3.38.25 \(\int \frac {2 e^x+e^{5 e^{-x} x} (5 x-5 x^2+2 e^x x^2)+2 e^x x^2 \log (x^2)}{e^{x+5 e^{-x} x} x+e^x x \log (x^2)} \, dx\)

Optimal. Leaf size=25 \[ -5+x^2+\log \left (-e^{5 e^{-x} x}-\log \left (x^2\right )\right ) \]

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Rubi [F]  time = 3.08, 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 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(5*x)/E^x + x^2 + 2*Defer[Int][1/(x*(E^((5*x)/E^x) + Log[x^2])), x] - 5*Defer[Int][Log[x^2]/(E^x*(E^((5*x)/E^x
) + Log[x^2])), x] + 5*Defer[Int][(x*Log[x^2])/(E^x*(E^((5*x)/E^x) + Log[x^2])), x]

Rubi steps

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

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Mathematica [A]  time = 0.96, size = 20, normalized size = 0.80 \begin {gather*} x^2+\log \left (e^{5 e^{-x} x}+\log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2 + Log[E^((5*x)/E^x) + Log[x^2]]

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fricas [A]  time = 0.74, size = 30, normalized size = 1.20 \begin {gather*} x^{2} - x + \log \left (e^{x} \log \left (x^{2}\right ) + e^{\left ({\left (x e^{x} + 5 \, x\right )} e^{\left (-x\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*log(x^2)+2*exp(x))/(x*exp(x)*exp(5*x/exp(x))+
x*exp(x)*log(x^2)),x, algorithm="fricas")

[Out]

x^2 - x + log(e^x*log(x^2) + e^((x*e^x + 5*x)*e^(-x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*log(x^2)+2*exp(x))/(x*exp(x)*exp(5*x/exp(x))+
x*exp(x)*log(x^2)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.40, size = 84, normalized size = 3.36

method	result	size
risch	\(x \left ({\mathrm e}^{x} x +5\right ) {\mathrm e}^{-x}-5 x \,{\mathrm e}^{-x}+\ln \left ({\mathrm e}^{5 x \,{\mathrm e}^{-x}}-\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )\right )}{2}\right )\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(exp(x)*x+5)*exp(-x)-5*x*exp(-x)+ln(exp(5*x*exp(-x))-1/2*I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I
*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x)))

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maxima [A]  time = 0.41, size = 18, normalized size = 0.72 \begin {gather*} x^{2} + \log \left (e^{\left (5 \, x e^{\left (-x\right )}\right )} + 2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*log(x^2)+2*exp(x))/(x*exp(x)*exp(5*x/exp(x))+
x*exp(x)*log(x^2)),x, algorithm="maxima")

[Out]

x^2 + log(e^(5*x*e^(-x)) + 2*log(x))

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mupad [B]  time = 2.23, size = 18, normalized size = 0.72 \begin {gather*} \ln \left (\ln \left (x^2\right )+{\mathrm {e}}^{5\,x\,{\mathrm {e}}^{-x}}\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x^2) + exp(5*x*exp(-x))) + x^2

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sympy [A]  time = 0.41, size = 17, normalized size = 0.68 \begin {gather*} x^{2} + \log {\left (e^{5 x e^{- x}} + \log {\left (x^{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x**2-5*x**2+5*x)*exp(5*x/exp(x))+2*x**2*exp(x)*ln(x**2)+2*exp(x))/(x*exp(x)*exp(5*x/exp(x
))+x*exp(x)*ln(x**2)),x)

[Out]

x**2 + log(exp(5*x*exp(-x)) + log(x**2))

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