3.97.83 \(\int \frac {-11552 x^3-4864 x^3 \log (x^2)-512 x^3 \log ^2(x^2)+e^{-\frac {1}{19+4 \log (x^2)}} (-353-152 \log (x^2)-16 \log ^2(x^2))}{361 x^2+152 x^2 \log (x^2)+16 x^2 \log ^2(x^2)} \, dx\)

Optimal. Leaf size=28 \[ 2+\frac {e^{\frac {x}{x-4 x \left (5+\log \left (x^2\right )\right )}}}{x}-16 x^2 \]

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Rubi [F]  time = 1.36, 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 {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{361 x^2+152 x^2 \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-11552*x^3 - 4864*x^3*Log[x^2] - 512*x^3*Log[x^2]^2 + (-353 - 152*Log[x^2] - 16*Log[x^2]^2)/E^(19 + 4*Log
[x^2])^(-1))/(361*x^2 + 152*x^2*Log[x^2] + 16*x^2*Log[x^2]^2),x]

[Out]

-16*x^2 - Defer[Int][E^(-19 - 4*Log[x^2])^(-1)/x^2, x] + 8*Defer[Int][E^(-19 - 4*Log[x^2])^(-1)/(x^2*(19 + 4*L
og[x^2])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-11552 x^3-4864 x^3 \log \left (x^2\right )-512 x^3 \log ^2\left (x^2\right )+e^{-\frac {1}{19+4 \log \left (x^2\right )}} \left (-353-152 \log \left (x^2\right )-16 \log ^2\left (x^2\right )\right )}{x^2 \left (19+4 \log \left (x^2\right )\right )^2} \, dx\\ &=\int \left (-32 x-\frac {e^{\frac {1}{-19-4 \log \left (x^2\right )}} \left (353+152 \log \left (x^2\right )+16 \log ^2\left (x^2\right )\right )}{x^2 \left (19+4 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=-16 x^2-\int \frac {e^{\frac {1}{-19-4 \log \left (x^2\right )}} \left (353+152 \log \left (x^2\right )+16 \log ^2\left (x^2\right )\right )}{x^2 \left (19+4 \log \left (x^2\right )\right )^2} \, dx\\ &=-16 x^2-\int \left (\frac {e^{\frac {1}{-19-4 \log \left (x^2\right )}}}{x^2}-\frac {8 e^{\frac {1}{-19-4 \log \left (x^2\right )}}}{x^2 \left (19+4 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=-16 x^2+8 \int \frac {e^{\frac {1}{-19-4 \log \left (x^2\right )}}}{x^2 \left (19+4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{\frac {1}{-19-4 \log \left (x^2\right )}}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 24, normalized size = 0.86 \begin {gather*} \frac {e^{-\frac {1}{19+4 \log \left (x^2\right )}}}{x}-16 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11552*x^3 - 4864*x^3*Log[x^2] - 512*x^3*Log[x^2]^2 + (-353 - 152*Log[x^2] - 16*Log[x^2]^2)/E^(19 +
4*Log[x^2])^(-1))/(361*x^2 + 152*x^2*Log[x^2] + 16*x^2*Log[x^2]^2),x]

[Out]

1/(E^(19 + 4*Log[x^2])^(-1)*x) - 16*x^2

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fricas [A]  time = 0.56, size = 26, normalized size = 0.93 \begin {gather*} -\frac {16 \, x^{3} - e^{\left (-\frac {1}{4 \, \log \left (x^{2}\right ) + 19}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(x^2)^2-152*log(x^2)-353)*exp(-1/(4*log(x^2)+19))-512*x^3*log(x^2)^2-4864*x^3*log(x^2)-1155
2*x^3)/(16*x^2*log(x^2)^2+152*x^2*log(x^2)+361*x^2),x, algorithm="fricas")

[Out]

