3.60.42 \(\int \frac {-8 x+8 x^2+(8 x-16 x^2) \log (x)+\frac {e^{-3+x^2} (-2 x+2 x^2+(6 x-4 x^2-8 x^3) \log (x))}{x}+(16 x-16 x \log (x)+\frac {e^{-3+x^2} (4 x+(4-4 x-8 x^2) \log (x))}{x}) \log (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x})+(8+\frac {2 e^{-3+x^2}}{x}) \log ^2(4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x})}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {2 \left (x-\left (x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )^2\right )}{\log (x)} \]

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Rubi [F]  time = 8.12, 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 {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8*x + 8*x^2 + (8*x - 16*x^2)*Log[x] + (E^(-3 + x^2)*(-2*x + 2*x^2 + (6*x - 4*x^2 - 8*x^3)*Log[x]))/x + (
16*x - 16*x*Log[x] + (E^(-3 + x^2)*(4*x + (4 - 4*x - 8*x^2)*Log[x]))/x)*Log[4*Log[5] + (E^(-3 + x^2)*Log[5])/x
] + (8 + (2*E^(-3 + x^2))/x)*Log[4*Log[5] + (E^(-3 + x^2)*Log[5])/x]^2)/(E^(-3 + x^2)*Log[x]^2 + 4*x*Log[x]^2)
,x]

[Out]

