\(\int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4))+(8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)) \log (\frac {e^{25 x^2}-x^2}{x})+(e^{25 x^2} x-x^3) \log ^2(\frac {e^{25 x^2}-x^2}{x})}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+(8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)) \log (\frac {e^{25 x^2}-x^2}{x})+(e^{25 x^2} x-x^3) \log ^2(\frac {e^{25 x^2}-x^2}{x})} \, dx\) [2272]

Optimal result

Integrand size = 248, antiderivative size = 32 \[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=x+\frac {2}{4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )} \] Output:

2/(ln(exp(25*x^2)/x-x)+64*x^3*ln(2)^4)+x
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=\frac {2+4 x^4 \log ^4(4)+x \log \left (\frac {e^{25 x^2}}{x}-x\right )}{4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )} \] Input:

Integrate[(2*x^2 + 24*x^5*Log[4]^4 - 16*x^9*Log[4]^8 + E^(25*x^2)*(2 - 100 
*x^2 - 24*x^3*Log[4]^4 + 16*x^7*Log[4]^8) + (8*E^(25*x^2)*x^4*Log[4]^4 - 8 
*x^6*Log[4]^4)*Log[(E^(25*x^2) - x^2)/x] + (E^(25*x^2)*x - x^3)*Log[(E^(25 
*x^2) - x^2)/x]^2)/(16*E^(25*x^2)*x^7*Log[4]^8 - 16*x^9*Log[4]^8 + (8*E^(2 
5*x^2)*x^4*Log[4]^4 - 8*x^6*Log[4]^4)*Log[(E^(25*x^2) - x^2)/x] + (E^(25*x 
^2)*x - x^3)*Log[(E^(25*x^2) - x^2)/x]^2),x]
 

Output:

(2 + 4*x^4*Log[4]^4 + x*Log[E^(25*x^2)/x - x])/(4*x^3*Log[4]^4 + Log[E^(25 
*x^2)/x - x])
 

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.

\(\displaystyle \int \frac {-16 x^9 \log ^8(4)+24 x^5 \log ^4(4)+2 x^2+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )+e^{25 x^2} \left (16 x^7 \log ^8(4)-24 x^3 \log ^4(4)-100 x^2+2\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )}{-16 x^9 \log ^8(4)+16 e^{25 x^2} x^7 \log ^8(4)+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-16 x^9 \log ^8(4)+24 x^5 \log ^4(4)+2 x^2+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )+e^{25 x^2} \left (16 x^7 \log ^8(4)-24 x^3 \log ^4(4)-100 x^2+2\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )}{x \left (e^{25 x^2}-x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {4 x \left (25 x^2-1\right )}{\left (x^2-e^{25 x^2}\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {16 x^7 \log ^8(4)-24 x^3 \log ^4(4)-100 x^2+x \log ^2\left (\frac {e^{25 x^2}}{x}-x\right )+8 x^4 \log ^4(4) \log \left (\frac {e^{25 x^2}}{x}-x\right )+2}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \int \frac {1}{x \left (4 \log ^4(4) x^3+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}dx-100 \int \frac {x}{\left (4 \log ^4(4) x^3+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}dx-24 \log ^4(4) \int \frac {x^2}{\left (4 \log ^4(4) x^3+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}dx-4 \int \frac {x}{\left (x^2-e^{25 x^2}\right ) \left (4 \log ^4(4) x^3+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}dx+100 \int \frac {x^3}{\left (x^2-e^{25 x^2}\right ) \left (4 \log ^4(4) x^3+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}dx+x\)

Input:

Int[(2*x^2 + 24*x^5*Log[4]^4 - 16*x^9*Log[4]^8 + E^(25*x^2)*(2 - 100*x^2 - 
 24*x^3*Log[4]^4 + 16*x^7*Log[4]^8) + (8*E^(25*x^2)*x^4*Log[4]^4 - 8*x^6*L 
og[4]^4)*Log[(E^(25*x^2) - x^2)/x] + (E^(25*x^2)*x - x^3)*Log[(E^(25*x^2) 
- x^2)/x]^2)/(16*E^(25*x^2)*x^7*Log[4]^8 - 16*x^9*Log[4]^8 + (8*E^(25*x^2) 
*x^4*Log[4]^4 - 8*x^6*Log[4]^4)*Log[(E^(25*x^2) - x^2)/x] + (E^(25*x^2)*x 
- x^3)*Log[(E^(25*x^2) - x^2)/x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 15.80 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.97

