Integrand size = 148, antiderivative size = 30 \[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx=e^{-4+x-\left (x+x \left (-\frac {e^x}{x}+\log (2)\right ) \log (5)\right )^2} (-1+x) \] Output:
(-1+x)*exp(x-4-((ln(2)-exp(x)/x)*x*ln(5)+x)^2)
\[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx=\int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx \] Input:
Integrate[E^(-4 + x - x^2 - 2*x^2*Log[2]*Log[5] - E^(2*x)*Log[5]^2 - x^2*L og[2]^2*Log[5]^2 + E^x*(2*x*Log[5] + 2*x*Log[2]*Log[5]^2))*(3*x - 2*x^2 + (4*x - 4*x^2)*Log[2]*Log[5] + E^(2*x)*(2 - 2*x)*Log[5]^2 + (2*x - 2*x^2)*L og[2]^2*Log[5]^2 + E^x*((-2 + 2*x^2)*Log[5] + (-2 + 2*x^2)*Log[2]*Log[5]^2 )),x]
Output:
Integrate[E^(-4 + x - x^2 - 2*x^2*Log[2]*Log[5] - E^(2*x)*Log[5]^2 - x^2*L og[2]^2*Log[5]^2 + E^x*(2*x*Log[5] + 2*x*Log[2]*Log[5]^2))*(3*x - 2*x^2 + (4*x - 4*x^2)*Log[2]*Log[5] + E^(2*x)*(2 - 2*x)*Log[5]^2 + (2*x - 2*x^2)*L og[2]^2*Log[5]^2 + E^x*((-2 + 2*x^2)*Log[5] + (-2 + 2*x^2)*Log[2]*Log[5]^2 )), 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.
| \(\displaystyle \int \left (-2 x^2+e^x \left (\left (2 x^2-2\right ) \log (2) \log ^2(5)+\left (2 x^2-2\right ) \log (5)\right )+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+\left (4 x-4 x^2\right ) \log (2) \log (5)+3 x+e^{2 x} (2-2 x) \log ^2(5)\right ) \exp \left (-x^2+x^2 \left (-\log ^2(2)\right ) \log ^2(5)-2 x^2 \log (2) \log (5)+x+e^x \left (2 x \log (2) \log ^2(5)+2 x \log (5)\right )-e^{2 x} \log ^2(5)-4\right ) \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \left (-2 x^2+e^x \left (\left (2 x^2-2\right ) \log (2) \log ^2(5)+\left (2 x^2-2\right ) \log (5)\right )+\left (4 x-4 x^2\right ) \log (2) \log (5) \left (1+\frac {1}{2} \log (2) \log (5)\right )+3 x+e^{2 x} (2-2 x) \log ^2(5)\right ) \exp \left (-\left (x^2 (1+\log (2) \log (5) (2+\log (2) \log (5)))\right )+x+e^x \left (2 x \log (2) \log ^2(5)+2 x \log (5)\right )-e^{2 x} \log ^2(5)-4\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-2 x^2 \exp \left (-\left (x^2 (1+\log (2) \log (5) (2+\log (2) \log (5)))\right )+x+e^x \left (2 x \log (2) \log ^2(5)+2 x \log (5)\right )-e^{2 x} \log ^2(5)-4\right )+3 x \exp \left (-\left (x^2 (1+\log (2) \log (5) (2+\log (2) \log (5)))\right )+x+e^x \left (2 x \log (2) \log ^2(5)+2 x \log (5)\right )-e^{2 x} \log ^2(5)-4\right )-2 (x-1) x \log (2) \log (5) (2+\log (2) \log (5)) \exp \left (-\left (x^2 (1+\log (2) \log (5) (2+\log (2) \log (5)))\right )+x+e^x \left (2 x \log (2) \log ^2(5)+2 x \log (5)\right )-e^{2 x} \log ^2(5)-4\right )+2 \left (x^2-1\right ) \log (5) (1+\log (2) \log (5)) \exp \left (-\left (x^2 (1+\log (2) \log (5) (2+\log (2) \log (5)))\right )+2 x+e^x \left (2 x \log (2) \log ^2(5)+2 x \log (5)\right )-e^{2 x} \log ^2(5)-4\right )-2 (x-1) \log ^2(5) \exp \left (-\left (x^2 (1+\log (2) \log (5) (2+\log (2) \log (5)))\right )+3 x+e^x \left (2 x \log (2) \log ^2(5)+2 x \log (5)\right )-e^{2 x} \log ^2(5)-4\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \log (5) (1+\log (2) \log (5)) \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+2 x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right )dx+2 \log ^2(5) \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+3 x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right )dx+2 \log (2) \log (5) (2+\log (2) \log (5)) \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right ) xdx+3 \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right ) xdx-2 \log ^2(5) \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+3 x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right ) xdx-2 \log (2) \log (5) (2+\log (2) \log (5)) \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right ) x^2dx-2 \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right ) x^2dx+2 \log (5) (1+\log (2) \log (5)) \int \exp \left (-\left ((1+\log (2) \log (5) (2+\log (2) \log (5))) x^2\right )+2 x+e^x \left (2 \log (2) \log ^2(5) x+2 \log (5) x\right )-e^{2 x} \log ^2(5)-4\right ) x^2dx\) |
Input:
Int[E^(-4 + x - x^2 - 2*x^2*Log[2]*Log[5] - E^(2*x)*Log[5]^2 - x^2*Log[2]^ 2*Log[5]^2 + E^x*(2*x*Log[5] + 2*x*Log[2]*Log[5]^2))*(3*x - 2*x^2 + (4*x - 4*x^2)*Log[2]*Log[5] + E^(2*x)*(2 - 2*x)*Log[5]^2 + (2*x - 2*x^2)*Log[2]^ 2*Log[5]^2 + E^x*((-2 + 2*x^2)*Log[5] + (-2 + 2*x^2)*Log[2]*Log[5]^2)),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(60\) vs. \(2(28)=56\).
