3.73.85 \(\int \frac {e^{-x} (e^x+e^{e^{-x} (e^{15} x^2+e^{15+x} x^2)} (2 e^{15+x} x^3+e^{15} (2 x^3-x^4))-e^x \log (x))}{x^2} \, dx\)

Optimal. Leaf size=24 \[ -1+e^{e^{15} x \left (x+e^{-x} x\right )}+\frac {\log (x)}{x} \]

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Rubi [F]  time = 1.69, 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} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E^15*(2*x^3 - x^4)) - E^x*Log[x])/(E^x*x^2)
,x]

[Out]

Log[x]/x + 2*Defer[Int][E^(15 + E^(15 - x)*(1 + E^x)*x^2)*x, x] + 2*Defer[Int][E^(15 - x + E^(15 - x)*(1 + E^x
)*x^2)*x, x] - Defer[Int][E^(15 - x + E^(15 - x)*(1 + E^x)*x^2)*x^2, x]

Rubi steps

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

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Mathematica [A]  time = 0.43, size = 25, normalized size = 1.04 \begin {gather*} e^{e^{15-x} \left (1+e^x\right ) x^2}+\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E^15*(2*x^3 - x^4)) - E^x*Log[x])/(E^
x*x^2),x]

[Out]

E^(E^(15 - x)*(1 + E^x)*x^2) + Log[x]/x

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fricas [A]  time = 0.58, size = 32, normalized size = 1.33 \begin {gather*} \frac {x e^{\left ({\left (x^{2} e^{30} + x^{2} e^{\left (x + 30\right )}\right )} e^{\left (-x - 15\right )}\right )} + \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="fricas")

[Out]

(x*e^((x^2*e^30 + x^2*e^(x + 30))*e^(-x - 15)) + log(x))/x

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giac [A]  time = 0.17, size = 25, normalized size = 1.04 \begin {gather*} \frac {\log \relax (x)}{x} + e^{\left (x^{2} e^{15} + x^{2} e^{\left (-x + 15\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.15, size = 24, normalized size = 1.00

method	result	size
risch	\(\frac {\ln \relax (x )}{x}+{\mathrm e}^{x^{2} \left ({\mathrm e}^{x +15}+{\mathrm e}^{15}\right ) {\mathrm e}^{-x}}\)	\(24\)
default	\({\mathrm e}^{x^{2} \left ({\mathrm e}^{15} {\mathrm e}^{x}+{\mathrm e}^{15}\right ) {\mathrm e}^{-x}}+\frac {\ln \relax (x )}{x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*ln(x)+exp
(x))/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

ln(x)/x+exp(x^2*(exp(x+15)+exp(15))*exp(-x))

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maxima [A]  time = 0.49, size = 34, normalized size = 1.42 \begin {gather*} \frac {x e^{\left (x^{2} e^{15} + x^{2} e^{\left (-x + 15\right )}\right )} + \log \relax (x) + 1}{x} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.46, size = 26, normalized size = 1.08 \begin {gather*} \frac {\ln \relax (x)}{x}+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{15}}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(x) - exp(x)*log(x) + exp(exp(-x)*(x^2*exp(15) + x^2*exp(15)*exp(x)))*(exp(15)*(2*x^3 - x^4)
+ 2*x^3*exp(15)*exp(x))))/x^2,x)

[Out]

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

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sympy [A]  time = 0.53, size = 26, normalized size = 1.08 \begin {gather*} e^{\left (x^{2} e^{15} e^{x} + x^{2} e^{15}\right ) e^{- x}} + \frac {\log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3*exp(15)*exp(x)+(-x**4+2*x**3)*exp(15))*exp((x**2*exp(15)*exp(x)+x**2*exp(15))/exp(x))-exp(x
)*ln(x)+exp(x))/exp(x)/x**2,x)

[Out]

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

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