method	result	size
parallelrisch	\(-\frac {-2-64 \ln \left (2\right )^{4} x^{4}-\ln \left (\frac {{\mathrm e}^{25 x^{2}}-x^{2}}{x}\right ) x}{64 x^{3} \ln \left (2\right )^{4}+\ln \left (\frac {{\mathrm e}^{25 x^{2}}-x^{2}}{x}\right )}\)	\(63\)
risch	\(x +\frac {4}{128 x^{3} \ln \left (2\right )^{4}-2 i \pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )+i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )}^{3}+2 i \pi -2 \ln \left (x \right )+2 \ln \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}\)	\(209\)

Input:

int(((x*exp(25*x^2)-x^3)*ln((exp(25*x^2)-x^2)/x)^2+(128*x^4*ln(2)^4*exp(25 
*x^2)-128*x^6*ln(2)^4)*ln((exp(25*x^2)-x^2)/x)+(4096*x^7*ln(2)^8-384*x^3*l 
n(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*ln(2)^8+384*x^5*ln(2)^4+2*x^2)/((x* 
exp(25*x^2)-x^3)*ln((exp(25*x^2)-x^2)/x)^2+(128*x^4*ln(2)^4*exp(25*x^2)-12 
8*x^6*ln(2)^4)*ln((exp(25*x^2)-x^2)/x)+4096*x^7*ln(2)^8*exp(25*x^2)-4096*x 
^9*ln(2)^8),x,method=_RETURNVERBOSE)
 

Output:

-(-2-64*ln(2)^4*x^4-ln((exp(25*x^2)-x^2)/x)*x)/(64*x^3*ln(2)^4+ln((exp(25* 
x^2)-x^2)/x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.94 \[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=\frac {64 \, x^{4} \log \left (2\right )^{4} + x \log \left (-\frac {x^{2} - e^{\left (25 \, x^{2}\right )}}{x}\right ) + 2}{64 \, x^{3} \log \left (2\right )^{4} + \log \left (-\frac {x^{2} - e^{\left (25 \, x^{2}\right )}}{x}\right )} \] Input:

integrate(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^ 
4*exp(25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^ 
8-384*x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(2) 
^4+2*x^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^ 
4*exp(25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8 
*exp(25*x^2)-4096*x^9*log(2)^8),x, algorithm="fricas")
 

Output:

(64*x^4*log(2)^4 + x*log(-(x^2 - e^(25*x^2))/x) + 2)/(64*x^3*log(2)^4 + lo 
g(-(x^2 - e^(25*x^2))/x))
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=x + \frac {2}{64 x^{3} \log {\left (2 \right )}^{4} + \log {\left (\frac {- x^{2} + e^{25 x^{2}}}{x} \right )}} \] Input:

integrate(((x*exp(25*x**2)-x**3)*ln((exp(25*x**2)-x**2)/x)**2+(128*x**4*ln 
(2)**4*exp(25*x**2)-128*x**6*ln(2)**4)*ln((exp(25*x**2)-x**2)/x)+(4096*x** 
7*ln(2)**8-384*x**3*ln(2)**4-100*x**2+2)*exp(25*x**2)-4096*x**9*ln(2)**8+3 
84*x**5*ln(2)**4+2*x**2)/((x*exp(25*x**2)-x**3)*ln((exp(25*x**2)-x**2)/x)* 
*2+(128*x**4*ln(2)**4*exp(25*x**2)-128*x**6*ln(2)**4)*ln((exp(25*x**2)-x** 
2)/x)+4096*x**7*ln(2)**8*exp(25*x**2)-4096*x**9*ln(2)**8),x)
 

Output:

x + 2/(64*x**3*log(2)**4 + log((-x**2 + exp(25*x**2))/x))
 

Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.91 \[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=\frac {64 \, x^{4} \log \left (2\right )^{4} + x \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - x \log \left (x\right ) + 2}{64 \, x^{3} \log \left (2\right )^{4} + \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - \log \left (x\right )} \] Input:

integrate(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^ 
4*exp(25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^ 
8-384*x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(2) 
^4+2*x^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^ 
4*exp(25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8 
*exp(25*x^2)-4096*x^9*log(2)^8),x, algorithm="maxima")
 