-8*ExpIntegralEi[3*Log[x]] + (2*(1 - x)*x)/Log[x] + 4*LogIntegral[x] - 4*Log[(4 + E^(-3 + x^2)/x)*Log[5]]*LogI
ntegral[x] - 16*E^3*Defer[Int][x/((E^x^2 + 4*E^3*x)*Log[x]), x] + 32*E^3*Defer[Int][x^3/((E^x^2 + 4*E^3*x)*Log
[x]), x] + 4*Defer[Int][Log[(4 + E^(-3 + x^2)/x)*Log[5]]/Log[x]^2, x] + 4*Defer[Int][Log[(4 + E^(-3 + x^2)/x)*
Log[5]]/(x*Log[x]), x] - 8*Defer[Int][(x*Log[(4 + E^(-3 + x^2)/x)*Log[5]])/Log[x], x] - 16*E^3*Defer[Int][Log[
(4 + E^(-3 + x^2)/x)*Log[5]]/((E^x^2 + 4*E^3*x)*Log[x]), x] + 32*E^3*Defer[Int][(x^2*Log[(4 + E^(-3 + x^2)/x)*
Log[5]])/((E^x^2 + 4*E^3*x)*Log[x]), x] + 2*Defer[Int][Log[(4 + E^(-3 + x^2)/x)*Log[5]]^2/(x*Log[x]^2), x] - 4
*Defer[Int][(E^x^2*LogIntegral[x])/(x*(E^x^2 + 4*E^3*x)), x] + 8*Defer[Int][(E^x^2*x*LogIntegral[x])/(E^x^2 +
4*E^3*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 \left (-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )\right )}{\left (e^{x^2}+4 e^3 x\right ) \log ^2(x)} \, dx\\ &=e^3 \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{\left (e^{x^2}+4 e^3 x\right ) \log ^2(x)} \, dx\\ &=e^3 \int \left (\frac {16 \left (-1+2 x^2\right ) \left (x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)}-\frac {2 \left (x-x^2-3 x \log (x)+2 x^2 \log (x)+4 x^3 \log (x)-2 x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )-2 \log (x) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )+2 x \log (x) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )+4 x^2 \log (x) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )-\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{e^3 x \log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {x-x^2-3 x \log (x)+2 x^2 \log (x)+4 x^3 \log (x)-2 x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )-2 \log (x) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )+2 x \log (x) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )+4 x^2 \log (x) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )-\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx\right )+\left (16 e^3\right ) \int \frac {\left (-1+2 x^2\right ) \left (x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=-\left (2 \int \frac {x-x^2-2 x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )-\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )+\log (x) \left (x \left (-3+2 x+4 x^2\right )+2 \left (-1+x+2 x^2\right ) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{x \log ^2(x)} \, dx\right )+\left (16 e^3\right ) \int \left (-\frac {x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)}+\frac {2 x^2 \left (x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)}\right ) \, dx\\ &=-\left (2 \int \left (\frac {1-x-3 \log (x)+2 x \log (x)+4 x^2 \log (x)}{\log ^2(x)}+\frac {2 \left (-x-\log (x)+x \log (x)+2 x^2 \log (x)\right ) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)}-\frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)}\right ) \, dx\right )-\left (16 e^3\right ) \int \frac {x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \left (x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=-\left (2 \int \frac {1-x-3 \log (x)+2 x \log (x)+4 x^2 \log (x)}{\log ^2(x)} \, dx\right )+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx-4 \int \frac {\left (-x-\log (x)+x \log (x)+2 x^2 \log (x)\right ) \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx-\left (16 e^3\right ) \int \left (\frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)}+\frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)}\right ) \, dx+\left (32 e^3\right ) \int \left (\frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)}+\frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)}\right ) \, dx\\ &=-\left (2 \int \left (\frac {1-x}{\log ^2(x)}+\frac {-3+2 x+4 x^2}{\log (x)}\right ) \, dx\right )+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx-4 \int \left (-\frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log ^2(x)}+\frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)}-\frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log (x)}+\frac {2 x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)}\right ) \, dx-\left (16 e^3\right ) \int \frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx-\left (16 e^3\right ) \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=-\left (2 \int \frac {1-x}{\log ^2(x)} \, dx\right )-2 \int \frac {-3+2 x+4 x^2}{\log (x)} \, dx+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log ^2(x)} \, dx-4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log (x)} \, dx-8 \int \frac {x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)} \, dx-\left (16 e^3\right ) \int \frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx-\left (16 e^3\right ) \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=\frac {2 (1-x) x}{\log (x)}-4 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right ) \text {li}(x)-2 \int \left (-\frac {3}{\log (x)}+\frac {2 x}{\log (x)}+\frac {4 x^2}{\log (x)}\right ) \, dx+2 \int \frac {1}{\log (x)} \, dx+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx-4 \int \frac {1-x}{\log (x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log (x)} \, dx+4 \int \frac {e^{x^2} \left (-1+2 x^2\right ) \text {li}(x)}{x \left (e^{x^2}+4 e^3 x\right )} \, dx-8 \int \frac {x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)} \, dx-\left (16 e^3\right ) \int \frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx-\left (16 e^3\right ) \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=\frac {2 (1-x) x}{\log (x)}+2 \text {li}(x)-4 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right ) \text {li}(x)+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx-4 \int \left (\frac {1}{\log (x)}-\frac {x}{\log (x)}\right ) \, dx-4 \int \frac {x}{\log (x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log (x)} \, dx+4 \int \left (-\frac {e^{x^2} \text {li}(x)}{x \left (e^{x^2}+4 e^3 x\right )}+\frac {2 e^{x^2} x \text {li}(x)}{e^{x^2}+4 e^3 x}\right ) \, dx+6 \int \frac {1}{\log (x)} \, dx-8 \int \frac {x^2}{\log (x)} \, dx-8 \int \frac {x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)} \, dx-\left (16 e^3\right ) \int \frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx-\left (16 e^3\right ) \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=\frac {2 (1-x) x}{\log (x)}+8 \text {li}(x)-4 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right ) \text {li}(x)+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx-4 \int \frac {1}{\log (x)} \, dx+4 \int \frac {x}{\log (x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log (x)} \, dx-4 \int \frac {e^{x^2} \text {li}(x)}{x \left (e^{x^2}+4 e^3 x\right )} \, dx-4 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-8 \int \frac {x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)} \, dx+8 \int \frac {e^{x^2} x \text {li}(x)}{e^{x^2}+4 e^3 x} \, dx-8 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-\left (16 e^3\right ) \int \frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx-\left (16 e^3\right ) \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=-4 \text {Ei}(2 \log (x))-8 \text {Ei}(3 \log (x))+\frac {2 (1-x) x}{\log (x)}+4 \text {li}(x)-4 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right ) \text {li}(x)+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log (x)} \, dx-4 \int \frac {e^{x^2} \text {li}(x)}{x \left (e^{x^2}+4 e^3 x\right )} \, dx+4 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-8 \int \frac {x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)} \, dx+8 \int \frac {e^{x^2} x \text {li}(x)}{e^{x^2}+4 e^3 x} \, dx-\left (16 e^3\right ) \int \frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx-\left (16 e^3\right ) \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ &=-8 \text {Ei}(3 \log (x))+\frac {2 (1-x) x}{\log (x)}+4 \text {li}(x)-4 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right ) \text {li}(x)+2 \int \frac {\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log ^2(x)} \, dx+4 \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{x \log (x)} \, dx-4 \int \frac {e^{x^2} \text {li}(x)}{x \left (e^{x^2}+4 e^3 x\right )} \, dx-8 \int \frac {x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\log (x)} \, dx+8 \int \frac {e^{x^2} x \text {li}(x)}{e^{x^2}+4 e^3 x} \, dx-\left (16 e^3\right ) \int \frac {x}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx-\left (16 e^3\right ) \int \frac {\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^3}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx+\left (32 e^3\right ) \int \frac {x^2 \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )}{\left (e^{x^2}+4 e^3 x\right ) \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 51, normalized size = 1.65 \begin {gather*} -\frac {2 \left ((-1+x) x+2 x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )+\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x + 8*x^2 + (8*x - 16*x^2)*Log[x] + (E^(-3 + x^2)*(-2*x + 2*x^2 + (6*x - 4*x^2 - 8*x^3)*Log[x]))
/x + (16*x - 16*x*Log[x] + (E^(-3 + x^2)*(4*x + (4 - 4*x - 8*x^2)*Log[x]))/x)*Log[4*Log[5] + (E^(-3 + x^2)*Log
[5])/x] + (8 + (2*E^(-3 + x^2))/x)*Log[4*Log[5] + (E^(-3 + x^2)*Log[5])/x]^2)/(E^(-3 + x^2)*Log[x]^2 + 4*x*Log
[x]^2),x]

