\(\int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2(\frac {2 x^2}{5})}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2(\frac {2 x^2}{5})}} (-6 x^2-12 x \log (4)-6 \log ^2(4)+(3 x^2+3 x \log (4)) \log (\frac {2 x^2}{5}))}{2 x \log ^3(\frac {2 x^2}{5})} \, dx\) [6512]

Optimal result

Integrand size = 111, antiderivative size = 27 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 e^{\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}} \]

[Out]

3*exp(1/4*exp((x+2*ln(2))^2/ln(2/5*x^2)^2))

Rubi [F]

\[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=\int \frac {\exp \left (\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx \]

[In]

Int[(E^(E^((x^2 + 2*x*Log[4] + Log[4]^2)/Log[(2*x^2)/5]^2)/4 + (x^2 + 2*x*Log[4] + Log[4]^2)/Log[(2*x^2)/5]^2)
*(-6*x^2 - 12*x*Log[4] - 6*Log[4]^2 + (3*x^2 + 3*x*Log[4])*Log[(2*x^2)/5]))/(2*x*Log[(2*x^2)/5]^3),x]

[Out]

-6*Log[4]*Defer[Int][E^(E^((x + Log[4])^2/Log[(2*x^2)/5]^2)/4 + (x + Log[4])^2/Log[(2*x^2)/5]^2)/Log[(2*x^2)/5
]^3, x] - 3*Log[4]^2*Defer[Int][E^(E^((x + Log[4])^2/Log[(2*x^2)/5]^2)/4 + (x + Log[4])^2/Log[(2*x^2)/5]^2)/(x
*Log[(2*x^2)/5]^3), x] - 3*Defer[Int][(E^(E^((x + Log[4])^2/Log[(2*x^2)/5]^2)/4 + (x + Log[4])^2/Log[(2*x^2)/5
]^2)*x)/Log[(2*x^2)/5]^3, x] + (3*Log[4]*Defer[Int][E^(E^((x + Log[4])^2/Log[(2*x^2)/5]^2)/4 + (x + Log[4])^2/
Log[(2*x^2)/5]^2)/Log[(2*x^2)/5]^2, x])/2 + (3*Defer[Int][(E^(E^((x + Log[4])^2/Log[(2*x^2)/5]^2)/4 + (x + Log
[4])^2/Log[(2*x^2)/5]^2)*x)/Log[(2*x^2)/5]^2, x])/2

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx \\ & = \frac {1}{2} \int \frac {3 \exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4)) \left (-2 (x+\log (4))+x \log \left (\frac {2 x^2}{5}\right )\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx \\ & = \frac {3}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4)) \left (-2 (x+\log (4))+x \log \left (\frac {2 x^2}{5}\right )\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx \\ & = \frac {3}{2} \int \left (-\frac {2 \exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))^2}{x \log ^3\left (\frac {2 x^2}{5}\right )}+\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \, dx \\ & = \frac {3}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))}{\log ^2\left (\frac {2 x^2}{5}\right )} \, dx-3 \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))^2}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx \\ & = \frac {3}{2} \int \left (\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^2\left (\frac {2 x^2}{5}\right )}+\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \log (4)}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \, dx-3 \int \left (\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^3\left (\frac {2 x^2}{5}\right )}+\frac {2 \exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \log (4)}{\log ^3\left (\frac {2 x^2}{5}\right )}+\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \log ^2(4)}{x \log ^3\left (\frac {2 x^2}{5}\right )}\right ) \, dx \\ & = \frac {3}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^2\left (\frac {2 x^2}{5}\right )} \, dx-3 \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^3\left (\frac {2 x^2}{5}\right )} \, dx+\frac {1}{2} (3 \log (4)) \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right )}{\log ^2\left (\frac {2 x^2}{5}\right )} \, dx-(6 \log (4)) \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right )}{\log ^3\left (\frac {2 x^2}{5}\right )} \, dx-\left (3 \log ^2(4)\right ) \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 4.86 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 e^{\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}} \]

[In]

Integrate[(E^(E^((x^2 + 2*x*Log[4] + Log[4]^2)/Log[(2*x^2)/5]^2)/4 + (x^2 + 2*x*Log[4] + Log[4]^2)/Log[(2*x^2)
/5]^2)*(-6*x^2 - 12*x*Log[4] - 6*Log[4]^2 + (3*x^2 + 3*x*Log[4])*Log[(2*x^2)/5]))/(2*x*Log[(2*x^2)/5]^3),x]

