3.10.15 \(\int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+(16-e^4-x+2 e^2 x-x^2) \log (-16 x+e^4 x+x^2-2 e^2 x^2+x^3) \log (\log ^2(-16 x+e^4 x+x^2-2 e^2 x^2+x^3))}{(-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4) \log (-16 x+e^4 x+x^2-2 e^2 x^2+x^3)} \, dx\)

Optimal. Leaf size=22 \[ \frac {\log \left (\log ^2\left (x \left (-16+\left (e^2-x\right )^2+x\right )\right )\right )}{x} \]

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Rubi [F]  time = 2.45, 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 {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-32 + 2*E^4 + 4*x - 8*E^2*x + 6*x^2 + (16 - E^4 - x + 2*E^2*x - x^2)*Log[-16*x + E^4*x + x^2 - 2*E^2*x^2
+ x^3]*Log[Log[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]^2])/((-16*x^2 + E^4*x^2 + x^3 - 2*E^2*x^3 + x^4)*Log[-16
*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]),x]

[Out]

(-2*(1 - 2*E^2)*(1 + (1 - 2*E^2)/Sqrt[65 - 4*E^2])*Defer[Int][1/((-1 + 2*E^2 - Sqrt[65 - 4*E^2] - 2*x)*Log[x*(
-16 + E^4 + (1 - 2*E^2)*x + x^2)]), x])/(16 - E^4) - (4*(33 - 4*E^2 + 2*E^4)*Defer[Int][1/((-1 + 2*E^2 + Sqrt[
65 - 4*E^2] - 2*x)*Log[x*(-16 + E^4 + (1 - 2*E^2)*x + x^2)]), x])/(Sqrt[65 - 4*E^2]*(16 - E^4)) - (2*(1 - 2*E^
2)*(1 - (1 - 2*E^2)/Sqrt[65 - 4*E^2])*Defer[Int][1/((-1 + 2*E^2 + Sqrt[65 - 4*E^2] - 2*x)*Log[x*(-16 + E^4 + (
1 - 2*E^2)*x + x^2)]), x])/(16 - E^4) + 2*Defer[Int][1/(x^2*Log[x*(-16 + E^4 + (1 - 2*E^2)*x + x^2)]), x] - (2
*(1 - 2*E^2)*Defer[Int][1/(x*Log[x*(-16 + E^4 + (1 - 2*E^2)*x + x^2)]), x])/(16 - E^4) - (4*(33 - 4*E^2 + 2*E^
4)*Defer[Int][1/((1 - 2*E^2 + Sqrt[65 - 4*E^2] + 2*x)*Log[x*(-16 + E^4 + (1 - 2*E^2)*x + x^2)]), x])/(Sqrt[65
- 4*E^2]*(16 - E^4)) - Defer[Int][Log[Log[x*(-16 + E^4 + (1 - 2*E^2)*x + x^2)]^2]/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (\left (-16+e^4\right ) x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx\\ &=\int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (\left (-16+e^4\right ) x^2+\left (1-2 e^2\right ) x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx\\ &=\int \frac {-32+2 e^4+\left (4-8 e^2\right ) x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (\left (-16+e^4\right ) x^2+\left (1-2 e^2\right ) x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx\\ &=\int \frac {-32+2 e^4+\left (4-8 e^2\right ) x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{x^2 \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx\\ &=\int \left (\frac {2 \left (16-e^4-2 \left (1-2 e^2\right ) x-3 x^2\right )}{x^2 \left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}-\frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {16-e^4-2 \left (1-2 e^2\right ) x-3 x^2}{x^2 \left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2} \, dx\\ &=2 \int \left (\frac {1}{x^2 \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}+\frac {-1+2 e^2}{(2-e) (2+e) \left (4+e^2\right ) x \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}+\frac {-33+4 e^2-2 e^4-\left (1-2 e^2\right ) x}{(2-e) (2+e) \left (4+e^2\right ) \left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}\right ) \, dx-\int \frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2} \, dx\\ &=2 \int \frac {1}{x^2 \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx+\frac {2 \int \frac {-33+4 e^2-2 e^4-\left (1-2 e^2\right ) x}{\left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\frac {\left (2 \left (1-2 e^2\right )\right ) \int \frac {1}{x \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\int \frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2} \, dx\\ &=2 \int \frac {1}{x^2 \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx+\frac {2 \int \left (\frac {4 e^2 \left (1-\frac {33+2 e^4}{4 e^2}\right )}{\left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}+\frac {\left (-1+2 e^2\right ) x}{\left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}\right ) \, dx}{16-e^4}-\frac {\left (2 \left (1-2 e^2\right )\right ) \int \frac {1}{x \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\int \frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2} \, dx\\ &=2 \int \frac {1}{x^2 \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx-\frac {\left (2 \left (1-2 e^2\right )\right ) \int \frac {1}{x \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\frac {\left (2 \left (1-2 e^2\right )\right ) \int \frac {x}{\left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\frac {\left (2 \left (33-4 e^2+2 e^4\right )\right ) \int \frac {1}{\left (16-e^4-\left (1-2 e^2\right ) x-x^2\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\int \frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2} \, dx\\ &=2 \int \frac {1}{x^2 \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx-\frac {\left (2 \left (1-2 e^2\right )\right ) \int \left (\frac {1+\frac {1-2 e^2}{\sqrt {65-4 e^2}}}{\left (-1+2 e^2-\sqrt {65-4 e^2}-2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}+\frac {1-\frac {1-2 e^2}{\sqrt {65-4 e^2}}}{\left (-1+2 e^2+\sqrt {65-4 e^2}-2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}\right ) \, dx}{16-e^4}-\frac {\left (2 \left (1-2 e^2\right )\right ) \int \frac {1}{x \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\frac {\left (2 \left (33-4 e^2+2 e^4\right )\right ) \int \left (\frac {2}{\sqrt {65-4 e^2} \left (-1+2 e^2+\sqrt {65-4 e^2}-2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}+\frac {2}{\sqrt {65-4 e^2} \left (1-2 e^2+\sqrt {65-4 e^2}+2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )}\right ) \, dx}{16-e^4}-\int \frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2} \, dx\\ &=2 \int \frac {1}{x^2 \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx-\frac {\left (2 \left (1-2 e^2\right )\right ) \int \frac {1}{x \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\frac {\left (4 \left (33-4 e^2+2 e^4\right )\right ) \int \frac {1}{\left (-1+2 e^2+\sqrt {65-4 e^2}-2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{\sqrt {65-4 e^2} \left (16-e^4\right )}-\frac {\left (4 \left (33-4 e^2+2 e^4\right )\right ) \int \frac {1}{\left (1-2 e^2+\sqrt {65-4 e^2}+2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{\sqrt {65-4 e^2} \left (16-e^4\right )}-\frac {\left (2 \left (1-2 e^2\right ) \left (1-\frac {1-2 e^2}{\sqrt {65-4 e^2}}\right )\right ) \int \frac {1}{\left (-1+2 e^2+\sqrt {65-4 e^2}-2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\frac {\left (2 \left (1-2 e^2\right ) \left (1+\frac {1-2 e^2}{\sqrt {65-4 e^2}}\right )\right ) \int \frac {1}{\left (-1+2 e^2-\sqrt {65-4 e^2}-2 x\right ) \log \left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )} \, dx}{16-e^4}-\int \frac {\log \left (\log ^2\left (x \left (-16+e^4+\left (1-2 e^2\right ) x+x^2\right )\right )\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 1.14 \begin {gather*} \frac {\log \left (\log ^2\left (x \left (-16+e^4+x-2 e^2 x+x^2\right )\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32 + 2*E^4 + 4*x - 8*E^2*x + 6*x^2 + (16 - E^4 - x + 2*E^2*x - x^2)*Log[-16*x + E^4*x + x^2 - 2*E^
2*x^2 + x^3]*Log[Log[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]^2])/((-16*x^2 + E^4*x^2 + x^3 - 2*E^2*x^3 + x^4)*L
og[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]),x]

[Out]

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

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fricas [A]  time = 0.52, size = 29, normalized size = 1.32 \begin {gather*} \frac {\log \left (\log \left (x^{3} - 2 \, x^{2} e^{2} + x^{2} + x e^{4} - 16 \, x\right )^{2}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)*log(log(x*exp(1)^4-2*
x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^
2)/log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm="fricas")

