3.8.58 \(\int \frac {-8 e^3 x-32 x^3+8 e^3 x \log (-x^2)+(-e^6+48 x^2-16 x^4+e^3 (-12-8 x^2)) \log ^2(-x^2)}{(e^6+8 e^3 x^2+16 x^4) \log ^2(-x^2)} \, dx\)

Optimal. Leaf size=31 \[ -x+\frac {-3+\frac {x}{\log \left (-x^2\right )}}{\frac {e^3}{4 x}+x} \]

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Rubi [F]  time = 0.51, 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 {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8*E^3*x - 32*x^3 + 8*E^3*x*Log[-x^2] + (-E^6 + 48*x^2 - 16*x^4 + E^3*(-12 - 8*x^2))*Log[-x^2]^2)/((E^6 +
8*E^3*x^2 + 16*x^4)*Log[-x^2]^2),x]

[Out]

-x - (12*x)/(E^3 + 4*x^2) - 4*Defer[Subst][Defer[Int][1/((E^3 + 4*x)*Log[-x]^2), x], x, x^2] + 4*E^3*Defer[Sub
st][Defer[Int][1/((E^3 + 4*x)^2*Log[-x]), x], x, x^2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=16 \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (4 e^3+16 x^2\right )^2 \log ^2\left (-x^2\right )} \, dx\\ &=16 \int \left (\frac {-e^3 \left (12+e^3\right )+8 \left (6-e^3\right ) x^2-16 x^4}{16 \left (e^3+4 x^2\right )^2}-\frac {x}{2 \left (e^3+4 x^2\right ) \log ^2\left (-x^2\right )}+\frac {e^3 x}{2 \left (e^3+4 x^2\right )^2 \log \left (-x^2\right )}\right ) \, dx\\ &=-\left (8 \int \frac {x}{\left (e^3+4 x^2\right ) \log ^2\left (-x^2\right )} \, dx\right )+\left (8 e^3\right ) \int \frac {x}{\left (e^3+4 x^2\right )^2 \log \left (-x^2\right )} \, dx+\int \frac {-e^3 \left (12+e^3\right )+8 \left (6-e^3\right ) x^2-16 x^4}{\left (e^3+4 x^2\right )^2} \, dx\\ &=-\frac {12 x}{e^3+4 x^2}-4 \operatorname {Subst}\left (\int \frac {1}{\left (e^3+4 x\right ) \log ^2(-x)} \, dx,x,x^2\right )-\frac {\int \frac {2 e^6+8 e^3 x^2}{e^3+4 x^2} \, dx}{2 e^3}+\left (4 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^3+4 x\right )^2 \log (-x)} \, dx,x,x^2\right )\\ &=-\frac {12 x}{e^3+4 x^2}-4 \operatorname {Subst}\left (\int \frac {1}{\left (e^3+4 x\right ) \log ^2(-x)} \, dx,x,x^2\right )+\left (4 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^3+4 x\right )^2 \log (-x)} \, dx,x,x^2\right )-\int 1 \, dx\\ &=-x-\frac {12 x}{e^3+4 x^2}-4 \operatorname {Subst}\left (\int \frac {1}{\left (e^3+4 x\right ) \log ^2(-x)} \, dx,x,x^2\right )+\left (4 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^3+4 x\right )^2 \log (-x)} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 39, normalized size = 1.26 \begin {gather*} x \left (-1-\frac {12}{e^3+4 x^2}+\frac {4 x}{\left (e^3+4 x^2\right ) \log \left (-x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E^3*x - 32*x^3 + 8*E^3*x*Log[-x^2] + (-E^6 + 48*x^2 - 16*x^4 + E^3*(-12 - 8*x^2))*Log[-x^2]^2)/(
(E^6 + 8*E^3*x^2 + 16*x^4)*Log[-x^2]^2),x]

[Out]

x*(-1 - 12/(E^3 + 4*x^2) + (4*x)/((E^3 + 4*x^2)*Log[-x^2]))

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fricas [A]  time = 0.56, size = 46, normalized size = 1.48 \begin {gather*} \frac {4 \, x^{2} - {\left (4 \, x^{3} + x e^{3} + 12 \, x\right )} \log \left (-x^{2}\right )}{{\left (4 \, x^{2} + e^{3}\right )} \log \left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*log(-x^2)^2+8*x*exp(3)*log(-x^2)-8*x*exp(3)-32*x^3)/(e
xp(3)^2+8*x^2*exp(3)+16*x^4)/log(-x^2)^2,x, algorithm="fricas")

[Out]

(4*x^2 - (4*x^3 + x*e^3 + 12*x)*log(-x^2))/((4*x^2 + e^3)*log(-x^2))

