3.35.26 \(\int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4))}} ((-8 x-4 x^2) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4))}{9 \log ^4(4)+e^x (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4))+e^{2 x} (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4))} \, dx\)

Optimal. Leaf size=32 \[ 2 \left (-e^{\frac {2}{3-e^x \left (\log (3)-\frac {x}{\log (4)}\right )^2}}+\log (2)\right ) \]

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Rubi [F]  time = 9.71, 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 (x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}\right ) \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x^2 - 2*x*Log[3]*Log[4] + Log[3]^2*Log[4]^2)))*((-8*x - 4*x^2)*Lo
g[4]^2 + (8 + 8*x)*Log[3]*Log[4]^3 - 4*Log[3]^2*Log[4]^4))/(9*Log[4]^4 + E^x*(-6*x^2*Log[4]^2 + 12*x*Log[3]*Lo
g[4]^3 - 6*Log[3]^2*Log[4]^4) + E^(2*x)*(x^4 - 4*x^3*Log[3]*Log[4] + 6*x^2*Log[3]^2*Log[4]^2 - 4*x*Log[3]^3*Lo
g[4]^3 + Log[3]^4*Log[4]^4)),x]

[Out]

4*Log[3]*Log[4]^3*(2 - Log[3]*Log[4])*Defer[Int][E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x - Log[3]*Log[4])^2)
)/(3*Log[4]^2 - E^x*(x - Log[3]*Log[4])^2)^2, x] - 4*Log[4]^2*(2 - Log[4]*Log[9])*Defer[Int][(E^(x - (2*Log[4]
^2)/(-3*Log[4]^2 + E^x*(x - Log[3]*Log[4])^2))*x)/(3*Log[4]^2 - E^x*(x - Log[3]*Log[4])^2)^2, x] - 4*Log[4]^2*
Defer[Int][(E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x - Log[3]*Log[4])^2))*x^2)/(3*Log[4]^2 - E^x*(x - Log[3]*
Log[4])^2)^2, x]

Rubi steps

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

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Mathematica [F]  time = 4.53, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x^2 - 2*x*Log[3]*Log[4] + Log[3]^2*Log[4]^2)))*((-8*x - 4*x
^2)*Log[4]^2 + (8 + 8*x)*Log[3]*Log[4]^3 - 4*Log[3]^2*Log[4]^4))/(9*Log[4]^4 + E^x*(-6*x^2*Log[4]^2 + 12*x*Log
[3]*Log[4]^3 - 6*Log[3]^2*Log[4]^4) + E^(2*x)*(x^4 - 4*x^3*Log[3]*Log[4] + 6*x^2*Log[3]^2*Log[4]^2 - 4*x*Log[3
]^3*Log[4]^3 + Log[3]^4*Log[4]^4)),x]

[Out]

Integrate[(E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x^2 - 2*x*Log[3]*Log[4] + Log[3]^2*Log[4]^2)))*((-8*x - 4*x
^2)*Log[4]^2 + (8 + 8*x)*Log[3]*Log[4]^3 - 4*Log[3]^2*Log[4]^4))/(9*Log[4]^4 + E^x*(-6*x^2*Log[4]^2 + 12*x*Log
[3]*Log[4]^3 - 6*Log[3]^2*Log[4]^4) + E^(2*x)*(x^4 - 4*x^3*Log[3]*Log[4] + 6*x^2*Log[3]^2*Log[4]^2 - 4*x*Log[3
]^3*Log[4]^3 + Log[3]^4*Log[4]^4)), x]

