3.44.14 \(\int \frac {e^4 (-8+2 x)+(16 x-8 x^2+x^3+e^4 (-16 x+8 x^2-x^3)) \log (x^2)-e^4 x \log (x^2) \log (\log (x^2))}{e^4 (16 x-8 x^2+x^3) \log (x^2)} \, dx\)

Optimal. Leaf size=20 \[ -x+\frac {x}{e^4}+\frac {\log \left (\log \left (x^2\right )\right )}{-4+x} \]

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Rubi [F]  time = 0.36, 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^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^4*(-8 + 2*x) + (16*x - 8*x^2 + x^3 + E^4*(-16*x + 8*x^2 - x^3))*Log[x^2] - E^4*x*Log[x^2]*Log[Log[x^2]]
)/(E^4*(16*x - 8*x^2 + x^3)*Log[x^2]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx}{e^4}\\ &=\frac {\int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \left (16-8 x+x^2\right ) \log \left (x^2\right )} \, dx}{e^4}\\ &=\frac {\int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(-4+x)^2 x \log \left (x^2\right )} \, dx}{e^4}\\ &=\frac {\int \left (1-e^4+\frac {2 e^4}{(-4+x) x \log \left (x^2\right )}-\frac {e^4 \log \left (\log \left (x^2\right )\right )}{(-4+x)^2}\right ) \, dx}{e^4}\\ &=-\left (\left (1-\frac {1}{e^4}\right ) x\right )+2 \int \frac {1}{(-4+x) x \log \left (x^2\right )} \, dx-\int \frac {\log \left (\log \left (x^2\right )\right )}{(-4+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 19, normalized size = 0.95 \begin {gather*} \left (-1+\frac {1}{e^4}\right ) x+\frac {\log \left (\log \left (x^2\right )\right )}{-4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(-8 + 2*x) + (16*x - 8*x^2 + x^3 + E^4*(-16*x + 8*x^2 - x^3))*Log[x^2] - E^4*x*Log[x^2]*Log[Log
[x^2]])/(E^4*(16*x - 8*x^2 + x^3)*Log[x^2]),x]

[Out]

(-1 + E^(-4))*x + Log[Log[x^2]]/(-4 + x)

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fricas [A]  time = 0.76, size = 34, normalized size = 1.70 \begin {gather*} \frac {{\left (x^{2} - {\left (x^{2} - 4 \, x\right )} e^{4} + e^{4} \log \left (\log \left (x^{2}\right )\right ) - 4 \, x\right )} e^{\left (-4\right )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))
/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.19, size = 38, normalized size = 1.90 \begin {gather*} -\frac {{\left (x^{2} e^{4} - x^{2} - 4 \, x e^{4} - e^{4} \log \left (\log \left (x^{2}\right )\right ) + 4 \, x\right )} e^{\left (-4\right )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))
/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (-x \,{\mathrm e}^{4} \ln \left (x^{2}\right ) \ln \left (\ln \left (x^{2}\right )\right )+\left (\left (-x^{3}+8 x^{2}-16 x \right ) {\mathrm e}^{4}+x^{3}-8 x^{2}+16 x \right ) \ln \left (x^{2}\right )+\left (2 x -8\right ) {\mathrm e}^{4}\right ) {\mathrm e}^{-4}}{\left (x^{3}-8 x^{2}+16 x \right ) \ln \left (x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*exp(4)*ln(x^2)*ln(ln(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*ln(x^2)+(2*x-8)*exp(4))/(x^3-8*x^
2+16*x)/exp(4)/ln(x^2),x)

[Out]

int((-x*exp(4)*ln(x^2)*ln(ln(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*ln(x^2)+(2*x-8)*exp(4))/(x^3-8*x^
2+16*x)/exp(4)/ln(x^2),x)

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maxima [A]  time = 0.48, size = 38, normalized size = 1.90 \begin {gather*} -\frac {{\left (x^{2} {\left (e^{4} - 1\right )} - 4 \, x {\left (e^{4} - 1\right )} - e^{4} \log \relax (2) - e^{4} \log \left (\log \relax (x)\right )\right )} e^{\left (-4\right )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))
/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, algorithm="maxima")

[Out]

-(x^2*(e^4 - 1) - 4*x*(e^4 - 1) - e^4*log(2) - e^4*log(log(x)))*e^(-4)/(x - 4)

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mupad [B]  time = 3.29, size = 18, normalized size = 0.90 \begin {gather*} \frac {\ln \left (\ln \left (x^2\right )\right )}{x-4}+x\,\left ({\mathrm {e}}^{-4}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4)*(log(x^2)*(16*x - exp(4)*(16*x - 8*x^2 + x^3) - 8*x^2 + x^3) + exp(4)*(2*x - 8) - x*log(x^2)*exp(
4)*log(log(x^2))))/(log(x^2)*(16*x - 8*x^2 + x^3)),x)

[Out]

log(log(x^2))/(x - 4) + x*(exp(-4) - 1)

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sympy [A]  time = 0.35, size = 19, normalized size = 0.95 \begin {gather*} \frac {x \left (1 - e^{4}\right )}{e^{4}} + \frac {\log {\left (\log {\left (x^{2} \right )} \right )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(4)*ln(x**2)*ln(ln(x**2))+((-x**3+8*x**2-16*x)*exp(4)+x**3-8*x**2+16*x)*ln(x**2)+(2*x-8)*exp(
4))/(x**3-8*x**2+16*x)/exp(4)/ln(x**2),x)

[Out]

x*(1 - exp(4))*exp(-4) + log(log(x**2))/(x - 4)

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