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

Optimal. Leaf size=17 \[ -2-\log \left (\log \left (-1+e^x-x (2+x)\right )\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6684} \begin {gather*} -\log \left (\log \left (-x^2-2 x+e^x-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[Log[-1 + E^x - 2*x - x^2]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 15, normalized size = 0.88 \begin {gather*} -\log \left (\log \left (e^x-(1+x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[Log[E^x - (1 + x)^2]]

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fricas [A]  time = 0.58, size = 16, normalized size = 0.94 \begin {gather*} -\log \left (\log \left (-x^{2} - 2 \, x + e^{x} - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(-x^2 - 2*x + e^x - 1))

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giac [A]  time = 0.27, size = 16, normalized size = 0.94 \begin {gather*} -\log \left (\log \left (-x^{2} - 2 \, x + e^{x} - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(-x^2 - 2*x + e^x - 1))

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maple [A]  time = 0.03, size = 17, normalized size = 1.00

method	result	size
norman	\(-\ln \left (\ln \left ({\mathrm e}^{x}-x^{2}-2 x -1\right )\right )\)	\(17\)
risch	\(-\ln \left (\ln \left ({\mathrm e}^{x}-x^{2}-2 x -1\right )\right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(ln(exp(x)-x^2-2*x-1))

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maxima [A]  time = 0.44, size = 16, normalized size = 0.94 \begin {gather*} -\log \left (\log \left (-x^{2} - 2 \, x + e^{x} - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(-x^2 - 2*x + e^x - 1))

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mupad [B]  time = 0.98, size = 16, normalized size = 0.94 \begin {gather*} -\ln \left (\ln \left ({\mathrm {e}}^x-2\,x-x^2-1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(exp(x) - 2*x - x^2 - 1))

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sympy [A]  time = 0.24, size = 15, normalized size = 0.88 \begin {gather*} - \log {\left (\log {\left (- x^{2} - 2 x + e^{x} - 1 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(-x**2 - 2*x + exp(x) - 1))

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