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giac [B]  time = 1.30, size = 61, normalized size = 1.97 \begin {gather*} -\frac {4 \, x^{3} \log \left (-x^{2}\right ) + x e^{3} \log \left (-x^{2}\right ) - 4 \, x^{2} + 24 \, x \log \left (-x^{2}\right )}{4 \, x^{2} \log \left (-x^{2}\right ) + e^{3} \log \left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*log(-x^2)^2+8*x*exp(3)*log(-x^2)-8*x*exp(3)-32*x^3)/(e
xp(3)^2+8*x^2*exp(3)+16*x^4)/log(-x^2)^2,x, algorithm="giac")

[Out]

-(4*x^3*log(-x^2) + x*e^3*log(-x^2) - 4*x^2 + 24*x*log(-x^2))/(4*x^2*log(-x^2) + e^3*log(-x^2))

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maple [A]  time = 0.14, size = 47, normalized size = 1.52

method	result	size
risch	\(-\frac {x \left (4 x^{2}+{\mathrm e}^{3}+12\right )}{{\mathrm e}^{3}+4 x^{2}}+\frac {4 x^{2}}{\left ({\mathrm e}^{3}+4 x^{2}\right ) \ln \left (-x^{2}\right )}\)	\(47\)
norman	\(\frac {\left (-{\mathrm e}^{3}-12\right ) x \ln \left (-x^{2}\right )+4 x^{2}-4 x^{3} \ln \left (-x^{2}\right )}{\left ({\mathrm e}^{3}+4 x^{2}\right ) \ln \left (-x^{2}\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*ln(-x^2)^2+8*x*exp(3)*ln(-x^2)-8*x*exp(3)-32*x^3)/(exp(3)^2+
8*x^2*exp(3)+16*x^4)/ln(-x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-x*(4*x^2+exp(3)+12)/(exp(3)+4*x^2)+4*x^2/(exp(3)+4*x^2)/ln(-x^2)

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maxima [C]  time = 0.55, size = 67, normalized size = 2.16 \begin {gather*} -\frac {4 i \, \pi x^{3} + {\left (12 i \, \pi + i \, \pi e^{3}\right )} x - 4 \, x^{2} + 2 \, {\left (4 \, x^{3} + x {\left (e^{3} + 12\right )}\right )} \log \relax (x)}{4 i \, \pi x^{2} + i \, \pi e^{3} + 2 \, {\left (4 \, x^{2} + e^{3}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*log(-x^2)^2+8*x*exp(3)*log(-x^2)-8*x*exp(3)-32*x^3)/(e
xp(3)^2+8*x^2*exp(3)+16*x^4)/log(-x^2)^2,x, algorithm="maxima")

[Out]

-(4*I*pi*x^3 + (12*I*pi + I*pi*e^3)*x - 4*x^2 + 2*(4*x^3 + x*(e^3 + 12))*log(x))/(4*I*pi*x^2 + I*pi*e^3 + 2*(4
*x^2 + e^3)*log(x))

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mupad [B]  time = 1.23, size = 53, normalized size = 1.71 \begin {gather*} -\frac {x\,\left (12\,\ln \left (-x^2\right )-4\,x+{\mathrm {e}}^3\,\ln \left (-x^2\right )+4\,x^2\,\ln \left (-x^2\right )\right )}{\ln \left (-x^2\right )\,\left (4\,x^2+{\mathrm {e}}^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x*exp(3) + 32*x^3 + log(-x^2)^2*(exp(6) + exp(3)*(8*x^2 + 12) - 48*x^2 + 16*x^4) - 8*x*exp(3)*log(-x^2
))/(log(-x^2)^2*(exp(6) + 8*x^2*exp(3) + 16*x^4)),x)

[Out]

-(x*(12*log(-x^2) - 4*x + exp(3)*log(-x^2) + 4*x^2*log(-x^2)))/(log(-x^2)*(exp(3) + 4*x^2))

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sympy [A]  time = 0.23, size = 32, normalized size = 1.03 \begin {gather*} \frac {4 x^{2}}{\left (4 x^{2} + e^{3}\right ) \log {\left (- x^{2} \right )}} - x - \frac {12 x}{4 x^{2} + e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(3)**2+(-8*x**2-12)*exp(3)-16*x**4+48*x**2)*ln(-x**2)**2+8*x*exp(3)*ln(-x**2)-8*x*exp(3)-32*x*
*3)/(exp(3)**2+8*x**2*exp(3)+16*x**4)/ln(-x**2)**2,x)

[Out]

4*x**2/((4*x**2 + exp(3))*log(-x**2)) - x - 12*x/(4*x**2 + exp(3))

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