\(\int \frac {2 x+16 x^2+2 x \log (\frac {5 e^{-4-8 x}}{x})+\log ^3(\frac {5 e^{-4-8 x}}{x})}{(-3+\log (5)) \log ^3(\frac {5 e^{-4-8 x}}{x})} \, dx\) [5876]

Optimal result

Integrand size = 62, antiderivative size = 28 \[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\frac {x+\frac {x^2}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )}}{-3+\log (5)} \]

[Out]

(x^2/ln(5/x/exp(4)/exp(x)^8)^2+x)/(ln(5)-3)

Rubi [F]

\[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx \]

[In]

Int[(2*x + 16*x^2 + 2*x*Log[(5*E^(-4 - 8*x))/x] + Log[(5*E^(-4 - 8*x))/x]^3)/((-3 + Log[5])*Log[(5*E^(-4 - 8*x
))/x]^3),x]

[Out]

-(x/(3 - Log[5])) - (2*Defer[Int][x/Log[(5*E^(-4 - 8*x))/x]^3, x])/(3 - Log[5]) - (16*Defer[Int][x^2/Log[(5*E^
(-4 - 8*x))/x]^3, x])/(3 - Log[5]) - (2*Defer[Int][x/Log[(5*E^(-4 - 8*x))/x]^2, x])/(3 - Log[5])

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx}{-3+\log (5)} \\ & = \frac {\int \left (1+\frac {2 x (1+8 x)}{\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}+\frac {2 x}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )}\right ) \, dx}{-3+\log (5)} \\ & = -\frac {x}{3-\log (5)}-\frac {2 \int \frac {x (1+8 x)}{\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx}{3-\log (5)}-\frac {2 \int \frac {x}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx}{3-\log (5)} \\ & = -\frac {x}{3-\log (5)}-\frac {2 \int \left (\frac {x}{\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}+\frac {8 x^2}{\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}\right ) \, dx}{3-\log (5)}-\frac {2 \int \frac {x}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx}{3-\log (5)} \\ & = -\frac {x}{3-\log (5)}-\frac {2 \int \frac {x}{\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx}{3-\log (5)}-\frac {2 \int \frac {x}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx}{3-\log (5)}-\frac {16 \int \frac {x^2}{\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx}{3-\log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\frac {x+\frac {x^2}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )}}{-3+\log (5)} \]

[In]

Integrate[(2*x + 16*x^2 + 2*x*Log[(5*E^(-4 - 8*x))/x] + Log[(5*E^(-4 - 8*x))/x]^3)/((-3 + Log[5])*Log[(5*E^(-4
- 8*x))/x]^3),x]

[Out]

(x + x^2/Log[(5*E^(-4 - 8*x))/x]^2)/(-3 + Log[5])

Maple [F(-1)]

Timed out.

\[\int \frac {\ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )^{3}+2 x \ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )+16 x^{2}+2 x}{\left (\ln \left (5\right )-3\right ) \ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )^{3}}d x\]

[In]

int((ln(5/x/exp(4)/exp(x)^8)^3+2*x*ln(5/x/exp(4)/exp(x)^8)+16*x^2+2*x)/(ln(5)-3)/ln(5/x/exp(4)/exp(x)^8)^3,x)

[Out]

int((ln(5/x/exp(4)/exp(x)^8)^3+2*x*ln(5/x/exp(4)/exp(x)^8)+16*x^2+2*x)/(ln(5)-3)/ln(5/x/exp(4)/exp(x)^8)^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\frac {x \log \left (\frac {5 \, e^{\left (-8 \, x - 4\right )}}{x}\right )^{2} + x^{2}}{{\left (\log \left (5\right ) - 3\right )} \log \left (\frac {5 \, e^{\left (-8 \, x - 4\right )}}{x}\right )^{2}} \]

