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

Optimal. Leaf size=34 \[ 2+\frac {1}{4} \left (2 x+\frac {e^3 \log \left (2 e^{-2 x}\right )}{e^2-\log ^2(x)}\right ) \]

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Rubi [F]  time = 0.81, 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^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x/2 + (E^(2 + E)*ExpIntegralEi[-E + Log[x]])/4 - (E^(2 - E)*ExpIntegralEi[E + Log[x]])/4 + (E^2*Defer[Int][Log
[2/E^(2*x)]/(x*(E - Log[x])^2), x])/8 - (E^2*Defer[Int][Log[2/E^(2*x)]/(x*(E + Log[x])^2), x])/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (e^4-e^5\right ) x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx\\ &=\int \frac {\left (e^4-e^5\right ) x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 x \left (e^2-\log ^2(x)\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {\left (e^4-e^5\right ) x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{x \left (e^2-\log ^2(x)\right )^2} \, dx\\ &=\frac {1}{2} \int \left (1+\frac {e^2 \log \left (2 e^{-2 x}\right )}{4 x (e-\log (x))^2}-\frac {e^2}{2 (e-\log (x))}-\frac {e^2 \log \left (2 e^{-2 x}\right )}{4 x (e+\log (x))^2}-\frac {e^2}{2 (e+\log (x))}\right ) \, dx\\ &=\frac {x}{2}+\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e-\log (x))^2} \, dx-\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e+\log (x))^2} \, dx-\frac {1}{4} e^2 \int \frac {1}{e-\log (x)} \, dx-\frac {1}{4} e^2 \int \frac {1}{e+\log (x)} \, dx\\ &=\frac {x}{2}+\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e-\log (x))^2} \, dx-\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e+\log (x))^2} \, dx-\frac {1}{4} e^2 \operatorname {Subst}\left (\int \frac {e^x}{e-x} \, dx,x,\log (x)\right )-\frac {1}{4} e^2 \operatorname {Subst}\left (\int \frac {e^x}{e+x} \, dx,x,\log (x)\right )\\ &=\frac {x}{2}+\frac {1}{4} e^{2+e} \text {Ei}(-e+\log (x))-\frac {1}{4} e^{2-e} \text {Ei}(e+\log (x))+\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e-\log (x))^2} \, dx-\frac {1}{8} e^2 \int \frac {\log \left (2 e^{-2 x}\right )}{x (e+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 33, normalized size = 0.97 \begin {gather*} \frac {1}{2} \left (x+\frac {e^3 \log \left (2 e^{-2 x}\right )}{2 \left (e^2-\log ^2(x)\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.27, size = 37, normalized size = 1.09 \begin {gather*} \frac {2 \, x \log \relax (x)^{2} + 2 \, x e^{3} - 2 \, x e^{2} - e^{3} \log \relax (2)}{4 \, {\left (\log \relax (x)^{2} - e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 37, normalized size = 1.09 \begin {gather*} \frac {2 \, x \log \relax (x)^{2} + 2 \, x e^{3} - 2 \, x e^{2} - e^{3} \log \relax (2)}{4 \, {\left (\log \relax (x)^{2} - e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.54, size = 59, normalized size = 1.74

method	result	size
default	\(\frac {\left ({\mathrm e}^{2}-{\mathrm e}^{3}\right ) x -x \ln \relax (x )^{2}-{\mathrm e}^{3} \left (\ln \left ({\mathrm e}^{x}\right )-x \right )+\frac {{\mathrm e}^{3} \left (\ln \left (2 \,{\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )\right )}{2}}{2 \,{\mathrm e}^{2}-2 \ln \relax (x )^{2}}\)	\(59\)
risch	\(-\frac {{\mathrm e}^{3} \ln \left ({\mathrm e}^{x}\right )}{2 \left ({\mathrm e}^{2}-\ln \relax (x )^{2}\right )}+\frac {-i \pi \,{\mathrm e}^{3} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+2 i \pi \,{\mathrm e}^{3} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-i \pi \,{\mathrm e}^{3} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 \,{\mathrm e}^{3} \ln \relax (2)-4 \,{\mathrm e}^{2} x +4 x \ln \relax (x )^{2}}{-8 \,{\mathrm e}^{2}+8 \ln \relax (x )^{2}}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 37, normalized size = 1.09 \begin {gather*} \frac {2 \, x \log \relax (x)^{2} + 2 \, x {\left (e^{3} - e^{2}\right )} - e^{3} \log \relax (2)}{4 \, {\left (\log \relax (x)^{2} - e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.07, size = 29, normalized size = 0.85 \begin {gather*} \frac {x}{2}-\frac {{\mathrm {e}}^3\,\left (2\,x-\ln \relax (2)\right )}{4\,\left ({\mathrm {e}}^2-{\ln \relax (x)}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x)^4 + x*exp(4) - x*exp(5) - log(x)^2*(2*x*exp(2) - x*exp(3)) + log(2*exp(-2*x))*exp(3)*log(x))/(2*
x*log(x)^4 + 2*x*exp(4) - 4*x*exp(2)*log(x)^2),x)

[Out]

x/2 - (exp(3)*(2*x - log(2)))/(4*(exp(2) - log(x)^2))

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sympy [A]  time = 0.30, size = 27, normalized size = 0.79 \begin {gather*} \frac {x}{2} + \frac {2 x e^{3} - e^{3} \log {\relax (2 )}}{4 \log {\relax (x )}^{2} - 4 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/2 + (2*x*exp(3) - exp(3)*log(2))/(4*log(x)**2 - 4*exp(2))

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