3.7.54 \(\int \frac {-2+e^{-4+x+x^2} (-2 x-4 x^2)}{-5 e^{-4+x+x^2} x-5 x \log (x)+(2 e^{-4+x+x^2} x+2 x \log (x)) \log (4 e^{-4+x+x^2}+4 \log (x))} \, dx\)

Optimal. Leaf size=23 \[ 5-\log \left (-\frac {5}{2}+\log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right ) \]

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Rubi [A]  time = 0.65, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6741, 12, 6684} \begin {gather*} -\log \left (5-2 \log \left (4 \left (e^{x^2+x-4}+\log (x)\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + E^(-4 + x + x^2)*(-2*x - 4*x^2))/(-5*E^(-4 + x + x^2)*x - 5*x*Log[x] + (2*E^(-4 + x + x^2)*x + 2*x*L
og[x])*Log[4*E^(-4 + x + x^2) + 4*Log[x]]),x]

[Out]

-Log[5 - 2*Log[4*(E^(-4 + x + x^2) + Log[x])]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6684

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

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 \left (2-e^{-4+x+x^2} \left (-2 x-4 x^2\right )\right )}{x \left (e^{x+x^2}+e^4 \log (x)\right ) \left (5-2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right )} \, dx\\ &=e^4 \int \frac {2-e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{x \left (e^{x+x^2}+e^4 \log (x)\right ) \left (5-2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right )} \, dx\\ &=-\log \left (5-2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 21, normalized size = 0.91 \begin {gather*} -\log \left (-5+2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + E^(-4 + x + x^2)*(-2*x - 4*x^2))/(-5*E^(-4 + x + x^2)*x - 5*x*Log[x] + (2*E^(-4 + x + x^2)*x +
2*x*Log[x])*Log[4*E^(-4 + x + x^2) + 4*Log[x]]),x]

[Out]

-Log[-5 + 2*Log[4*(E^(-4 + x + x^2) + Log[x])]]

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fricas [A]  time = 0.64, size = 22, normalized size = 0.96 \begin {gather*} -\log \left (2 \, \log \left (4 \, e^{\left (x^{2} + x - 4\right )} + 4 \, \log \relax (x)\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*log(x)+2*x*exp(x^2+x-4))*log(4*log(x)+4*exp(x^2+x-4))-5*x*log(x)
-5*x*exp(x^2+x-4)),x, algorithm="fricas")

[Out]

-log(2*log(4*e^(x^2 + x - 4) + 4*log(x)) - 5)

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giac [A]  time = 0.45, size = 23, normalized size = 1.00 \begin {gather*} -\log \left (2 \, \log \left (4 \, e^{4} \log \relax (x) + 4 \, e^{\left (x^{2} + x\right )}\right ) - 13\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*log(x)+2*x*exp(x^2+x-4))*log(4*log(x)+4*exp(x^2+x-4))-5*x*log(x)
-5*x*exp(x^2+x-4)),x, algorithm="giac")

[Out]

-log(2*log(4*e^4*log(x) + 4*e^(x^2 + x)) - 13)

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maple [A]  time = 0.04, size = 21, normalized size = 0.91

method	result	size
risch	\(-\ln \left (\ln \left (4 \ln \relax (x )+4 \,{\mathrm e}^{x^{2}+x -4}\right )-\frac {5}{2}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*ln(x)+2*x*exp(x^2+x-4))*ln(4*ln(x)+4*exp(x^2+x-4))-5*x*ln(x)-5*x*exp(x
^2+x-4)),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(4*ln(x)+4*exp(x^2+x-4))-5/2)

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maxima [A]  time = 0.66, size = 22, normalized size = 0.96 \begin {gather*} -\log \left (2 \, \log \relax (2) + \log \left (e^{4} \log \relax (x) + e^{\left (x^{2} + x\right )}\right ) - \frac {13}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*log(x)+2*x*exp(x^2+x-4))*log(4*log(x)+4*exp(x^2+x-4))-5*x*log(x)
-5*x*exp(x^2+x-4)),x, algorithm="maxima")

[Out]

-log(2*log(2) + log(e^4*log(x) + e^(x^2 + x)) - 13/2)

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mupad [B]  time = 0.93, size = 21, normalized size = 0.91 \begin {gather*} -\ln \left (\ln \left (4\,\ln \relax (x)+4\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^x\right )-\frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + x^2 - 4)*(2*x + 4*x^2) + 2)/(5*x*log(x) - log(4*exp(x + x^2 - 4) + 4*log(x))*(2*x*log(x) + 2*x*ex
p(x + x^2 - 4)) + 5*x*exp(x + x^2 - 4)),x)

[Out]

-log(log(4*log(x) + 4*exp(x^2)*exp(-4)*exp(x)) - 5/2)

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sympy [A]  time = 0.92, size = 22, normalized size = 0.96 \begin {gather*} - \log {\left (\log {\left (4 e^{x^{2} + x - 4} + 4 \log {\relax (x )} \right )} - \frac {5}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2-2*x)*exp(x**2+x-4)-2)/((2*x*ln(x)+2*x*exp(x**2+x-4))*ln(4*ln(x)+4*exp(x**2+x-4))-5*x*ln(x)
-5*x*exp(x**2+x-4)),x)

[Out]

-log(log(4*exp(x**2 + x - 4) + 4*log(x)) - 5/2)

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