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

Optimal. Leaf size=19 \[ \frac {1}{2 \log \left (2 \left (x+\log \left (e^x x^2\right )\right )\right )} \]

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Rubi [A]  time = 0.10, antiderivative size = 21, normalized size of antiderivative = 1.11, number of steps used = 1, number of rules used = 1, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6686} \begin {gather*} \frac {1}{2 \log \left (2 \log \left (e^x x^2\right )+2 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*Log[2*x + 2*Log[E^x*x^2]])

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{2 \log \left (2 \left (x+\log \left (e^x x^2\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*Log[2*(x + Log[E^x*x^2])])

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fricas [A]  time = 0.72, size = 18, normalized size = 0.95 \begin {gather*} \frac {1}{2 \, \log \left (2 \, x + 2 \, \log \left (x^{2} e^{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/log(2*x + 2*log(x^2*e^x))

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giac [A]  time = 0.33, size = 15, normalized size = 0.79 \begin {gather*} \frac {1}{2 \, \log \left (4 \, x + 2 \, \log \left (x^{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/log(4*x + 2*log(x^2))

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maple [C]  time = 0.10, size = 97, normalized size = 5.11

method	result	size
risch	\(\frac {1}{2 \ln \left (4 \ln \relax (x )+2 \ln \left ({\mathrm e}^{x}\right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}-i \pi \,\mathrm {csgn}\left (i x^{2} {\mathrm e}^{x}\right ) \left (-\mathrm {csgn}\left (i x^{2} {\mathrm e}^{x}\right )+\mathrm {csgn}\left (i x^{2}\right )\right ) \left (-\mathrm {csgn}\left (i x^{2} {\mathrm e}^{x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x}\right )\right )+2 x \right )}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{2 \, {\left (2 \, \log \relax (2) + \log \left (x + \log \relax (x)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(2*log(2) + log(x + log(x)))

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mupad [B]  time = 1.21, size = 13, normalized size = 0.68 \begin {gather*} \frac {1}{2\,\ln \left (4\,x+\ln \left (x^4\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2*log(4*x + log(x^4)))

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sympy [A]  time = 0.25, size = 17, normalized size = 0.89 \begin {gather*} \frac {1}{2 \log {\left (2 x + 2 \log {\left (x^{2} e^{x} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2*log(2*x + 2*log(x**2*exp(x))))

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