3.19.19 \(\int \frac {-4 x+x^3+e^x (4-x^2)+(-4+x^2) \log (\frac {1}{x})+(2 x-2 x^2+2 e^x x^2) \log (e^x-x-\log (\frac {1}{x}))}{-e^x x^2+x^3+x^2 \log (\frac {1}{x})} \, dx\)

Optimal. Leaf size=29 \[ -2-\frac {-4+x}{x}+x-\log ^2\left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \]

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Rubi [F]  time = 1.31, 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 {-4 x+x^3+e^x \left (4-x^2\right )+\left (-4+x^2\right ) \log \left (\frac {1}{x}\right )+\left (2 x-2 x^2+2 e^x x^2\right ) \log \left (e^x-x-\log \left (\frac {1}{x}\right )\right )}{-e^x x^2+x^3+x^2 \log \left (\frac {1}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 25, normalized size = 0.86 \begin {gather*} \frac {4}{x}+x-\log ^2\left (e^x-x-\log \left (\frac {1}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x + x^3 + E^x*(4 - x^2) + (-4 + x^2)*Log[x^(-1)] + (2*x - 2*x^2 + 2*E^x*x^2)*Log[E^x - x - Log[x
^(-1)]])/(-(E^x*x^2) + x^3 + x^2*Log[x^(-1)]),x]

[Out]

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

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fricas [A]  time = 0.79, size = 29, normalized size = 1.00 \begin {gather*} -\frac {x \log \left (-x + e^{x} - \log \left (\frac {1}{x}\right )\right )^{2} - x^{2} - 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-2*x^2+2*x)*log(-log(1/x)+exp(x)-x)+(x^2-4)*log(1/x)+(-x^2+4)*exp(x)+x^3-4*x)/(x^2*log
(1/x)-exp(x)*x^2+x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 25, normalized size = 0.86 \begin {gather*} -\frac {x \log \left (-x + e^{x} + \log \relax (x)\right )^{2} - x^{2} - 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-2*x^2+2*x)*log(-log(1/x)+exp(x)-x)+(x^2-4)*log(1/x)+(-x^2+4)*exp(x)+x^3-4*x)/(x^2*log
(1/x)-exp(x)*x^2+x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 24, normalized size = 0.83

method	result	size
risch	\(-\ln \left (\ln \relax (x )+{\mathrm e}^{x}-x \right )^{2}+\frac {x^{2}+4}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x^2-2*x^2+2*x)*ln(-ln(1/x)+exp(x)-x)+(x^2-4)*ln(1/x)+(-x^2+4)*exp(x)+x^3-4*x)/(x^2*ln(1/x)-exp(
x)*x^2+x^3),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(x)+exp(x)-x)^2+1/x*(x^2+4)

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maxima [A]  time = 0.53, size = 25, normalized size = 0.86 \begin {gather*} -\frac {x \log \left (-x + e^{x} + \log \relax (x)\right )^{2} - x^{2} - 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x^2-2*x^2+2*x)*log(-log(1/x)+exp(x)-x)+(x^2-4)*log(1/x)+(-x^2+4)*exp(x)+x^3-4*x)/(x^2*log
(1/x)-exp(x)*x^2+x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.34, size = 24, normalized size = 0.83 \begin {gather*} x-{\ln \left ({\mathrm {e}}^x-\ln \left (\frac {1}{x}\right )-x\right )}^2+\frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(x) - log(1/x) - x)*(2*x + 2*x^2*exp(x) - 2*x^2) - 4*x - exp(x)*(x^2 - 4) + x^3 + log(1/x)*(x^2 -
4))/(x^2*log(1/x) - x^2*exp(x) + x^3),x)

[Out]

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

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sympy [A]  time = 0.71, size = 17, normalized size = 0.59 \begin {gather*} x - \log {\left (- x + e^{x} - \log {\left (\frac {1}{x} \right )} \right )}^{2} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x**2-2*x**2+2*x)*ln(-ln(1/x)+exp(x)-x)+(x**2-4)*ln(1/x)+(-x**2+4)*exp(x)+x**3-4*x)/(x**2*
ln(1/x)-exp(x)*x**2+x**3),x)

[Out]

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

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