3.100.24 \(\int \frac {e^{2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}} (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} (3-6 \log (x)+3 \log ^2(x)))}{3-6 \log (x)+3 \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ e^{2 e^{-\frac {8}{3-3 \log (x)}} x^4} x \]

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Rubi [F]  time = 4.57, 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 {\exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3-6 \log (x)+3 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*E^(8/(-3 + 3*Log[x]))*x^4 + 8/(-3 + 3*Log[x]))*(8*x^4 - 48*x^4*Log[x] + 24*x^4*Log[x]^2 + (3 - 6*Log
[x] + 3*Log[x]^2)/E^(8/(-3 + 3*Log[x]))))/(3 - 6*Log[x] + 3*Log[x]^2),x]

[Out]

Defer[Int][E^(2*E^(8/(3*(-1 + Log[x])))*x^4), x] + 8*Defer[Int][E^(2*E^(8/(-3 + 3*Log[x]))*x^4 + 8/(-3 + 3*Log
[x]))*x^4, x] - (16*Defer[Int][(E^(2*E^(8/(-3 + 3*Log[x]))*x^4 + 8/(-3 + 3*Log[x]))*x^4)/(-1 + Log[x])^2, x])/
3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{3 (1-\log (x))^2} \, dx\\ &=\frac {1}{3} \int \frac {\exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) \left (8 x^4-48 x^4 \log (x)+24 x^4 \log ^2(x)+e^{-\frac {8}{-3+3 \log (x)}} \left (3-6 \log (x)+3 \log ^2(x)\right )\right )}{(1-\log (x))^2} \, dx\\ &=\frac {1}{3} \int \left (3 \exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4-\frac {8}{3 (-1+\log (x))}+\frac {8}{-3+3 \log (x)}\right )+\frac {8 \exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) x^4 \left (1-6 \log (x)+3 \log ^2(x)\right )}{(-1+\log (x))^2}\right ) \, dx\\ &=\frac {8}{3} \int \frac {\exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) x^4 \left (1-6 \log (x)+3 \log ^2(x)\right )}{(-1+\log (x))^2} \, dx+\int \exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4-\frac {8}{3 (-1+\log (x))}+\frac {8}{-3+3 \log (x)}\right ) \, dx\\ &=\frac {8}{3} \int \left (3 \exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) x^4-\frac {2 \exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) x^4}{(-1+\log (x))^2}\right ) \, dx+\int e^{2 e^{\frac {8}{3 (-1+\log (x))}} x^4} \, dx\\ &=-\left (\frac {16}{3} \int \frac {\exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) x^4}{(-1+\log (x))^2} \, dx\right )+8 \int \exp \left (2 e^{\frac {8}{-3+3 \log (x)}} x^4+\frac {8}{-3+3 \log (x)}\right ) x^4 \, dx+\int e^{2 e^{\frac {8}{3 (-1+\log (x))}} x^4} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.21, size = 21, normalized size = 1.00 \begin {gather*} e^{2 e^{\frac {8}{3 (-1+\log (x))}} x^4} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^(8/(-3 + 3*Log[x]))*x^4 + 8/(-3 + 3*Log[x]))*(8*x^4 - 48*x^4*Log[x] + 24*x^4*Log[x]^2 + (3 -
6*Log[x] + 3*Log[x]^2)/E^(8/(-3 + 3*Log[x]))))/(3 - 6*Log[x] + 3*Log[x]^2),x]

[Out]

E^(2*E^(8/(3*(-1 + Log[x])))*x^4)*x

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fricas [B]  time = 2.33, size = 45, normalized size = 2.14 \begin {gather*} x e^{\left (\frac {2 \, {\left (3 \, {\left (x^{4} \log \relax (x) - x^{4}\right )} e^{\left (\frac {8}{3 \, {\left (\log \relax (x) - 1\right )}}\right )} + 4\right )}}{3 \, {\left (\log \relax (x) - 1\right )}} - \frac {8}{3 \, {\left (\log \relax (x) - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)^2-6*log(x)+3)*exp(-4/(3*log(x)-3))^2+24*x^4*log(x)^2-48*x^4*log(x)+8*x^4)*exp(2*x^4/exp(-
4/(3*log(x)-3))^2)/(3*log(x)^2-6*log(x)+3)/exp(-4/(3*log(x)-3))^2,x, algorithm="fricas")

