3.95.22 \(\int \frac {2 x^2+8 x^3+8 x^4+e^x (-2-8 x-8 x^2)+(e^x (-2-4 x)+2 x+4 x^2) \log (-\frac {4 e^x x}{e^x-x})}{(e^x x^3-x^4) \log ^3(-\frac {4 e^x x}{e^x-x})} \, dx\)

Optimal. Leaf size=26 \[ \frac {\left (2+\frac {1}{x}\right )^2}{\log ^2\left (\frac {4 e^x x}{-e^x+x}\right )} \]

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Rubi [F]  time = 3.22, 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 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x^2 + 8*x^3 + 8*x^4 + E^x*(-2 - 8*x - 8*x^2) + (E^x*(-2 - 4*x) + 2*x + 4*x^2)*Log[(-4*E^x*x)/(E^x - x)]
)/((E^x*x^3 - x^4)*Log[(-4*E^x*x)/(E^x - x)]^3),x]

[Out]

-2*Defer[Int][1/(x^3*Log[(-4*E^x*x)/(E^x - x)]^3), x] - 8*Defer[Int][1/(x^2*Log[(-4*E^x*x)/(E^x - x)]^3), x] -
2*Defer[Int][1/((E^x - x)*x^2*Log[(-4*E^x*x)/(E^x - x)]^3), x] - 8*Defer[Int][1/(x*Log[(-4*E^x*x)/(E^x - x)]^
3), x] - 6*Defer[Int][1/((E^x - x)*x*Log[(-4*E^x*x)/(E^x - x)]^3), x] + 8*Defer[Int][x/((E^x - x)*Log[(-4*E^x*
x)/(E^x - x)]^3), x] - 2*Defer[Int][1/(x^3*Log[(-4*E^x*x)/(E^x - x)]^2), x] - 4*Defer[Int][1/(x^2*Log[(-4*E^x*
x)/(E^x - x)]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 (1+2 x) \left ((1+2 x) \left (-e^x+x^2\right )+\left (-e^x+x\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )\right )}{\left (e^x-x\right ) x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\\ &=2 \int \frac {(1+2 x) \left ((1+2 x) \left (-e^x+x^2\right )+\left (-e^x+x\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )\right )}{\left (e^x-x\right ) x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\\ &=2 \int \left (\frac {(-1+x) (1+2 x)^2}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}-\frac {(1+2 x) \left (1+2 x+\log \left (-\frac {4 e^x x}{e^x-x}\right )\right )}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}\right ) \, dx\\ &=2 \int \frac {(-1+x) (1+2 x)^2}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-2 \int \frac {(1+2 x) \left (1+2 x+\log \left (-\frac {4 e^x x}{e^x-x}\right )\right )}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\\ &=2 \int \left (-\frac {1}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}-\frac {3}{\left (e^x-x\right ) x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}+\frac {4 x}{\left (e^x-x\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}\right ) \, dx-2 \int \left (\frac {(1+2 x)^2}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}+\frac {1+2 x}{x^3 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\right )-2 \int \frac {(1+2 x)^2}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-2 \int \frac {1+2 x}{x^3 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-6 \int \frac {1}{\left (e^x-x\right ) x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx+8 \int \frac {x}{\left (e^x-x\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\\ &=-\left (2 \int \left (\frac {1}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}+\frac {4}{x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}+\frac {4}{x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}\right ) \, dx\right )-2 \int \left (\frac {1}{x^3 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )}+\frac {2}{x^2 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )}\right ) \, dx-2 \int \frac {1}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-6 \int \frac {1}{\left (e^x-x\right ) x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx+8 \int \frac {x}{\left (e^x-x\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\\ &=-\left (2 \int \frac {1}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\right )-2 \int \frac {1}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-2 \int \frac {1}{x^3 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-4 \int \frac {1}{x^2 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-6 \int \frac {1}{\left (e^x-x\right ) x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-8 \int \frac {1}{x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx-8 \int \frac {1}{x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx+8 \int \frac {x}{\left (e^x-x\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 29, normalized size = 1.12 \begin {gather*} \frac {(1+2 x)^2}{x^2 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + 8*x^3 + 8*x^4 + E^x*(-2 - 8*x - 8*x^2) + (E^x*(-2 - 4*x) + 2*x + 4*x^2)*Log[(-4*E^x*x)/(E^x
- x)])/((E^x*x^3 - x^4)*Log[(-4*E^x*x)/(E^x - x)]^3),x]

[Out]

(1 + 2*x)^2/(x^2*Log[(-4*E^x*x)/(E^x - x)]^2)

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fricas [A]  time = 1.12, size = 30, normalized size = 1.15 \begin {gather*} \frac {4 \, x^{2} + 4 \, x + 1}{x^{2} \log \left (\frac {4 \, x e^{x}}{x - e^{x}}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-2)*exp(x)+4*x^2+2*x)*log(-4*x*exp(x)/(exp(x)-x))+(-8*x^2-8*x-2)*exp(x)+8*x^4+8*x^3+2*x^2)/(e
xp(x)*x^3-x^4)/log(-4*x*exp(x)/(exp(x)-x))^3,x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-2)*exp(x)+4*x^2+2*x)*log(-4*x*exp(x)/(exp(x)-x))+(-8*x^2-8*x-2)*exp(x)+8*x^4+8*x^3+2*x^2)/(e
xp(x)*x^3-x^4)/log(-4*x*exp(x)/(exp(x)-x))^3,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.24, size = 267, normalized size = 10.27

