3.101.39 \(\int \frac {(-20+12 x+(4-2 x) \log (e^{8+x})) \log (\frac {1}{5} e^{-x} (-10 x^2+2 x^2 \log (e^{8+x})))}{-5 x+x \log (e^{8+x})} \, dx\)

Optimal. Leaf size=23 \[ \log ^2\left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right ) \]

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Rubi [F]  time = 1.16, 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 {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-20 + 12*x + (4 - 2*x)*Log[E^(8 + x)])*Log[(-10*x^2 + 2*x^2*Log[E^(8 + x)])/(5*E^x)])/(-5*x + x*Log[E^(8
+ x)]),x]

[Out]

12*Defer[Int][Log[(2*x^2*(-5 + Log[E^(8 + x)]))/(5*E^x)]/(-5 + Log[E^(8 + x)]), x] - 20*Defer[Int][Log[(2*x^2*
(-5 + Log[E^(8 + x)]))/(5*E^x)]/(x*(-5 + Log[E^(8 + x)])), x] - 2*Defer[Int][(Log[E^(8 + x)]*Log[(2*x^2*(-5 +
Log[E^(8 + x)]))/(5*E^x)])/(-5 + Log[E^(8 + x)]), x] + 4*Defer[Int][(Log[E^(8 + x)]*Log[(2*x^2*(-5 + Log[E^(8
+ x)]))/(5*E^x)])/(x*(-5 + Log[E^(8 + x)])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (20-12 x-(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{5 x-x \log \left (e^{8+x}\right )} \, dx\\ &=\int \left (\frac {12 \log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{-5+\log \left (e^{8+x}\right )}-\frac {20 \log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{x \left (-5+\log \left (e^{8+x}\right )\right )}-\frac {2 \log \left (e^{8+x}\right ) \log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{-5+\log \left (e^{8+x}\right )}+\frac {4 \log \left (e^{8+x}\right ) \log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{x \left (-5+\log \left (e^{8+x}\right )\right )}\right ) \, dx\\ &=-\left (2 \int \frac {\log \left (e^{8+x}\right ) \log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{-5+\log \left (e^{8+x}\right )} \, dx\right )+4 \int \frac {\log \left (e^{8+x}\right ) \log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{x \left (-5+\log \left (e^{8+x}\right )\right )} \, dx+12 \int \frac {\log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{-5+\log \left (e^{8+x}\right )} \, dx-20 \int \frac {\log \left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right )}{x \left (-5+\log \left (e^{8+x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 23, normalized size = 1.00 \begin {gather*} \log ^2\left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-20 + 12*x + (4 - 2*x)*Log[E^(8 + x)])*Log[(-10*x^2 + 2*x^2*Log[E^(8 + x)])/(5*E^x)])/(-5*x + x*Lo
g[E^(8 + x)]),x]

[Out]

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

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fricas [A]  time = 1.59, size = 18, normalized size = 0.78 \begin {gather*} \log \left (\frac {2}{5} \, {\left (x^{3} + 3 \, x^{2}\right )} e^{\left (-x\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-2*x)*log(exp(3)*exp(5+x))+12*x-20)*log(1/5*(2*x^2*log(exp(3)*exp(5+x))-10*x^2)/exp(x))/(x*log(ex
p(3)*exp(5+x))-5*x),x, algorithm="fricas")

[Out]

log(2/5*(x^3 + 3*x^2)*e^(-x))^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left ({\left (x - 2\right )} \log \left (e^{\left (x + 8\right )}\right ) - 6 \, x + 10\right )} \log \left (\frac {2}{5} \, {\left (x^{2} \log \left (e^{\left (x + 8\right )}\right ) - 5 \, x^{2}\right )} e^{\left (-x\right )}\right )}{x \log \left (e^{\left (x + 8\right )}\right ) - 5 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-2*x)*log(exp(3)*exp(5+x))+12*x-20)*log(1/5*(2*x^2*log(exp(3)*exp(5+x))-10*x^2)/exp(x))/(x*log(ex
p(3)*exp(5+x))-5*x),x, algorithm="giac")

[Out]

integrate(-2*((x - 2)*log(e^(x + 8)) - 6*x + 10)*log(2/5*(x^2*log(e^(x + 8)) - 5*x^2)*e^(-x))/(x*log(e^(x + 8)
) - 5*x), x)