[Out]

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

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fricas [A]  time = 0.76, size = 56, normalized size = 1.81 \begin {gather*} -\frac {2 \, {\left (x^{2} + 2 \, x \log \left (e^{\left (x^{2} - \log \relax (x) - 3\right )} \log \relax (5) + 4 \, \log \relax (5)\right ) + \log \left (e^{\left (x^{2} - \log \relax (x) - 3\right )} \log \relax (5) + 4 \, \log \relax (5)\right )^{2} - x\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(-log(x)+x^2-3)+8)*log(log(5)*exp(-log(x)+x^2-3)+4*log(5))^2+(((-8*x^2-4*x+4)*log(x)+4*x)*exp
(-log(x)+x^2-3)-16*x*log(x)+16*x)*log(log(5)*exp(-log(x)+x^2-3)+4*log(5))+((-8*x^3-4*x^2+6*x)*log(x)+2*x^2-2*x
)*exp(-log(x)+x^2-3)+(-16*x^2+8*x)*log(x)+8*x^2-8*x)/(x*log(x)^2*exp(-log(x)+x^2-3)+4*x*log(x)^2),x, algorithm
="fricas")

[Out]

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

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giac [B]  time = 0.53, size = 93, normalized size = 3.00 \begin {gather*} -\frac {2 \, {\left (x^{2} + 2 \, x \log \left (4 \, x e^{3} \log \relax (5) + e^{\left (x^{2}\right )} \log \relax (5)\right ) + \log \left (4 \, x e^{3} \log \relax (5) + e^{\left (x^{2}\right )} \log \relax (5)\right )^{2} - 2 \, x \log \relax (x) - 2 \, \log \left (4 \, x e^{3} + e^{\left (x^{2}\right )}\right ) \log \relax (x) + \log \relax (x)^{2} - 7 \, x - 6 \, \log \left (4 \, x e^{3} \log \relax (5) + e^{\left (x^{2}\right )} \log \relax (5)\right ) + 9\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(-log(x)+x^2-3)+8)*log(log(5)*exp(-log(x)+x^2-3)+4*log(5))^2+(((-8*x^2-4*x+4)*log(x)+4*x)*exp
(-log(x)+x^2-3)-16*x*log(x)+16*x)*log(log(5)*exp(-log(x)+x^2-3)+4*log(5))+((-8*x^3-4*x^2+6*x)*log(x)+2*x^2-2*x
)*exp(-log(x)+x^2-3)+(-16*x^2+8*x)*log(x)+8*x^2-8*x)/(x*log(x)^2*exp(-log(x)+x^2-3)+4*x*log(x)^2),x, algorithm
="giac")

[Out]

-2*(x^2 + 2*x*log(4*x*e^3*log(5) + e^(x^2)*log(5)) + log(4*x*e^3*log(5) + e^(x^2)*log(5))^2 - 2*x*log(x) - 2*l
og(4*x*e^3 + e^(x^2))*log(x) + log(x)^2 - 7*x - 6*log(4*x*e^3*log(5) + e^(x^2)*log(5)) + 9)/log(x)

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maple [B]  time = 0.06, size = 63, normalized size = 2.03

