\(\int \frac {-4 x+8 e^{8-2 x} x+(-8+e^{8-2 x}+x) \log (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x))}{(-8+e^{8-2 x}+x) \log ^3(64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x))} \, dx\) [2749]

Optimal result

Integrand size = 98, antiderivative size = 20 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {\log \left (e^x\right )}{\log ^2\left (\left (-8+e^{8-2 x}+x\right )^2\right )} \]

[Out]

ln(exp(x))/ln((exp(-x+4)^2-8+x)^2)^2

Rubi [F]

\[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx \]

[In]

Int[(-4*x + 8*E^(8 - 2*x)*x + (-8 + E^(8 - 2*x) + x)*Log[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-16 + 2
*x)])/((-8 + E^(8 - 2*x) + x)*Log[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-16 + 2*x)]^3),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 x-8 e^{8-2 x} x-\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (8-e^{8-2 x}-x\right ) \log ^3\left (e^{-4 x} \left (e^8-8 e^{2 x}+e^{2 x} x\right )^2\right )} \, dx \\ & = \int \left (\frac {4 e^8 x (-15+2 x)}{(-8+x) \left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}+\frac {-4 x-8 \log \left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )+x \log \left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}{(-8+x) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}\right ) \, dx \\ & = \left (4 e^8\right ) \int \frac {x (-15+2 x)}{(-8+x) \left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\int \frac {-4 x-8 \log \left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )+x \log \left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}{(-8+x) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx \\ & = \left (4 e^8\right ) \int \left (\frac {1}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}+\frac {8}{(-8+x) \left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}+\frac {2 x}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}\right ) \, dx+\int \frac {-\frac {4 x}{-8+x}+\log \left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}{\log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx \\ & = \left (4 e^8\right ) \int \frac {1}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (8 e^8\right ) \int \frac {x}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (32 e^8\right ) \int \frac {1}{(-8+x) \left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\int \left (-\frac {4 x}{(-8+x) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}+\frac {1}{\log ^2\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}\right ) \, dx \\ & = -\left (4 \int \frac {x}{(-8+x) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx\right )+\left (4 e^8\right ) \int \frac {1}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (8 e^8\right ) \int \frac {x}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (32 e^8\right ) \int \frac {1}{(-8+x) \left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\int \frac {1}{\log ^2\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx \\ & = -\left (4 \int \left (\frac {1}{\log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}+\frac {8}{(-8+x) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )}\right ) \, dx\right )+\left (4 e^8\right ) \int \frac {1}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (8 e^8\right ) \int \frac {x}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (32 e^8\right ) \int \frac {1}{(-8+x) \left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\int \frac {1}{\log ^2\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx \\ & = -\left (4 \int \frac {1}{\log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx\right )-32 \int \frac {1}{(-8+x) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (4 e^8\right ) \int \frac {1}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (8 e^8\right ) \int \frac {x}{\left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\left (32 e^8\right ) \int \frac {1}{(-8+x) \left (e^8-8 e^{2 x}+e^{2 x} x\right ) \log ^3\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx+\int \frac {1}{\log ^2\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\log ^2\left (e^{-4 x} \left (e^8+e^{2 x} (-8+x)\right )^2\right )} \]

[In]

Integrate[(-4*x + 8*E^(8 - 2*x)*x + (-8 + E^(8 - 2*x) + x)*Log[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-
16 + 2*x)])/((-8 + E^(8 - 2*x) + x)*Log[64 + E^(16 - 4*x) - 16*x + x^2 + E^(8 - 2*x)*(-16 + 2*x)]^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80

method	result	size
parallelrisch	\(\frac {x}{\ln \left ({\mathrm e}^{-4 x +16}+\left (2 x -16\right ) {\mathrm e}^{-2 x +8}+x^{2}-16 x +64\right )^{2}}\)	\(36\)
risch	\(-\frac {4 x}{{\left (\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )^{2}\right )}^{2}+\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{-2 x +8}-8+x \right )^{2}\right )}^{3}+4 i \ln \left ({\mathrm e}^{-2 x +8}-8+x \right )\right )}^{2}}\)	\(104\)

[In]

int(((exp(-x+4)^2-8+x)*ln(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)+8*x*exp(-x+4)^2-4*x)/(exp(-x+4)^2-8+x)
/ln(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)^3,x,method=_RETURNVERBOSE)

[Out]

x/ln(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\log \left (x^{2} + 2 \, {\left (x - 8\right )} e^{\left (-2 \, x + 8\right )} - 16 \, x + e^{\left (-4 \, x + 16\right )} + 64\right )^{2}} \]

[In]

integrate(((exp(-x+4)^2-8+x)*log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)+8*x*exp(-x+4)^2-4*x)/(exp(-x+4)
^2-8+x)/log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)^3,x, algorithm="fricas")

[Out]

x/log(x^2 + 2*(x - 8)*e^(-2*x + 8) - 16*x + e^(-4*x + 16) + 64)^2

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\log {\left (x^{2} - 16 x + \left (2 x - 16\right ) e^{8 - 2 x} + e^{16 - 4 x} + 64 \right )}^{2}} \]

[In]

integrate(((exp(-x+4)**2-8+x)*ln(exp(-x+4)**4+(2*x-16)*exp(-x+4)**2+x**2-16*x+64)+8*x*exp(-x+4)**2-4*x)/(exp(-
x+4)**2-8+x)/ln(exp(-x+4)**4+(2*x-16)*exp(-x+4)**2+x**2-16*x+64)**3,x)

[Out]

x/log(x**2 - 16*x + (2*x - 16)*exp(8 - 2*x) + exp(16 - 4*x) + 64)**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (18) = 36\).

Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.00 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{4 \, {\left (4 \, x^{2} - 4 \, x \log \left ({\left (x - 8\right )} e^{\left (2 \, x\right )} + e^{8}\right ) + \log \left ({\left (x - 8\right )} e^{\left (2 \, x\right )} + e^{8}\right )^{2}\right )}} \]

[In]

integrate(((exp(-x+4)^2-8+x)*log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)+8*x*exp(-x+4)^2-4*x)/(exp(-x+4)
^2-8+x)/log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.92 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {x}{\log \left (x^{2} + 2 \, x e^{\left (-2 \, x + 8\right )} - 16 \, x - 16 \, e^{\left (-2 \, x + 8\right )} + e^{\left (-4 \, x + 16\right )} + 64\right )^{2}} \]

[In]

integrate(((exp(-x+4)^2-8+x)*log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)+8*x*exp(-x+4)^2-4*x)/(exp(-x+4)
^2-8+x)/log(exp(-x+4)^4+(2*x-16)*exp(-x+4)^2+x^2-16*x+64)^3,x, algorithm="giac")

[Out]

x/log(x^2 + 2*x*e^(-2*x + 8) - 16*x - 16*e^(-2*x + 8) + e^(-4*x + 16) + 64)^2

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 297, normalized size of antiderivative = 14.85 \[ \int \frac {-4 x+8 e^{8-2 x} x+\left (-8+e^{8-2 x}+x\right ) \log \left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )}{\left (-8+e^{8-2 x}+x\right ) \log ^3\left (64+e^{16-4 x}-16 x+x^2+e^{8-2 x} (-16+2 x)\right )} \, dx=\frac {\frac {x}{8}-\frac {7}{8}}{2\,{\mathrm {e}}^{8-2\,x}-1}-\frac {\frac {x+{\mathrm {e}}^{8-2\,x}-8}{4\,\left (2\,{\mathrm {e}}^{8-2\,x}-1\right )}-\frac {\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )\,\left (x+{\mathrm {e}}^{8-2\,x}-8\right )\,\left (28\,{\mathrm {e}}^{8-2\,x}-4\,x\,{\mathrm {e}}^{8-2\,x}+1\right )}{8\,{\left (2\,{\mathrm {e}}^{8-2\,x}-1\right )}^3}}{\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )}+\frac {\frac {x^2}{4}-\frac {15\,x}{4}+\frac {225}{16}}{6\,{\mathrm {e}}^{8-2\,x}-12\,{\mathrm {e}}^{16-4\,x}+8\,{\mathrm {e}}^{24-6\,x}-1}+\frac {\frac {x^2}{4}-\frac {7\,x}{2}+\frac {195}{16}}{4\,{\mathrm {e}}^{16-4\,x}-4\,{\mathrm {e}}^{8-2\,x}+1}+\frac {x+\frac {\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )\,\left (x+{\mathrm {e}}^{8-2\,x}-8\right )}{4\,\left (2\,{\mathrm {e}}^{8-2\,x}-1\right )}}{{\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{16}-16\,x+x^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8\,\left (2\,x-16\right )+64\right )}^2} \]

[In]

int((log(exp(16 - 4*x) - 16*x + exp(8 - 2*x)*(2*x - 16) + x^2 + 64)*(x + exp(8 - 2*x) - 8) - 4*x + 8*x*exp(8 -
2*x))/(log(exp(16 - 4*x) - 16*x + exp(8 - 2*x)*(2*x - 16) + x^2 + 64)^3*(x + exp(8 - 2*x) - 8)),x)

[Out]

(x/8 - 7/8)/(2*exp(8 - 2*x) - 1) - ((x + exp(8 - 2*x) - 8)/(4*(2*exp(8 - 2*x) - 1)) - (log(exp(-4*x)*exp(16) -
16*x + x^2 + exp(-2*x)*exp(8)*(2*x - 16) + 64)*(x + exp(8 - 2*x) - 8)*(28*exp(8 - 2*x) - 4*x*exp(8 - 2*x) + 1
))/(8*(2*exp(8 - 2*x) - 1)^3))/log(exp(-4*x)*exp(16) - 16*x + x^2 + exp(-2*x)*exp(8)*(2*x - 16) + 64) + (x^2/4
- (15*x)/4 + 225/16)/(6*exp(8 - 2*x) - 12*exp(16 - 4*x) + 8*exp(24 - 6*x) - 1) + (x^2/4 - (7*x)/2 + 195/16)/(
4*exp(16 - 4*x) - 4*exp(8 - 2*x) + 1) + (x + (log(exp(-4*x)*exp(16) - 16*x + x^2 + exp(-2*x)*exp(8)*(2*x - 16)
+ 64)*(x + exp(8 - 2*x) - 8))/(4*(2*exp(8 - 2*x) - 1)))/log(exp(-4*x)*exp(16) - 16*x + x^2 + exp(-2*x)*exp(8)
*(2*x - 16) + 64)^2