\(\int \frac {(2+e^x+x)^{\frac {1}{2 \log (\frac {1}{4} (-12+x))}} ((-12+e^x (-12+x)+x) \log (\frac {1}{4} (-12+x))+(-2-e^x-x) \log (2+e^x+x))}{(-48-20 x+2 x^2+e^x (-24+2 x)) \log ^2(\frac {1}{4} (-12+x))} \, dx\) [4866]

Optimal result

Integrand size = 91, antiderivative size = 21 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\left (2+e^x+x\right )^{\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \]

[Out]

exp(1/2*ln(exp(x)+2+x)/ln(1/4*x-3))

Rubi [F]

\[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx \]

[In]

Int[((2 + E^x + x)^(1/(2*Log[(-12 + x)/4]))*((-12 + E^x*(-12 + x) + x)*Log[(-12 + x)/4] + (-2 - E^x - x)*Log[2
+ E^x + x]))/((-48 - 20*x + 2*x^2 + E^x*(-24 + 2*x))*Log[(-12 + x)/4]^2),x]

[Out]

Defer[Int][(2 + E^x + x)^(-1 + 1/(2*Log[-3 + x/4]))/Log[-3 + x/4], x]/2 + Defer[Int][(E^x*(2 + E^x + x)^(-1 +
1/(2*Log[-3 + x/4])))/Log[-3 + x/4], x]/2 - Defer[Int][((2 + E^x + x)^(-1 + 1/(2*Log[-3 + x/4]))*Log[2 + E^x +
x])/Log[-3 + x/4]^2, x]/2 - 7*Defer[Int][((2 + E^x + x)^(-1 + 1/(2*Log[-3 + x/4]))*Log[2 + E^x + x])/((-12 +
x)*Log[-3 + x/4]^2), x] - Defer[Int][(E^x*(2 + E^x + x)^(-1 + 1/(2*Log[-3 + x/4]))*Log[2 + E^x + x])/((-12 + x
)*Log[-3 + x/4]^2), x]/2

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \left (-\left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )\right )-\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{2 (12-x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx \\ & = \frac {1}{2} \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \left (-\left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )\right )-\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{(12-x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx \\ & = \frac {1}{2} \int \left (\frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \left (-12 \log \left (-3+\frac {x}{4}\right )+x \log \left (-3+\frac {x}{4}\right )-\log \left (2+e^x+x\right )\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )}+\frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \left (-12 \log \left (-3+\frac {x}{4}\right )+x \log \left (-3+\frac {x}{4}\right )-2 \log \left (2+e^x+x\right )-x \log \left (2+e^x+x\right )\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \left (-12 \log \left (-3+\frac {x}{4}\right )+x \log \left (-3+\frac {x}{4}\right )-\log \left (2+e^x+x\right )\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx+\frac {1}{2} \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \left (-12 \log \left (-3+\frac {x}{4}\right )+x \log \left (-3+\frac {x}{4}\right )-2 \log \left (2+e^x+x\right )-x \log \left (2+e^x+x\right )\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx \\ & = \frac {1}{2} \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \left (-\left ((-12+x) \log \left (-3+\frac {x}{4}\right )\right )+(2+x) \log \left (2+e^x+x\right )\right )}{(12-x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx+\frac {1}{2} \int \left (\frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )}-\frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )} \, dx-\frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx+\frac {1}{2} \int \left (\frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )}-\frac {(2+x) \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )} \, dx+\frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )} \, dx-\frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx-\frac {1}{2} \int \frac {(2+x) \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx \\ & = \frac {1}{2} \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )} \, dx+\frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )} \, dx-\frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx-\frac {1}{2} \int \left (\frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{\log ^2\left (-3+\frac {x}{4}\right )}+\frac {14 \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )} \, dx+\frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}}}{\log \left (-3+\frac {x}{4}\right )} \, dx-\frac {1}{2} \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{\log ^2\left (-3+\frac {x}{4}\right )} \, dx-\frac {1}{2} \int \frac {e^x \left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx-7 \int \frac {\left (2+e^x+x\right )^{-1+\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \log \left (2+e^x+x\right )}{(-12+x) \log ^2\left (-3+\frac {x}{4}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\left (2+e^x+x\right )^{\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \]

[In]

Integrate[((2 + E^x + x)^(1/(2*Log[(-12 + x)/4]))*((-12 + E^x*(-12 + x) + x)*Log[(-12 + x)/4] + (-2 - E^x - x)
*Log[2 + E^x + x]))/((-48 - 20*x + 2*x^2 + E^x*(-24 + 2*x))*Log[(-12 + x)/4]^2),x]

