\(\int \frac {(\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))})^x (5-x+e^x (-1-\log (4))+(-5+2 x+e^x (1+x+(1+x) \log (4))) \log (x \log (4))+(-5+x+e^x (1+\log (4))) \log (x \log (4)) \log (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}))}{(-5+x+e^x (1+\log (4))) \log (x \log (4))} \, dx\) [7440]

Optimal result

Integrand size = 130, antiderivative size = 23 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\left (\frac {x \left (-5+e^x+x+e^x \log (4)\right )}{\log (x \log (4))}\right )^x \]

[Out]

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

Rubi [F]

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

[In]

Int[(((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x*(5 - x + E^x*(-1 - Log[4]) + (-5 + 2*x + E^x*(1 + x +
(1 + x)*Log[4]))*Log[x*Log[4]] + (-5 + x + E^x*(1 + Log[4]))*Log[x*Log[4]]*Log[(-5*x + x^2 + E^x*(x + x*Log[4
]))/Log[x*Log[4]]]))/((-5 + x + E^x*(1 + Log[4]))*Log[x*Log[4]]),x]

[Out]

Defer[Int][((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x, x] + Defer[Int][x*((-5*x + x^2 + E^x*(x + x*Lo
g[4]))/Log[x*Log[4]])^x, x] + Defer[Int][(x^2*((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x)/(5 - x - E^
x*(1 + Log[4])), x] + 6*Defer[Int][(x*((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x)/(-5 + x + E^x*(1 +
Log[4])), x] - Defer[Int][((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x/Log[x*Log[4]], x] + Defer[Int][(
(-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x*Log[(x*(-5 + x + E^x*(1 + Log[4])))/Log[x*Log[4]]], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-6+x) x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{5-x-e^x (1+\log (4))}+\frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (-1+\log (x \log (4))+x \log (x \log (4))+\log (x \log (4)) \log \left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right )\right )}{\log (x \log (4))}\right ) \, dx \\ & = \int \frac {(-6+x) x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{5-x-e^x (1+\log (4))} \, dx+\int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (-1+\log (x \log (4))+x \log (x \log (4))+\log (x \log (4)) \log \left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right )\right )}{\log (x \log (4))} \, dx \\ & = \int \left (\frac {x^2 \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{5-x-e^x (1+\log (4))}+\frac {6 x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{-5+x+e^x (1+\log (4))}\right ) \, dx+\int \left (\frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x (-1+\log (x \log (4))+x \log (x \log (4)))}{\log (x \log (4))}+\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \log \left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right )\right ) \, dx \\ & = 6 \int \frac {x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{-5+x+e^x (1+\log (4))} \, dx+\int \frac {x^2 \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{5-x-e^x (1+\log (4))} \, dx+\int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x (-1+\log (x \log (4))+x \log (x \log (4)))}{\log (x \log (4))} \, dx+\int \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \log \left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right ) \, dx \\ & = 6 \int \frac {x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{-5+x+e^x (1+\log (4))} \, dx+\int \left (\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x+x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x-\frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{\log (x \log (4))}\right ) \, dx+\int \frac {x^2 \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{5-x-e^x (1+\log (4))} \, dx+\int \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \log \left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right ) \, dx \\ & = 6 \int \frac {x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{-5+x+e^x (1+\log (4))} \, dx+\int \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \, dx+\int x \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \, dx+\int \frac {x^2 \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{5-x-e^x (1+\log (4))} \, dx-\int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{\log (x \log (4))} \, dx+\int \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \log \left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right ) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right )^x \]

[In]

Integrate[(((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x*(5 - x + E^x*(-1 - Log[4]) + (-5 + 2*x + E^x*(1
+ x + (1 + x)*Log[4]))*Log[x*Log[4]] + (-5 + x + E^x*(1 + Log[4]))*Log[x*Log[4]]*Log[(-5*x + x^2 + E^x*(x + x
*Log[4]))/Log[x*Log[4]]]))/((-5 + x + E^x*(1 + Log[4]))*Log[x*Log[4]]),x]

[Out]

((x*(-5 + x + E^x*(1 + Log[4])))/Log[x*Log[4]])^x

Maple [A] (verified)

Time = 113.98 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13

method	result	size
parallelrisch	\({\mathrm e}^{\ln \left (\frac {x \left ({\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (2\right )-5+x \right )}{\ln \left (2 x \ln \left (2\right )\right )}\right ) x}\)	\(26\)

