Integrand size = 137, antiderivative size = 28 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=\frac {1}{2} x \left (x+\log \left (\frac {1}{16} \left (-5+\left (e^2-x\right )^2\right )^2 x^2\right )\right ) \] Output:
1/2*(ln(1/16*x^2*((exp(2)-x)^2-5)^2)+x)*x
Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=\frac {1}{2} x \left (x+\log \left (\frac {1}{16} x^2 \left (-5+e^4-2 e^2 x+x^2\right )^2\right )\right ) \] Input:
Integrate[(-10 - 10*x + 6*x^2 + 2*x^3 + E^4*(2 + 2*x) + E^2*(-8*x - 4*x^2) + (-5 + E^4 - 2*E^2*x + x^2)*Log[(25*x^2 + E^8*x^2 - 4*E^6*x^3 - 10*x^4 + x^6 + E^4*(-10*x^2 + 6*x^4) + E^2*(20*x^3 - 4*x^5))/16])/(-10 + 2*E^4 - 4 *E^2*x + 2*x^2),x]
Output:
(x*(x + Log[(x^2*(-5 + E^4 - 2*E^2*x + x^2)^2)/16]))/2
Time = 0.58 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {7239, 2009}
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.
| \(\displaystyle \int \frac {2 x^3+6 x^2+e^2 \left (-4 x^2-8 x\right )+\left (x^2-2 e^2 x+e^4-5\right ) \log \left (\frac {1}{16} \left (x^6-10 x^4-4 e^6 x^3+e^8 x^2+25 x^2+e^2 \left (20 x^3-4 x^5\right )+e^4 \left (6 x^4-10 x^2\right )\right )\right )-10 x+e^4 (2 x+2)-10}{2 x^2-4 e^2 x+2 e^4-10} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (\frac {1}{2} \log \left (\frac {1}{16} x^2 \left (x^2-2 e^2 x+e^4-5\right )^2\right )+\frac {x^3+\left (3-2 e^2\right ) x^2-\left (5+4 e^2-e^4\right ) x+e^4-5}{x^2-2 e^2 x+e^4-5}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (-x^2+2 e^2 x-e^4+5\right )^2\right )\) |
Input:
Int[(-10 - 10*x + 6*x^2 + 2*x^3 + E^4*(2 + 2*x) + E^2*(-8*x - 4*x^2) + (-5 + E^4 - 2*E^2*x + x^2)*Log[(25*x^2 + E^8*x^2 - 4*E^6*x^3 - 10*x^4 + x^6 + E^4*(-10*x^2 + 6*x^4) + E^2*(20*x^3 - 4*x^5))/16])/(-10 + 2*E^4 - 4*E^2*x + 2*x^2),x]
Output:
x^2/2 + (x*Log[(x^2*(5 - E^4 + 2*E^2*x - x^2)^2)/16])/2
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(70\) vs. \(2(23)=46\).
Time = 6.87 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.54
| method | result | size |
| risch | \(\frac {x^{2}}{2}+\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{8}}{16}-\frac {x^{3} {\mathrm e}^{6}}{4}+\frac {\left (6 x^{4}-10 x^{2}\right ) {\mathrm e}^{4}}{16}+\frac {\left (-4 x^{5}+20 x^{3}\right ) {\mathrm e}^{2}}{16}+\frac {x^{6}}{16}-\frac {5 x^{4}}{8}+\frac {25 x^{2}}{16}\right )}{2}\) | \(71\) |
| norman | \(\frac {x^{2}}{2}+\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{8}}{16}-\frac {x^{3} {\mathrm e}^{6}}{4}+\frac {\left (6 x^{4}-10 x^{2}\right ) {\mathrm e}^{4}}{16}+\frac {\left (-4 x^{5}+20 x^{3}\right ) {\mathrm e}^{2}}{16}+\frac {x^{6}}{16}-\frac {5 x^{4}}{8}+\frac {25 x^{2}}{16}\right )}{2}\) | \(77\) |
| default | \(\frac {x^{2}}{2}-2 x \ln \left (2\right )+\frac {x \ln \left (x^{2} {\mathrm e}^{8}-4 x^{3} {\mathrm e}^{6}+6 x^{4} {\mathrm e}^{4}-10 x^{2} {\mathrm e}^{4}-4 \,{\mathrm e}^{2} x^{5}+20 x^{3} {\mathrm e}^{2}+x^{6}-10 x^{4}+25 x^{2}\right )}{2}\) | \(79\) |
