3.57.19 \(\int \frac {e^{\frac {64 x^2+32 x^3+16 x^4+(32 x+16 x^2+8 x^3) \log (x)+(4+2 x+x^2) \log ^2(x)}{(32 x+16 x^2) (i \pi +\log (5-\log (5)))^2}} (32 x+96 x^2+80 x^3+68 x^4+16 x^5+(8+8 x+4 x^2+17 x^3+4 x^4) \log (x)+(-4-4 x) \log ^2(x))}{(32 x^2+32 x^3+8 x^4) (i \pi +\log (5-\log (5)))^2} \, dx\)

Optimal. Leaf size=40 \[ e^{\frac {\left (x+\frac {4}{2+x}\right ) \left (x+\frac {\log (x)}{4}\right )^2}{x (i \pi +\log (5-\log (5)))^2}} \]

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Rubi [F]  time = 18.59, 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 {64 x^2+32 x^3+16 x^4+\left (32 x+16 x^2+8 x^3\right ) \log (x)+\left (4+2 x+x^2\right ) \log ^2(x)}{\left (32 x+16 x^2\right ) (i \pi +\log (5-\log (5)))^2}\right ) \left (32 x+96 x^2+80 x^3+68 x^4+16 x^5+\left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)+(-4-4 x) \log ^2(x)\right )}{\left (32 x^2+32 x^3+8 x^4\right ) (i \pi +\log (5-\log (5)))^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((64*x^2 + 32*x^3 + 16*x^4 + (32*x + 16*x^2 + 8*x^3)*Log[x] + (4 + 2*x + x^2)*Log[x]^2)/((32*x + 16*x^2
)*(I*Pi + Log[5 - Log[5]])^2))*(32*x + 96*x^2 + 80*x^3 + 68*x^4 + 16*x^5 + (8 + 8*x + 4*x^2 + 17*x^3 + 4*x^4)*
Log[x] + (-4 - 4*x)*Log[x]^2))/((32*x^2 + 32*x^3 + 8*x^4)*(I*Pi + Log[5 - Log[5]])^2),x]

[Out]

