\(\int \frac {(-20+12 x+(4-2 x) \log (e^{8+x})) \log (\frac {1}{5} e^{-x} (-10 x^2+2 x^2 \log (e^{8+x})))}{-5 x+x \log (e^{8+x})} \, dx\) [10039]

Optimal result

Integrand size = 59, antiderivative size = 23 \[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx=\log ^2\left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx=\int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx=\log ^2\left (\frac {2}{5} e^{-x} x^2 \left (-5+\log \left (e^{8+x}\right )\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\({\ln \left (\frac {2 \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right ) x^{2} {\mathrm e}^{-x}}{5}\right )}^{2}\)	\(23\)
default	\(4 \ln \left (\frac {\left (2 x^{2} \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-10 x^{2}\right ) {\mathrm e}^{-x}}{5}\right ) \ln \left (x \right )+2 \ln \left (\frac {\left (2 x^{2} \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-10 x^{2}\right ) {\mathrm e}^{-x}}{5}\right ) \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )-2 \ln \left (\frac {\left (2 x^{2} \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-10 x^{2}\right ) {\mathrm e}^{-x}}{5}\right ) x -x^{2}-2 \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x \right ) \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )+4 x \ln \left (x \right )-4 \ln \left (x \right )^{2}-4 \operatorname {dilog}\left (\frac {\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )-4 \ln \left (x \right ) \ln \left (\frac {\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )+2 \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right ) \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )-2 \ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )+10+2 x -{\ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right )}^{2}-4 \operatorname {dilog}\left (-\frac {x}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )-4 \ln \left (\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5\right ) \ln \left (-\frac {x}{\ln \left ({\mathrm e}^{3} {\mathrm e}^{5+x}\right )-5-x}\right )\)	\(307\)

[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]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx=\log \left (\frac {2}{5} \, {\left (x^{3} + 3 \, x^{2}\right )} e^{\left (-x\right )}\right )^{2} \]

[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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx=\text {Timed out} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (19) = 38\).

Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.39 \[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx=-x^{2} - 2 \, {\left (x - \log \left (x + 3\right ) - 2 \, \log \left (x\right )\right )} \log \left (\frac {2}{5} \, {\left ({\left (x + 8\right )} x^{2} - 5 \, x^{2}\right )} e^{\left (-x\right )}\right ) + 2 \, {\left (x - 2 \, \log \left (x\right ) + 3\right )} \log \left (x + 3\right ) - \log \left (x + 3\right )^{2} + 4 \, x \log \left (x\right ) - 4 \, \log \left (x\right )^{2} - 6 \, \log \left (x + 3\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx=\log \left (\frac {2}{5} \, x^{3} e^{\left (-x\right )} + \frac {6}{5} \, x^{2} e^{\left (-x\right )}\right )^{2} \]

[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]

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

Mupad [B] (verification not implemented)

Time = 18.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (-20+12 x+(4-2 x) \log \left (e^{8+x}\right )\right ) \log \left (\frac {1}{5} e^{-x} \left (-10 x^2+2 x^2 \log \left (e^{8+x}\right )\right )\right )}{-5 x+x \log \left (e^{8+x}\right )} \, dx={\left (x-\ln \left (\frac {2\,x^2\,\left (x+8\right )}{5}-2\,x^2\right )\right )}^2 \]

[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