-(16*x^3 - e^(-1/(4*log(x^2) + 19)))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {512 \, x^{3} \log \left (x^{2}\right )^{2} + 4864 \, x^{3} \log \left (x^{2}\right ) + 11552 \, x^{3} + {\left (16 \, \log \left (x^{2}\right )^{2} + 152 \, \log \left (x^{2}\right ) + 353\right )} e^{\left (-\frac {1}{4 \, \log \left (x^{2}\right ) + 19}\right )}}{16 \, x^{2} \log \left (x^{2}\right )^{2} + 152 \, x^{2} \log \left (x^{2}\right ) + 361 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(x^2)^2-152*log(x^2)-353)*exp(-1/(4*log(x^2)+19))-512*x^3*log(x^2)^2-4864*x^3*log(x^2)-1155
2*x^3)/(16*x^2*log(x^2)^2+152*x^2*log(x^2)+361*x^2),x, algorithm="giac")

[Out]

integrate(-(512*x^3*log(x^2)^2 + 4864*x^3*log(x^2) + 11552*x^3 + (16*log(x^2)^2 + 152*log(x^2) + 353)*e^(-1/(4
*log(x^2) + 19)))/(16*x^2*log(x^2)^2 + 152*x^2*log(x^2) + 361*x^2), x)

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maple [A]  time = 0.16, size = 24, normalized size = 0.86

method	result	size
risch	\(-16 x^{2}+\frac {{\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}}{x}\)	\(24\)
default	\(-16 x^{2}+\frac {\left (4 \ln \left (x^{2}\right )-8 \ln \relax (x )+19\right ) {\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}+8 \ln \relax (x ) {\mathrm e}^{-\frac {1}{4 \ln \left (x^{2}\right )+19}}}{x \left (4 \ln \left (x^{2}\right )+19\right )}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*ln(x^2)^2-152*ln(x^2)-353)*exp(-1/(4*ln(x^2)+19))-512*x^3*ln(x^2)^2-4864*x^3*ln(x^2)-11552*x^3)/(16*
x^2*ln(x^2)^2+152*x^2*ln(x^2)+361*x^2),x,method=_RETURNVERBOSE)

[Out]

-16*x^2+1/x*exp(-1/(4*ln(x^2)+19))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -16 \, x^{2} - \int \frac {{\left (64 \, \log \relax (x)^{2} + 304 \, \log \relax (x) + 353\right )} e^{\left (-\frac {1}{8 \, \log \relax (x) + 19}\right )}}{64 \, x^{2} \log \relax (x)^{2} + 304 \, x^{2} \log \relax (x) + 361 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(x^2)^2-152*log(x^2)-353)*exp(-1/(4*log(x^2)+19))-512*x^3*log(x^2)^2-4864*x^3*log(x^2)-1155
2*x^3)/(16*x^2*log(x^2)^2+152*x^2*log(x^2)+361*x^2),x, algorithm="maxima")

[Out]

-16*x^2 - integrate((64*log(x)^2 + 304*log(x) + 353)*e^(-1/(8*log(x) + 19))/(64*x^2*log(x)^2 + 304*x^2*log(x)
+ 361*x^2), x)

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mupad [B]  time = 6.06, size = 21, normalized size = 0.75 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {1}{\ln \left (x^8\right )+19}}}{x}-16\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4864*x^3*log(x^2) + exp(-1/(4*log(x^2) + 19))*(152*log(x^2) + 16*log(x^2)^2 + 353) + 11552*x^3 + 512*x^3
*log(x^2)^2)/(152*x^2*log(x^2) + 361*x^2 + 16*x^2*log(x^2)^2),x)

[Out]

exp(-1/(log(x^8) + 19))/x - 16*x^2

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sympy [A]  time = 0.55, size = 19, normalized size = 0.68 \begin {gather*} - 16 x^{2} + \frac {e^{- \frac {1}{4 \log {\left (x^{2} \right )} + 19}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*ln(x**2)**2-152*ln(x**2)-353)*exp(-1/(4*ln(x**2)+19))-512*x**3*ln(x**2)**2-4864*x**3*ln(x**2)-
11552*x**3)/(16*x**2*ln(x**2)**2+152*x**2*ln(x**2)+361*x**2),x)

[Out]

-16*x**2 + exp(-1/(4*log(x**2) + 19))/x

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