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

Optimal. Leaf size=23 \[ \left (\frac {x \left (-5+e^x+x+e^x \log (4)\right )}{\log (x \log (4))}\right )^x \]

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Rubi [F]  time = 3.72, 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 (\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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 22, normalized size = 0.96 \begin {gather*} \left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right )^x \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.94, size = 28, normalized size = 1.22 \begin {gather*} \left (\frac {x^{2} + {\left (2 \, x \log \relax (2) + x\right )} e^{x} - 5 \, x}{\log \left (2 \, x \log \relax (2)\right )}\right )^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)*(x+1)+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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left ({\left (2 \, \log \relax (2) + 1\right )} e^{x} + x - 5\right )} \log \left (2 \, x \log \relax (2)\right ) \log \left (\frac {x^{2} + {\left (2 \, x \log \relax (2) + x\right )} e^{x} - 5 \, x}{\log \left (2 \, x \log \relax (2)\right )}\right ) - {\left (2 \, \log \relax (2) + 1\right )} e^{x} + {\left ({\left (2 \, {\left (x + 1\right )} \log \relax (2) + x + 1\right )} e^{x} + 2 \, x - 5\right )} \log \left (2 \, x \log \relax (2)\right ) - x + 5\right )} \left (\frac {x^{2} + {\left (2 \, x \log \relax (2) + x\right )} e^{x} - 5 \, x}{\log \left (2 \, x \log \relax (2)\right )}\right )^{x}}{{\left ({\left (2 \, \log \relax (2) + 1\right )} e^{x} + x - 5\right )} \log \left (2 \, x \log \relax (2)\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)*(x+1)+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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (1+2 \ln \relax (2)\right ) {\mathrm e}^{x}+x -5\right ) \ln \left (2 x \ln \relax (2)\right ) \ln \left (\frac {\left (x +2 x \ln \relax (2)\right ) {\mathrm e}^{x}+x^{2}-5 x}{\ln \left (2 x \ln \relax (2)\right )}\right )+\left (\left (2 \ln \relax (2) \left (x +1\right )+x +1\right ) {\mathrm e}^{x}+2 x -5\right ) \ln \left (2 x \ln \relax (2)\right )+\left (-1-2 \ln \relax (2)\right ) {\mathrm e}^{x}+5-x \right ) {\mathrm e}^{x \ln \left (\frac {\left (x +2 x \ln \relax (2)\right ) {\mathrm e}^{x}+x^{2}-5 x}{\ln \left (2 x \ln \relax (2)\right )}\right )}}{\left (\left (1+2 \ln \relax (2)\right ) {\mathrm e}^{x}+x -5\right ) \ln \left (2 x \ln \relax (2)\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)*(x+1)+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]

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)*(x+1)+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)

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maxima [A]  time = 0.58, size = 33, normalized size = 1.43 \begin {gather*} e^{\left (x \log \left ({\left (2 \, \log \relax (2) + 1\right )} e^{x} + x - 5\right ) + x \log \relax (x) - x \log \left (\log \relax (2) + \log \relax (x) + \log \left (\log \relax (2)\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)*(x+1)+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))))

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mupad [B]  time = 5.51, size = 31, normalized size = 1.35 \begin {gather*} {\left (\frac {x\,{\mathrm {e}}^x-5\,x+x^2+2\,x\,{\mathrm {e}}^x\,\ln \relax (2)}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )+\ln \relax (x)}\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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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((((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)*
(x+1)+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

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