3.75.39 \(\int \frac {e^{\frac {-x \log (9)+(-5+e^{5 e^4 x} x) \log (x)}{\log (x)}} (\log (9)-\log (9) \log (x)+e^{5 e^4 x} (1+5 e^4 x) \log ^2(x))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ e^{-5+e^{5 e^4 x} x-\frac {x \log (9)}{\log (x)}} \]

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Rubi [F]  time = 1.58, 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 {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right ) \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-(x*Log[9]) + (-5 + E^(5*E^4*x)*x)*Log[x])/Log[x])*(Log[9] - Log[9]*Log[x] + E^(5*E^4*x)*(1 + 5*E^4*x
)*Log[x]^2))/Log[x]^2,x]

[Out]

Defer[Int][E^((-(x*Log[9]) - 5*Log[x] + 5*E^4*x*Log[x] + E^(5*E^4*x)*x*Log[x])/Log[x]), x] + 5*Defer[Int][E^((
-(x*Log[9]) - Log[x] + 5*E^4*x*Log[x] + E^(5*E^4*x)*x*Log[x])/Log[x])*x, x] + Log[9]*Defer[Int][E^((-(x*Log[9]
) + (-5 + E^(5*E^4*x)*x)*Log[x])/Log[x])/Log[x]^2, x] - Log[9]*Defer[Int][E^((-(x*Log[9]) + (-5 + E^(5*E^4*x)*
x)*Log[x])/Log[x])/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\exp \left (5 e^4 x+\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right ) \left (1+5 e^4 x\right )-\frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right ) \log (9) (-1+\log (x))}{\log ^2(x)}\right ) \, dx\\ &=-\left (\log (9) \int \frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right ) (-1+\log (x))}{\log ^2(x)} \, dx\right )+\int \exp \left (5 e^4 x+\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right ) \left (1+5 e^4 x\right ) \, dx\\ &=-\left (\log (9) \int \left (-\frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log ^2(x)}+\frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log (x)}\right ) \, dx\right )+\int \exp \left (\frac {-x \log (9)-5 \log (x)+5 e^4 x \log (x)+e^{5 e^4 x} x \log (x)}{\log (x)}\right ) \left (1+5 e^4 x\right ) \, dx\\ &=\log (9) \int \frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log ^2(x)} \, dx-\log (9) \int \frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log (x)} \, dx+\int \left (\exp \left (\frac {-x \log (9)-5 \log (x)+5 e^4 x \log (x)+e^{5 e^4 x} x \log (x)}{\log (x)}\right )+5 \exp \left (4+\frac {-x \log (9)-5 \log (x)+5 e^4 x \log (x)+e^{5 e^4 x} x \log (x)}{\log (x)}\right ) x\right ) \, dx\\ &=5 \int \exp \left (4+\frac {-x \log (9)-5 \log (x)+5 e^4 x \log (x)+e^{5 e^4 x} x \log (x)}{\log (x)}\right ) x \, dx+\log (9) \int \frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log ^2(x)} \, dx-\log (9) \int \frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log (x)} \, dx+\int \exp \left (\frac {-x \log (9)-5 \log (x)+5 e^4 x \log (x)+e^{5 e^4 x} x \log (x)}{\log (x)}\right ) \, dx\\ &=5 \int \exp \left (\frac {-x \log (9)-\log (x)+5 e^4 x \log (x)+e^{5 e^4 x} x \log (x)}{\log (x)}\right ) x \, dx+\log (9) \int \frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log ^2(x)} \, dx-\log (9) \int \frac {\exp \left (\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}\right )}{\log (x)} \, dx+\int \exp \left (\frac {-x \log (9)-5 \log (x)+5 e^4 x \log (x)+e^{5 e^4 x} x \log (x)}{\log (x)}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 24, normalized size = 1.04 \begin {gather*} 9^{-\frac {x}{\log (x)}} e^{-5+e^{5 e^4 x} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-(x*Log[9]) + (-5 + E^(5*E^4*x)*x)*Log[x])/Log[x])*(Log[9] - Log[9]*Log[x] + E^(5*E^4*x)*(1 + 5
*E^4*x)*Log[x]^2))/Log[x]^2,x]

