3.83.38 \(\int \frac {e^2 (48 x^2-18 x^3-4 x^4)+4 e^2 x^3 \log (x)}{64-64 x-16 x^2+16 x^3+4 x^4+(16 x-8 x^2-4 x^3) \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^2 x^2}{-3-x+\frac {4+x}{x}+\frac {\log (x)}{2}} \]

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Rubi [F]  time = 0.83, 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^2 \left (48 x^2-18 x^3-4 x^4\right )+4 e^2 x^3 \log (x)}{64-64 x-16 x^2+16 x^3+4 x^4+\left (16 x-8 x^2-4 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^2*(48*x^2 - 18*x^3 - 4*x^4) + 4*E^2*x^3*Log[x])/(64 - 64*x - 16*x^2 + 16*x^3 + 4*x^4 + (16*x - 8*x^2 -
4*x^3)*Log[x] + x^2*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^2 x^2 \left (24-9 x-2 x^2+2 x \log (x)\right )}{\left (8-4 x-2 x^2+x \log (x)\right )^2} \, dx\\ &=\left (2 e^2\right ) \int \frac {x^2 \left (24-9 x-2 x^2+2 x \log (x)\right )}{\left (8-4 x-2 x^2+x \log (x)\right )^2} \, dx\\ &=\left (2 e^2\right ) \int \left (\frac {x^2 \left (8-x+2 x^2\right )}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}-\frac {2 x^2}{-8+4 x+2 x^2-x \log (x)}\right ) \, dx\\ &=\left (2 e^2\right ) \int \frac {x^2 \left (8-x+2 x^2\right )}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx-\left (4 e^2\right ) \int \frac {x^2}{-8+4 x+2 x^2-x \log (x)} \, dx\\ &=\left (2 e^2\right ) \int \left (\frac {8 x^2}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}-\frac {x^3}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}+\frac {2 x^4}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}\right ) \, dx-\left (4 e^2\right ) \int \frac {x^2}{-8+4 x+2 x^2-x \log (x)} \, dx\\ &=-\left (\left (2 e^2\right ) \int \frac {x^3}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx\right )+\left (4 e^2\right ) \int \frac {x^4}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx-\left (4 e^2\right ) \int \frac {x^2}{-8+4 x+2 x^2-x \log (x)} \, dx+\left (16 e^2\right ) \int \frac {x^2}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 e^2 x^3}{8-4 x-2 x^2+x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(48*x^2 - 18*x^3 - 4*x^4) + 4*E^2*x^3*Log[x])/(64 - 64*x - 16*x^2 + 16*x^3 + 4*x^4 + (16*x - 8*
x^2 - 4*x^3)*Log[x] + x^2*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.58, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{3} e^{2}}{2 \, x^{2} - x \log \relax (x) + 4 \, x - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3*exp(2)*log(x)+(-4*x^4-18*x^3+48*x^2)*exp(2))/(x^2*log(x)^2+(-4*x^3-8*x^2+16*x)*log(x)+4*x^4+1
6*x^3-16*x^2-64*x+64),x, algorithm="fricas")

[Out]

-2*x^3*e^2/(2*x^2 - x*log(x) + 4*x - 8)

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giac [A]  time = 0.21, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{3} e^{2}}{2 \, x^{2} - x \log \relax (x) + 4 \, x - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3*exp(2)*log(x)+(-4*x^4-18*x^3+48*x^2)*exp(2))/(x^2*log(x)^2+(-4*x^3-8*x^2+16*x)*log(x)+4*x^4+1
6*x^3-16*x^2-64*x+64),x, algorithm="giac")

[Out]

-2*x^3*e^2/(2*x^2 - x*log(x) + 4*x - 8)

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maple [A]  time = 0.08, size = 25, normalized size = 0.93

method	result	size
norman	\(-\frac {2 x^{3} {\mathrm e}^{2}}{2 x^{2}-x \ln \relax (x )+4 x -8}\)	\(25\)
risch	\(-\frac {2 x^{3} {\mathrm e}^{2}}{2 x^{2}-x \ln \relax (x )+4 x -8}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3*exp(2)*ln(x)+(-4*x^4-18*x^3+48*x^2)*exp(2))/(x^2*ln(x)^2+(-4*x^3-8*x^2+16*x)*ln(x)+4*x^4+16*x^3-16*
x^2-64*x+64),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{3} e^{2}}{2 \, x^{2} - x \log \relax (x) + 4 \, x - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3*exp(2)*log(x)+(-4*x^4-18*x^3+48*x^2)*exp(2))/(x^2*log(x)^2+(-4*x^3-8*x^2+16*x)*log(x)+4*x^4+1
6*x^3-16*x^2-64*x+64),x, algorithm="maxima")

[Out]

-2*x^3*e^2/(2*x^2 - x*log(x) + 4*x - 8)

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mupad [B]  time = 4.98, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2\,x^3\,{\mathrm {e}}^2}{4\,x-x\,\ln \relax (x)+2\,x^2-8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*(18*x^3 - 48*x^2 + 4*x^4) - 4*x^3*exp(2)*log(x))/(x^2*log(x)^2 - 64*x - 16*x^2 + 16*x^3 + 4*x^4 -
log(x)*(8*x^2 - 16*x + 4*x^3) + 64),x)

[Out]

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

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sympy [A]  time = 0.15, size = 22, normalized size = 0.81 \begin {gather*} \frac {2 x^{3} e^{2}}{- 2 x^{2} + x \log {\relax (x )} - 4 x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**3*exp(2)*ln(x)+(-4*x**4-18*x**3+48*x**2)*exp(2))/(x**2*ln(x)**2+(-4*x**3-8*x**2+16*x)*ln(x)+4*
x**4+16*x**3-16*x**2-64*x+64),x)

[Out]

2*x**3*exp(2)/(-2*x**2 + x*log(x) - 4*x + 8)

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