Time = 0.59 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03
| method | result | size |
| risch | \(\left (-1+x \right ) \left (\frac {1}{4}\right )^{x^{2} \ln \left (5\right )} 4^{\ln \left (5\right )^{2} {\mathrm e}^{x} x} 25^{{\mathrm e}^{x} x} {\mathrm e}^{-x^{2} \ln \left (2\right )^{2} \ln \left (5\right )^{2}-4-\ln \left (5\right )^{2} {\mathrm e}^{2 x}-x^{2}+x}\) | \(61\) |
| norman | \(x \,{\mathrm e}^{-\ln \left (5\right )^{2} {\mathrm e}^{2 x}+\left (2 x \ln \left (2\right ) \ln \left (5\right )^{2}+2 x \ln \left (5\right )\right ) {\mathrm e}^{x}-x^{2} \ln \left (2\right )^{2} \ln \left (5\right )^{2}-2 x^{2} \ln \left (2\right ) \ln \left (5\right )-x^{2}+x -4}-{\mathrm e}^{-\ln \left (5\right )^{2} {\mathrm e}^{2 x}+\left (2 x \ln \left (2\right ) \ln \left (5\right )^{2}+2 x \ln \left (5\right )\right ) {\mathrm e}^{x}-x^{2} \ln \left (2\right )^{2} \ln \left (5\right )^{2}-2 x^{2} \ln \left (2\right ) \ln \left (5\right )-x^{2}+x -4}\) | \(124\) |
| parallelrisch | \(x \,{\mathrm e}^{-\ln \left (5\right )^{2} {\mathrm e}^{2 x}+\left (2 x \ln \left (2\right ) \ln \left (5\right )^{2}+2 x \ln \left (5\right )\right ) {\mathrm e}^{x}-x^{2} \ln \left (2\right )^{2} \ln \left (5\right )^{2}-2 x^{2} \ln \left (2\right ) \ln \left (5\right )-x^{2}+x -4}-{\mathrm e}^{-\ln \left (5\right )^{2} {\mathrm e}^{2 x}+\left (2 x \ln \left (2\right ) \ln \left (5\right )^{2}+2 x \ln \left (5\right )\right ) {\mathrm e}^{x}-x^{2} \ln \left (2\right )^{2} \ln \left (5\right )^{2}-2 x^{2} \ln \left (2\right ) \ln \left (5\right )-x^{2}+x -4}\) | \(124\) |
Input:
int(((2-2*x)*ln(5)^2*exp(x)^2+((2*x^2-2)*ln(2)*ln(5)^2+(2*x^2-2)*ln(5))*ex p(x)+(-2*x^2+2*x)*ln(2)^2*ln(5)^2+(-4*x^2+4*x)*ln(2)*ln(5)-2*x^2+3*x)*exp( -ln(5)^2*exp(x)^2+(2*x*ln(2)*ln(5)^2+2*x*ln(5))*exp(x)-x^2*ln(2)^2*ln(5)^2 -2*x^2*ln(2)*ln(5)-x^2+x-4),x,method=_RETURNVERBOSE)
Output:
(-1+x)*(1/4)^(x^2*ln(5))*4^(ln(5)^2*exp(x)*x)*25^(exp(x)*x)*exp(-x^2*ln(2) ^2*ln(5)^2-4-ln(5)^2*exp(2*x)-x^2+x)
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.07 \[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx={\left (x - 1\right )} e^{\left (-x^{2} \log \left (5\right )^{2} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (5\right ) \log \left (2\right ) - e^{\left (2 \, x\right )} \log \left (5\right )^{2} - x^{2} + 2 \, {\left (x \log \left (5\right )^{2} \log \left (2\right ) + x \log \left (5\right )\right )} e^{x} + x - 4\right )} \] Input:
integrate(((2-2*x)*log(5)^2*exp(x)^2+((2*x^2-2)*log(2)*log(5)^2+(2*x^2-2)* log(5))*exp(x)+(-2*x^2+2*x)*log(2)^2*log(5)^2+(-4*x^2+4*x)*log(2)*log(5)-2 *x^2+3*x)*exp(-exp(x)^2*log(5)^2+(2*x*log(2)*log(5)^2+2*x*log(5))*exp(x)-x ^2*log(2)^2*log(5)^2-2*x^2*log(2)*log(5)-x^2+x-4),x, algorithm="fricas")
Output:
(x - 1)*e^(-x^2*log(5)^2*log(2)^2 - 2*x^2*log(5)*log(2) - e^(2*x)*log(5)^2 - x^2 + 2*(x*log(5)^2*log(2) + x*log(5))*e^x + x - 4)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (24) = 48\).
Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx=\left (x - 1\right ) e^{- 2 x^{2} \log {\left (2 \right )} \log {\left (5 \right )} - x^{2} \log {\left (2 \right )}^{2} \log {\left (5 \right )}^{2} - x^{2} + x + \left (2 x \log {\left (5 \right )} + 2 x \log {\left (2 \right )} \log {\left (5 \right )}^{2}\right ) e^{x} - e^{2 x} \log {\left (5 \right )}^{2} - 4} \] Input:
integrate(((2-2*x)*ln(5)**2*exp(x)**2+((2*x**2-2)*ln(2)*ln(5)**2+(2*x**2-2 )*ln(5))*exp(x)+(-2*x**2+2*x)*ln(2)**2*ln(5)**2+(-4*x**2+4*x)*ln(2)*ln(5)- 2*x**2+3*x)*exp(-exp(x)**2*ln(5)**2+(2*x*ln(2)*ln(5)**2+2*x*ln(5))*exp(x)- x**2*ln(2)**2*ln(5)**2-2*x**2*ln(2)*ln(5)-x**2+x-4),x)
Output:
(x - 1)*exp(-2*x**2*log(2)*log(5) - x**2*log(2)**2*log(5)**2 - x**2 + x + (2*x*log(5) + 2*x*log(2)*log(5)**2)*exp(x) - exp(2*x)*log(5)**2 - 4)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (31) = 62\).
Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx={\left (x - 1\right )} e^{\left (-x^{2} \log \left (5\right )^{2} \log \left (2\right )^{2} + 2 \, x e^{x} \log \left (5\right )^{2} \log \left (2\right ) - 2 \, x^{2} \log \left (5\right ) \log \left (2\right ) + 2 \, x e^{x} \log \left (5\right ) - e^{\left (2 \, x\right )} \log \left (5\right )^{2} - x^{2} + x - 4\right )} \] Input:
integrate(((2-2*x)*log(5)^2*exp(x)^2+((2*x^2-2)*log(2)*log(5)^2+(2*x^2-2)* log(5))*exp(x)+(-2*x^2+2*x)*log(2)^2*log(5)^2+(-4*x^2+4*x)*log(2)*log(5)-2 *x^2+3*x)*exp(-exp(x)^2*log(5)^2+(2*x*log(2)*log(5)^2+2*x*log(5))*exp(x)-x ^2*log(2)^2*log(5)^2-2*x^2*log(2)*log(5)-x^2+x-4),x, algorithm="maxima")
Output:
(x - 1)*e^(-x^2*log(5)^2*log(2)^2 + 2*x*e^x*log(5)^2*log(2) - 2*x^2*log(5) *log(2) + 2*x*e^x*log(5) - e^(2*x)*log(5)^2 - x^2 + x - 4)
Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (31) = 62\).
