3.38.42 \(\int e^{-x+e^{-x} (-5 x+e^x (1+x^3))+\frac {3-e^{e^{-x} (-5 x+e^x (1+x^3))} \log (3)}{\log (3)}} (5-5 x-3 e^x x^2) \, dx\)

Optimal. Leaf size=28 \[ e^{-e^{1+x \left (-5 e^{-x}+x^2\right )}+\frac {3}{\log (3)}} \]

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Rubi [F]  time = 1.99, 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 \exp \left (-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}\right ) \left (5-5 x-3 e^x x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[E^(-x + (-5*x + E^x*(1 + x^3))/E^x + (3 - E^((-5*x + E^x*(1 + x^3))/E^x)*Log[3])/Log[3])*(5 - 5*x - 3*E^x*
x^2),x]

[Out]

5*Defer[Int][E^(1 - E^(1 - (5*x)/E^x + x^3) - x - (5*x)/E^x + x^3 + 3/Log[3]), x] - 5*Defer[Int][E^(1 - E^(1 -
(5*x)/E^x + x^3) - x - (5*x)/E^x + x^3 + 3/Log[3])*x, x] - 3*Defer[Int][E^(1 - E^(1 - (5*x)/E^x + x^3) - (5*x
)/E^x + x^3 + 3/Log[3])*x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) \left (5-5 x-3 e^x x^2\right ) \, dx\\ &=\int \left (5 \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right )-5 \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x-3 \exp \left (1-e^{1-5 e^{-x} x+x^3}-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x^2\right ) \, dx\\ &=-\left (3 \int \exp \left (1-e^{1-5 e^{-x} x+x^3}-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x^2 \, dx\right )+5 \int \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) \, dx-5 \int \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.21, size = 26, normalized size = 0.93 \begin {gather*} e^{-e^{1-5 e^{-x} x+x^3}+\frac {3}{\log (3)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-x + (-5*x + E^x*(1 + x^3))/E^x + (3 - E^((-5*x + E^x*(1 + x^3))/E^x)*Log[3])/Log[3])*(5 - 5*x -
3*E^x*x^2),x]

[Out]

E^(-E^(1 - (5*x)/E^x + x^3) + 3/Log[3])

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fricas [B]  time = 0.69, size = 76, normalized size = 2.71 \begin {gather*} e^{\left (-{\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} + x + \frac {{\left ({\left ({\left (x^{3} - x + 1\right )} \log \relax (3) + 3\right )} e^{x} - 5 \, x \log \relax (3) - e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} + x\right )} \log \relax (3)\right )} e^{\left (-x\right )}}{\log \relax (3)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-((x^3 + 1)*e^x - 5*x)*e^(-x) + x + (((x^3 - x + 1)*log(3) + 3)*e^x - 5*x*log(3) - e^(((x^3 + 1)*e^x - 5*x)
*e^(-x) + x)*log(3))*e^(-x)/log(3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -{\left (3 \, x^{2} e^{x} + 5 \, x - 5\right )} e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} - x - \frac {e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )}\right )} \log \relax (3) - 3}{\log \relax (3)}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(3*x^2*e^x + 5*x - 5)*e^(((x^3 + 1)*e^x - 5*x)*e^(-x) - x - (e^(((x^3 + 1)*e^x - 5*x)*e^(-x))*log(3
) - 3)/log(3)), x)

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maple [A]  time = 0.10, size = 31, normalized size = 1.11

method	result	size
norman	\({\mathrm e}^{\frac {-\ln \relax (3) {\mathrm e}^{\left (\left (x^{3}+1\right ) {\mathrm e}^{x}-5 x \right ) {\mathrm e}^{-x}}+3}{\ln \relax (3)}}\)	\(31\)
risch	\({\mathrm e}^{-\frac {\ln \relax (3) {\mathrm e}^{\left ({\mathrm e}^{x} x^{3}+{\mathrm e}^{x}-5 x \right ) {\mathrm e}^{-x}}-3}{\ln \relax (3)}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((-ln(3)*exp(((x^3+1)*exp(x)-5*x)/exp(x))+3)/ln(3))

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maxima [A]  time = 0.93, size = 23, normalized size = 0.82 \begin {gather*} e^{\left (\frac {3}{\log \relax (3)} - e^{\left (x^{3} - 5 \, x e^{\left (-x\right )} + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(3/log(3) - e^(x^3 - 5*x*e^(-x) + 1))

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mupad [B]  time = 2.40, size = 25, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{\frac {3}{\ln \relax (3)}}\,{\mathrm {e}}^{-{\mathrm {e}}^{x^3}\,\mathrm {e}\,{\mathrm {e}}^{-5\,x\,{\mathrm {e}}^{-x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(3/log(3))*exp(-exp(x^3)*exp(1)*exp(-5*x*exp(-x)))

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sympy [A]  time = 0.54, size = 26, normalized size = 0.93 \begin {gather*} e^{\frac {- e^{\left (- 5 x + \left (x^{3} + 1\right ) e^{x}\right ) e^{- x}} \log {\relax (3 )} + 3}{\log {\relax (3 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((-exp((-5*x + (x**3 + 1)*exp(x))*exp(-x))*log(3) + 3)/log(3))

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