3.60.92 \(\int \frac {e^x (-10+10 x)+(-5 e^x+5 x) \log (\frac {e^{2 x}-2 e^x x+x^2}{x^2})+(6 e^x x-6 x^2) \log ^2(\frac {e^{2 x}-2 e^x x+x^2}{x^2})}{(e^x-x) \log ^2(\frac {e^{2 x}-2 e^x x+x^2}{x^2})} \, dx\)

Optimal. Leaf size=24 \[ x \left (3 x-\frac {5}{\log \left (\frac {\left (-e^x+x\right )^2}{x^2}\right )}\right ) \]

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Rubi [F]  time = 0.78, 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 (-10+10 x)+\left (-5 e^x+5 x\right ) \log \left (\frac {e^{2 x}-2 e^x x+x^2}{x^2}\right )+\left (6 e^x x-6 x^2\right ) \log ^2\left (\frac {e^{2 x}-2 e^x x+x^2}{x^2}\right )}{\left (e^x-x\right ) \log ^2\left (\frac {e^{2 x}-2 e^x x+x^2}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-10 + 10*x) + (-5*E^x + 5*x)*Log[(E^(2*x) - 2*E^x*x + x^2)/x^2] + (6*E^x*x - 6*x^2)*Log[(E^(2*x) - 2
*E^x*x + x^2)/x^2]^2)/((E^x - x)*Log[(E^(2*x) - 2*E^x*x + x^2)/x^2]^2),x]

[Out]

3*x^2 - 10*Defer[Int][E^x/((E^x - x)*Log[(E^x - x)^2/x^2]^2), x] + 10*Defer[Int][(E^x*x)/((E^x - x)*Log[(E^x -
x)^2/x^2]^2), x] - 5*Defer[Int][Log[(E^x - x)^2/x^2]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (6 x+\frac {10 e^x (-1+x)}{\left (e^x-x\right ) \log ^2\left (\frac {\left (e^x-x\right )^2}{x^2}\right )}-\frac {5}{\log \left (\frac {\left (e^x-x\right )^2}{x^2}\right )}\right ) \, dx\\ &=3 x^2-5 \int \frac {1}{\log \left (\frac {\left (e^x-x\right )^2}{x^2}\right )} \, dx+10 \int \frac {e^x (-1+x)}{\left (e^x-x\right ) \log ^2\left (\frac {\left (e^x-x\right )^2}{x^2}\right )} \, dx\\ &=3 x^2-5 \int \frac {1}{\log \left (\frac {\left (e^x-x\right )^2}{x^2}\right )} \, dx+10 \int \left (-\frac {e^x}{\left (e^x-x\right ) \log ^2\left (\frac {\left (e^x-x\right )^2}{x^2}\right )}+\frac {e^x x}{\left (e^x-x\right ) \log ^2\left (\frac {\left (e^x-x\right )^2}{x^2}\right )}\right ) \, dx\\ &=3 x^2-5 \int \frac {1}{\log \left (\frac {\left (e^x-x\right )^2}{x^2}\right )} \, dx-10 \int \frac {e^x}{\left (e^x-x\right ) \log ^2\left (\frac {\left (e^x-x\right )^2}{x^2}\right )} \, dx+10 \int \frac {e^x x}{\left (e^x-x\right ) \log ^2\left (\frac {\left (e^x-x\right )^2}{x^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 25, normalized size = 1.04 \begin {gather*} 3 x^2-\frac {5 x}{\log \left (\frac {\left (e^x-x\right )^2}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-10 + 10*x) + (-5*E^x + 5*x)*Log[(E^(2*x) - 2*E^x*x + x^2)/x^2] + (6*E^x*x - 6*x^2)*Log[(E^(2*
x) - 2*E^x*x + x^2)/x^2]^2)/((E^x - x)*Log[(E^(2*x) - 2*E^x*x + x^2)/x^2]^2),x]

[Out]

3*x^2 - (5*x)/Log[(E^x - x)^2/x^2]