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fricas [B]  time = 0.62, size = 82, normalized size = 2.56 \begin {gather*} -2 \, e^{\left (-x - \frac {4 \, {\left (3 \, x + 2\right )} \log \relax (2)^{2} - {\left (4 \, x \log \relax (3)^{2} \log \relax (2)^{2} - 4 \, x^{2} \log \relax (3) \log \relax (2) + x^{3}\right )} e^{x}}{{\left (4 \, \log \relax (3)^{2} \log \relax (2)^{2} - 4 \, x \log \relax (3) \log \relax (2) + x^{2}\right )} e^{x} - 12 \, \log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*log(3)^2*log(2)^4+8*(8*x+8)*log(3)*log(2)^3+4*(-4*x^2-8*x)*log(2)^2)*exp(x)*exp(-8*log(2)^2/((4
*log(3)^2*log(2)^2-4*x*log(2)*log(3)+x^2)*exp(x)-12*log(2)^2))/((16*log(3)^4*log(2)^4-32*x*log(3)^3*log(2)^3+2
4*x^2*log(3)^2*log(2)^2-8*x^3*log(3)*log(2)+x^4)*exp(x)^2+(-96*log(3)^2*log(2)^4+96*x*log(3)*log(2)^3-24*x^2*l
og(2)^2)*exp(x)+144*log(2)^4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {16 \, {\left (4 \, \log \relax (3)^{2} \log \relax (2)^{4} - 4 \, {\left (x + 1\right )} \log \relax (3) \log \relax (2)^{3} + {\left (x^{2} + 2 \, x\right )} \log \relax (2)^{2}\right )} e^{\left (x - \frac {8 \, \log \relax (2)^{2}}{{\left (4 \, \log \relax (3)^{2} \log \relax (2)^{2} - 4 \, x \log \relax (3) \log \relax (2) + x^{2}\right )} e^{x} - 12 \, \log \relax (2)^{2}}\right )}}{144 \, \log \relax (2)^{4} + {\left (16 \, \log \relax (3)^{4} \log \relax (2)^{4} - 32 \, x \log \relax (3)^{3} \log \relax (2)^{3} + 24 \, x^{2} \log \relax (3)^{2} \log \relax (2)^{2} - 8 \, x^{3} \log \relax (3) \log \relax (2) + x^{4}\right )} e^{\left (2 \, x\right )} - 24 \, {\left (4 \, \log \relax (3)^{2} \log \relax (2)^{4} - 4 \, x \log \relax (3) \log \relax (2)^{3} + x^{2} \log \relax (2)^{2}\right )} e^{x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*log(3)^2*log(2)^4+8*(8*x+8)*log(3)*log(2)^3+4*(-4*x^2-8*x)*log(2)^2)*exp(x)*exp(-8*log(2)^2/((4
*log(3)^2*log(2)^2-4*x*log(2)*log(3)+x^2)*exp(x)-12*log(2)^2))/((16*log(3)^4*log(2)^4-32*x*log(3)^3*log(2)^3+2
4*x^2*log(3)^2*log(2)^2-8*x^3*log(3)*log(2)+x^4)*exp(x)^2+(-96*log(3)^2*log(2)^4+96*x*log(3)*log(2)^3-24*x^2*l
og(2)^2)*exp(x)+144*log(2)^4),x, algorithm="giac")

[Out]

integrate(-16*(4*log(3)^2*log(2)^4 - 4*(x + 1)*log(3)*log(2)^3 + (x^2 + 2*x)*log(2)^2)*e^(x - 8*log(2)^2/((4*l
og(3)^2*log(2)^2 - 4*x*log(3)*log(2) + x^2)*e^x - 12*log(2)^2))/(144*log(2)^4 + (16*log(3)^4*log(2)^4 - 32*x*l
og(3)^3*log(2)^3 + 24*x^2*log(3)^2*log(2)^2 - 8*x^3*log(3)*log(2) + x^4)*e^(2*x) - 24*(4*log(3)^2*log(2)^4 - 4
*x*log(3)*log(2)^3 + x^2*log(2)^2)*e^x), x)

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maple [A]  time = 0.68, size = 46, normalized size = 1.44

method	result	size
risch	\(-2 \,{\mathrm e}^{-\frac {8 \ln \relax (2)^{2}}{4 \ln \relax (2)^{2} {\mathrm e}^{x} \ln \relax (3)^{2}-4 \ln \relax (2) {\mathrm e}^{x} \ln \relax (3) x +{\mathrm e}^{x} x^{2}-12 \ln \relax (2)^{2}}}\)	\(46\)
norman	\(\frac {24 \ln \relax (2)^{2} {\mathrm e}^{-\frac {8 \ln \relax (2)^{2}}{\left (4 \ln \relax (3)^{2} \ln \relax (2)^{2}-4 x \ln \relax (2) \ln \relax (3)+x^{2}\right ) {\mathrm e}^{x}-12 \ln \relax (2)^{2}}}-2 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{-\frac {8 \ln \relax (2)^{2}}{\left (4 \ln \relax (3)^{2} \ln \relax (2)^{2}-4 x \ln \relax (2) \ln \relax (3)+x^{2}\right ) {\mathrm e}^{x}-12 \ln \relax (2)^{2}}}-8 \ln \relax (2)^{2} {\mathrm e}^{x} \ln \relax (3)^{2} {\mathrm e}^{-\frac {8 \ln \relax (2)^{2}}{\left (4 \ln \relax (3)^{2} \ln \relax (2)^{2}-4 x \ln \relax (2) \ln \relax (3)+x^{2}\right ) {\mathrm e}^{x}-12 \ln \relax (2)^{2}}}+8 \ln \relax (2) {\mathrm e}^{x} \ln \relax (3) x \,{\mathrm e}^{-\frac {8 \ln \relax (2)^{2}}{\left (4 \ln \relax (3)^{2} \ln \relax (2)^{2}-4 x \ln \relax (2) \ln \relax (3)+x^{2}\right ) {\mathrm e}^{x}-12 \ln \relax (2)^{2}}}}{4 \ln \relax (2)^{2} {\mathrm e}^{x} \ln \relax (3)^{2}-4 \ln \relax (2) {\mathrm e}^{x} \ln \relax (3) x +{\mathrm e}^{x} x^{2}-12 \ln \relax (2)^{2}}\)	\(233\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-64*ln(3)^2*ln(2)^4+8*(8*x+8)*ln(3)*ln(2)^3+4*(-4*x^2-8*x)*ln(2)^2)*exp(x)*exp(-8*ln(2)^2/((4*ln(3)^2*ln(
2)^2-4*x*ln(2)*ln(3)+x^2)*exp(x)-12*ln(2)^2))/((16*ln(3)^4*ln(2)^4-32*x*ln(3)^3*ln(2)^3+24*x^2*ln(3)^2*ln(2)^2
-8*x^3*ln(3)*ln(2)+x^4)*exp(x)^2+(-96*ln(3)^2*ln(2)^4+96*x*ln(3)*ln(2)^3-24*x^2*ln(2)^2)*exp(x)+144*ln(2)^4),x
,method=_RETURNVERBOSE)

