\(\int \frac {e^{\frac {32 x+(-4+8 x+12 x^2) \log (\log (5))}{(1-2 x+x^2) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{(-1+3 x-3 x^2+x^3) \log (\log (5))} \, dx\) [5809]

Optimal result

Integrand size = 68, antiderivative size = 32 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=-3+e^{4 \left (-1+\frac {4 \left (x+\frac {2}{\log (\log (5))}\right )}{-4+\frac {(1+x)^2}{x}}\right )} \]

[Out]

exp(16*(x+2/ln(ln(5)))/((1+x)^2/x-4)-4)-3

Rubi [F]

\[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=\int \frac {\exp \left (\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx \]

[In]

Int[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Log[Log[5]]))*(-32 - 32*x - 32*x*Log[Log[5]]
))/((-1 + 3*x - 3*x^2 + x^3)*Log[Log[5]]),x]

[Out]

(-32*(2 + Log[Log[5]])*Defer[Int][1/(E^((4*(Log[Log[5]] - 3*x^2*Log[Log[5]] - 2*x*(4 + Log[Log[5]])))/((-1 + x
)^2*Log[Log[5]]))*(-1 + x)^3), x])/Log[Log[5]] - 32*(1 + Log[Log[5]]^(-1))*Defer[Int][1/(E^((4*(Log[Log[5]] -
3*x^2*Log[Log[5]] - 2*x*(4 + Log[Log[5]])))/((-1 + x)^2*Log[Log[5]]))*(-1 + x)^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (-32+x (-32-32 \log (\log (5))))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx \\ & = \frac {\int \frac {\exp \left (\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (-32+x (-32-32 \log (\log (5))))}{-1+3 x-3 x^2+x^3} \, dx}{\log (\log (5))} \\ & = \frac {\int \frac {\exp \left (\frac {-4 \log (\log (5))+12 x^2 \log (\log (5))+8 x (4+\log (\log (5)))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (32-x (-32-32 \log (\log (5))))}{1-3 x+3 x^2-x^3} \, dx}{\log (\log (5))} \\ & = \frac {\int \frac {32 \exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (1+x (1+\log (\log (5))))}{(1-x)^3} \, dx}{\log (\log (5))} \\ & = \frac {32 \int \frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (1+x (1+\log (\log (5))))}{(1-x)^3} \, dx}{\log (\log (5))} \\ & = \frac {32 \int \left (\frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (-2-\log (\log (5)))}{(-1+x)^3}+\frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (-1-\log (\log (5)))}{(-1+x)^2}\right ) \, dx}{\log (\log (5))} \\ & = \frac {(32 (-2-\log (\log (5)))) \int \frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right )}{(-1+x)^3} \, dx}{\log (\log (5))}+\frac {(32 (-1-\log (\log (5)))) \int \frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right )}{(-1+x)^2} \, dx}{\log (\log (5))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\frac {-4 \log (\log (5))+12 x^2 \log (\log (5))+8 x (4+\log (\log (5)))}{(-1+x)^2 \log (\log (5))}} \]

[In]

Integrate[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Log[Log[5]]))*(-32 - 32*x - 32*x*Log[L
og[5]]))/((-1 + 3*x - 3*x^2 + x^3)*Log[Log[5]]),x]

[Out]

E^((-4*Log[Log[5]] + 12*x^2*Log[Log[5]] + 8*x*(4 + Log[Log[5]]))/((-1 + x)^2*Log[Log[5]]))

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}\)	\(36\)
risch	\({\mathrm e}^{\frac {12 \ln \left (\ln \left (5\right )\right ) x^{2}+8 x \ln \left (\ln \left (5\right )\right )-4 \ln \left (\ln \left (5\right )\right )+32 x}{\ln \left (\ln \left (5\right )\right ) \left (-1+x \right )^{2}}}\)	\(37\)
gosper	\({\mathrm e}^{\frac {12 \ln \left (\ln \left (5\right )\right ) x^{2}+8 x \ln \left (\ln \left (5\right )\right )-4 \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}\)	\(42\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}-2 x \,{\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}+{\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}}{\left (-1+x \right )^{2}}\)	\(120\)
default	\(\frac {-\frac {4 i {\mathrm e}^{12} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}+64 \,{\mathrm e}^{12} \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )-\frac {4 i {\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}+32 \,{\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\ln \left (\ln \left (5\right )\right )}\)	\(502\)
derivativedivides	\(-\frac {32 \left (\frac {i {\mathrm e}^{12} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}-2 \,{\mathrm e}^{12} \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )+\frac {i {\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}-{\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )\right )}{\ln \left (\ln \left (5\right )\right )}\)	\(503\)