[Out]

E^(-5 + E^(5*E^4*x)*x)/9^(x/Log[x])

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fricas [A]  time = 1.11, size = 27, normalized size = 1.17 \begin {gather*} e^{\left (-\frac {2 \, x \log \relax (3) - {\left (x e^{\left (5 \, x e^{4}\right )} - 5\right )} \log \relax (x)}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*log(x)^2-2*log(3)*log(x)+2*log(3))*exp(((x*exp(5*x*exp(4))-5)*log(x)
-2*x*log(3))/log(x))/log(x)^2,x, algorithm="fricas")

[Out]

e^(-(2*x*log(3) - (x*e^(5*x*e^4) - 5)*log(x))/log(x))

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giac [A]  time = 0.23, size = 20, normalized size = 0.87 \begin {gather*} e^{\left (x e^{\left (5 \, x e^{4}\right )} - \frac {2 \, x \log \relax (3)}{\log \relax (x)} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*log(x)^2-2*log(3)*log(x)+2*log(3))*exp(((x*exp(5*x*exp(4))-5)*log(x)
-2*x*log(3))/log(x))/log(x)^2,x, algorithm="giac")

[Out]

e^(x*e^(5*x*e^4) - 2*x*log(3)/log(x) - 5)

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maple [A]  time = 0.05, size = 21, normalized size = 0.91

method	result	size
risch	\(\left (\frac {1}{9}\right )^{\frac {x}{\ln \relax (x )}} {\mathrm e}^{x \,{\mathrm e}^{5 x \,{\mathrm e}^{4}}-5}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x*exp(4)+1)*exp(5*x*exp(4))*ln(x)^2-2*ln(3)*ln(x)+2*ln(3))*exp(((x*exp(5*x*exp(4))-5)*ln(x)-2*x*ln(3))
/ln(x))/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

(1/9)^(x/ln(x))*exp(x*exp(5*x*exp(4))-5)

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maxima [A]  time = 0.56, size = 20, normalized size = 0.87 \begin {gather*} e^{\left (x e^{\left (5 \, x e^{4}\right )} - \frac {2 \, x \log \relax (3)}{\log \relax (x)} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*log(x)^2-2*log(3)*log(x)+2*log(3))*exp(((x*exp(5*x*exp(4))-5)*log(x)
-2*x*log(3))/log(x))/log(x)^2,x, algorithm="maxima")

[Out]

e^(x*e^(5*x*e^4) - 2*x*log(3)/log(x) - 5)

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mupad [B]  time = 6.73, size = 23, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{5\,x\,{\mathrm {e}}^4}}}{3^{\frac {2\,x}{\ln \relax (x)}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(x)*(x*exp(5*x*exp(4)) - 5) - 2*x*log(3))/log(x))*(2*log(3) - 2*log(3)*log(x) + exp(5*x*exp(4))*l
og(x)^2*(5*x*exp(4) + 1)))/log(x)^2,x)

[Out]

(exp(-5)*exp(x*exp(5*x*exp(4))))/3^((2*x)/log(x))

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sympy [A]  time = 0.60, size = 26, normalized size = 1.13 \begin {gather*} e^{\frac {- 2 x \log {\relax (3 )} + \left (x e^{5 x e^{4}} - 5\right ) \log {\relax (x )}}{\log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*ln(x)**2-2*ln(3)*ln(x)+2*ln(3))*exp(((x*exp(5*x*exp(4))-5)*ln(x)-2*x
*ln(3))/ln(x))/ln(x)**2,x)

[Out]

exp((-2*x*log(3) + (x*exp(5*x*exp(4)) - 5)*log(x))/log(x))

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