[Out]

-2*exp(-8*ln(2)^2/(4*ln(2)^2*exp(x)*ln(3)^2-4*ln(2)*exp(x)*ln(3)*x+exp(x)*x^2-12*ln(2)^2))

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maxima [A]  time = 1.00, size = 42, normalized size = 1.31 \begin {gather*} -2 \, e^{\left (-\frac {8 \, \log \relax (2)^{2}}{{\left (4 \, \log \relax (3)^{2} \log \relax (2)^{2} - 4 \, x \log \relax (3) \log \relax (2) + x^{2}\right )} e^{x} - 12 \, \log \relax (2)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*log(3)^2*log(2)^4+8*(8*x+8)*log(3)*log(2)^3+4*(-4*x^2-8*x)*log(2)^2)*exp(x)*exp(-8*log(2)^2/((4
*log(3)^2*log(2)^2-4*x*log(2)*log(3)+x^2)*exp(x)-12*log(2)^2))/((16*log(3)^4*log(2)^4-32*x*log(3)^3*log(2)^3+2
4*x^2*log(3)^2*log(2)^2-8*x^3*log(3)*log(2)+x^4)*exp(x)^2+(-96*log(3)^2*log(2)^4+96*x*log(3)*log(2)^3-24*x^2*l
og(2)^2)*exp(x)+144*log(2)^4),x, algorithm="maxima")

[Out]

-2*e^(-8*log(2)^2/((4*log(3)^2*log(2)^2 - 4*x*log(3)*log(2) + x^2)*e^x - 12*log(2)^2))

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mupad [B]  time = 3.00, size = 45, normalized size = 1.41 \begin {gather*} -2\,{\mathrm {e}}^{-\frac {8\,{\ln \relax (2)}^2}{x^2\,{\mathrm {e}}^x-12\,{\ln \relax (2)}^2+4\,{\mathrm {e}}^x\,{\ln \relax (2)}^2\,{\ln \relax (3)}^2-4\,x\,{\mathrm {e}}^x\,\ln \relax (2)\,\ln \relax (3)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(8*log(2)^2)/(exp(x)*(x^2 + 4*log(2)^2*log(3)^2 - 4*x*log(2)*log(3)) - 12*log(2)^2))*exp(x)*(4*log(
2)^2*(8*x + 4*x^2) + 64*log(2)^4*log(3)^2 - 8*log(2)^3*log(3)*(8*x + 8)))/(exp(2*x)*(x^4 + 16*log(2)^4*log(3)^
4 - 8*x^3*log(2)*log(3) - 32*x*log(2)^3*log(3)^3 + 24*x^2*log(2)^2*log(3)^2) - exp(x)*(24*x^2*log(2)^2 + 96*lo
g(2)^4*log(3)^2 - 96*x*log(2)^3*log(3)) + 144*log(2)^4),x)

[Out]

-2*exp(-(8*log(2)^2)/(x^2*exp(x) - 12*log(2)^2 + 4*exp(x)*log(2)^2*log(3)^2 - 4*x*exp(x)*log(2)*log(3)))

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sympy [A]  time = 0.82, size = 46, normalized size = 1.44 \begin {gather*} - 2 e^{- \frac {8 \log {\relax (2 )}^{2}}{\left (x^{2} - 4 x \log {\relax (2 )} \log {\relax (3 )} + 4 \log {\relax (2 )}^{2} \log {\relax (3 )}^{2}\right ) e^{x} - 12 \log {\relax (2 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*ln(3)**2*ln(2)**4+8*(8*x+8)*ln(3)*ln(2)**3+4*(-4*x**2-8*x)*ln(2)**2)*exp(x)*exp(-8*ln(2)**2/((4
*ln(3)**2*ln(2)**2-4*x*ln(2)*ln(3)+x**2)*exp(x)-12*ln(2)**2))/((16*ln(3)**4*ln(2)**4-32*x*ln(3)**3*ln(2)**3+24
*x**2*ln(3)**2*ln(2)**2-8*x**3*ln(3)*ln(2)+x**4)*exp(x)**2+(-96*ln(3)**2*ln(2)**4+96*x*ln(3)*ln(2)**3-24*x**2*
ln(2)**2)*exp(x)+144*ln(2)**4),x)

[Out]

-2*exp(-8*log(2)**2/((x**2 - 4*x*log(2)*log(3) + 4*log(2)**2*log(3)**2)*exp(x) - 12*log(2)**2))

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