[Out]

(2 + E^x + x)^(1/(2*Log[-3 + x/4]))

Maple [A] (verified)

Time = 26.41 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\ln \left ({\mathrm e}^{x}+2+x \right )}{2 \ln \left (\frac {x}{4}-3\right )}}\)	\(18\)
risch	\(\left ({\mathrm e}^{x}+2+x \right )^{\frac {1}{-4 \ln \left (2\right )+2 \ln \left (x -12\right )}}\)	\(20\)

[In]

int(((-exp(x)-x-2)*ln(exp(x)+2+x)+((x-12)*exp(x)+x-12)*ln(1/4*x-3))*exp(1/2*ln(exp(x)+2+x)/ln(1/4*x-3))/((2*x-
24)*exp(x)+2*x^2-20*x-48)/ln(1/4*x-3)^2,x,method=_RETURNVERBOSE)

[Out]

exp(1/2*ln(exp(x)+2+x)/ln(1/4*x-3))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx={\left (x + e^{x} + 2\right )}^{\frac {1}{2 \, \log \left (\frac {1}{4} \, x - 3\right )}} \]

[In]

integrate(((-exp(x)-x-2)*log(exp(x)+2+x)+((x-12)*exp(x)+x-12)*log(1/4*x-3))*exp(1/2*log(exp(x)+2+x)/log(1/4*x-
3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/log(1/4*x-3)^2,x, algorithm="fricas")

[Out]

(x + e^x + 2)^(1/2/log(1/4*x - 3))

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\text {Timed out} \]

[In]

integrate(((-exp(x)-x-2)*ln(exp(x)+2+x)+((x-12)*exp(x)+x-12)*ln(1/4*x-3))*exp(1/2*ln(exp(x)+2+x)/ln(1/4*x-3))/
((2*x-24)*exp(x)+2*x**2-20*x-48)/ln(1/4*x-3)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\frac {1}{{\left (x + e^{x} + 2\right )}^{\frac {1}{2 \, {\left (2 \, \log \left (2\right ) - \log \left (x - 12\right )\right )}}}} \]

[In]

integrate(((-exp(x)-x-2)*log(exp(x)+2+x)+((x-12)*exp(x)+x-12)*log(1/4*x-3))*exp(1/2*log(exp(x)+2+x)/log(1/4*x-
3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/log(1/4*x-3)^2,x, algorithm="maxima")

[Out]

1/((x + e^x + 2)^(1/2/(2*log(2) - log(x - 12))))

Giac [F]

\[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\int { -\frac {{\left ({\left (x + e^{x} + 2\right )} \log \left (x + e^{x} + 2\right ) - {\left ({\left (x - 12\right )} e^{x} + x - 12\right )} \log \left (\frac {1}{4} \, x - 3\right )\right )} {\left (x + e^{x} + 2\right )}^{\frac {1}{2 \, \log \left (\frac {1}{4} \, x - 3\right )}}}{2 \, {\left (x^{2} + {\left (x - 12\right )} e^{x} - 10 \, x - 24\right )} \log \left (\frac {1}{4} \, x - 3\right )^{2}} \,d x } \]

[In]

integrate(((-exp(x)-x-2)*log(exp(x)+2+x)+((x-12)*exp(x)+x-12)*log(1/4*x-3))*exp(1/2*log(exp(x)+2+x)/log(1/4*x-
3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/log(1/4*x-3)^2,x, algorithm="giac")

[Out]

integrate(-1/2*((x + e^x + 2)*log(x + e^x + 2) - ((x - 12)*e^x + x - 12)*log(1/4*x - 3))*(x + e^x + 2)^(1/2/lo
g(1/4*x - 3))/((x^2 + (x - 12)*e^x - 10*x - 24)*log(1/4*x - 3)^2), x)

Mupad [B] (verification not implemented)

Time = 11.73 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx={\mathrm {e}}^{\frac {\ln \left (x+{\mathrm {e}}^x+2\right )}{2\,\ln \left (\frac {x}{4}-3\right )}} \]

[In]

int((exp(log(x + exp(x) + 2)/(2*log(x/4 - 3)))*(log(x + exp(x) + 2)*(x + exp(x) + 2) - log(x/4 - 3)*(x + exp(x
)*(x - 12) - 12)))/(log(x/4 - 3)^2*(20*x - exp(x)*(2*x - 24) - 2*x^2 + 48)),x)

[Out]

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