[In]

int((-32*x*ln(ln(5))-32*x-32)*exp(((12*x^2+8*x-4)*ln(ln(5))+32*x)/(x^2-2*x+1)/ln(ln(5)))/(x^3-3*x^2+3*x-1)/ln(
ln(5)),x,method=_RETURNVERBOSE)

[Out]

exp(((12*x^2+8*x-4)*ln(ln(5))+32*x)/(x^2-2*x+1)/ln(ln(5)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {4 \, {\left ({\left (3 \, x^{2} + 2 \, x - 1\right )} \log \left (\log \left (5\right )\right ) + 8 \, x\right )}}{{\left (x^{2} - 2 \, x + 1\right )} \log \left (\log \left (5\right )\right )}\right )} \]

[In]

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^
2+3*x-1)/log(log(5)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\frac {32 x + \left (12 x^{2} + 8 x - 4\right ) \log {\left (\log {\left (5 \right )} \right )}}{\left (x^{2} - 2 x + 1\right ) \log {\left (\log {\left (5 \right )} \right )}}} \]

[In]

integrate((-32*x*ln(ln(5))-32*x-32)*exp(((12*x**2+8*x-4)*ln(ln(5))+32*x)/(x**2-2*x+1)/ln(ln(5)))/(x**3-3*x**2+
3*x-1)/ln(ln(5)),x)

[Out]

exp((32*x + (12*x**2 + 8*x - 4)*log(log(5)))/((x**2 - 2*x + 1)*log(log(5))))

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {32}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {16}{x^{2} - 2 \, x + 1} + \frac {32}{x \log \left (\log \left (5\right )\right ) - \log \left (\log \left (5\right )\right )} + \frac {32}{x - 1} + 12\right )} \]

[In]

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^
2+3*x-1)/log(log(5)),x, algorithm="maxima")

[Out]

e^(32/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 16/(x^2 - 2*x + 1) + 32/(x*log(log(5)) - log(log(5))
) + 32/(x - 1) + 12)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (29) = 58\).

Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.12 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {12 \, x^{2} \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {8 \, x \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {32 \, x}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} - \frac {4 \, \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )}\right )} \]

[In]

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^
2+3*x-1)/log(log(5)),x, algorithm="giac")

[Out]

e^(12*x^2*log(log(5))/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 8*x*log(log(5))/(x^2*log(log(5)) - 2
*x*log(log(5)) + log(log(5))) + 32*x/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) - 4*log(log(5))/(x^2*lo
g(log(5)) - 2*x*log(log(5)) + log(log(5))))

Mupad [B] (verification not implemented)

Time = 13.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx={\mathrm {e}}^{\frac {32\,x}{\ln \left (\ln \left (5\right )\right )\,x^2-2\,\ln \left (\ln \left (5\right )\right )\,x+\ln \left (\ln \left (5\right )\right )}}\,{\ln \left (5\right )}^{\frac {12\,x^2+8\,x-4}{\ln \left ({\ln \left (5\right )}^{x^2-2\,x+1}\right )}} \]

[In]

int(-(exp((32*x + log(log(5))*(8*x + 12*x^2 - 4))/(log(log(5))*(x^2 - 2*x + 1)))*(32*x + 32*x*log(log(5)) + 32
))/(log(log(5))*(3*x - 3*x^2 + x^3 - 1)),x)

[Out]

exp((32*x)/(log(log(5)) + x^2*log(log(5)) - 2*x*log(log(5))))*log(5)^((8*x + 12*x^2 - 4)/log(log(5)^(x^2 - 2*x
+ 1)))