3.59.9 \(\int \frac {e^{\frac {32 x+(-4+8 x+12 x^2) \log (\log (5))}{(1-2 x+x^2) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{(-1+3 x-3 x^2+x^3) \log (\log (5))} \, dx\) [5809]

Optimal. Leaf size=32 \[ -3+e^{4 \left (-1+\frac {4 \left (x+\frac {2}{\log (\log (5))}\right )}{-4+\frac {(1+x)^2}{x}}\right )} \]

[Out]

exp(16*(x+2/ln(ln(5)))/((1+x)^2/x-4)-4)-3

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Rubi [F]
time = 1.32, 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 (\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Log[Log[5]]))*(-32 - 32*x - 32*x*Log[Log[5]]
))/((-1 + 3*x - 3*x^2 + x^3)*Log[Log[5]]),x]

[Out]

(-32*(2 + Log[Log[5]])*Defer[Int][1/(E^((4*(Log[Log[5]] - 3*x^2*Log[Log[5]] - 2*x*(4 + Log[Log[5]])))/((-1 + x
)^2*Log[Log[5]]))*(-1 + x)^3), x])/Log[Log[5]] - 32*(1 + Log[Log[5]]^(-1))*Defer[Int][1/(E^((4*(Log[Log[5]] -
3*x^2*Log[Log[5]] - 2*x*(4 + Log[Log[5]])))/((-1 + x)^2*Log[Log[5]]))*(-1 + x)^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (-32+x (-32-32 \log (\log (5))))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx\\ &=\frac {\int \frac {\exp \left (\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (-32+x (-32-32 \log (\log (5))))}{-1+3 x-3 x^2+x^3} \, dx}{\log (\log (5))}\\ &=\frac {\int \frac {\exp \left (\frac {-4 \log (\log (5))+12 x^2 \log (\log (5))+8 x (4+\log (\log (5)))}{\left (1-2 x+x^2\right ) \log (\log (5))}\right ) (32-x (-32-32 \log (\log (5))))}{1-3 x+3 x^2-x^3} \, dx}{\log (\log (5))}\\ &=\frac {\int \frac {32 \exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (1+x (1+\log (\log (5))))}{(1-x)^3} \, dx}{\log (\log (5))}\\ &=\frac {32 \int \frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (1+x (1+\log (\log (5))))}{(1-x)^3} \, dx}{\log (\log (5))}\\ &=\frac {32 \int \left (\frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (-2-\log (\log (5)))}{(-1+x)^3}+\frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right ) (-1-\log (\log (5)))}{(-1+x)^2}\right ) \, dx}{\log (\log (5))}\\ &=\frac {(32 (-2-\log (\log (5)))) \int \frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right )}{(-1+x)^3} \, dx}{\log (\log (5))}+\frac {(32 (-1-\log (\log (5)))) \int \frac {\exp \left (-\frac {4 \left (\log (\log (5))-3 x^2 \log (\log (5))-2 x (4+\log (\log (5)))\right )}{(-1+x)^2 \log (\log (5))}\right )}{(-1+x)^2} \, dx}{\log (\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.17, size = 35, normalized size = 1.09 \begin {gather*} e^{\frac {-4 \log (\log (5))+12 x^2 \log (\log (5))+8 x (4+\log (\log (5)))}{(-1+x)^2 \log (\log (5))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Log[Log[5]]))*(-32 - 32*x - 32*x*Log[L
og[5]]))/((-1 + 3*x - 3*x^2 + x^3)*Log[Log[5]]),x]

[Out]

E^((-4*Log[Log[5]] + 12*x^2*Log[Log[5]] + 8*x*(4 + Log[Log[5]]))/((-1 + x)^2*Log[Log[5]]))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.72, size = 587, normalized size = 18.34 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*x*ln(ln(5))-32*x-32)*exp(((12*x^2+8*x-4)*ln(ln(5))+32*x)/(x^2-2*x+1)/ln(ln(5)))/(x^3-3*x^2+3*x-1)/ln(
ln(5)),x,method=_RETURNVERBOSE)

[Out]

