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

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

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Rubi [F]  time = 6.91, 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+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 26, normalized size = 1.00 \begin {gather*} \frac {2-x}{x \log \left (e^4+e^x-x-\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 23, normalized size = 0.88 \begin {gather*} -\frac {x - 2}{x \log \left (-x + e^{4} + e^{x} - \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Sign error %%%{ln(` w`),0%%%}
Simplificat

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maple [A]  time = 0.04, size = 24, normalized size = 0.92

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x-2)/x/ln(-ln(x)+exp(x)+exp(4)-x)

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maxima [A]  time = 0.44, size = 23, normalized size = 0.88 \begin {gather*} -\frac {x - 2}{x \log \left (-x + e^{4} + e^{x} - \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.98, size = 140, normalized size = 5.38 \begin {gather*} \frac {2\,\left (2\,x+x\,{\mathrm {e}}^4+x^2\,{\mathrm {e}}^4+x^3\,{\mathrm {e}}^4+x^2-x^4+1\right )}{x^2\,\left (x-x\,{\mathrm {e}}^x+1\right )\,\left (x^2+x+1\right )}-\frac {2}{x^2}-\frac {2\,\ln \relax (x)}{x\,\left (x-x\,{\mathrm {e}}^x+1\right )}-\frac {\frac {x-2}{x}-\frac {2\,\ln \left ({\mathrm {e}}^4-x+{\mathrm {e}}^x-\ln \relax (x)\right )\,\left (x-{\mathrm {e}}^4-{\mathrm {e}}^x+\ln \relax (x)\right )}{x\,\left (x-x\,{\mathrm {e}}^x+1\right )}}{\ln \left ({\mathrm {e}}^4-x+{\mathrm {e}}^x-\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.92, size = 17, normalized size = 0.65 \begin {gather*} \frac {2 - x}{x \log {\left (- x + e^{x} - \log {\relax (x )} + e^{4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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