| parts | \(\frac {x^{2}}{2}-2 x \ln \left (2\right )+\frac {x \ln \left (x^{2} {\mathrm e}^{8}-4 x^{3} {\mathrm e}^{6}+6 x^{4} {\mathrm e}^{4}-10 x^{2} {\mathrm e}^{4}-4 \,{\mathrm e}^{2} x^{5}+20 x^{3} {\mathrm e}^{2}+x^{6}-10 x^{4}+25 x^{2}\right )}{2}\) | \(79\) |
| parallelrisch | \(-3 \,{\mathrm e}^{4}+5+\frac {x^{2}}{2}+\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{8}}{16}-\frac {x^{3} {\mathrm e}^{6}}{4}+\frac {\left (6 x^{4}-10 x^{2}\right ) {\mathrm e}^{4}}{16}+\frac {\left (-4 x^{5}+20 x^{3}\right ) {\mathrm e}^{2}}{16}+\frac {x^{6}}{16}-\frac {5 x^{4}}{8}+\frac {25 x^{2}}{16}\right )}{2}\) | \(84\) |
Input:
int(((exp(2)^2-2*exp(2)*x+x^2-5)*ln(1/16*x^2*exp(2)^4-1/4*x^3*exp(2)^3+1/1 6*(6*x^4-10*x^2)*exp(2)^2+1/16*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8*x^4+25/ 16*x^2)+(2+2*x)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)/(2*exp(2 )^2-4*exp(2)*x+2*x^2-10),x,method=_RETURNVERBOSE)
Output:
1/2*x^2+1/2*x*ln(1/16*x^2*exp(8)-1/4*x^3*exp(6)+1/16*(6*x^4-10*x^2)*exp(4) +1/16*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8*x^4+25/16*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.43 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{2} \, x \log \left (\frac {1}{16} \, x^{6} - \frac {5}{8} \, x^{4} - \frac {1}{4} \, x^{3} e^{6} + \frac {1}{16} \, x^{2} e^{8} + \frac {25}{16} \, x^{2} + \frac {1}{8} \, {\left (3 \, x^{4} - 5 \, x^{2}\right )} e^{4} - \frac {1}{4} \, {\left (x^{5} - 5 \, x^{3}\right )} e^{2}\right ) \] Input:
integrate(((exp(2)^2-2*exp(2)*x+x^2-5)*log(1/16*x^2*exp(2)^4-1/4*x^3*exp(2 )^3+1/16*(6*x^4-10*x^2)*exp(2)^2+1/16*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8* x^4+25/16*x^2)+(2+2*x)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)/( 2*exp(2)^2-4*exp(2)*x+2*x^2-10),x, algorithm="fricas")
Output:
1/2*x^2 + 1/2*x*log(1/16*x^6 - 5/8*x^4 - 1/4*x^3*e^6 + 1/16*x^2*e^8 + 25/1 6*x^2 + 1/8*(3*x^4 - 5*x^2)*e^4 - 1/4*(x^5 - 5*x^3)*e^2)
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (22) = 44\).
Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.71 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=\frac {x^{2}}{2} + \frac {x \log {\left (\frac {x^{6}}{16} - \frac {5 x^{4}}{8} - \frac {x^{3} e^{6}}{4} + \frac {25 x^{2}}{16} + \frac {x^{2} e^{8}}{16} + \left (\frac {3 x^{4}}{8} - \frac {5 x^{2}}{8}\right ) e^{4} + \left (- \frac {x^{5}}{4} + \frac {5 x^{3}}{4}\right ) e^{2} \right )}}{2} \] Input:
integrate(((exp(2)**2-2*exp(2)*x+x**2-5)*ln(1/16*x**2*exp(2)**4-1/4*x**3*e xp(2)**3+1/16*(6*x**4-10*x**2)*exp(2)**2+1/16*(-4*x**5+20*x**3)*exp(2)+1/1 6*x**6-5/8*x**4+25/16*x**2)+(2+2*x)*exp(2)**2+(-4*x**2-8*x)*exp(2)+2*x**3+ 6*x**2-10*x-10)/(2*exp(2)**2-4*exp(2)*x+2*x**2-10),x)
Output:
x**2/2 + x*log(x**6/16 - 5*x**4/8 - x**3*exp(6)/4 + 25*x**2/16 + x**2*exp( 8)/16 + (3*x**4/8 - 5*x**2/8)*exp(4) + (-x**5/4 + 5*x**3/4)*exp(2))/2
Leaf count of result is larger than twice the leaf count of optimal. 417 vs. \(2 (23) = 46\).