[In]

integrate((log(5/x/exp(4)/exp(x)^8)^3+2*x*log(5/x/exp(4)/exp(x)^8)+16*x^2+2*x)/(log(5)-3)/log(5/x/exp(4)/exp(x
)^8)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\frac {x^{2}}{\left (-3 + \log {\left (5 \right )}\right ) \log {\left (\frac {5 e^{- 8 x}}{x e^{4}} \right )}^{2}} + \frac {x}{-3 + \log {\left (5 \right )}} \]

[In]

integrate((ln(5/x/exp(4)/exp(x)**8)**3+2*x*ln(5/x/exp(4)/exp(x)**8)+16*x**2+2*x)/(ln(5)-3)/ln(5/x/exp(4)/exp(x
)**8)**3,x)

[Out]

x**2/((-3 + log(5))*log(5*exp(-4)*exp(-8*x)/x)**2) + x/(-3 + log(5))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.93 \[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\frac {x + \frac {x^{2}}{64 \, x^{2} - 16 \, x {\left (\log \left (5\right ) - 4\right )} + \log \left (5\right )^{2} + 2 \, {\left (8 \, x - \log \left (5\right ) + 4\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 8 \, \log \left (5\right ) + 16}}{\log \left (5\right ) - 3} \]

[In]

integrate((log(5/x/exp(4)/exp(x)^8)^3+2*x*log(5/x/exp(4)/exp(x)^8)+16*x^2+2*x)/(log(5)-3)/log(5/x/exp(4)/exp(x
)^8)^3,x, algorithm="maxima")

[Out]

(x + x^2/(64*x^2 - 16*x*(log(5) - 4) + log(5)^2 + 2*(8*x - log(5) + 4)*log(x) + log(x)^2 - 8*log(5) + 16))/(lo
g(5) - 3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (27) = 54\).

Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.68 \[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\frac {x + \frac {8 \, x^{3} + x^{2}}{512 \, x^{3} - 128 \, x^{2} \log \left (5\right ) + 8 \, x \log \left (5\right )^{2} + 128 \, x^{2} \log \left (x\right ) - 16 \, x \log \left (5\right ) \log \left (x\right ) + 8 \, x \log \left (x\right )^{2} + 576 \, x^{2} - 80 \, x \log \left (5\right ) + \log \left (5\right )^{2} + 80 \, x \log \left (x\right ) - 2 \, \log \left (5\right ) \log \left (x\right ) + \log \left (x\right )^{2} + 192 \, x - 8 \, \log \left (5\right ) + 8 \, \log \left (x\right ) + 16}}{\log \left (5\right ) - 3} \]

[In]

integrate((log(5/x/exp(4)/exp(x)^8)^3+2*x*log(5/x/exp(4)/exp(x)^8)+16*x^2+2*x)/(log(5)-3)/log(5/x/exp(4)/exp(x
)^8)^3,x, algorithm="giac")

[Out]

(x + (8*x^3 + x^2)/(512*x^3 - 128*x^2*log(5) + 8*x*log(5)^2 + 128*x^2*log(x) - 16*x*log(5)*log(x) + 8*x*log(x)
^2 + 576*x^2 - 80*x*log(5) + log(5)^2 + 80*x*log(x) - 2*log(5)*log(x) + log(x)^2 + 192*x - 8*log(5) + 8*log(x)
+ 16))/(log(5) - 3)

Mupad [B] (verification not implemented)

Time = 13.41 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {2 x+16 x^2+2 x \log \left (\frac {5 e^{-4-8 x}}{x}\right )+\log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )}{(-3+\log (5)) \log ^3\left (\frac {5 e^{-4-8 x}}{x}\right )} \, dx=\frac {x}{\ln \left (5\right )-3}+\frac {x^2}{{\ln \left (\frac {5\,{\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^{-4}}{x}\right )}^2\,\left (\ln \left (5\right )-3\right )} \]

[In]

int((2*x + 2*x*log((5*exp(-8*x)*exp(-4))/x) + log((5*exp(-8*x)*exp(-4))/x)^3 + 16*x^2)/(log((5*exp(-8*x)*exp(-
4))/x)^3*(log(5) - 3)),x)

[Out]

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