3.9.43 \(\int \frac {e^{-3-e^x} (-2+4 \log (2)-4 e^x x \log (2)+(-1+e^x x) \log (x^2))}{\log (2)} \, dx\)

Optimal. Leaf size=25 \[ e^{-3-e^x} \left (4 x-\frac {x \log \left (x^2\right )}{\log (2)}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 0.45, 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^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-3 - E^x)*(-2 + 4*Log[2] - 4*E^x*x*Log[2] + (-1 + E^x*x)*Log[x^2]))/Log[2],x]

[Out]

(2*ExpIntegralEi[-E^x])/(E^3*Log[2]) - (2*ExpIntegralEi[-E^x]*(1 - Log[4]))/(E^3*Log[2]) - (E^(-3 - E^x)*x*Log
[x^2])/Log[2] - 4*Defer[Int][E^(-3 - E^x + x)*x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right ) \, dx}{\log (2)}\\ &=\frac {\int e^{-3-e^x} \left (-4 e^x x \log (2)-2 (1-\log (4))+\left (-1+e^x x\right ) \log \left (x^2\right )\right ) \, dx}{\log (2)}\\ &=\frac {\int \left (-4 e^{-3-e^x+x} x \log (2)-2 e^{-3-e^x} (1-\log (4))+e^{-3-e^x} \left (-1+e^x x\right ) \log \left (x^2\right )\right ) \, dx}{\log (2)}\\ &=-\left (4 \int e^{-3-e^x+x} x \, dx\right )+\frac {\int e^{-3-e^x} \left (-1+e^x x\right ) \log \left (x^2\right ) \, dx}{\log (2)}-\frac {(2 (1-\log (4))) \int e^{-3-e^x} \, dx}{\log (2)}\\ &=-\frac {e^{-3-e^x} x \log \left (x^2\right )}{\log (2)}-4 \int e^{-3-e^x+x} x \, dx-\frac {\int -2 e^{-3-e^x} \, dx}{\log (2)}-\frac {(2 (1-\log (4))) \operatorname {Subst}\left (\int \frac {e^{-3-x}}{x} \, dx,x,e^x\right )}{\log (2)}\\ &=-\frac {2 \text {Ei}\left (-e^x\right ) (1-\log (4))}{e^3 \log (2)}-\frac {e^{-3-e^x} x \log \left (x^2\right )}{\log (2)}-4 \int e^{-3-e^x+x} x \, dx+\frac {2 \int e^{-3-e^x} \, dx}{\log (2)}\\ &=-\frac {2 \text {Ei}\left (-e^x\right ) (1-\log (4))}{e^3 \log (2)}-\frac {e^{-3-e^x} x \log \left (x^2\right )}{\log (2)}-4 \int e^{-3-e^x+x} x \, dx+\frac {2 \operatorname {Subst}\left (\int \frac {e^{-3-x}}{x} \, dx,x,e^x\right )}{\log (2)}\\ &=\frac {2 \text {Ei}\left (-e^x\right )}{e^3 \log (2)}-\frac {2 \text {Ei}\left (-e^x\right ) (1-\log (4))}{e^3 \log (2)}-\frac {e^{-3-e^x} x \log \left (x^2\right )}{\log (2)}-4 \int e^{-3-e^x+x} x \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [C]  time = 0.28, size = 60, normalized size = 2.40 \begin {gather*} -4 \left (-e^{-3-e^x} x+\frac {\text {Ei}\left (-e^x\right )}{e^3}\right )+\frac {\frac {\text {Ei}\left (-e^x\right ) \log (16)}{e^3}-e^{-3-e^x} x \log \left (x^2\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 - E^x)*(-2 + 4*Log[2] - 4*E^x*x*Log[2] + (-1 + E^x*x)*Log[x^2]))/Log[2],x]

[Out]

-4*(-(E^(-3 - E^x)*x) + ExpIntegralEi[-E^x]/E^3) + ((ExpIntegralEi[-E^x]*Log[16])/E^3 - E^(-3 - E^x)*x*Log[x^2
])/Log[2]

________________________________________________________________________________________

fricas [A]  time = 1.25, size = 25, normalized size = 1.00 \begin {gather*} \frac {{\left (4 \, x \log \relax (2) - x \log \left (x^{2}\right )\right )} e^{\left (-e^{x} - 3\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.39, size = 38, normalized size = 1.52 \begin {gather*} \frac {{\left (4 \, x e^{\left (x - e^{x}\right )} \log \relax (2) - x e^{\left (x - e^{x}\right )} \log \left (x^{2}\right )\right )} e^{\left (-x - 3\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.24, size = 24, normalized size = 0.96

method	result	size
norman	\(\left (4 x -\frac {x \ln \left (x^{2}\right )}{\ln \relax (2)}\right ) {\mathrm e}^{-{\mathrm e}^{x}-3}\)	\(24\)
default	\(\frac {\left (\left (4 \ln \relax (2)-\ln \left (x^{2}\right )+2 \ln \relax (x )\right ) x -2 x \ln \relax (x )\right ) {\mathrm e}^{-{\mathrm e}^{x}-3}}{\ln \relax (2)}\)	\(36\)
risch	\(\frac {\left (\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+4 x \ln \relax (2)-2 x \ln \relax (x )\right ) {\mathrm e}^{-{\mathrm e}^{x}-3}}{\ln \relax (2)}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x-x/ln(2)*ln(x^2))/exp(3+exp(x))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*Ei(-e^x)*e^(-3)*log(2) - Ei(-e^x)*e^(-3) + (2*x*log(2) - x*log(x))*e^(-e^x - 3) - (2*log(2) - 1)*integrat
e(e^(-e^x - 3), x))/log(2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-{\mathrm {e}}^x-3}\,\left (4\,\ln \relax (2)+\ln \left (x^2\right )\,\left (x\,{\mathrm {e}}^x-1\right )-4\,x\,{\mathrm {e}}^x\,\ln \relax (2)-2\right )}{\ln \relax (2)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 2.27, size = 24, normalized size = 0.96 \begin {gather*} \frac {\left (- x \log {\left (x^{2} \right )} + 4 x \log {\relax (2 )}\right ) e^{- e^{x} - 3}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________