3.88.72 \(\int \frac {2+e^{-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x} (5 x+25 x^2-375 e^{3 x} x^2)}{5 x} \, dx\)

Optimal. Leaf size=30 \[ e^{e^{5 \left (-5 \left (2+e^{3 x}\right )+x\right )} x}-\frac {1}{5} \log \left (\frac {100}{x^2}\right ) \]

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Rubi [F]  time = 0.74, 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 {2+\exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (5 x+25 x^2-375 e^{3 x} x^2\right )}{5 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + E^(-50 - 25*E^(3*x) + 5*x + E^(-50 - 25*E^(3*x) + 5*x)*x)*(5*x + 25*x^2 - 375*E^(3*x)*x^2))/(5*x),x]

[Out]

(2*Log[x])/5 + Defer[Int][E^(-50 - 25*E^(3*x) + 5*x + E^(-50 - 25*E^(3*x) + 5*x)*x), x] + 5*Defer[Int][E^(-50
- 25*E^(3*x) + 5*x + E^(-50 - 25*E^(3*x) + 5*x)*x)*x, x] - 75*Defer[Int][E^(-50 - 25*E^(3*x) + 8*x + E^(-50 -
25*E^(3*x) + 5*x)*x)*x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {2+\exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (5 x+25 x^2-375 e^{3 x} x^2\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (\frac {2}{x}-5 \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (-1-5 x+75 e^{3 x} x\right )\right ) \, dx\\ &=\frac {2 \log (x)}{5}-\int \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \left (-1-5 x+75 e^{3 x} x\right ) \, dx\\ &=\frac {2 \log (x)}{5}-\int \left (-\exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right )-5 \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) x+75 \exp \left (-50-25 e^{3 x}+8 x+e^{-50-25 e^{3 x}+5 x} x\right ) x\right ) \, dx\\ &=\frac {2 \log (x)}{5}+5 \int \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) x \, dx-75 \int \exp \left (-50-25 e^{3 x}+8 x+e^{-50-25 e^{3 x}+5 x} x\right ) x \, dx+\int \exp \left (-50-25 e^{3 x}+5 x+e^{-50-25 e^{3 x}+5 x} x\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.44, size = 25, normalized size = 0.83 \begin {gather*} e^{e^{-50-25 e^{3 x}+5 x} x}+\frac {2 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + E^(-50 - 25*E^(3*x) + 5*x + E^(-50 - 25*E^(3*x) + 5*x)*x)*(5*x + 25*x^2 - 375*E^(3*x)*x^2))/(5*
x),x]

[Out]

E^(E^(-50 - 25*E^(3*x) + 5*x)*x) + (2*Log[x])/5

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fricas [B]  time = 0.56, size = 59, normalized size = 1.97 \begin {gather*} \frac {1}{5} \, {\left (2 \, e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} \log \relax (x) + 5 \, e^{\left (x e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} + 5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )}\right )} e^{\left (-5 \, x + 25 \, e^{\left (3 \, x\right )} + 50\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-375*x^2*exp(3*x)+25*x^2+5*x)*exp(-25*exp(3*x)+5*x-50)*exp(x*exp(-25*exp(3*x)+5*x-50))+2)/x,x,
algorithm="fricas")

[Out]

1/5*(2*e^(5*x - 25*e^(3*x) - 50)*log(x) + 5*e^(x*e^(5*x - 25*e^(3*x) - 50) + 5*x - 25*e^(3*x) - 50))*e^(-5*x +
25*e^(3*x) + 50)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {5 \, {\left (75 \, x^{2} e^{\left (3 \, x\right )} - 5 \, x^{2} - x\right )} e^{\left (x e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} + 5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )} - 2}{5 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-375*x^2*exp(3*x)+25*x^2+5*x)*exp(-25*exp(3*x)+5*x-50)*exp(x*exp(-25*exp(3*x)+5*x-50))+2)/x,x,
algorithm="giac")

[Out]

integrate(-1/5*(5*(75*x^2*e^(3*x) - 5*x^2 - x)*e^(x*e^(5*x - 25*e^(3*x) - 50) + 5*x - 25*e^(3*x) - 50) - 2)/x,
x)

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maple [A]  time = 0.08, size = 21, normalized size = 0.70

method	result	size
norman	\({\mathrm e}^{x \,{\mathrm e}^{-25 \,{\mathrm e}^{3 x}+5 x -50}}+\frac {2 \ln \relax (x )}{5}\)	\(21\)
risch	\({\mathrm e}^{x \,{\mathrm e}^{-25 \,{\mathrm e}^{3 x}+5 x -50}}+\frac {2 \ln \relax (x )}{5}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-375*x^2*exp(3*x)+25*x^2+5*x)*exp(-25*exp(3*x)+5*x-50)*exp(x*exp(-25*exp(3*x)+5*x-50))+2)/x,x,method
=_RETURNVERBOSE)

[Out]

exp(x*exp(-25*exp(3*x)+5*x-50))+2/5*ln(x)

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maxima [A]  time = 0.66, size = 20, normalized size = 0.67 \begin {gather*} e^{\left (x e^{\left (5 \, x - 25 \, e^{\left (3 \, x\right )} - 50\right )}\right )} + \frac {2}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-375*x^2*exp(3*x)+25*x^2+5*x)*exp(-25*exp(3*x)+5*x-50)*exp(x*exp(-25*exp(3*x)+5*x-50))+2)/x,x,
algorithm="maxima")

[Out]

e^(x*e^(5*x - 25*e^(3*x) - 50)) + 2/5*log(x)

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mupad [B]  time = 5.59, size = 21, normalized size = 0.70 \begin {gather*} {\mathrm {e}}^{x\,{\mathrm {e}}^{-25\,{\mathrm {e}}^{3\,x}}\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{-50}}+\frac {2\,\ln \relax (x)}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(5*x - 25*exp(3*x) - 50)*exp(x*exp(5*x - 25*exp(3*x) - 50))*(5*x - 375*x^2*exp(3*x) + 25*x^2))/5 + 2/
5)/x,x)

[Out]

exp(x*exp(-25*exp(3*x))*exp(5*x)*exp(-50)) + (2*log(x))/5

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sympy [A]  time = 0.41, size = 22, normalized size = 0.73 \begin {gather*} e^{x e^{5 x - 25 e^{3 x} - 50}} + \frac {2 \log {\relax (x )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-375*x**2*exp(3*x)+25*x**2+5*x)*exp(-25*exp(3*x)+5*x-50)*exp(x*exp(-25*exp(3*x)+5*x-50))+2)/x,
x)

[Out]

exp(x*exp(5*x - 25*exp(3*x) - 50)) + 2*log(x)/5

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