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

Optimal. Leaf size=28 \[ \frac {x+\frac {x^2}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )}}{-3+\log (5)} \]

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Rubi [F]  time = 0.28, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 28, normalized size = 1.00 \begin {gather*} \frac {x+\frac {x^2}{\log ^2\left (\frac {5 e^{-4-8 x}}{x}\right )}}{-3+\log (5)} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.63, size = 41, normalized size = 1.46 \begin {gather*} \frac {x \log \left (\frac {5 \, e^{\left (-8 \, x - 4\right )}}{x}\right )^{2} + x^{2}}{{\left (\log \relax (5) - 3\right )} \log \left (\frac {5 \, e^{\left (-8 \, x - 4\right )}}{x}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.21, size = 103, normalized size = 3.68 \begin {gather*} \frac {x + \frac {8 \, x^{3} + x^{2}}{512 \, x^{3} - 128 \, x^{2} \log \relax (5) + 8 \, x \log \relax (5)^{2} + 128 \, x^{2} \log \relax (x) - 16 \, x \log \relax (5) \log \relax (x) + 8 \, x \log \relax (x)^{2} + 576 \, x^{2} - 80 \, x \log \relax (5) + \log \relax (5)^{2} + 80 \, x \log \relax (x) - 2 \, \log \relax (5) \log \relax (x) + \log \relax (x)^{2} + 192 \, x - 8 \, \log \relax (5) + 8 \, \log \relax (x) + 16}}{\log \relax (5) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 10.36, size = 169, normalized size = 6.04

method	result	size
default	\(\frac {x \ln \relax (x )^{2}+\left (1-16 \ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )-16 \ln \relax (x )-128 x \right ) x^{2}+\left (\left (\ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )+\ln \relax (x )+8 \ln \left ({\mathrm e}^{x}\right )\right )^{2}-16 \left (\ln \left ({\mathrm e}^{x}\right )-x \right ) \left (\ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )+\ln \relax (x )+8 \ln \left ({\mathrm e}^{x}\right )\right )+64 \left (\ln \left ({\mathrm e}^{x}\right )-x \right )^{2}\right ) x +\left (-2 \ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )-2 \ln \relax (x )-16 x \right ) x \ln \relax (x )+64 x^{3}+16 x^{2} \ln \relax (x )}{\left (\ln \relax (5)-3\right ) \ln \left (\frac {5 \,{\mathrm e}^{-4} {\mathrm e}^{-8 x}}{x}\right )^{2}}\)	\(169\)
risch	\(\frac {x}{\ln \relax (5)-3}-\frac {4 x^{2}}{\left (\ln \relax (5)-3\right ) \left (2 i \ln \relax (x )-2 i \ln \relax (5)+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{6 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{7 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{5 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-8 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-8 x}}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{7 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{8 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{3 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{3 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{6 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+\pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-\pi \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\pi \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{3 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{5 x}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-8 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-8 x}}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-8 x}}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{6 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{7 x}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{7 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{7 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{8 x}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{8 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\pi \mathrm {csgn}\left (i {\mathrm e}^{5 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{6 x}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{6 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+8 i+16 i \ln \left ({\mathrm e}^{x}\right )+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{5 x}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{6 x}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{7 x}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{8 x}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right )^{3}-\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-8 x}}{x}\right )^{3}\right )^{2}}\)	\(637\)

Verification of antiderivative is not currently implemented for this CAS.

[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,m
ethod=_RETURNVERBOSE)

[Out]

1/(ln(5)-3)*(x*ln(x)^2+(1-16*ln(5/x/exp(4)/exp(x)^8)-16*ln(x)-128*x)*x^2+((ln(5/x/exp(4)/exp(x)^8)+ln(x)+8*ln(
exp(x)))^2-16*(ln(exp(x))-x)*(ln(5/x/exp(4)/exp(x)^8)+ln(x)+8*ln(exp(x)))+64*(ln(exp(x))-x)^2)*x+(-2*ln(5/x/ex
p(4)/exp(x)^8)-2*ln(x)-16*x)*x*ln(x)+64*x^3+16*x^2*ln(x))/ln(5/x/exp(4)/exp(x)^8)^2

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maxima [A]  time = 0.50, size = 54, normalized size = 1.93 \begin {gather*} \frac {x + \frac {x^{2}}{64 \, x^{2} - 16 \, x {\left (\log \relax (5) - 4\right )} + \log \relax (5)^{2} + 2 \, {\left (8 \, x - \log \relax (5) + 4\right )} \log \relax (x) + \log \relax (x)^{2} - 8 \, \log \relax (5) + 16}}{\log \relax (5) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.45, size = 33, normalized size = 1.18 \begin {gather*} \frac {x}{\ln \relax (5)-3}+\frac {x^2}{{\ln \left (\frac {5\,{\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^{-4}}{x}\right )}^2\,\left (\ln \relax (5)-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.23, size = 29, normalized size = 1.04 \begin {gather*} \frac {x^{2}}{\left (-3 + \log {\relax (5 )}\right ) \log {\left (\frac {5 e^{- 8 x}}{x e^{4}} \right )}^{2}} + \frac {x}{-3 + \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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