3.26.9 \(\int \frac {e^{-5 x} (e^{2 e^{-5 x}} (-2 e^{5 x}-10 x)-e^{5 x} x^3 \log (4))}{x^3 \log (4)} \, dx\)

Optimal. Leaf size=22 \[ -5-x+\frac {e^{2 e^{-5 x}}}{x^2 \log (4)} \]

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Rubi [F]  time = 0.46, 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^{-5 x} \left (e^{2 e^{-5 x}} \left (-2 e^{5 x}-10 x\right )-e^{5 x} x^3 \log (4)\right )}{x^3 \log (4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 25, normalized size = 1.14 \begin {gather*} -\frac {-\frac {e^{2 e^{-5 x}}}{x^2}+x \log (4)}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 26, normalized size = 1.18 \begin {gather*} -\frac {2 \, x^{3} \log \relax (2) - e^{\left (2 \, e^{\left (-5 \, x\right )}\right )}}{2 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*log(2)*exp(5*x))/x^3/log(2)/exp(5*x),x, algorithm="fri
cas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*log(2)*exp(5*x))/x^3/log(2)/exp(5*x),x, algorithm="gia
c")

[Out]

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

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maple [A]  time = 0.04, size = 21, normalized size = 0.95

method	result	size
risch	\(-x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{-5 x}}}{2 x^{2} \ln \relax (2)}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*ln(2)*exp(5*x))/x^3/ln(2)/exp(5*x),x,method=_RETURNVERBOSE)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x^3*log(2)*exp(5*x))/x^3/log(2)/exp(5*x),x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 1.49, size = 20, normalized size = 0.91 \begin {gather*} \frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{-5\,x}}}{2\,x^2\,\ln \relax (2)}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*exp(-5*x))/(2*x^2*log(2)) - x

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sympy [A]  time = 0.15, size = 17, normalized size = 0.77 \begin {gather*} - x + \frac {e^{2 e^{- 5 x}}}{2 x^{2} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*exp(5*x)-10*x)*exp(2/exp(5*x))-2*x**3*ln(2)*exp(5*x))/x**3/ln(2)/exp(5*x),x)

[Out]

-x + exp(2*exp(-5*x))/(2*x**2*log(2))

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