[Out]

x*e^(2/3*(3*(x^4*log(x) - x^4)*e^(8/3/(log(x) - 1)) + 4)/(log(x) - 1) - 8/3/(log(x) - 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (24 \, x^{4} \log \relax (x)^{2} - 48 \, x^{4} \log \relax (x) + 8 \, x^{4} + 3 \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} e^{\left (-\frac {8}{3 \, {\left (\log \relax (x) - 1\right )}}\right )}\right )} e^{\left (2 \, x^{4} e^{\left (\frac {8}{3 \, {\left (\log \relax (x) - 1\right )}}\right )} + \frac {8}{3 \, {\left (\log \relax (x) - 1\right )}}\right )}}{3 \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)^2-6*log(x)+3)*exp(-4/(3*log(x)-3))^2+24*x^4*log(x)^2-48*x^4*log(x)+8*x^4)*exp(2*x^4/exp(-
4/(3*log(x)-3))^2)/(3*log(x)^2-6*log(x)+3)/exp(-4/(3*log(x)-3))^2,x, algorithm="giac")

[Out]

integrate(1/3*(24*x^4*log(x)^2 - 48*x^4*log(x) + 8*x^4 + 3*(log(x)^2 - 2*log(x) + 1)*e^(-8/3/(log(x) - 1)))*e^
(2*x^4*e^(8/3/(log(x) - 1)) + 8/3/(log(x) - 1))/(log(x)^2 - 2*log(x) + 1), x)

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maple [A]  time = 0.12, size = 18, normalized size = 0.86

method	result	size
risch	\(x \,{\mathrm e}^{2 x^{4} {\mathrm e}^{\frac {8}{3 \left (\ln \relax (x )-1\right )}}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*ln(x)^2-6*ln(x)+3)*exp(-4/(3*ln(x)-3))^2+24*x^4*ln(x)^2-48*x^4*ln(x)+8*x^4)*exp(2*x^4/exp(-4/(3*ln(x)-
3))^2)/(3*ln(x)^2-6*ln(x)+3)/exp(-4/(3*ln(x)-3))^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(2*x^4*exp(8/3/(ln(x)-1)))

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maxima [A]  time = 0.40, size = 17, normalized size = 0.81 \begin {gather*} x e^{\left (2 \, x^{4} e^{\left (\frac {8}{3 \, {\left (\log \relax (x) - 1\right )}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)^2-6*log(x)+3)*exp(-4/(3*log(x)-3))^2+24*x^4*log(x)^2-48*x^4*log(x)+8*x^4)*exp(2*x^4/exp(-
4/(3*log(x)-3))^2)/(3*log(x)^2-6*log(x)+3)/exp(-4/(3*log(x)-3))^2,x, algorithm="maxima")

[Out]

x*e^(2*x^4*e^(8/3/(log(x) - 1)))

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mupad [B]  time = 7.27, size = 19, normalized size = 0.90 \begin {gather*} x\,{\mathrm {e}}^{2\,x^4\,{\mathrm {e}}^{\frac {8}{3\,\ln \relax (x)-3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x^4*exp(8/(3*log(x) - 3)))*exp(8/(3*log(x) - 3))*(exp(-8/(3*log(x) - 3))*(3*log(x)^2 - 6*log(x) + 3
) - 48*x^4*log(x) + 24*x^4*log(x)^2 + 8*x^4))/(3*log(x)^2 - 6*log(x) + 3),x)

[Out]

x*exp(2*x^4*exp(8/(3*log(x) - 3)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*ln(x)**2-6*ln(x)+3)*exp(-4/(3*ln(x)-3))**2+24*x**4*ln(x)**2-48*x**4*ln(x)+8*x**4)*exp(2*x**4/exp
(-4/(3*ln(x)-3))**2)/(3*ln(x)**2-6*ln(x)+3)/exp(-4/(3*ln(x)-3))**2,x)

[Out]

Timed out

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