\(\int \frac {-4 x-4 e^x x+(4 e^x+4 x+4 \log (4)) \log (e^x+x+\log (4))+(8 e^x+8 x+8 \log (4)) \log ^2(e^x+x+\log (4))}{(e^x+x+\log (4)) \log ^2(e^x+x+\log (4))} \, dx\) [7180]

Optimal result

Integrand size = 76, antiderivative size = 19 \[ \int \frac {-4 x-4 e^x x+\left (4 e^x+4 x+4 \log (4)\right ) \log \left (e^x+x+\log (4)\right )+\left (8 e^x+8 x+8 \log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )}{\left (e^x+x+\log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )} \, dx=4 \left (5+2 x+\frac {x}{\log \left (e^x+x+\log (4)\right )}\right ) \]

[Out]

20+4*x/ln(exp(x)+x+2*ln(2))+8*x

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-4 x-4 e^x x+\left (4 e^x+4 x+4 \log (4)\right ) \log \left (e^x+x+\log (4)\right )+\left (8 e^x+8 x+8 \log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )}{\left (e^x+x+\log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )} \, dx=8 x+\frac {4 x}{\log \left (e^x+x+\log (4)\right )} \]

[In]

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

[Out]

8*x + (4*x)/Log[E^x + x + Log[4]]

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method	result	size
risch	\(8 x +\frac {4 x}{\ln \left ({\mathrm e}^{x}+x +2 \ln \left (2\right )\right )}\)	\(19\)
norman	\(\frac {4 x +8 \ln \left ({\mathrm e}^{x}+x +2 \ln \left (2\right )\right ) x}{\ln \left ({\mathrm e}^{x}+x +2 \ln \left (2\right )\right )}\)	\(29\)
parallelrisch	\(\frac {4 x +8 \ln \left ({\mathrm e}^{x}+x +2 \ln \left (2\right )\right ) x}{\ln \left ({\mathrm e}^{x}+x +2 \ln \left (2\right )\right )}\)	\(29\)

[In]

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

[Out]

8*x+4*x/ln(exp(x)+x+2*ln(2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-4 x-4 e^x x+\left (4 e^x+4 x+4 \log (4)\right ) \log \left (e^x+x+\log (4)\right )+\left (8 e^x+8 x+8 \log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )}{\left (e^x+x+\log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )} \, dx=\frac {4 \, {\left (2 \, x \log \left (x + e^{x} + 2 \, \log \left (2\right )\right ) + x\right )}}{\log \left (x + e^{x} + 2 \, \log \left (2\right )\right )} \]

[In]

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

[Out]

4*(2*x*log(x + e^x + 2*log(2)) + x)/log(x + e^x + 2*log(2))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-4 x-4 e^x x+\left (4 e^x+4 x+4 \log (4)\right ) \log \left (e^x+x+\log (4)\right )+\left (8 e^x+8 x+8 \log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )}{\left (e^x+x+\log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )} \, dx=8 x + \frac {4 x}{\log {\left (x + e^{x} + 2 \log {\left (2 \right )} \right )}} \]

[In]

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

[Out]

8*x + 4*x/log(x + exp(x) + 2*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-4 x-4 e^x x+\left (4 e^x+4 x+4 \log (4)\right ) \log \left (e^x+x+\log (4)\right )+\left (8 e^x+8 x+8 \log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )}{\left (e^x+x+\log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )} \, dx=\frac {4 \, {\left (2 \, x \log \left (x + e^{x} + 2 \, \log \left (2\right )\right ) + x\right )}}{\log \left (x + e^{x} + 2 \, \log \left (2\right )\right )} \]

[In]

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

[Out]

4*(2*x*log(x + e^x + 2*log(2)) + x)/log(x + e^x + 2*log(2))

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-4 x-4 e^x x+\left (4 e^x+4 x+4 \log (4)\right ) \log \left (e^x+x+\log (4)\right )+\left (8 e^x+8 x+8 \log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )}{\left (e^x+x+\log (4)\right ) \log ^2\left (e^x+x+\log (4)\right )} \, dx=\frac {4 \, {\left (2 \, x \log \left (x + e^{x} + 2 \, \log \left (2\right )\right ) + x\right )}}{\log \left (x + e^{x} + 2 \, \log \left (2\right )\right )} \]

[In]

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

[Out]

4*(2*x*log(x + e^x + 2*log(2)) + x)/log(x + e^x + 2*log(2))

Mupad [F(-1)]

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

[In]

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

[Out]

\text{Hanged}