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maple [C]  time = 0.09, size = 307, normalized size = 13.35

method	result	size
default	\(4 \ln \left (\frac {\left (2 x^{2} \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-10 x^{2}\right ) {\mathrm e}^{-x}}{5}\right ) \ln \relax (x )+2 \ln \left (\frac {\left (2 x^{2} \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-10 x^{2}\right ) {\mathrm e}^{-x}}{5}\right ) \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )-2 \ln \left (\frac {\left (2 x^{2} \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-10 x^{2}\right ) {\mathrm e}^{-x}}{5}\right ) x -x^{2}-2 \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x \right ) \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )+4 x \ln \relax (x )-4 \ln \relax (x )^{2}-4 \dilog \left (\frac {\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )-4 \ln \relax (x ) \ln \left (\frac {\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )+2 \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right ) \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )-2 \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )+10+2 x -\ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )^{2}-4 \dilog \left (-\frac {x}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )-4 \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right ) \ln \left (-\frac {x}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )\)	\(307\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4-2*x)*ln(exp(3)*exp(5+x))+12*x-20)*ln(1/5*(2*x^2*ln(exp(3)*exp(5+x))-10*x^2)/exp(x))/(x*ln(exp(3)*exp(5
+x))-5*x),x,method=_RETURNVERBOSE)

[Out]

4*ln(1/5*(2*x^2*ln(exp(3)*exp(5+x))-10*x^2)/exp(x))*ln(x)+2*ln(1/5*(2*x^2*ln(exp(3)*exp(5+x))-10*x^2)/exp(x))*
ln(ln(exp(3)*exp(5+x))-5)-2*ln(1/5*(2*x^2*ln(exp(3)*exp(5+x))-10*x^2)/exp(x))*x-x^2-2*(ln(exp(3)*exp(5+x))-5-x
)*ln(ln(exp(3)*exp(5+x))-5)+4*x*ln(x)-4*ln(x)^2-4*dilog((ln(exp(3)*exp(5+x))-5)/(ln(exp(3)*exp(5+x))-5-x))-4*l
n(x)*ln((ln(exp(3)*exp(5+x))-5)/(ln(exp(3)*exp(5+x))-5-x))+2*(ln(exp(3)*exp(5+x))-5)*ln(ln(exp(3)*exp(5+x))-5)
-2*ln(exp(3)*exp(5+x))+10+2*x-ln(ln(exp(3)*exp(5+x))-5)^2-4*dilog(-x/(ln(exp(3)*exp(5+x))-5-x))-4*ln(ln(exp(3)
*exp(5+x))-5)*ln(-x/(ln(exp(3)*exp(5+x))-5-x))

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maxima [B]  time = 0.39, size = 78, normalized size = 3.39 \begin {gather*} -x^{2} - 2 \, {\left (x - \log \left (x + 3\right ) - 2 \, \log \relax (x)\right )} \log \left (\frac {2}{5} \, {\left ({\left (x + 8\right )} x^{2} - 5 \, x^{2}\right )} e^{\left (-x\right )}\right ) + 2 \, {\left (x - 2 \, \log \relax (x) + 3\right )} \log \left (x + 3\right ) - \log \left (x + 3\right )^{2} + 4 \, x \log \relax (x) - 4 \, \log \relax (x)^{2} - 6 \, \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-2*x)*log(exp(3)*exp(5+x))+12*x-20)*log(1/5*(2*x^2*log(exp(3)*exp(5+x))-10*x^2)/exp(x))/(x*log(ex
p(3)*exp(5+x))-5*x),x, algorithm="maxima")

[Out]

-x^2 - 2*(x - log(x + 3) - 2*log(x))*log(2/5*((x + 8)*x^2 - 5*x^2)*e^(-x)) + 2*(x - 2*log(x) + 3)*log(x + 3) -
log(x + 3)^2 + 4*x*log(x) - 4*log(x)^2 - 6*log(x + 3)

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mupad [B]  time = 7.19, size = 21, normalized size = 0.91 \begin {gather*} {\left (x-\ln \left (\frac {2\,x^2\,\left (x+8\right )}{5}-2\,x^2\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(-x)*((2*x^2*log(exp(x + 5)*exp(3)))/5 - 2*x^2))*(log(exp(x + 5)*exp(3))*(2*x - 4) - 12*x + 20))/(
5*x - x*log(exp(x + 5)*exp(3))),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-2*x)*ln(exp(3)*exp(5+x))+12*x-20)*ln(1/5*(2*x**2*ln(exp(3)*exp(5+x))-10*x**2)/exp(x))/(x*ln(exp(
3)*exp(5+x))-5*x),x)

[Out]

Timed out

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