Defer[Int][E^(((4 + 2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2)), x]/(2*(I*Pi + Lo
g[5 - Log[5]])^2) + Defer[Int][E^(((4 + 2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2
))/x, x]/(I*Pi + Log[5 - Log[5]])^2 + (2*Defer[Int][E^(((4 + 2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi
+ Log[5 - Log[5]])^2))*x, x])/(I*Pi + Log[5 - Log[5]])^2 + (8*Defer[Int][E^(((4 + 2*x + x^2)*(4*x + Log[x])^2
)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2))/(2 + x)^2, x])/(I*Pi + Log[5 - Log[5]])^2 - Defer[Int][E^(((4 +
2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2))/(2 + x), x]/(I*Pi + Log[5 - Log[5]])^
2 + Defer[Int][E^(((4 + 2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2))*Log[x], x]/(2
*(I*Pi + Log[5 - Log[5]])^2) + Defer[Int][(E^(((4 + 2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5
- Log[5]])^2))*Log[x])/x^2, x]/(4*(I*Pi + Log[5 - Log[5]])^2) - (2*Defer[Int][(E^(((4 + 2*x + x^2)*(4*x + Log[
x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2))*Log[x])/(2 + x)^2, x])/(I*Pi + Log[5 - Log[5]])^2 + Defer[I
nt][(E^(((4 + 2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2))*Log[x])/(2 + x), x]/(8*
(I*Pi + Log[5 - Log[5]])^2) - Defer[Int][(E^(((4 + 2*x + x^2)*(4*x + Log[x])^2)/(x*(32 + 16*x)*(I*Pi + Log[5 -
Log[5]])^2))*Log[x]^2)/x^2, x]/(8*(I*Pi + Log[5 - Log[5]])^2) + Defer[Int][(E^(((4 + 2*x + x^2)*(4*x + Log[x]
)^2)/(x*(32 + 16*x)*(I*Pi + Log[5 - Log[5]])^2))*Log[x]^2)/(2 + x)^2, x]/(8*(I*Pi + Log[5 - Log[5]])^2)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (\frac {64 x^2+32 x^3+16 x^4+\left (32 x+16 x^2+8 x^3\right ) \log (x)+\left (4+2 x+x^2\right ) \log ^2(x)}{\left (32 x+16 x^2\right ) (i \pi +\log (5-\log (5)))^2}\right ) \left (32 x+96 x^2+80 x^3+68 x^4+16 x^5+\left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)+(-4-4 x) \log ^2(x)\right )}{32 x^2+32 x^3+8 x^4} \, dx}{(i \pi +\log (5-\log (5)))^2}\\ &=\frac {\int \frac {\exp \left (\frac {64 x^2+32 x^3+16 x^4+\left (32 x+16 x^2+8 x^3\right ) \log (x)+\left (4+2 x+x^2\right ) \log ^2(x)}{\left (32 x+16 x^2\right ) (i \pi +\log (5-\log (5)))^2}\right ) \left (32 x+96 x^2+80 x^3+68 x^4+16 x^5+\left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)+(-4-4 x) \log ^2(x)\right )}{x^2 \left (32+32 x+8 x^2\right )} \, dx}{(i \pi +\log (5-\log (5)))^2}\\ &=\frac {\int \frac {\exp \left (\frac {64 x^2+32 x^3+16 x^4+\left (32 x+16 x^2+8 x^3\right ) \log (x)+\left (4+2 x+x^2\right ) \log ^2(x)}{\left (32 x+16 x^2\right ) (i \pi +\log (5-\log (5)))^2}\right ) \left (32 x+96 x^2+80 x^3+68 x^4+16 x^5+\left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)+(-4-4 x) \log ^2(x)\right )}{8 x^2 (2+x)^2} \, dx}{(i \pi +\log (5-\log (5)))^2}\\ &=\frac {\int \frac {\exp \left (\frac {64 x^2+32 x^3+16 x^4+\left (32 x+16 x^2+8 x^3\right ) \log (x)+\left (4+2 x+x^2\right ) \log ^2(x)}{\left (32 x+16 x^2\right ) (i \pi +\log (5-\log (5)))^2}\right ) \left (32 x+96 x^2+80 x^3+68 x^4+16 x^5+\left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)+(-4-4 x) \log ^2(x)\right )}{x^2 (2+x)^2} \, dx}{8 (i \pi +\log (5-\log (5)))^2}\\ &=\frac {\int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) \left (32 x+96 x^2+80 x^3+68 x^4+16 x^5+\left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)+(-4-4 x) \log ^2(x)\right )}{x^2 (2+x)^2} \, dx}{8 (i \pi +\log (5-\log (5)))^2}\\ &=\frac {\int \left (\frac {96 \exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right )}{(2+x)^2}+\frac {32 \exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right )}{x (2+x)^2}+\frac {80 \exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) x}{(2+x)^2}+\frac {68 \exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) x^2}{(2+x)^2}+\frac {16 \exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) x^3}{(2+x)^2}+\frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) \left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)}{x^2 (2+x)^2}-\frac {4 \exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) (1+x) \log ^2(x)}{x^2 (2+x)^2}\right ) \, dx}{8 (i \pi +\log (5-\log (5)))^2}\\ &=\frac {\int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) \left (8+8 x+4 x^2+17 x^3+4 x^4\right ) \log (x)}{x^2 (2+x)^2} \, dx}{8 (i \pi +\log (5-\log (5)))^2}-\frac {\int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) (1+x) \log ^2(x)}{x^2 (2+x)^2} \, dx}{2 (i \pi +\log (5-\log (5)))^2}+\frac {2 \int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) x^3}{(2+x)^2} \, dx}{(i \pi +\log (5-\log (5)))^2}+\frac {4 \int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right )}{x (2+x)^2} \, dx}{(i \pi +\log (5-\log (5)))^2}+\frac {17 \int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) x^2}{(2+x)^2} \, dx}{2 (i \pi +\log (5-\log (5)))^2}+\frac {10 \int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right ) x}{(2+x)^2} \, dx}{(i \pi +\log (5-\log (5)))^2}+\frac {12 \int \frac {\exp \left (\frac {\left (4+2 x+x^2\right ) (4 x+\log (x))^2}{x (32+16 x) (i \pi +\log (5-\log (5)))^2}\right )}{(2+x)^2} \, dx}{(i \pi +\log (5-\log (5)))^2}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.50, size = 82, normalized size = 2.05 \begin {gather*} e^{-\frac {\left (4+2 x+x^2\right ) \left (16 x^2+\log ^2(x)\right )}{16 x (2+x) (\pi -i \log (5-\log (5)))^2}} x^{-\frac {4+2 x+x^2}{2 (2+x) (\pi -i \log (5-\log (5)))^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((64*x^2 + 32*x^3 + 16*x^4 + (32*x + 16*x^2 + 8*x^3)*Log[x] + (4 + 2*x + x^2)*Log[x]^2)/((32*x +
16*x^2)*(I*Pi + Log[5 - Log[5]])^2))*(32*x + 96*x^2 + 80*x^3 + 68*x^4 + 16*x^5 + (8 + 8*x + 4*x^2 + 17*x^3 + 4
*x^4)*Log[x] + (-4 - 4*x)*Log[x]^2))/((32*x^2 + 32*x^3 + 8*x^4)*(I*Pi + Log[5 - Log[5]])^2),x]

