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

Optimal. Leaf size=26 \[ \frac {e^{4-e^x-4 e^{-x} x} x}{5 \log (5)} \]

________________________________________________________________________________________

Rubi [F]  time = 0.89, 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 {e^{-x-e^{-x} \left (-4 e^x+e^{2 x}+4 x\right )} \left (e^x-4 x-e^{2 x} x+4 x^2\right )}{5 \log (5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{4-e^x-4 e^{-x} x} x}{5 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(4 - E^x - (4*x)/E^x)*x)/(5*Log[5])

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 30, normalized size = 1.15 \begin {gather*} \frac {x e^{\left (-{\left ({\left (x - 4\right )} e^{x} + 4 \, x + e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} + x\right )}}{5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.18, size = 24, normalized size = 0.92 \begin {gather*} \frac {x e^{\left (-{\left (4 \, x + e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} + 4\right )}}{5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 28, normalized size = 1.08

method	result	size
norman	\(\frac {x \,{\mathrm e}^{-\left ({\mathrm e}^{2 x}-4 \,{\mathrm e}^{x}+4 x \right ) {\mathrm e}^{-x}}}{5 \ln \relax (5)}\)	\(28\)
risch	\(\frac {x \,{\mathrm e}^{\left (-{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x}-4 x \right ) {\mathrm e}^{-x}}}{5 \ln \relax (5)}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.46, size = 21, normalized size = 0.81 \begin {gather*} \frac {x e^{\left (-4 \, x e^{\left (-x\right )} - e^{x} + 4\right )}}{5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 7.94, size = 22, normalized size = 0.85 \begin {gather*} \frac {x\,{\mathrm {e}}^4\,{\mathrm {e}}^{-4\,x\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{-{\mathrm {e}}^x}}{5\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.32, size = 24, normalized size = 0.92 \begin {gather*} \frac {x e^{- \left (4 x + e^{2 x} - 4 e^{x}\right ) e^{- x}}}{5 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________