3.49.75 \(\int e^{-x-e^{-x} (10 x^2+2 x^3) \log ^2(5)} (e^x+(-20 x^2+4 x^3+2 x^4) \log ^2(5)) \, dx\)

Optimal. Leaf size=21 \[ e^{-2 e^{-x} x^2 (5+x) \log ^2(5)} x \]

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Rubi [F]  time = 1.41, 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 e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[E^(-x - ((10*x^2 + 2*x^3)*Log[5]^2)/E^x)*(E^x + (-20*x^2 + 4*x^3 + 2*x^4)*Log[5]^2),x]

[Out]

Defer[Int][E^((-2*x^2*(5 + x)*Log[5]^2)/E^x), x] - 20*Log[5]^2*Defer[Int][x^2/E^((x*(E^x + 10*x*Log[5]^2 + 2*x
^2*Log[5]^2))/E^x), x] + 4*Log[5]^2*Defer[Int][x^3/E^((x*(E^x + 10*x*Log[5]^2 + 2*x^2*Log[5]^2))/E^x), x] + 2*
Log[5]^2*Defer[Int][x^4/E^((x*(E^x + 10*x*Log[5]^2 + 2*x^2*Log[5]^2))/E^x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx\\ &=\int \left (\exp \left (x-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right )+2 \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^2 \left (-10+2 x+x^2\right ) \log ^2(5)\right ) \, dx\\ &=\left (2 \log ^2(5)\right ) \int \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^2 \left (-10+2 x+x^2\right ) \, dx+\int \exp \left (x-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) \, dx\\ &=\left (2 \log ^2(5)\right ) \int \left (-10 \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^2+2 \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^3+\exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^4\right ) \, dx+\int e^{-2 e^{-x} x^2 (5+x) \log ^2(5)} \, dx\\ &=\left (2 \log ^2(5)\right ) \int \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^4 \, dx+\left (4 \log ^2(5)\right ) \int \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^3 \, dx-\left (20 \log ^2(5)\right ) \int \exp \left (-e^{-x} x \left (e^x+10 x \log ^2(5)+2 x^2 \log ^2(5)\right )\right ) x^2 \, dx+\int e^{-2 e^{-x} x^2 (5+x) \log ^2(5)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 21, normalized size = 1.00 \begin {gather*} e^{-2 e^{-x} x^2 (5+x) \log ^2(5)} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-x - ((10*x^2 + 2*x^3)*Log[5]^2)/E^x)*(E^x + (-20*x^2 + 4*x^3 + 2*x^4)*Log[5]^2),x]

[Out]

x/E^((2*x^2*(5 + x)*Log[5]^2)/E^x)

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fricas [A]  time = 0.96, size = 31, normalized size = 1.48 \begin {gather*} x e^{\left (-{\left (2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \relax (5)^{2} + x e^{x}\right )} e^{\left (-x\right )} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+(2*x^4+4*x^3-20*x^2)*log(5)^2)/exp(x)/exp((2*x^3+10*x^2)*log(5)^2/exp(x)),x, algorithm="fric
as")

[Out]

x*e^(-(2*(x^3 + 5*x^2)*log(5)^2 + x*e^x)*e^(-x) + x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (2 \, {\left (x^{4} + 2 \, x^{3} - 10 \, x^{2}\right )} \log \relax (5)^{2} + e^{x}\right )} e^{\left (-2 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (-x\right )} \log \relax (5)^{2} - x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+(2*x^4+4*x^3-20*x^2)*log(5)^2)/exp(x)/exp((2*x^3+10*x^2)*log(5)^2/exp(x)),x, algorithm="giac
")

[Out]

integrate((2*(x^4 + 2*x^3 - 10*x^2)*log(5)^2 + e^x)*e^(-2*(x^3 + 5*x^2)*e^(-x)*log(5)^2 - x), x)

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maple [A]  time = 0.09, size = 20, normalized size = 0.95

method	result	size
risch	\(x \,{\mathrm e}^{-2 x^{2} \left (5+x \right ) \ln \relax (5)^{2} {\mathrm e}^{-x}}\)	\(20\)
norman	\(x \,{\mathrm e}^{-\left (2 x^{3}+10 x^{2}\right ) \ln \relax (5)^{2} {\mathrm e}^{-x}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)+(2*x^4+4*x^3-20*x^2)*ln(5)^2)/exp(x)/exp((2*x^3+10*x^2)*ln(5)^2/exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (2 \, {\left (x^{4} + 2 \, x^{3} - 10 \, x^{2}\right )} \log \relax (5)^{2} + e^{x}\right )} e^{\left (-2 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (-x\right )} \log \relax (5)^{2} - x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+(2*x^4+4*x^3-20*x^2)*log(5)^2)/exp(x)/exp((2*x^3+10*x^2)*log(5)^2/exp(x)),x, algorithm="maxi
ma")

[Out]

integrate((2*(x^4 + 2*x^3 - 10*x^2)*log(5)^2 + e^x)*e^(-2*(x^3 + 5*x^2)*e^(-x)*log(5)^2 - x), x)

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mupad [B]  time = 3.43, size = 30, normalized size = 1.43 \begin {gather*} x\,{\mathrm {e}}^{-2\,x^3\,{\mathrm {e}}^{-x}\,{\ln \relax (5)}^2}\,{\mathrm {e}}^{-10\,x^2\,{\mathrm {e}}^{-x}\,{\ln \relax (5)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-x)*exp(-exp(-x)*log(5)^2*(10*x^2 + 2*x^3))*(exp(x) + log(5)^2*(4*x^3 - 20*x^2 + 2*x^4)),x)

[Out]

x*exp(-2*x^3*exp(-x)*log(5)^2)*exp(-10*x^2*exp(-x)*log(5)^2)

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sympy [A]  time = 0.37, size = 20, normalized size = 0.95 \begin {gather*} x e^{- \left (2 x^{3} + 10 x^{2}\right ) e^{- x} \log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+(2*x**4+4*x**3-20*x**2)*ln(5)**2)/exp(x)/exp((2*x**3+10*x**2)*ln(5)**2/exp(x)),x)

[Out]

x*exp(-(2*x**3 + 10*x**2)*exp(-x)*log(5)**2)

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