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

Optimal. Leaf size=21 \[ e^{x+\frac {1}{2} e^4 \left (x+\frac {5 e x}{\log (x)}\right )} \]

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Rubi [F]  time = 0.62, 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 {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \left (-5 e^5+5 e^5 \log (x)+\left (2+e^4\right ) \log ^2(x)\right )}{2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

((2 + E^4)*Defer[Int][E^((5*E^5*x + (2*x + E^4*x)*Log[x])/(2*Log[x])), x])/2 - (5*Defer[Int][E^((5*E^5*x + 10*
Log[x] + 2*(1 + E^4/2)*x*Log[x])/(2*Log[x]))/Log[x]^2, x])/2 + (5*Defer[Int][E^((5*E^5*x + 10*Log[x] + 2*(1 +
E^4/2)*x*Log[x])/(2*Log[x]))/Log[x], x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {\exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \left (-5 e^5+5 e^5 \log (x)+\left (2+e^4\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (2 \exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \left (1+\frac {e^4}{2}\right )-\frac {5 \exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log ^2(x)}+\frac {5 \exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log (x)}\right ) \, dx\\ &=-\left (\frac {5}{2} \int \frac {\exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log ^2(x)} \, dx\right )+\frac {5}{2} \int \frac {\exp \left (5+\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right )}{\log (x)} \, dx+\frac {1}{2} \left (2+e^4\right ) \int \exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \, dx\\ &=-\left (\frac {5}{2} \int \frac {\exp \left (\frac {5 e^5 x+10 \log (x)+2 \left (1+\frac {e^4}{2}\right ) x \log (x)}{2 \log (x)}\right )}{\log ^2(x)} \, dx\right )+\frac {5}{2} \int \frac {\exp \left (\frac {5 e^5 x+10 \log (x)+2 \left (1+\frac {e^4}{2}\right ) x \log (x)}{2 \log (x)}\right )}{\log (x)} \, dx+\frac {1}{2} \left (2+e^4\right ) \int \exp \left (\frac {5 e^5 x+\left (2 x+e^4 x\right ) \log (x)}{2 \log (x)}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 21, normalized size = 1.00 \begin {gather*} e^{\frac {1}{2} x \left (2+e^4+\frac {5 e^5}{\log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^((x*(2 + E^4 + (5*E^5)/Log[x]))/2)

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fricas [A]  time = 0.86, size = 24, normalized size = 1.14 \begin {gather*} e^{\left (\frac {5 \, x e^{5} + {\left (x e^{4} + 2 \, x\right )} \log \relax (x)}{2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(1/2*(5*x*e^5 + (x*e^4 + 2*x)*log(x))/log(x))

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giac [A]  time = 0.16, size = 17, normalized size = 0.81 \begin {gather*} e^{\left (\frac {1}{2} \, x e^{4} + x + \frac {5 \, x e^{5}}{2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(1/2*x*e^4 + x + 5/2*x*e^5/log(x))

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maple [A]  time = 0.05, size = 23, normalized size = 1.10

method	result	size
risch	\({\mathrm e}^{\frac {x \left ({\mathrm e}^{4} \ln \relax (x )+5 \,{\mathrm e}^{5}+2 \ln \relax (x )\right )}{2 \ln \relax (x )}}\)	\(23\)
norman	\({\mathrm e}^{\frac {\left (x \,{\mathrm e} \,{\mathrm e}^{3}+2 x \right ) \ln \relax (x )+5 x \,{\mathrm e}^{2} {\mathrm e}^{3}}{2 \ln \relax (x )}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(1/2*x*(exp(4)*ln(x)+5*exp(5)+2*ln(x))/ln(x))

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maxima [A]  time = 0.40, size = 17, normalized size = 0.81 \begin {gather*} e^{\left (\frac {1}{2} \, x e^{4} + x + \frac {5 \, x e^{5}}{2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(1/2*x*e^4 + x + 5/2*x*e^5/log(x))

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mupad [B]  time = 2.58, size = 19, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{\frac {x\,{\mathrm {e}}^4}{2}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^5}{2\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((x*exp(4))/2)*exp(x)*exp((5*x*exp(5))/(2*log(x)))

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sympy [A]  time = 0.38, size = 26, normalized size = 1.24 \begin {gather*} e^{\frac {\frac {5 x e^{5}}{2} + \frac {\left (2 x + x e^{4}\right ) \log {\relax (x )}}{2}}{\log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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