Integrand size = 174, antiderivative size = 31 \[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=\frac {2 \left (x-\left (x+\log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )^2\right )}{\log (x)} \] Output:
2*(x-(ln((4+exp(-ln(x)+x^2-3))*ln(5))+x)^2)/ln(x)
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=-\frac {2 \left ((-1+x) x+2 x \log \left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )+\log ^2\left (\left (4+\frac {e^{-3+x^2}}{x}\right ) \log (5)\right )\right )}{\log (x)} \] Input:
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]
Output:
(-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]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {8 x^2+\left (\frac {2 e^{x^2-3}}{x}+8\right ) \log ^2\left (\frac {e^{x^2-3} \log (5)}{x}+4 \log (5)\right )+\left (8 x-16 x^2\right ) \log (x)+\left (\frac {e^{x^2-3} \left (\left (-8 x^2-4 x+4\right ) \log (x)+4 x\right )}{x}+16 x-16 x \log (x)\right ) \log \left (\frac {e^{x^2-3} \log (5)}{x}+4 \log (5)\right )+\frac {e^{x^2-3} \left (2 x^2+\left (-8 x^3-4 x^2+6 x\right ) \log (x)-2 x\right )}{x}-8 x}{e^{x^2-3} \log ^2(x)+4 x \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^3 \left (8 x^2+\left (\frac {2 e^{x^2-3}}{x}+8\right ) \log ^2\left (\frac {e^{x^2-3} \log (5)}{x}+4 \log (5)\right )+\left (8 x-16 x^2\right ) \log (x)+\left (\frac {e^{x^2-3} \left (\left (-8 x^2-4 x+4\right ) \log (x)+4 x\right )}{x}+16 x-16 x \log (x)\right ) \log \left (\frac {e^{x^2-3} \log (5)}{x}+4 \log (5)\right )+\frac {e^{x^2-3} \left (2 x^2+\left (-8 x^3-4 x^2+6 x\right ) \log (x)-2 x\right )}{x}-8 x\right )}{\left (e^{x^2}+4 e^3 x\right ) \log ^2(x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^3 \int -\frac {2 \left (-4 x^2+4 x-\left (4+\frac {e^{x^2-3}}{x}\right ) \log ^2\left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )-4 \left (x-2 x^2\right ) \log (x)-2 \left (-4 \log (x) x+4 x+\frac {e^{x^2-3} \left (x+\left (-2 x^2-x+1\right ) \log (x)\right )}{x}\right ) \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )+\frac {e^{x^2-3} \left (-x^2+x-\left (-4 x^3-2 x^2+3 x\right ) \log (x)\right )}{x}\right )}{\left (4 e^3 x+e^{x^2}\right ) \log ^2(x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -2 e^3 \int \frac {-4 x^2+4 x-\left (4+\frac {e^{x^2-3}}{x}\right ) \log ^2\left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )-4 \left (x-2 x^2\right ) \log (x)-2 \left (-4 \log (x) x+4 x+\frac {e^{x^2-3} \left (x+\left (-2 x^2-x+1\right ) \log (x)\right )}{x}\right ) \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )+\frac {e^{x^2-3} \left (-x^2+x-\left (-4 x^3-2 x^2+3 x\right ) \log (x)\right )}{x}}{\left (4 e^3 x+e^{x^2}\right ) \log ^2(x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -2 e^3 \int \left (\frac {4 \log (x) x^3+2 \log (x) x^2+4 \log (x) \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right ) x^2-x^2-3 \log (x) x+2 \log (x) \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right ) x-2 \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right ) x+x-\log ^2\left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )-2 \log (x) \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )}{e^3 x \log ^2(x)}-\frac {8 \left (2 x^2-1\right ) \left (x+\log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )\right )}{\left (4 e^3 x+e^{x^2}\right ) \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 e^3 \left (\frac {2 \log (5) \int \frac {e^{x^2} \operatorname {LogIntegral}(x)}{x \left (e^3 \log (625) x+e^{x^2} \log (5)\right )}dx}{e^3}-\frac {4 \log (5) \int \frac {e^{x^2} x \operatorname {LogIntegral}(x)}{e^3 \log (625) x+e^{x^2} \log (5)}dx}{e^3}-\frac {2 \int \frac {\log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )}{\log ^2(x)}dx}{e^3}-\frac {\int \frac {\log ^2\left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )}{x \log ^2(x)}dx}{e^3}+8 \int \frac {x}{\left (4 e^3 x+e^{x^2}\right ) \log (x)}dx-\frac {2 \int \frac {\log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )}{x \log (x)}dx}{e^3}+\frac {4 \int \frac {x \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )}{\log (x)}dx}{e^3}+8 \int \frac {\log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )}{\left (4 e^3 x+e^{x^2}\right ) \log (x)}dx-16 \int \frac {x^2 \log \left (\log (625)+\frac {e^{x^2-3} \log (5)}{x}\right )}{\left (4 e^3 x+e^{x^2}\right ) \log (x)}dx-16 \int \frac {x^3}{\left (4 e^3 x+e^{x^2}\right ) \log (x)}dx+\frac {4 \operatorname {ExpIntegralEi}(3 \log (x))}{e^3}+\frac {2 \operatorname {LogIntegral}(x) \log \left (\frac {e^{x^2-3} \log (5)}{x}+\log (625)\right )}{e^3}-\frac {2 \operatorname {LogIntegral}(x)}{e^3}-\frac {(1-x) x}{e^3 \log (x)}\right )\) |
Input:
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]
Output:
$Aborted
Time = 0.68 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77
| method | result | size |
| parallelrisch | \(-\frac {16 x^{2}+32 \ln \left (\left (4+{\mathrm e}^{-\ln \left (x \right )+x^{2}-3}\right ) \ln \left (5\right )\right ) x +16 {\ln \left (\left (4+{\mathrm e}^{-\ln \left (x \right )+x^{2}-3}\right ) \ln \left (5\right )\right )}^{2}-16 x}{8 \ln \left (x \right )}\) | \(55\) |
| risch | \(-\frac {2 \ln \left (\frac {\ln \left (5\right ) {\mathrm e}^{x^{2}-3}}{x}+4 \ln \left (5\right )\right )^{2}}{\ln \left (x \right )}-\frac {4 x \ln \left (\frac {\ln \left (5\right ) {\mathrm e}^{x^{2}-3}}{x}+4 \ln \left (5\right )\right )}{\ln \left (x \right )}-\frac {2 x \left (-1+x \right )}{\ln \left (x \right )}\) | \(63\) |
Input:
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)
Output:
-1/8*(16*x^2+32*ln((4+exp(-ln(x)+x^2-3))*ln(5))*x+16*ln((4+exp(-ln(x)+x^2- 3))*ln(5))^2-16*x)/ln(x)
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=-\frac {2 \, {\left (x^{2} + 2 \, x \log \left (e^{\left (x^{2} - \log \left (x\right ) - 3\right )} \log \left (5\right ) + 4 \, \log \left (5\right )\right ) + \log \left (e^{\left (x^{2} - \log \left (x\right ) - 3\right )} \log \left (5\right ) + 4 \, \log \left (5\right )\right )^{2} - x\right )}}{\log \left (x\right )} \] Input:
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")
Output:
-2*(x^2 + 2*x*log(e^(x^2 - log(x) - 3)*log(5) + 4*log(5)) + log(e^(x^2 - l og(x) - 3)*log(5) + 4*log(5))^2 - x)/log(x)
Exception generated. \[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
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)*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)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (30) = 60\).
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.42 \[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=-\frac {2 \, {\left (x^{2} + x {\left (2 \, \log \left (\log \left (5\right )\right ) - 7\right )} + 2 \, {\left (x - \log \left (x\right ) + \log \left (\log \left (5\right )\right ) - 3\right )} \log \left (4 \, x e^{3} + e^{\left (x^{2}\right )}\right ) + \log \left (4 \, x e^{3} + e^{\left (x^{2}\right )}\right )^{2} - 2 \, x \log \left (x\right ) + \log \left (x\right )^{2} + \log \left (\log \left (5\right )\right )^{2} - 6 \, \log \left (\log \left (5\right )\right ) + 9\right )}}{\log \left (x\right )} \] Input:
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")
Output:
-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(l og(5))^2 - 6*log(log(5)) + 9)/log(x)
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (30) = 60\).
Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.00 \[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=-\frac {2 \, {\left (x^{2} + 2 \, x \log \left (4 \, x e^{3} \log \left (5\right ) + e^{\left (x^{2}\right )} \log \left (5\right )\right ) + \log \left (4 \, x e^{3} \log \left (5\right ) + e^{\left (x^{2}\right )} \log \left (5\right )\right )^{2} - 2 \, x \log \left (x\right ) - 2 \, \log \left (4 \, x e^{3} + e^{\left (x^{2}\right )}\right ) \log \left (x\right ) + \log \left (x\right )^{2} - 7 \, x - 6 \, \log \left (4 \, x e^{3} \log \left (5\right ) + e^{\left (x^{2}\right )} \log \left (5\right )\right ) + 9\right )}}{\log \left (x\right )} \] Input:
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")
Output:
-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*log(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)
Time = 2.81 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.32 \[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=\frac {2\,x}{\ln \left (x\right )}-\frac {2\,x^2}{\ln \left (x\right )}-\frac {2\,{\ln \left (\frac {4\,x\,\ln \left (5\right )+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}\,\ln \left (5\right )}{x}\right )}^2}{\ln \left (x\right )}-\frac {4\,x\,\ln \left (\frac {4\,x\,\ln \left (5\right )+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}\,\ln \left (5\right )}{x}\right )}{\ln \left (x\right )} \] Input:
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)*lo g(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)
Output:
(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)
\[ \int \frac {-8 x+8 x^2+\left (8 x-16 x^2\right ) \log (x)+\frac {e^{-3+x^2} \left (-2 x+2 x^2+\left (6 x-4 x^2-8 x^3\right ) \log (x)\right )}{x}+\left (16 x-16 x \log (x)+\frac {e^{-3+x^2} \left (4 x+\left (4-4 x-8 x^2\right ) \log (x)\right )}{x}\right ) \log \left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )+\left (8+\frac {2 e^{-3+x^2}}{x}\right ) \log ^2\left (4 \log (5)+\frac {e^{-3+x^2} \log (5)}{x}\right )}{e^{-3+x^2} \log ^2(x)+4 x \log ^2(x)} \, dx=\int \frac {\left (2 \,{\mathrm e}^{-\mathrm {log}\left (x \right )+x^{2}-3}+8\right ) \mathrm {log}\left (\mathrm {log}\left (5\right ) {\mathrm e}^{-\mathrm {log}\left (x \right )+x^{2}-3}+4 \,\mathrm {log}\left (5\right )\right )^{2}+\left (\left (\left (-8 x^{2}-4 x +4\right ) \mathrm {log}\left (x \right )+4 x \right ) {\mathrm e}^{-\mathrm {log}\left (x \right )+x^{2}-3}-16 \,\mathrm {log}\left (x \right ) x +16 x \right ) \mathrm {log}\left (\mathrm {log}\left (5\right ) {\mathrm e}^{-\mathrm {log}\left (x \right )+x^{2}-3}+4 \,\mathrm {log}\left (5\right )\right )+\left (\left (-8 x^{3}-4 x^{2}+6 x \right ) \mathrm {log}\left (x \right )+2 x^{2}-2 x \right ) {\mathrm e}^{-\mathrm {log}\left (x \right )+x^{2}-3}+\left (-16 x^{2}+8 x \right ) \mathrm {log}\left (x \right )+8 x^{2}-8 x}{x \mathrm {log}\left (x \right )^{2} {\mathrm e}^{-\mathrm {log}\left (x \right )+x^{2}-3}+4 \mathrm {log}\left (x \right )^{2} x}d x \] Input:
int(((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(-l og(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)
Output:
int(((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(-l og(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)