method	result	size
risch	\(-\frac {4 \left (4 x^{2}+4 x +1\right )}{x^{2} \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{3}+\pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{3}-2 i \ln \left (x -{\mathrm e}^{x}\right )+4 i \ln \relax (2)+2 i \ln \left ({\mathrm e}^{x}\right )+2 i \ln \relax (x )\right )^{2}}\)	\(267\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x-2)*exp(x)+4*x^2+2*x)*ln(-4*x*exp(x)/(exp(x)-x))+(-8*x^2-8*x-2)*exp(x)+8*x^4+8*x^3+2*x^2)/(exp(x)*x
^3-x^4)/ln(-4*x*exp(x)/(exp(x)-x))^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.51, size = 95, normalized size = 3.65 \begin {gather*} \frac {4 \, x^{2} + 4 \, x + 1}{x^{4} + 4 \, x^{3} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} + x^{2} \log \left (x - e^{x}\right )^{2} + x^{2} \log \relax (x)^{2} - 2 \, {\left (x^{3} + 2 \, x^{2} \log \relax (2) + x^{2} \log \relax (x)\right )} \log \left (x - e^{x}\right ) + 2 \, {\left (x^{3} + 2 \, x^{2} \log \relax (2)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-2)*exp(x)+4*x^2+2*x)*log(-4*x*exp(x)/(exp(x)-x))+(-8*x^2-8*x-2)*exp(x)+8*x^4+8*x^3+2*x^2)/(e
xp(x)*x^3-x^4)/log(-4*x*exp(x)/(exp(x)-x))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.39, size = 377, normalized size = 14.50 \begin {gather*} \frac {\frac {\left (x-{\mathrm {e}}^x\right )\,\left (2\,x+1\right )}{x^2\,\left ({\mathrm {e}}^x-x^2\right )}+\frac {\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )\,\left (x-{\mathrm {e}}^x\right )\,\left (2\,{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^{2\,x}-5\,x^2\,{\mathrm {e}}^x-7\,x^3\,{\mathrm {e}}^x+2\,x^4\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x+3\,x^3+4\,x^4\right )}{x^2\,{\left ({\mathrm {e}}^x-x^2\right )}^3}}{\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )}+\frac {\frac {{\left (2\,x+1\right )}^2}{x^2}-\frac {\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )\,\left (x-{\mathrm {e}}^x\right )\,\left (2\,x+1\right )}{x^2\,\left ({\mathrm {e}}^x-x^2\right )}}{{\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )}^2}+\frac {2\,x+2}{x^2}+\frac {-4\,x^6+18\,x^5-23\,x^4+4\,x^3+3\,x^2+2\,x}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{2\,x}-2\,x^2\,{\mathrm {e}}^x+x^4\right )}-\frac {2\,x^4-5\,x^3+x^2-x+6}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^x-x^2\right )}-\frac {2\,x^8-11\,x^7+20\,x^6-11\,x^5-4\,x^4+4\,x^3}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{3\,x}+3\,x^4\,{\mathrm {e}}^x-3\,x^2\,{\mathrm {e}}^{2\,x}-x^6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((4*x*exp(x))/(x - exp(x)))*(2*x - exp(x)*(4*x + 2) + 4*x^2) - exp(x)*(8*x + 8*x^2 + 2) + 2*x^2 + 8*x^
3 + 8*x^4)/(log((4*x*exp(x))/(x - exp(x)))^3*(x^3*exp(x) - x^4)),x)

[Out]

(((x - exp(x))*(2*x + 1))/(x^2*(exp(x) - x^2)) + (log((4*x*exp(x))/(x - exp(x)))*(x - exp(x))*(2*exp(2*x) + 2*
x*exp(2*x) - 5*x^2*exp(x) - 7*x^3*exp(x) + 2*x^4*exp(x) - x*exp(x) + 3*x^3 + 4*x^4))/(x^2*(exp(x) - x^2)^3))/l
og((4*x*exp(x))/(x - exp(x))) + ((2*x + 1)^2/x^2 - (log((4*x*exp(x))/(x - exp(x)))*(x - exp(x))*(2*x + 1))/(x^
2*(exp(x) - x^2)))/log((4*x*exp(x))/(x - exp(x)))^2 + (2*x + 2)/x^2 + (2*x + 3*x^2 + 4*x^3 - 23*x^4 + 18*x^5 -
4*x^6)/((2*x - x^2)*(exp(2*x) - 2*x^2*exp(x) + x^4)) - (x^2 - x - 5*x^3 + 2*x^4 + 6)/((2*x - x^2)*(exp(x) - x
^2)) - (4*x^3 - 4*x^4 - 11*x^5 + 20*x^6 - 11*x^7 + 2*x^8)/((2*x - x^2)*(exp(3*x) + 3*x^4*exp(x) - 3*x^2*exp(2*
x) - x^6))

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sympy [A]  time = 0.27, size = 29, normalized size = 1.12 \begin {gather*} \frac {4 x^{2} + 4 x + 1}{x^{2} \log {\left (- \frac {4 x e^{x}}{- x + e^{x}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-2)*exp(x)+4*x**2+2*x)*ln(-4*x*exp(x)/(exp(x)-x))+(-8*x**2-8*x-2)*exp(x)+8*x**4+8*x**3+2*x**2
)/(exp(x)*x**3-x**4)/ln(-4*x*exp(x)/(exp(x)-x))**3,x)

[Out]

(4*x**2 + 4*x + 1)/(x**2*log(-4*x*exp(x)/(-x + exp(x)))**2)

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