3.93.19 \(\int \frac {e^2 x+32 e^4 \log (-2 x)}{2 x+x^2+16 e^2 x \log ^2(-2 x)} \, dx\)

Optimal. Leaf size=19 \[ e^2 \log \left (2+x+16 e^2 \log ^2(-2 x)\right ) \]

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Rubi [F]  time = 0.39, 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 {e^2 x+32 e^4 \log (-2 x)}{2 x+x^2+16 e^2 x \log ^2(-2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^2*x + 32*E^4*Log[-2*x])/(2*x + x^2 + 16*E^2*x*Log[-2*x]^2),x]

[Out]

E^2*Defer[Int][(2 + x + 16*E^2*Log[-2*x]^2)^(-1), x] + 32*E^4*Defer[Int][Log[-2*x]/(x*(2 + x + 16*E^2*Log[-2*x
]^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 \left (x+32 e^2 \log (-2 x)\right )}{2 x+x^2+16 e^2 x \log ^2(-2 x)} \, dx\\ &=e^2 \int \frac {x+32 e^2 \log (-2 x)}{2 x+x^2+16 e^2 x \log ^2(-2 x)} \, dx\\ &=e^2 \int \left (\frac {1}{2+x+16 e^2 \log ^2(-2 x)}+\frac {32 e^2 \log (-2 x)}{x \left (2+x+16 e^2 \log ^2(-2 x)\right )}\right ) \, dx\\ &=e^2 \int \frac {1}{2+x+16 e^2 \log ^2(-2 x)} \, dx+\left (32 e^4\right ) \int \frac {\log (-2 x)}{x \left (2+x+16 e^2 \log ^2(-2 x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 21, normalized size = 1.11 \begin {gather*} e^2 \log \left (4+2 x+32 e^2 \log ^2(-2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*x + 32*E^4*Log[-2*x])/(2*x + x^2 + 16*E^2*x*Log[-2*x]^2),x]

[Out]

E^2*Log[4 + 2*x + 32*E^2*Log[-2*x]^2]

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fricas [A]  time = 0.61, size = 17, normalized size = 0.89 \begin {gather*} e^{2} \log \left (16 \, e^{2} \log \left (-2 \, x\right )^{2} + x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*exp(1)^2*exp(2)*log(-2*x)+exp(2)*x)/(16*x*exp(1)^2*log(-2*x)^2+x^2+2*x),x, algorithm="fricas")

[Out]

e^2*log(16*e^2*log(-2*x)^2 + x + 2)

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giac [A]  time = 0.23, size = 17, normalized size = 0.89 \begin {gather*} e^{2} \log \left (16 \, e^{2} \log \left (-2 \, x\right )^{2} + x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*exp(1)^2*exp(2)*log(-2*x)+exp(2)*x)/(16*x*exp(1)^2*log(-2*x)^2+x^2+2*x),x, algorithm="giac")

[Out]

e^2*log(16*e^2*log(-2*x)^2 + x + 2)

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maple [A]  time = 0.08, size = 19, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{2} \ln \left (\ln \left (-2 x \right )^{2}+\frac {\left (2+x \right ) {\mathrm e}^{-2}}{16}\right )\)	\(19\)
norman	\({\mathrm e}^{2} \ln \left (2+16 \,{\mathrm e}^{2} \ln \left (-2 x \right )^{2}+x \right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*exp(1)^2*exp(2)*ln(-2*x)+exp(2)*x)/(16*x*exp(1)^2*ln(-2*x)^2+x^2+2*x),x,method=_RETURNVERBOSE)

[Out]

exp(2)*ln(ln(-2*x)^2+1/16*(2+x)*exp(-2))

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maxima [B]  time = 0.49, size = 39, normalized size = 2.05 \begin {gather*} e^{2} \log \left (\frac {1}{16} \, {\left (16 \, e^{2} \log \relax (2)^{2} + 32 \, e^{2} \log \relax (2) \log \left (-x\right ) + 16 \, e^{2} \log \left (-x\right )^{2} + x + 2\right )} e^{\left (-2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*exp(1)^2*exp(2)*log(-2*x)+exp(2)*x)/(16*x*exp(1)^2*log(-2*x)^2+x^2+2*x),x, algorithm="maxima")

[Out]

e^2*log(1/16*(16*e^2*log(2)^2 + 32*e^2*log(2)*log(-x) + 16*e^2*log(-x)^2 + x + 2)*e^(-2))

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mupad [B]  time = 7.29, size = 17, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^2\,\ln \left (16\,{\mathrm {e}}^2\,{\ln \left (-2\,x\right )}^2+x+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*log(-2*x)*exp(4) + x*exp(2))/(2*x + x^2 + 16*x*log(-2*x)^2*exp(2)),x)

[Out]

exp(2)*log(x + 16*log(-2*x)^2*exp(2) + 2)

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sympy [A]  time = 0.19, size = 20, normalized size = 1.05 \begin {gather*} e^{2} \log {\left (\frac {x + 2}{16 e^{2}} + \log {\left (- 2 x \right )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*exp(1)**2*exp(2)*ln(-2*x)+exp(2)*x)/(16*x*exp(1)**2*ln(-2*x)**2+x**2+2*x),x)

[Out]

exp(2)*log((x + 2)*exp(-2)/16 + log(-2*x)**2)

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