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.13 \[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx={\left (x e^{\left (-x^{2} \log \left (5\right )^{2} \log \left (2\right )^{2} + 2 \, x e^{x} \log \left (5\right )^{2} \log \left (2\right ) - 2 \, x^{2} \log \left (5\right ) \log \left (2\right ) + 2 \, x e^{x} \log \left (5\right ) - e^{\left (2 \, x\right )} \log \left (5\right )^{2} - x^{2} + x\right )} - e^{\left (-x^{2} \log \left (5\right )^{2} \log \left (2\right )^{2} + 2 \, x e^{x} \log \left (5\right )^{2} \log \left (2\right ) - 2 \, x^{2} \log \left (5\right ) \log \left (2\right ) + 2 \, x e^{x} \log \left (5\right ) - e^{\left (2 \, x\right )} \log \left (5\right )^{2} - x^{2} + x\right )}\right )} e^{\left (-4\right )} \] Input:
integrate(((2-2*x)*log(5)^2*exp(x)^2+((2*x^2-2)*log(2)*log(5)^2+(2*x^2-2)* log(5))*exp(x)+(-2*x^2+2*x)*log(2)^2*log(5)^2+(-4*x^2+4*x)*log(2)*log(5)-2 *x^2+3*x)*exp(-exp(x)^2*log(5)^2+(2*x*log(2)*log(5)^2+2*x*log(5))*exp(x)-x ^2*log(2)^2*log(5)^2-2*x^2*log(2)*log(5)-x^2+x-4),x, algorithm="giac")
Output:
(x*e^(-x^2*log(5)^2*log(2)^2 + 2*x*e^x*log(5)^2*log(2) - 2*x^2*log(5)*log( 2) + 2*x*e^x*log(5) - e^(2*x)*log(5)^2 - x^2 + x) - e^(-x^2*log(5)^2*log(2 )^2 + 2*x*e^x*log(5)^2*log(2) - 2*x^2*log(5)*log(2) + 2*x*e^x*log(5) - e^( 2*x)*log(5)^2 - x^2 + x))*e^(-4)
Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx=\frac {2^{2\,x\,{\mathrm {e}}^x\,{\ln \left (5\right )}^2}\,5^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x-{\mathrm {e}}^{2\,x}\,{\ln \left (5\right )}^2-x^2-x^2\,{\ln \left (2\right )}^2\,{\ln \left (5\right )}^2-4}\,\left (x-1\right )}{2^{2\,x^2\,\ln \left (5\right )}} \] Input:
int(exp(x - exp(2*x)*log(5)^2 - x^2 + exp(x)*(2*x*log(5) + 2*x*log(2)*log( 5)^2) - 2*x^2*log(2)*log(5) - x^2*log(2)^2*log(5)^2 - 4)*(3*x + exp(x)*(lo g(5)*(2*x^2 - 2) + log(2)*log(5)^2*(2*x^2 - 2)) - 2*x^2 + log(2)^2*log(5)^ 2*(2*x - 2*x^2) - exp(2*x)*log(5)^2*(2*x - 2) + log(2)*log(5)*(4*x - 4*x^2 )),x)
Output:
(2^(2*x*exp(x)*log(5)^2)*5^(2*x*exp(x))*exp(x - exp(2*x)*log(5)^2 - x^2 - x^2*log(2)^2*log(5)^2 - 4)*(x - 1))/2^(2*x^2*log(5))
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40 \[ \int e^{-4+x-x^2-2 x^2 \log (2) \log (5)-e^{2 x} \log ^2(5)-x^2 \log ^2(2) \log ^2(5)+e^x \left (2 x \log (5)+2 x \log (2) \log ^2(5)\right )} \left (3 x-2 x^2+\left (4 x-4 x^2\right ) \log (2) \log (5)+e^{2 x} (2-2 x) \log ^2(5)+\left (2 x-2 x^2\right ) \log ^2(2) \log ^2(5)+e^x \left (\left (-2+2 x^2\right ) \log (5)+\left (-2+2 x^2\right ) \log (2) \log ^2(5)\right )\right ) \, dx=\frac {e^{2 e^{x} \mathrm {log}\left (5\right )^{2} \mathrm {log}\left (2\right ) x +2 e^{x} \mathrm {log}\left (5\right ) x +x} \left (x -1\right )}{e^{e^{2 x} \mathrm {log}\left (5\right )^{2}+\mathrm {log}\left (5\right )^{2} \mathrm {log}\left (2\right )^{2} x^{2}+x^{2}} 5^{2 \,\mathrm {log}\left (2\right ) x^{2}} e^{4}} \] Input:
int(((2-2*x)*log(5)^2*exp(x)^2+((2*x^2-2)*log(2)*log(5)^2+(2*x^2-2)*log(5) )*exp(x)+(-2*x^2+2*x)*log(2)^2*log(5)^2+(-4*x^2+4*x)*log(2)*log(5)-2*x^2+3 *x)*exp(-exp(x)^2*log(5)^2+(2*x*log(2)*log(5)^2+2*x*log(5))*exp(x)-x^2*log (2)^2*log(5)^2-2*x^2*log(2)*log(5)-x^2+x-4),x)
Output:
(e**(2*e**x*log(5)**2*log(2)*x + 2*e**x*log(5)*x + x)*(x - 1))/(e**(e**(2* x)*log(5)**2 + log(5)**2*log(2)**2*x**2 + x**2)*5**(2*log(2)*x**2)*e**4)