\right )}^{18}}{16}+\frac {{\mathrm {e}}^x\,{\ln \left (-\frac {1}{x^2+x}\right )}^9}{8}+2\,x-\frac {{\mathrm {e}}^{2\,x}}{16} \] Input:

int((log(-1/(x + x^2))^18*(18*x + 9) - log(-1/(x + x^2))^9*(exp(x)*(18*x + 
 9) - exp(x)*log(-1/(x + x^2))*(x + x^2)) + log(-1/(x + x^2))*(16*x - exp( 
2*x)*(x + x^2) + 16*x^2))/(log(-1/(x + x^2))*(8*x + 8*x^2)),x)
 

Output:

2*x - exp(2*x)/16 - log(-1/(x + x^2))^18/16 + (exp(x)*log(-1/(x + x^2))^9) 
/8
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx=-\frac {e^{2 x}}{16}+\frac {e^{x} \mathrm {log}\left (-\frac {1}{x^{2}+x}\right )^{9}}{8}-\frac {\mathrm {log}\left (-\frac {1}{x^{2}+x}\right )^{18}}{16}+2 x \] Input:

int(((18*x+9)*log(-1/(x^2+x))^18+((x^2+x)*exp(x)*log(-1/(x^2+x))+(-18*x-9) 
*exp(x))*log(-1/(x^2+x))^9+((-x^2-x)*exp(x)^2+16*x^2+16*x)*log(-1/(x^2+x)) 
)/(8*x^2+8*x)/log(-1/(x^2+x)),x)
 

Output:

( - e**(2*x) + 2*e**x*log(( - 1)/(x**2 + x))**9 - log(( - 1)/(x**2 + x))** 
18 + 32*x)/16