[Out]

1/(E^(((4 + 2*x + x^2)*(16*x^2 + Log[x]^2))/(16*x*(2 + x)*(Pi - I*Log[5 - Log[5]])^2))*x^((4 + 2*x + x^2)/(2*(
2 + x)*(Pi - I*Log[5 - Log[5]])^2)))

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fricas [B]  time = 1.01, size = 64, normalized size = 1.60 \begin {gather*} e^{\left (\frac {16 \, x^{4} + 32 \, x^{3} + {\left (x^{2} + 2 \, x + 4\right )} \log \relax (x)^{2} + 64 \, x^{2} + 8 \, {\left (x^{3} + 2 \, x^{2} + 4 \, x\right )} \log \relax (x)}{16 \, {\left (x^{2} + 2 \, x\right )} \log \left (\log \relax (5) - 5\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-4)*log(x)^2+(4*x^4+17*x^3+4*x^2+8*x+8)*log(x)+16*x^5+68*x^4+80*x^3+96*x^2+32*x)*exp(((x^2+2*x
+4)*log(x)^2+(8*x^3+16*x^2+32*x)*log(x)+16*x^4+32*x^3+64*x^2)/(16*x^2+32*x)/log(log(5)-5)^2)/(8*x^4+32*x^3+32*
x^2)/log(log(5)-5)^2,x, algorithm="fricas")

[Out]

e^(1/16*(16*x^4 + 32*x^3 + (x^2 + 2*x + 4)*log(x)^2 + 64*x^2 + 8*(x^3 + 2*x^2 + 4*x)*log(x))/((x^2 + 2*x)*log(
log(5) - 5)^2))

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giac [B]  time = 1.62, size = 272, normalized size = 6.80 \begin {gather*} e^{\left (\frac {x^{4}}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {x^{3} \log \relax (x)}{2 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}} + \frac {x^{2} \log \relax (x)^{2}}{16 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}} + \frac {2 \, x^{3}}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {x^{2} \log \relax (x)}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {x \log \relax (x)^{2}}{8 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}} + \frac {4 \, x^{2}}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {2 \, x \log \relax (x)}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {\log \relax (x)^{2}}{4 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-4)*log(x)^2+(4*x^4+17*x^3+4*x^2+8*x+8)*log(x)+16*x^5+68*x^4+80*x^3+96*x^2+32*x)*exp(((x^2+2*x
+4)*log(x)^2+(8*x^3+16*x^2+32*x)*log(x)+16*x^4+32*x^3+64*x^2)/(16*x^2+32*x)/log(log(5)-5)^2)/(8*x^4+32*x^3+32*
x^2)/log(log(5)-5)^2,x, algorithm="giac")