Output:

(64*x^4*log(2)^4 + x*log(-x^2 + e^(25*x^2)) - x*log(x) + 2)/(64*x^3*log(2) 
^4 + log(-x^2 + e^(25*x^2)) - log(x))
 

Giac [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.91 \[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=\frac {64 \, x^{4} \log \left (2\right )^{4} + x \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - x \log \left (x\right ) + 2}{64 \, x^{3} \log \left (2\right )^{4} + \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - \log \left (x\right )} \] Input:

integrate(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^ 
4*exp(25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^ 
8-384*x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(2) 
^4+2*x^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^ 
4*exp(25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8 
*exp(25*x^2)-4096*x^9*log(2)^8),x, algorithm="giac")
 

Output:

(64*x^4*log(2)^4 + x*log(-x^2 + e^(25*x^2)) - x*log(x) + 2)/(64*x^3*log(2) 
^4 + log(-x^2 + e^(25*x^2)) - log(x))
 

Mupad [B] (verification not implemented)

Time = 2.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=x+\frac {2}{\ln \left (\frac {{\mathrm {e}}^{25\,x^2}-x^2}{x}\right )+64\,x^3\,{\ln \left (2\right )}^4} \] Input:

int(-(384*x^5*log(2)^4 - 4096*x^9*log(2)^8 - log((exp(25*x^2) - x^2)/x)*(1 
28*x^6*log(2)^4 - 128*x^4*exp(25*x^2)*log(2)^4) - exp(25*x^2)*(384*x^3*log 
(2)^4 - 4096*x^7*log(2)^8 + 100*x^2 - 2) + log((exp(25*x^2) - x^2)/x)^2*(x 
*exp(25*x^2) - x^3) + 2*x^2)/(4096*x^9*log(2)^8 + log((exp(25*x^2) - x^2)/ 
x)*(128*x^6*log(2)^4 - 128*x^4*exp(25*x^2)*log(2)^4) - log((exp(25*x^2) - 
x^2)/x)^2*(x*exp(25*x^2) - x^3) - 4096*x^7*exp(25*x^2)*log(2)^8),x)
 

Output:

x + 2/(log((exp(25*x^2) - x^2)/x) + 64*x^3*log(2)^4)
 

Reduce [F]

\[ \int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx=\int \frac {\left (x \,{\mathrm e}^{25 x^{2}}-x^{3}\right ) \mathrm {log}\left (\frac {{\mathrm e}^{25 x^{2}}-x^{2}}{x}\right )^{2}+\left (128 x^{4} \mathrm {log}\left (2\right )^{4} {\mathrm e}^{25 x^{2}}-128 x^{6} \mathrm {log}\left (2\right )^{4}\right ) \mathrm {log}\left (\frac {{\mathrm e}^{25 x^{2}}-x^{2}}{x}\right )+\left (4096 x^{7} \mathrm {log}\left (2\right )^{8}-384 x^{3} \mathrm {log}\left (2\right )^{4}-100 x^{2}+2\right ) {\mathrm e}^{25 x^{2}}-4096 x^{9} \mathrm {log}\left (2\right )^{8}+384 x^{5} \mathrm {log}\left (2\right )^{4}+2 x^{2}}{\left (x \,{\mathrm e}^{25 x^{2}}-x^{3}\right ) \mathrm {log}\left (\frac {{\mathrm e}^{25 x^{2}}-x^{2}}{x}\right )^{2}+\left (128 x^{4} \mathrm {log}\left (2\right )^{4} {\mathrm e}^{25 x^{2}}-128 x^{6} \mathrm {log}\left (2\right )^{4}\right ) \mathrm {log}\left (\frac {{\mathrm e}^{25 x^{2}}-x^{2}}{x}\right )+4096 x^{7} \mathrm {log}\left (2\right )^{8} {\mathrm e}^{25 x^{2}}-4096 x^{9} \mathrm {log}\left (2\right )^{8}}d x \] Input:

int(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp( 
25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^8-384* 
x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(2)^4+2*x 
^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp( 
25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8*exp(2 
5*x^2)-4096*x^9*log(2)^8),x)
 

Output:

int(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp( 
25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^8-384* 
x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(2)^4+2*x 
^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp( 
25*x^2)-128*x^6*log(2)^4)*log((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8*exp(2 
5*x^2)-4096*x^9*log(2)^8),x)