3.76.79 \(\int \frac {e^{-x+\frac {e^{-x} x}{15}} (15 e^x+e^{3+3 x} (-x-45 e^x x+x^2)+(x-x^2) \log (x))}{15 x \log (2)} \, dx\)

Optimal. Leaf size=31 \[ 5+\frac {e^{\frac {e^{-x} x}{15}} \left (-e^{3+3 x}+\log (x)\right )}{\log (2)} \]

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Rubi [F]  time = 1.06, 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+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x + x/(15*E^x))*(15*E^x + E^(3 + 3*x)*(-x - 45*E^x*x + x^2) + (x - x^2)*Log[x]))/(15*x*Log[2]),x]

[Out]

(Log[x]*Defer[Int][E^(-x + x/(15*E^x)), x])/(15*Log[2]) - Defer[Int][E^(3 + 2*x + x/(15*E^x)), x]/(15*Log[2])
- (3*Defer[Int][E^(3 + 3*x + x/(15*E^x)), x])/Log[2] + Defer[Int][E^(x/(15*E^x))/x, x]/Log[2] - (Log[x]*Defer[
Int][E^(-x + x/(15*E^x))*x, x])/(15*Log[2]) + Defer[Int][E^(3 + 2*x + x/(15*E^x))*x, x]/(15*Log[2]) - Defer[In
t][Defer[Int][E^(((-15 + E^(-x))*x)/15), x]/x, x]/(15*Log[2]) + Defer[Int][Defer[Int][E^(((-15 + E^(-x))*x)/15
)*x, x]/x, x]/(15*Log[2])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{x} \, dx}{15 \log (2)}\\ &=\frac {\int \left (-45 e^{3+3 x+\frac {e^{-x} x}{15}}+e^{3+2 x+\frac {e^{-x} x}{15}} (-1+x)+\frac {15 e^{\frac {e^{-x} x}{15}}}{x}-e^{-x+\frac {e^{-x} x}{15}} (-1+x) \log (x)\right ) \, dx}{15 \log (2)}\\ &=\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} (-1+x) \, dx}{15 \log (2)}-\frac {\int e^{-x+\frac {e^{-x} x}{15}} (-1+x) \log (x) \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)}\\ &=\frac {\int \left (-e^{3+2 x+\frac {e^{-x} x}{15}}+e^{3+2 x+\frac {e^{-x} x}{15}} x\right ) \, dx}{15 \log (2)}+\frac {\int \frac {-\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} \, dx+\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} x \, dx}{x} \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)}+\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}-\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)}\\ &=-\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}+\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)}+\frac {\int \left (-\frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} \, dx}{x}+\frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} x \, dx}{x}\right ) \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)}+\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}-\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)}\\ &=-\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}+\frac {\int e^{3+2 x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)}-\frac {\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} \, dx}{x} \, dx}{15 \log (2)}+\frac {\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} x \, dx}{x} \, dx}{15 \log (2)}+\frac {\int \frac {e^{\frac {e^{-x} x}{15}}}{x} \, dx}{\log (2)}-\frac {3 \int e^{3+3 x+\frac {e^{-x} x}{15}} \, dx}{\log (2)}+\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} \, dx}{15 \log (2)}-\frac {\log (x) \int e^{-x+\frac {e^{-x} x}{15}} x \, dx}{15 \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.57, size = 34, normalized size = 1.10 \begin {gather*} \frac {e^{\frac {e^{-x} x}{15}} \left (-15 e^{3+3 x}+15 \log (x)\right )}{15 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + x/(15*E^x))*(15*E^x + E^(3 + 3*x)*(-x - 45*E^x*x + x^2) + (x - x^2)*Log[x]))/(15*x*Log[2]),
x]

[Out]

(E^(x/(15*E^x))*(-15*E^(3 + 3*x) + 15*Log[x]))/(15*Log[2])

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fricas [A]  time = 1.57, size = 35, normalized size = 1.13 \begin {gather*} \frac {{\left (e^{x} \log \relax (x) - e^{\left (4 \, x + 3\right )}\right )} e^{\left (-\frac {1}{15} \, {\left (15 \, x e^{x} - x\right )} e^{\left (-x\right )}\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x))*exp(1/15*x/exp(x))/x/log(2)/exp(x),
x, algorithm="fricas")

[Out]

(e^x*log(x) - e^(4*x + 3))*e^(-1/15*(15*x*e^x - x)*e^(-x))/log(2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x^{2} - 45 \, x e^{x} - x\right )} e^{\left (3 \, x + 3\right )} - {\left (x^{2} - x\right )} \log \relax (x) + 15 \, e^{x}\right )} e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} - x\right )}}{15 \, x \log \relax (2)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x))*exp(1/15*x/exp(x))/x/log(2)/exp(x),
x, algorithm="giac")

[Out]

integrate(1/15*((x^2 - 45*x*e^x - x)*e^(3*x + 3) - (x^2 - x)*log(x) + 15*e^x)*e^(1/15*x*e^(-x) - x)/(x*log(2))
, x)

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maple [A]  time = 0.08, size = 28, normalized size = 0.90

method	result	size
risch	\(\frac {\left (-15 \,{\mathrm e}^{3 x +3}+15 \ln \relax (x )\right ) {\mathrm e}^{\frac {x \,{\mathrm e}^{-x}}{15}}}{15 \ln \relax (2)}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*((-x^2+x)*ln(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x))*exp(1/15*x/exp(x))/x/ln(2)/exp(x),x,method
=_RETURNVERBOSE)

[Out]

1/15/ln(2)*(-15*exp(3*x+3)+15*ln(x))*exp(1/15*x*exp(-x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {-15 \, e^{\left (\frac {1}{15} \, x e^{\left (-x\right )}\right )} \log \relax (x) + 15 \, e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} + 3 \, x + 3\right )}}{15 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x))*exp(1/15*x/exp(x))/x/log(2)/exp(x),
x, algorithm="maxima")

[Out]

-1/15*(15*e^(1/15*x*e^(-x) + 3*x + 3) - integrate(-((x^2 - x)*log(x) - 15*e^x)*e^(1/15*x*e^(-x) - x)/x, x))/lo
g(2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-x}}{15}-x}\,\left (15\,{\mathrm {e}}^x+\ln \relax (x)\,\left (x-x^2\right )-{\mathrm {e}}^{3\,x+3}\,\left (x+45\,x\,{\mathrm {e}}^x-x^2\right )\right )}{15\,x\,\ln \relax (2)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*exp((x*exp(-x))/15)*(15*exp(x) + log(x)*(x - x^2) - exp(3*x + 3)*(x + 45*x*exp(x) - x^2)))/(15*x*
log(2)),x)

[Out]

int((exp((x*exp(-x))/15 - x)*(15*exp(x) + log(x)*(x - x^2) - exp(3*x + 3)*(x + 45*x*exp(x) - x^2)))/(15*x*log(
2)), x)

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sympy [A]  time = 0.40, size = 27, normalized size = 0.87 \begin {gather*} \frac {\left (- e^{3} + e^{- 3 x} \log {\relax (x )}\right ) e^{3 x} e^{\frac {x e^{- x}}{15}}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((-x**2+x)*ln(x)+(-45*exp(x)*x+x**2-x)*exp(3*x+3)+15*exp(x))*exp(1/15*x/exp(x))/x/ln(2)/exp(x),
x)

[Out]

(-exp(3) + exp(-3*x)*log(x))*exp(3*x)*exp(x*exp(-x)/15)/log(2)

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