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fricas [B]  time = 0.65, size = 48, normalized size = 2.00 \begin {gather*} \frac {3 \, x^{2} \log \left (\frac {x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}{x^{2}}\right ) - 5 \, x}{\log \left (\frac {x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(x)*x-6*x^2)*log((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2+(-5*exp(x)+5*x)*log((exp(x)^2-2*exp(x)*x+x^
2)/x^2)+(10*x-10)*exp(x))/(exp(x)-x)/log((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 2.32, size = 48, normalized size = 2.00 \begin {gather*} \frac {3 \, x^{2} \log \left (\frac {x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}{x^{2}}\right ) - 5 \, x}{\log \left (\frac {x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(x)*x-6*x^2)*log((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2+(-5*exp(x)+5*x)*log((exp(x)^2-2*exp(x)*x+x^
2)/x^2)+(10*x-10)*exp(x))/(exp(x)-x)/log((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.21, size = 49, normalized size = 2.04

method	result	size
norman	\(\frac {-5 x +3 x^{2} \ln \left (\frac {{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}{x^{2}}\right )}{\ln \left (\frac {{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}{x^{2}}\right )}\)	\(49\)
risch	\(3 x^{2}-\frac {10 i x}{\pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-x \right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-x \right )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-x \right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-x \right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-x \right )^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-x \right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x \right )^{2}}{x^{2}}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-x \right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x \right )^{2}}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right )+\pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x \right )^{2}}{x^{2}}\right )^{3}-\pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x \right )^{2}}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )-\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-4 i \ln \relax (x )+4 i \ln \left (x -{\mathrm e}^{x}\right )}\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*exp(x)*x-6*x^2)*ln((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2+(-5*exp(x)+5*x)*ln((exp(x)^2-2*exp(x)*x+x^2)/x^2)+
(10*x-10)*exp(x))/(exp(x)-x)/ln((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 39, normalized size = 1.62 \begin {gather*} \frac {6 \, x^{2} \log \relax (x) - 6 \, x^{2} \log \left (-x + e^{x}\right ) + 5 \, x}{2 \, {\left (\log \relax (x) - \log \left (-x + e^{x}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(x)*x-6*x^2)*log((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2+(-5*exp(x)+5*x)*log((exp(x)^2-2*exp(x)*x+x^
2)/x^2)+(10*x-10)*exp(x))/(exp(x)-x)/log((exp(x)^2-2*exp(x)*x+x^2)/x^2)^2,x, algorithm="maxima")

[Out]

1/2*(6*x^2*log(x) - 6*x^2*log(-x + e^x) + 5*x)/(log(x) - log(-x + e^x))

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mupad [B]  time = 4.94, size = 30, normalized size = 1.25 \begin {gather*} 3\,x^2-\frac {5\,x}{\ln \left (\frac {1}{x^2}\right )+\ln \left ({\mathrm {e}}^{2\,x}-2\,x\,{\mathrm {e}}^x+x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(10*x - 10) + log((exp(2*x) - 2*x*exp(x) + x^2)/x^2)*(5*x - 5*exp(x)) + log((exp(2*x) - 2*x*exp(x
) + x^2)/x^2)^2*(6*x*exp(x) - 6*x^2))/(log((exp(2*x) - 2*x*exp(x) + x^2)/x^2)^2*(x - exp(x))),x)

[Out]

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

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sympy [A]  time = 0.21, size = 27, normalized size = 1.12 \begin {gather*} 3 x^{2} - \frac {5 x}{\log {\left (\frac {x^{2} - 2 x e^{x} + e^{2 x}}{x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(x)*x-6*x**2)*ln((exp(x)**2-2*exp(x)*x+x**2)/x**2)**2+(-5*exp(x)+5*x)*ln((exp(x)**2-2*exp(x)*
x+x**2)/x**2)+(10*x-10)*exp(x))/(exp(x)-x)/ln((exp(x)**2-2*exp(x)*x+x**2)/x**2)**2,x)

[Out]

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

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