[In]

int((((1+2*ln(2))*exp(x)+x-5)*ln(2*x*ln(2))*ln(((x+2*x*ln(2))*exp(x)+x^2-5*x)/ln(2*x*ln(2)))+((2*ln(2)*(1+x)+x
+1)*exp(x)+2*x-5)*ln(2*x*ln(2))+(-1-2*ln(2))*exp(x)+5-x)*exp(x*ln(((x+2*x*ln(2))*exp(x)+x^2-5*x)/ln(2*x*ln(2))
))/((1+2*ln(2))*exp(x)+x-5)/ln(2*x*ln(2)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\left (\frac {x^{2} + {\left (2 \, x \log \left (2\right ) + x\right )} e^{x} - 5 \, x}{\log \left (2 \, x \log \left (2\right )\right )}\right )^{x} \]

[In]

integrate((((1+2*log(2))*exp(x)+x-5)*log(2*x*log(2))*log(((x+2*x*log(2))*exp(x)+x^2-5*x)/log(2*x*log(2)))+((2*
log(2)*(1+x)+x+1)*exp(x)+2*x-5)*log(2*x*log(2))+(-1-2*log(2))*exp(x)+5-x)*exp(x*log(((x+2*x*log(2))*exp(x)+x^2
-5*x)/log(2*x*log(2))))/((1+2*log(2))*exp(x)+x-5)/log(2*x*log(2)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

integrate((((1+2*ln(2))*exp(x)+x-5)*ln(2*x*ln(2))*ln(((x+2*x*ln(2))*exp(x)+x**2-5*x)/ln(2*x*ln(2)))+((2*ln(2)*
(1+x)+x+1)*exp(x)+2*x-5)*ln(2*x*ln(2))+(-1-2*ln(2))*exp(x)+5-x)*exp(x*ln(((x+2*x*ln(2))*exp(x)+x**2-5*x)/ln(2*
x*ln(2))))/((1+2*ln(2))*exp(x)+x-5)/ln(2*x*ln(2)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=e^{\left (x \log \left ({\left (2 \, \log \left (2\right ) + 1\right )} e^{x} + x - 5\right ) + x \log \left (x\right ) - x \log \left (\log \left (2\right ) + \log \left (x\right ) + \log \left (\log \left (2\right )\right )\right )\right )} \]

[In]

integrate((((1+2*log(2))*exp(x)+x-5)*log(2*x*log(2))*log(((x+2*x*log(2))*exp(x)+x^2-5*x)/log(2*x*log(2)))+((2*
log(2)*(1+x)+x+1)*exp(x)+2*x-5)*log(2*x*log(2))+(-1-2*log(2))*exp(x)+5-x)*exp(x*log(((x+2*x*log(2))*exp(x)+x^2
-5*x)/log(2*x*log(2))))/((1+2*log(2))*exp(x)+x-5)/log(2*x*log(2)),x, algorithm="maxima")

[Out]

e^(x*log((2*log(2) + 1)*e^x + x - 5) + x*log(x) - x*log(log(2) + log(x) + log(log(2))))

Giac [F]

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

[In]

integrate((((1+2*log(2))*exp(x)+x-5)*log(2*x*log(2))*log(((x+2*x*log(2))*exp(x)+x^2-5*x)/log(2*x*log(2)))+((2*
log(2)*(1+x)+x+1)*exp(x)+2*x-5)*log(2*x*log(2))+(-1-2*log(2))*exp(x)+5-x)*exp(x*log(((x+2*x*log(2))*exp(x)+x^2
-5*x)/log(2*x*log(2))))/((1+2*log(2))*exp(x)+x-5)/log(2*x*log(2)),x, algorithm="giac")

[Out]

integrate((((2*log(2) + 1)*e^x + x - 5)*log(2*x*log(2))*log((x^2 + (2*x*log(2) + x)*e^x - 5*x)/log(2*x*log(2))
) - (2*log(2) + 1)*e^x + ((2*(x + 1)*log(2) + x + 1)*e^x + 2*x - 5)*log(2*x*log(2)) - x + 5)*((x^2 + (2*x*log(
2) + x)*e^x - 5*x)/log(2*x*log(2)))^x/(((2*log(2) + 1)*e^x + x - 5)*log(2*x*log(2))), x)

Mupad [B] (verification not implemented)

Time = 12.90 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx={\left (\frac {x\,{\mathrm {e}}^x-5\,x+x^2+2\,x\,{\mathrm {e}}^x\,\ln \left (2\right )}{\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )}\right )}^x \]

[In]

int((exp(x*log((exp(x)*(x + 2*x*log(2)) - 5*x + x^2)/log(2*x*log(2))))*(log(2*x*log(2))*(2*x + exp(x)*(x + 2*l
og(2)*(x + 1) + 1) - 5) - exp(x)*(2*log(2) + 1) - x + log((exp(x)*(x + 2*x*log(2)) - 5*x + x^2)/log(2*x*log(2)
))*log(2*x*log(2))*(x + exp(x)*(2*log(2) + 1) - 5) + 5))/(log(2*x*log(2))*(x + exp(x)*(2*log(2) + 1) - 5)),x)

[Out]

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