[Out]

3*E^(E^((x + Log[4])^2/Log[(2*x^2)/5]^2)/4)

Maple [A] (verified)

Time = 24.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
risch	\(3 \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {\left (x +2 \ln \left (2\right )\right )^{2}}{\ln \left (\frac {2 x^{2}}{5}\right )^{2}}}}{4}}\)	\(24\)
parallelrisch	\(3 \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {4 \ln \left (2\right )^{2}+4 x \ln \left (2\right )+x^{2}}{\ln \left (\frac {2 x^{2}}{5}\right )^{2}}}}{4}}\)	\(31\)

[In]

int(1/2*((6*x*ln(2)+3*x^2)*ln(2/5*x^2)-24*ln(2)^2-24*x*ln(2)-6*x^2)*exp((4*ln(2)^2+4*x*ln(2)+x^2)/ln(2/5*x^2)^
2)*exp(1/4*exp((4*ln(2)^2+4*x*ln(2)+x^2)/ln(2/5*x^2)^2))/x/ln(2/5*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

3*exp(1/4*exp((x+2*ln(2))^2/ln(2/5*x^2)^2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (23) = 46\).

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.33 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 \, e^{\left (\frac {e^{\left (\frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )} \log \left (\frac {2}{5} \, x^{2}\right )^{2} + 4 \, x^{2} + 16 \, x \log \left (2\right ) + 16 \, \log \left (2\right )^{2}}{4 \, \log \left (\frac {2}{5} \, x^{2}\right )^{2}} - \frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )} \]

[In]

integrate(1/2*((6*x*log(2)+3*x^2)*log(2/5*x^2)-24*log(2)^2-24*x*log(2)-6*x^2)*exp((4*log(2)^2+4*x*log(2)+x^2)/
log(2/5*x^2)^2)*exp(1/4*exp((4*log(2)^2+4*x*log(2)+x^2)/log(2/5*x^2)^2))/x/log(2/5*x^2)^3,x, algorithm="fricas
")

[Out]

3*e^(1/4*(e^((x^2 + 4*x*log(2) + 4*log(2)^2)/log(2/5*x^2)^2)*log(2/5*x^2)^2 + 4*x^2 + 16*x*log(2) + 16*log(2)^
2)/log(2/5*x^2)^2 - (x^2 + 4*x*log(2) + 4*log(2)^2)/log(2/5*x^2)^2)

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 e^{\frac {e^{\frac {x^{2} + 4 x \log {\left (2 \right )} + 4 \log {\left (2 \right )}^{2}}{\log {\left (\frac {2 x^{2}}{5} \right )}^{2}}}}{4}} \]

[In]

integrate(1/2*((6*x*ln(2)+3*x**2)*ln(2/5*x**2)-24*ln(2)**2-24*x*ln(2)-6*x**2)*exp((4*ln(2)**2+4*x*ln(2)+x**2)/
ln(2/5*x**2)**2)*exp(1/4*exp((4*ln(2)**2+4*x*ln(2)+x**2)/ln(2/5*x**2)**2))/x/ln(2/5*x**2)**3,x)

[Out]

3*exp(exp((x**2 + 4*x*log(2) + 4*log(2)**2)/log(2*x**2/5)**2)/4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (23) = 46\).

Time = 0.54 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.59 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 \, e^{\left (\frac {1}{4} \, e^{\left (\frac {x^{2}}{\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, \log \left (x\right )^{2}} + \frac {4 \, x \log \left (2\right )}{\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, \log \left (x\right )^{2}} + \frac {4 \, \log \left (2\right )^{2}}{\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, \log \left (x\right )^{2}}\right )}\right )} \]

[In]

integrate(1/2*((6*x*log(2)+3*x^2)*log(2/5*x^2)-24*log(2)^2-24*x*log(2)-6*x^2)*exp((4*log(2)^2+4*x*log(2)+x^2)/
log(2/5*x^2)^2)*exp(1/4*exp((4*log(2)^2+4*x*log(2)+x^2)/log(2/5*x^2)^2))/x/log(2/5*x^2)^3,x, algorithm="maxima
")

[Out]