method	result	size
risch	\(-\frac {2 \ln \left (\frac {\ln \relax (5) {\mathrm e}^{x^{2}-3}}{x}+4 \ln \relax (5)\right )^{2}}{\ln \relax (x )}-\frac {4 x \ln \left (\frac {\ln \relax (5) {\mathrm e}^{x^{2}-3}}{x}+4 \ln \relax (5)\right )}{\ln \relax (x )}-\frac {2 x \left (x -1\right )}{\ln \relax (x )}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(-ln(x)+x^2-3)+8)*ln(ln(5)*exp(-ln(x)+x^2-3)+4*ln(5))^2+(((-8*x^2-4*x+4)*ln(x)+4*x)*exp(-ln(x)+x^2-
3)-16*x*ln(x)+16*x)*ln(ln(5)*exp(-ln(x)+x^2-3)+4*ln(5))+((-8*x^3-4*x^2+6*x)*ln(x)+2*x^2-2*x)*exp(-ln(x)+x^2-3)
+(-16*x^2+8*x)*ln(x)+8*x^2-8*x)/(x*ln(x)^2*exp(-ln(x)+x^2-3)+4*x*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.49, size = 75, normalized size = 2.42 \begin {gather*} -\frac {2 \, {\left (x^{2} + x {\left (2 \, \log \left (\log \relax (5)\right ) - 7\right )} + 2 \, {\left (x - \log \relax (x) + \log \left (\log \relax (5)\right ) - 3\right )} \log \left (4 \, x e^{3} + e^{\left (x^{2}\right )}\right ) + \log \left (4 \, x e^{3} + e^{\left (x^{2}\right )}\right )^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2} + \log \left (\log \relax (5)\right )^{2} - 6 \, \log \left (\log \relax (5)\right ) + 9\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(-log(x)+x^2-3)+8)*log(log(5)*exp(-log(x)+x^2-3)+4*log(5))^2+(((-8*x^2-4*x+4)*log(x)+4*x)*exp
(-log(x)+x^2-3)-16*x*log(x)+16*x)*log(log(5)*exp(-log(x)+x^2-3)+4*log(5))+((-8*x^3-4*x^2+6*x)*log(x)+2*x^2-2*x
)*exp(-log(x)+x^2-3)+(-16*x^2+8*x)*log(x)+8*x^2-8*x)/(x*log(x)^2*exp(-log(x)+x^2-3)+4*x*log(x)^2),x, algorithm
="maxima")

[Out]

-2*(x^2 + x*(2*log(log(5)) - 7) + 2*(x - log(x) + log(log(5)) - 3)*log(4*x*e^3 + e^(x^2)) + log(4*x*e^3 + e^(x
^2))^2 - 2*x*log(x) + log(x)^2 + log(log(5))^2 - 6*log(log(5)) + 9)/log(x)

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mupad [B]  time = 4.73, size = 72, normalized size = 2.32 \begin {gather*} \frac {2\,x}{\ln \relax (x)}-\frac {2\,x^2}{\ln \relax (x)}-\frac {2\,{\ln \left (\frac {4\,x\,\ln \relax (5)+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}\,\ln \relax (5)}{x}\right )}^2}{\ln \relax (x)}-\frac {4\,x\,\ln \left (\frac {4\,x\,\ln \relax (5)+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}\,\ln \relax (5)}{x}\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*log(5) + exp(x^2 - log(x) - 3)*log(5))^2*(2*exp(x^2 - log(x) - 3) + 8) - exp(x^2 - log(x) - 3)*(2*x
- 2*x^2 + log(x)*(4*x^2 - 6*x + 8*x^3)) - 8*x + log(x)*(8*x - 16*x^2) + log(4*log(5) + exp(x^2 - log(x) - 3)*
log(5))*(16*x + exp(x^2 - log(x) - 3)*(4*x - log(x)*(4*x + 8*x^2 - 4)) - 16*x*log(x)) + 8*x^2)/(4*x*log(x)^2 +
x*exp(x^2 - log(x) - 3)*log(x)^2),x)

[Out]

(2*x)/log(x) - (2*x^2)/log(x) - (2*log((4*x*log(5) + exp(x^2)*exp(-3)*log(5))/x)^2)/log(x) - (4*x*log((4*x*log
(5) + exp(x^2)*exp(-3)*log(5))/x))/log(x)

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sympy [B]  time = 0.76, size = 61, normalized size = 1.97 \begin {gather*} - \frac {4 x \log {\left (4 \log {\relax (5 )} + \frac {e^{x^{2} - 3} \log {\relax (5 )}}{x} \right )}}{\log {\relax (x )}} + \frac {- 2 x^{2} + 2 x}{\log {\relax (x )}} - \frac {2 \log {\left (4 \log {\relax (5 )} + \frac {e^{x^{2} - 3} \log {\relax (5 )}}{x} \right )}^{2}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(-ln(x)+x**2-3)+8)*ln(ln(5)*exp(-ln(x)+x**2-3)+4*ln(5))**2+(((-8*x**2-4*x+4)*ln(x)+4*x)*exp(-
ln(x)+x**2-3)-16*x*ln(x)+16*x)*ln(ln(5)*exp(-ln(x)+x**2-3)+4*ln(5))+((-8*x**3-4*x**2+6*x)*ln(x)+2*x**2-2*x)*ex
p(-ln(x)+x**2-3)+(-16*x**2+8*x)*ln(x)+8*x**2-8*x)/(x*ln(x)**2*exp(-ln(x)+x**2-3)+4*x*ln(x)**2),x)

[Out]

-4*x*log(4*log(5) + exp(x**2 - 3)*log(5)/x)/log(x) + (-2*x**2 + 2*x)/log(x) - 2*log(4*log(5) + exp(x**2 - 3)*l
og(5)/x)**2/log(x)

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