[Out]

log(log(x^3 - 2*x^2*e^2 + x^2 + x*e^4 - 16*x)^2)/x

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giac [A]  time = 18.90, size = 29, normalized size = 1.32 \begin {gather*} \frac {\log \left (\log \left (x^{3} - 2 \, x^{2} e^{2} + x^{2} + x e^{4} - 16 \, x\right )^{2}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)*log(log(x*exp(1)^4-2*
x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^
2)/log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm="giac")

[Out]

log(log(x^3 - 2*x^2*e^2 + x^2 + x*e^4 - 16*x)^2)/x

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (-{\mathrm e}^{4}+2 \,{\mathrm e}^{2} x -x^{2}-x +16\right ) \ln \left (x \,{\mathrm e}^{4}-2 x^{2} {\mathrm e}^{2}+x^{3}+x^{2}-16 x \right ) \ln \left (\ln \left (x \,{\mathrm e}^{4}-2 x^{2} {\mathrm e}^{2}+x^{3}+x^{2}-16 x \right )^{2}\right )+2 \,{\mathrm e}^{4}-8 \,{\mathrm e}^{2} x +6 x^{2}+4 x -32}{\left (x^{2} {\mathrm e}^{4}-2 x^{3} {\mathrm e}^{2}+x^{4}+x^{3}-16 x^{2}\right ) \ln \left (x \,{\mathrm e}^{4}-2 x^{2} {\mathrm e}^{2}+x^{3}+x^{2}-16 x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*ln(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)*ln(ln(x*exp(1)^4-2*x^2*exp(1
)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2)/ln(x*e
xp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x)

[Out]

int(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*ln(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)*ln(ln(x*exp(1)^4-2*x^2*exp(1
)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2)/ln(x*e
xp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x)

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maxima [A]  time = 0.62, size = 26, normalized size = 1.18 \begin {gather*} \frac {2 \, \log \left (\log \left (x^{2} - x {\left (2 \, e^{2} - 1\right )} + e^{4} - 16\right ) + \log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)*log(log(x*exp(1)^4-2*
x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^
2)/log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm="maxima")

[Out]

2*log(log(x^2 - x*(2*e^2 - 1) + e^4 - 16) + log(x))/x

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mupad [B]  time = 1.99, size = 29, normalized size = 1.32 \begin {gather*} \frac {\ln \left ({\ln \left (x\,{\mathrm {e}}^4-16\,x-2\,x^2\,{\mathrm {e}}^2+x^2+x^3\right )}^2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + 2*exp(4) - 8*x*exp(2) + 6*x^2 - log(x*exp(4) - 16*x - 2*x^2*exp(2) + x^2 + x^3)*log(log(x*exp(4) -
16*x - 2*x^2*exp(2) + x^2 + x^3)^2)*(x + exp(4) - 2*x*exp(2) + x^2 - 16) - 32)/(log(x*exp(4) - 16*x - 2*x^2*ex
p(2) + x^2 + x^3)*(x^2*exp(4) - 2*x^3*exp(2) - 16*x^2 + x^3 + x^4)),x)

[Out]

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

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sympy [A]  time = 0.66, size = 29, normalized size = 1.32 \begin {gather*} \frac {\log {\left (\log {\left (x^{3} - 2 x^{2} e^{2} + x^{2} - 16 x + x e^{4} \right )}^{2} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(1)**4+2*x*exp(1)**2-x**2-x+16)*ln(x*exp(1)**4-2*x**2*exp(1)**2+x**3+x**2-16*x)*ln(ln(x*exp(1)
**4-2*x**2*exp(1)**2+x**3+x**2-16*x)**2)+2*exp(1)**4-8*x*exp(1)**2+6*x**2+4*x-32)/(x**2*exp(1)**4-2*x**3*exp(1
)**2+x**4+x**3-16*x**2)/ln(x*exp(1)**4-2*x**2*exp(1)**2+x**3+x**2-16*x),x)

[Out]

log(log(x**3 - 2*x**2*exp(2) + x**2 - 16*x + x*exp(4))**2)/x

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