32/ln(ln(5))*(-1/8*I*Pi^(1/2)*exp(12-1/4*(32+32/ln(ln(5)))^2/(16+32/ln(ln(5))))/(1+2/ln(ln(5)))^(1/2)*erf(4*I*
(1+2/ln(ln(5)))^(1/2)/(x-1)+1/8*I*(32+32/ln(ln(5)))/(1+2/ln(ln(5)))^(1/2))+1/(16+32/ln(ln(5)))*exp(12+(16+32/l
n(ln(5)))/(x-1)^2+(32+32/ln(ln(5)))/(x-1))+6*I/(16+32/ln(ln(5)))*Pi^(1/2)*exp(12-1/4*(32+32/ln(ln(5)))^2/(16+3
2/ln(ln(5))))/(1+2/ln(ln(5)))^(1/2)*erf(4*I*(1+2/ln(ln(5)))^(1/2)/(x-1)+1/8*I*(32+32/ln(ln(5)))/(1+2/ln(ln(5))
)^(1/2))+4*I/(16+32/ln(ln(5)))*Pi^(1/2)*exp(12-1/4*(32+32/ln(ln(5)))^2/(16+32/ln(ln(5))))/(1+2/ln(ln(5)))^(1/2
)*erf(4*I*(1+2/ln(ln(5)))^(1/2)/(x-1)+1/8*I*(32+32/ln(ln(5)))/(1+2/ln(ln(5)))^(1/2))/ln(ln(5))-1/8*I*ln(ln(5))
*Pi^(1/2)*exp(12-1/4*(32+32/ln(ln(5)))^2/(16+32/ln(ln(5))))/(1+2/ln(ln(5)))^(1/2)*erf(4*I*(1+2/ln(ln(5)))^(1/2
)/(x-1)+1/8*I*(32+32/ln(ln(5)))/(1+2/ln(ln(5)))^(1/2))+1/2*ln(ln(5))/(16+32/ln(ln(5)))*exp(12+(16+32/ln(ln(5))
)/(x-1)^2+(32+32/ln(ln(5)))/(x-1))+2*I*ln(ln(5))/(16+32/ln(ln(5)))*Pi^(1/2)*exp(12-1/4*(32+32/ln(ln(5)))^2/(16
+32/ln(ln(5))))/(1+2/ln(ln(5)))^(1/2)*erf(4*I*(1+2/ln(ln(5)))^(1/2)/(x-1)+1/8*I*(32+32/ln(ln(5)))/(1+2/ln(ln(5
)))^(1/2)))

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Maxima [A]
time = 0.71, size = 58, normalized size = 1.81 \begin {gather*} e^{\left (\frac {32}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {16}{x^{2} - 2 \, x + 1} + \frac {32}{x \log \left (\log \left (5\right )\right ) - \log \left (\log \left (5\right )\right )} + \frac {32}{x - 1} + 12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^
2+3*x-1)/log(log(5)),x, algorithm="maxima")

[Out]

e^(32/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 16/(x^2 - 2*x + 1) + 32/(x*log(log(5)) - log(log(5))
) + 32/(x - 1) + 12)

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Fricas [A]
time = 0.36, size = 36, normalized size = 1.12 \begin {gather*} e^{\left (\frac {4 \, {\left ({\left (3 \, x^{2} + 2 \, x - 1\right )} \log \left (\log \left (5\right )\right ) + 8 \, x\right )}}{{\left (x^{2} - 2 \, x + 1\right )} \log \left (\log \left (5\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^
2+3*x-1)/log(log(5)),x, algorithm="fricas")

[Out]

e^(4*((3*x^2 + 2*x - 1)*log(log(5)) + 8*x)/((x^2 - 2*x + 1)*log(log(5))))

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Sympy [A]
time = 0.23, size = 32, normalized size = 1.00 \begin {gather*} e^{\frac {32 x + \left (12 x^{2} + 8 x - 4\right ) \log {\left (\log {\left (5 \right )} \right )}}{\left (x^{2} - 2 x + 1\right ) \log {\left (\log {\left (5 \right )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*ln(ln(5))-32*x-32)*exp(((12*x**2+8*x-4)*ln(ln(5))+32*x)/(x**2-2*x+1)/ln(ln(5)))/(x**3-3*x**2+
3*x-1)/ln(ln(5)),x)

[Out]

exp((32*x + (12*x**2 + 8*x - 4)*log(log(5)))/((x**2 - 2*x + 1)*log(log(5))))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (29) = 58\).
time = 0.40, size = 100, normalized size = 3.12 \begin {gather*} e^{\left (\frac {12 \, x^{2} \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {8 \, x \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {32 \, x}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} - \frac {4 \, \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^
2+3*x-1)/log(log(5)),x, algorithm="giac")

[Out]

e^(12*x^2*log(log(5))/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 8*x*log(log(5))/(x^2*log(log(5)) - 2
*x*log(log(5)) + log(log(5))) + 32*x/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) - 4*log(log(5))/(x^2*lo
g(log(5)) - 2*x*log(log(5)) + log(log(5))))

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Mupad [B]
time = 5.77, size = 52, normalized size = 1.62 \begin {gather*} {\mathrm {e}}^{\frac {32\,x}{\ln \left (\ln \left (5\right )\right )\,x^2-2\,\ln \left (\ln \left (5\right )\right )\,x+\ln \left (\ln \left (5\right )\right )}}\,{\ln \left (5\right )}^{\frac {12\,x^2+8\,x-4}{\ln \left ({\ln \left (5\right )}^{x^2-2\,x+1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((32*x + log(log(5))*(8*x + 12*x^2 - 4))/(log(log(5))*(x^2 - 2*x + 1)))*(32*x + 32*x*log(log(5)) + 32
))/(log(log(5))*(3*x - 3*x^2 + x^3 - 1)),x)

[Out]

exp((32*x)/(log(log(5)) + x^2*log(log(5)) - 2*x*log(log(5))))*log(5)^((8*x + 12*x^2 - 4)/log(log(5)^(x^2 - 2*x
+ 1)))

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