[Out]

e^(x^4/(x^2*log(log(5) - 5)^2 + 2*x*log(log(5) - 5)^2) + 1/2*x^3*log(x)/(x^2*log(log(5) - 5)^2 + 2*x*log(log(5
) - 5)^2) + 1/16*x^2*log(x)^2/(x^2*log(log(5) - 5)^2 + 2*x*log(log(5) - 5)^2) + 2*x^3/(x^2*log(log(5) - 5)^2 +
2*x*log(log(5) - 5)^2) + x^2*log(x)/(x^2*log(log(5) - 5)^2 + 2*x*log(log(5) - 5)^2) + 1/8*x*log(x)^2/(x^2*log
(log(5) - 5)^2 + 2*x*log(log(5) - 5)^2) + 4*x^2/(x^2*log(log(5) - 5)^2 + 2*x*log(log(5) - 5)^2) + 2*x*log(x)/(
x^2*log(log(5) - 5)^2 + 2*x*log(log(5) - 5)^2) + 1/4*log(x)^2/(x^2*log(log(5) - 5)^2 + 2*x*log(log(5) - 5)^2))

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maple [A]  time = 0.34, size = 35, normalized size = 0.88

method	result	size
risch	\({\mathrm e}^{\frac {\left (x^{2}+2 x +4\right ) \left (4 x +\ln \relax (x )\right )^{2}}{16 x \left (2+x \right ) \ln \left (\ln \relax (5)-5\right )^{2}}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-4)*ln(x)^2+(4*x^4+17*x^3+4*x^2+8*x+8)*ln(x)+16*x^5+68*x^4+80*x^3+96*x^2+32*x)*exp(((x^2+2*x+4)*ln(x
)^2+(8*x^3+16*x^2+32*x)*ln(x)+16*x^4+32*x^3+64*x^2)/(16*x^2+32*x)/ln(ln(5)-5)^2)/(8*x^4+32*x^3+32*x^2)/ln(ln(5
)-5)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {8 \, e^{\left (\frac {x^{4}}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {x^{3} \log \relax (x)}{2 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}} + \frac {x^{2} \log \relax (x)^{2}}{16 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}} + \frac {2 \, x^{3}}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {x^{2} \log \relax (x)}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {x \log \relax (x)^{2}}{8 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}} + \frac {4 \, x^{2}}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {2 \, x \log \relax (x)}{x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}} + \frac {\log \relax (x)^{2}}{4 \, {\left (x^{2} \log \left (\log \relax (5) - 5\right )^{2} + 2 \, x \log \left (\log \relax (5) - 5\right )^{2}\right )}}\right )} \log \left (\log \relax (5) - 5\right )^{2}}{8 \, \log \left (\log \relax (5) - 5\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-4)*log(x)^2+(4*x^4+17*x^3+4*x^2+8*x+8)*log(x)+16*x^5+68*x^4+80*x^3+96*x^2+32*x)*exp(((x^2+2*x
+4)*log(x)^2+(8*x^3+16*x^2+32*x)*log(x)+16*x^4+32*x^3+64*x^2)/(16*x^2+32*x)/log(log(5)-5)^2)/(8*x^4+32*x^3+32*
x^2)/log(log(5)-5)^2,x, algorithm="maxima")

[Out]

1/8*integrate((16*x^5 + 68*x^4 + 80*x^3 - 4*(x + 1)*log(x)^2 + 96*x^2 + (4*x^4 + 17*x^3 + 4*x^2 + 8*x + 8)*log
(x) + 32*x)*e^(1/16*(16*x^4 + 32*x^3 + (x^2 + 2*x + 4)*log(x)^2 + 64*x^2 + 8*(x^3 + 2*x^2 + 4*x)*log(x))/((x^2
+ 2*x)*log(log(5) - 5)^2))/(x^4 + 4*x^3 + 4*x^2), x)/log(log(5) - 5)^2