Time = 0.20 (sec) , antiderivative size = 417, normalized size of antiderivative = 14.89 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=\frac {1}{10} \, \sqrt {5} {\left (e^{6} + 15 \, e^{2}\right )} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + \frac {3}{10} \, \sqrt {5} {\left (e^{4} + 5\right )} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + \frac {1}{10} \, \sqrt {5} e^{4} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) - \frac {1}{2} \, \sqrt {5} e^{2} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + \frac {1}{2} \, x^{2} - x {\left (2 \, \log \left (2\right ) + 3\right )} + \frac {1}{10} \, {\left (\sqrt {5} e^{2} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 5 \, \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right )\right )} e^{4} - \frac {1}{5} \, {\left (\sqrt {5} {\left (e^{4} + 5\right )} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 10 \, e^{2} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + 10 \, x\right )} e^{2} - \frac {2}{5} \, {\left (\sqrt {5} e^{2} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 5 \, \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right )\right )} e^{2} + 2 \, x e^{2} + {\left (x - e^{2}\right )} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + \frac {1}{2} \, {\left (3 \, e^{4} + 5\right )} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + 3 \, e^{2} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + x \log \left (x\right ) - \frac {3}{2} \, \sqrt {5} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 3 \, x - \frac {5}{2} \, \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) \] Input:
integrate(((exp(2)^2-2*exp(2)*x+x^2-5)*log(1/16*x^2*exp(2)^4-1/4*x^3*exp(2 )^3+1/16*(6*x^4-10*x^2)*exp(2)^2+1/16*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8* x^4+25/16*x^2)+(2+2*x)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)/( 2*exp(2)^2-4*exp(2)*x+2*x^2-10),x, algorithm="maxima")
Output:
1/10*sqrt(5)*(e^6 + 15*e^2)*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 3/10*sqrt(5)*(e^4 + 5)*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 1/1 0*sqrt(5)*e^4*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) - 1/2*sqrt(5)*e ^2*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 1/2*x^2 - x*(2*log(2) + 3) + 1/10*(sqrt(5)*e^2*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 5*lo g(x^2 - 2*x*e^2 + e^4 - 5))*e^4 - 1/5*(sqrt(5)*(e^4 + 5)*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 10*e^2*log(x^2 - 2*x*e^2 + e^4 - 5) + 10*x)* e^2 - 2/5*(sqrt(5)*e^2*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 5*lo g(x^2 - 2*x*e^2 + e^4 - 5))*e^2 + 2*x*e^2 + (x - e^2)*log(x^2 - 2*x*e^2 + e^4 - 5) + 1/2*(3*e^4 + 5)*log(x^2 - 2*x*e^2 + e^4 - 5) + 3*e^2*log(x^2 - 2*x*e^2 + e^4 - 5) + x*log(x) - 3/2*sqrt(5)*log((x - sqrt(5) - e^2)/(x + s qrt(5) - e^2)) + 3*x - 5/2*log(x^2 - 2*x*e^2 + e^4 - 5)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.43 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{2} \, x \log \left (\frac {1}{16} \, x^{6} - \frac {1}{4} \, x^{5} e^{2} + \frac {3}{8} \, x^{4} e^{4} - \frac {5}{8} \, x^{4} - \frac {1}{4} \, x^{3} e^{6} + \frac {5}{4} \, x^{3} e^{2} + \frac {1}{16} \, x^{2} e^{8} - \frac {5}{8} \, x^{2} e^{4} + \frac {25}{16} \, x^{2}\right ) \] Input:
integrate(((exp(2)^2-2*exp(2)*x+x^2-5)*log(1/16*x^2*exp(2)^4-1/4*x^3*exp(2 )^3+1/16*(6*x^4-10*x^2)*exp(2)^2+1/16*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8* x^4+25/16*x^2)+(2+2*x)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)/( 2*exp(2)^2-4*exp(2)*x+2*x^2-10),x, algorithm="giac")