3*e^(1/4*e^(x^2/(log(5)^2 - 2*log(5)*log(2) + log(2)^2 - 4*(log(5) - log(2))*log(x) + 4*log(x)^2) + 4*x*log(2)
/(log(5)^2 - 2*log(5)*log(2) + log(2)^2 - 4*(log(5) - log(2))*log(x) + 4*log(x)^2) + 4*log(2)^2/(log(5)^2 - 2*
log(5)*log(2) + log(2)^2 - 4*(log(5) - log(2))*log(x) + 4*log(x)^2)))

Giac [F]

\[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=\int { -\frac {3 \, {\left (2 \, x^{2} + 8 \, x \log \left (2\right ) + 8 \, \log \left (2\right )^{2} - {\left (x^{2} + 2 \, x \log \left (2\right )\right )} \log \left (\frac {2}{5} \, x^{2}\right )\right )} e^{\left (\frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}} + \frac {1}{4} \, e^{\left (\frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )}\right )}}{2 \, x \log \left (\frac {2}{5} \, x^{2}\right )^{3}} \,d x } \]

[In]

integrate(1/2*((6*x*log(2)+3*x^2)*log(2/5*x^2)-24*log(2)^2-24*x*log(2)-6*x^2)*exp((4*log(2)^2+4*x*log(2)+x^2)/
log(2/5*x^2)^2)*exp(1/4*exp((4*log(2)^2+4*x*log(2)+x^2)/log(2/5*x^2)^2))/x/log(2/5*x^2)^3,x, algorithm="giac")

[Out]

integrate(-3/2*(2*x^2 + 8*x*log(2) + 8*log(2)^2 - (x^2 + 2*x*log(2))*log(2/5*x^2))*e^((x^2 + 4*x*log(2) + 4*lo
g(2)^2)/log(2/5*x^2)^2 + 1/4*e^((x^2 + 4*x*log(2) + 4*log(2)^2)/log(2/5*x^2)^2))/(x*log(2/5*x^2)^3), x)

Mupad [B] (verification not implemented)

Time = 13.73 (sec) , antiderivative size = 139, normalized size of antiderivative = 5.15 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3\,{\mathrm {e}}^{\frac {2^{\frac {4\,x}{2\,\ln \left (x^2\right )\,\ln \left (2\right )-2\,\ln \left (x^2\right )\,\ln \left (5\right )-2\,\ln \left (2\right )\,\ln \left (5\right )+{\ln \left (x^2\right )}^2+{\ln \left (2\right )}^2+{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {4\,{\ln \left (2\right )}^2}{2\,\ln \left (x^2\right )\,\ln \left (2\right )-2\,\ln \left (x^2\right )\,\ln \left (5\right )-2\,\ln \left (2\right )\,\ln \left (5\right )+{\ln \left (x^2\right )}^2+{\ln \left (2\right )}^2+{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {x^2}{2\,\ln \left (x^2\right )\,\ln \left (2\right )-2\,\ln \left (x^2\right )\,\ln \left (5\right )-2\,\ln \left (2\right )\,\ln \left (5\right )+{\ln \left (x^2\right )}^2+{\ln \left (2\right )}^2+{\ln \left (5\right )}^2}}}{4}} \]

[In]

int(-(exp(exp((4*x*log(2) + 4*log(2)^2 + x^2)/log((2*x^2)/5)^2)/4)*exp((4*x*log(2) + 4*log(2)^2 + x^2)/log((2*
x^2)/5)^2)*(24*x*log(2) + 24*log(2)^2 + 6*x^2 - log((2*x^2)/5)*(6*x*log(2) + 3*x^2)))/(2*x*log((2*x^2)/5)^3),x
)

[Out]

3*exp((2^((4*x)/(2*log(x^2)*log(2) - 2*log(x^2)*log(5) - 2*log(2)*log(5) + log(x^2)^2 + log(2)^2 + log(5)^2))*
exp((4*log(2)^2)/(2*log(x^2)*log(2) - 2*log(x^2)*log(5) - 2*log(2)*log(5) + log(x^2)^2 + log(2)^2 + log(5)^2))
*exp(x^2/(2*log(x^2)*log(2) - 2*log(x^2)*log(5) - 2*log(2)*log(5) + log(x^2)^2 + log(2)^2 + log(5)^2)))/4)