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mupad [B]  time = 3.93, size = 254, normalized size = 6.35 \begin {gather*} x^{\frac {16\,x^2}{16\,{\ln \left (\ln \relax (5)-5\right )}^2\,x^2+32\,{\ln \left (\ln \relax (5)-5\right )}^2\,x}+\frac {x^2+4}{2\,\left (x\,{\ln \left (\ln \relax (5)-5\right )}^2+2\,{\ln \left (\ln \relax (5)-5\right )}^2\right )}}\,{\mathrm {e}}^{\frac {4\,{\ln \relax (x)}^2}{16\,{\ln \left (\ln \relax (5)-5\right )}^2\,x^2+32\,{\ln \left (\ln \relax (5)-5\right )}^2\,x}}\,{\mathrm {e}}^{\frac {2\,x\,{\ln \relax (x)}^2}{16\,{\ln \left (\ln \relax (5)-5\right )}^2\,x^2+32\,{\ln \left (\ln \relax (5)-5\right )}^2\,x}}\,{\mathrm {e}}^{\frac {16\,x^4}{16\,{\ln \left (\ln \relax (5)-5\right )}^2\,x^2+32\,{\ln \left (\ln \relax (5)-5\right )}^2\,x}}\,{\mathrm {e}}^{\frac {32\,x^3}{16\,{\ln \left (\ln \relax (5)-5\right )}^2\,x^2+32\,{\ln \left (\ln \relax (5)-5\right )}^2\,x}}\,{\mathrm {e}}^{\frac {64\,x^2}{16\,{\ln \left (\ln \relax (5)-5\right )}^2\,x^2+32\,{\ln \left (\ln \relax (5)-5\right )}^2\,x}}\,{\mathrm {e}}^{\frac {x^2\,{\ln \relax (x)}^2}{16\,{\ln \left (\ln \relax (5)-5\right )}^2\,x^2+32\,{\ln \left (\ln \relax (5)-5\right )}^2\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(x)^2*(2*x + x^2 + 4) + 64*x^2 + 32*x^3 + 16*x^4 + log(x)*(32*x + 16*x^2 + 8*x^3))/(log(log(5) -
5)^2*(32*x + 16*x^2)))*(32*x + log(x)*(8*x + 4*x^2 + 17*x^3 + 4*x^4 + 8) + 96*x^2 + 80*x^3 + 68*x^4 + 16*x^5 -
log(x)^2*(4*x + 4)))/(log(log(5) - 5)^2*(32*x^2 + 32*x^3 + 8*x^4)),x)

[Out]

x^((16*x^2)/(32*x*log(log(5) - 5)^2 + 16*x^2*log(log(5) - 5)^2) + (x^2 + 4)/(2*(x*log(log(5) - 5)^2 + 2*log(lo
g(5) - 5)^2)))*exp((4*log(x)^2)/(32*x*log(log(5) - 5)^2 + 16*x^2*log(log(5) - 5)^2))*exp((2*x*log(x)^2)/(32*x*
log(log(5) - 5)^2 + 16*x^2*log(log(5) - 5)^2))*exp((16*x^4)/(32*x*log(log(5) - 5)^2 + 16*x^2*log(log(5) - 5)^2
))*exp((32*x^3)/(32*x*log(log(5) - 5)^2 + 16*x^2*log(log(5) - 5)^2))*exp((64*x^2)/(32*x*log(log(5) - 5)^2 + 16
*x^2*log(log(5) - 5)^2))*exp((x^2*log(x)^2)/(32*x*log(log(5) - 5)^2 + 16*x^2*log(log(5) - 5)^2))

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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(((-4*x-4)*ln(x)**2+(4*x**4+17*x**3+4*x**2+8*x+8)*ln(x)+16*x**5+68*x**4+80*x**3+96*x**2+32*x)*exp(((x
**2+2*x+4)*ln(x)**2+(8*x**3+16*x**2+32*x)*ln(x)+16*x**4+32*x**3+64*x**2)/(16*x**2+32*x)/ln(ln(5)-5)**2)/(8*x**
4+32*x**3+32*x**2)/ln(ln(5)-5)**2,x)

[Out]

Timed out

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