Output:
1/2*x^2 + 1/2*x*log(1/16*x^6 - 1/4*x^5*e^2 + 3/8*x^4*e^4 - 5/8*x^4 - 1/4*x ^3*e^6 + 5/4*x^3*e^2 + 1/16*x^2*e^8 - 5/8*x^2*e^4 + 25/16*x^2)
Time = 3.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=\frac {x\,\left (x+\ln \left (\frac {{\mathrm {e}}^2\,\left (20\,x^3-4\,x^5\right )}{16}-\frac {{\mathrm {e}}^4\,\left (10\,x^2-6\,x^4\right )}{16}-\frac {x^3\,{\mathrm {e}}^6}{4}+\frac {x^2\,{\mathrm {e}}^8}{16}+\frac {25\,x^2}{16}-\frac {5\,x^4}{8}+\frac {x^6}{16}\right )\right )}{2} \] Input:
int((log((exp(2)*(20*x^3 - 4*x^5))/16 - (exp(4)*(10*x^2 - 6*x^4))/16 - (x^ 3*exp(6))/4 + (x^2*exp(8))/16 + (25*x^2)/16 - (5*x^4)/8 + x^6/16)*(exp(4) - 2*x*exp(2) + x^2 - 5) - 10*x - exp(2)*(8*x + 4*x^2) + 6*x^2 + 2*x^3 + ex p(4)*(2*x + 2) - 10)/(2*exp(4) - 4*x*exp(2) + 2*x^2 - 10),x)
Output:
(x*(x + log((exp(2)*(20*x^3 - 4*x^5))/16 - (exp(4)*(10*x^2 - 6*x^4))/16 - (x^3*exp(6))/4 + (x^2*exp(8))/16 + (25*x^2)/16 - (5*x^4)/8 + x^6/16)))/2
Time = 0.22 (sec) , antiderivative size = 285, normalized size of antiderivative = 10.18 \[ \int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 \left (-8 x-4 x^2\right )+\left (-5+e^4-2 e^2 x+x^2\right ) \log \left (\frac {1}{16} \left (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 \left (-10 x^2+6 x^4\right )+e^2 \left (20 x^3-4 x^5\right )\right )\right )}{-10+2 e^4-4 e^2 x+2 x^2} \, dx=-\sqrt {5}\, \mathrm {log}\left (-\sqrt {5}-e^{2}+x \right )-\sqrt {5}\, \mathrm {log}\left (\sqrt {5}-e^{2}+x \right )+\frac {\sqrt {5}\, \mathrm {log}\left (\frac {1}{16} e^{8} x^{2}-\frac {1}{4} e^{6} x^{3}+\frac {3}{8} e^{4} x^{4}-\frac {1}{4} e^{2} x^{5}-\frac {5}{8} e^{4} x^{2}+\frac {1}{16} x^{6}+\frac {5}{4} e^{2} x^{3}-\frac {5}{8} x^{4}+\frac {25}{16} x^{2}\right )}{2}-\sqrt {5}\, \mathrm {log}\left (x \right )+\mathrm {log}\left (-\sqrt {5}-e^{2}+x \right ) e^{2}+\mathrm {log}\left (\sqrt {5}-e^{2}+x \right ) e^{2}-\frac {\mathrm {log}\left (\frac {1}{16} e^{8} x^{2}-\frac {1}{4} e^{6} x^{3}+\frac {3}{8} e^{4} x^{4}-\frac {1}{4} e^{2} x^{5}-\frac {5}{8} e^{4} x^{2}+\frac {1}{16} x^{6}+\frac {5}{4} e^{2} x^{3}-\frac {5}{8} x^{4}+\frac {25}{16} x^{2}\right ) e^{2}}{2}+\frac {\mathrm {log}\left (\frac {1}{16} e^{8} x^{2}-\frac {1}{4} e^{6} x^{3}+\frac {3}{8} e^{4} x^{4}-\frac {1}{4} e^{2} x^{5}-\frac {5}{8} e^{4} x^{2}+\frac {1}{16} x^{6}+\frac {5}{4} e^{2} x^{3}-\frac {5}{8} x^{4}+\frac {25}{16} x^{2}\right ) x}{2}+\mathrm {log}\left (x \right ) e^{2}+\frac {x^{2}}{2} \] Input:
int(((exp(2)^2-2*exp(2)*x+x^2-5)*log(1/16*x^2*exp(2)^4-1/4*x^3*exp(2)^3+1/ 16*(6*x^4-10*x^2)*exp(2)^2+1/16*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8*x^4+25 /16*x^2)+(2+2*x)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)/(2*exp( 2)^2-4*exp(2)*x+2*x^2-10),x)
Output:
( - 2*sqrt(5)*log( - sqrt(5) - e**2 + x) - 2*sqrt(5)*log(sqrt(5) - e**2 + x) + sqrt(5)*log((e**8*x**2 - 4*e**6*x**3 + 6*e**4*x**4 - 10*e**4*x**2 - 4 *e**2*x**5 + 20*e**2*x**3 + x**6 - 10*x**4 + 25*x**2)/16) - 2*sqrt(5)*log( x) + 2*log( - sqrt(5) - e**2 + x)*e**2 + 2*log(sqrt(5) - e**2 + x)*e**2 - log((e**8*x**2 - 4*e**6*x**3 + 6*e**4*x**4 - 10*e**4*x**2 - 4*e**2*x**5 + 20*e**2*x**3 + x**6 - 10*x**4 + 25*x**2)/16)*e**2 + log((e**8*x**2 - 4*e** 6*x**3 + 6*e**4*x**4 - 10*e**4*x**2 - 4*e**2*x**5 + 20*e**2*x**3 + x**6 - 10*x**4 + 25*x**2)/16)*x + 2*log(x)*e**2 + x**2)/2