3.58.86 \(\int \frac {10 x}{\log ^2(\frac {1}{e^4 x^2})+2 \log (\frac {1}{e^4 x^2}) \log (4 x^2)+\log ^2(4 x^2)} \, dx\)

Optimal. Leaf size=24 \[ 16+\frac {5 x^2}{\left (\log \left (\frac {1}{e^4 x^2}\right )+\log \left (4 x^2\right )\right )^2} \]

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Rubi [F]  time = 0.09, 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 {10 x}{\log ^2\left (\frac {1}{e^4 x^2}\right )+2 \log \left (\frac {1}{e^4 x^2}\right ) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10*x)/(Log[1/(E^4*x^2)]^2 + 2*Log[1/(E^4*x^2)]*Log[4*x^2] + Log[4*x^2]^2),x]

[Out]

5*Defer[Subst][Defer[Int][(4 - Log[x^(-1)] - Log[4*x])^(-2), x], x, x^2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=10 \int \frac {x}{\log ^2\left (\frac {1}{e^4 x^2}\right )+2 \log \left (\frac {1}{e^4 x^2}\right ) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )} \, dx\\ &=10 \int \frac {x}{\left (4-\log \left (\frac {1}{x^2}\right )-\log \left (4 x^2\right )\right )^2} \, dx\\ &=5 \operatorname {Subst}\left (\int \frac {1}{\left (4-\log \left (\frac {1}{x}\right )-\log (4 x)\right )^2} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.79 \begin {gather*} \frac {5 x^2}{\left (-4+\log \left (\frac {1}{x^2}\right )+\log \left (4 x^2\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x)/(Log[1/(E^4*x^2)]^2 + 2*Log[1/(E^4*x^2)]*Log[4*x^2] + Log[4*x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.57, size = 17, normalized size = 0.71 \begin {gather*} \frac {5 \, x^{2}}{4 \, {\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x/(log(4*x^2)^2+2*log(1/x^2/exp(4))*log(4*x^2)+log(1/x^2/exp(4))^2),x, algorithm="fricas")

[Out]

5/4*x^2/(log(2)^2 - 4*log(2) + 4)

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giac [A]  time = 0.16, size = 17, normalized size = 0.71 \begin {gather*} \frac {5 \, x^{2}}{4 \, {\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x/(log(4*x^2)^2+2*log(1/x^2/exp(4))*log(4*x^2)+log(1/x^2/exp(4))^2),x, algorithm="giac")

[Out]

5/4*x^2/(log(2)^2 - 4*log(2) + 4)

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maple [A]  time = 0.50, size = 20, normalized size = 0.83

method	result	size
risch	\(-\frac {20 x^{2}}{-64+64 \ln \relax (2)-16 \ln \relax (2)^{2}}\)	\(20\)
gosper	\(\frac {5 x^{2}}{\ln \left (4 x^{2}\right )^{2}+2 \ln \left (\frac {{\mathrm e}^{-4}}{x^{2}}\right ) \ln \left (4 x^{2}\right )+\ln \left (\frac {{\mathrm e}^{-4}}{x^{2}}\right )^{2}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*x/(ln(4*x^2)^2+2*ln(1/x^2/exp(4))*ln(4*x^2)+ln(1/x^2/exp(4))^2),x,method=_RETURNVERBOSE)

[Out]

-20*x^2/(-64+64*ln(2)-16*ln(2)^2)

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maxima [A]  time = 0.45, size = 17, normalized size = 0.71 \begin {gather*} \frac {5 \, x^{2}}{4 \, {\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x/(log(4*x^2)^2+2*log(1/x^2/exp(4))*log(4*x^2)+log(1/x^2/exp(4))^2),x, algorithm="maxima")

[Out]

5/4*x^2/(log(2)^2 - 4*log(2) + 4)

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mupad [B]  time = 4.18, size = 11, normalized size = 0.46 \begin {gather*} \frac {5\,x^2}{{\left (\ln \relax (4)-4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x)/(log(4*x^2)^2 + 2*log(exp(-4)/x^2)*log(4*x^2) + log(exp(-4)/x^2)^2),x)

[Out]

(5*x^2)/(log(4) - 4)^2

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sympy [A]  time = 0.21, size = 17, normalized size = 0.71 \begin {gather*} \frac {5 x^{2}}{- 16 \log {\relax (2 )} + 4 \log {\relax (2 )}^{2} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x/(ln(4*x**2)**2+2*ln(1/x**2/exp(4))*ln(4*x**2)+ln(1/x**2/exp(4))**2),x)

[Out]

5*x**2/(-16*log(2) + 4*log(2)**2 + 16)

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