Integrand size = 177, antiderivative size = 26 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=\left (e^3+e^{\frac {e^2}{x}}+x+x \left (e^x+x\right )\right )^2 \log (5) \] Output:
(exp(exp(2)/x)+exp(3)+(exp(x)+x)*x+x)^2*ln(5)
Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=\left (e^{\frac {e^2}{x}}+x+e^x x+x^2\right ) \left (2 e^3+e^{\frac {e^2}{x}}+x+e^x x+x^2\right ) \log (5) \] Input:
Integrate[(-2*E^(2 + (2*E^2)/x)*Log[5] + E^(2*x)*(2*x^3 + 2*x^4)*Log[5] + E^x*(4*x^3 + 8*x^4 + 2*x^5 + E^3*(2*x^2 + 2*x^3))*Log[5] + (2*x^3 + 6*x^4 + 4*x^5 + E^3*(2*x^2 + 4*x^3))*Log[5] + E^(E^2/x)*(E^x*(-2*E^2*x + 2*x^2 + 2*x^3)*Log[5] + (-2*E^5 + 2*x^2 + 4*x^3 + E^2*(-2*x - 2*x^2))*Log[5]))/x^ 2,x]
Output:
(E^(E^2/x) + x + E^x*x + x^2)*(2*E^3 + E^(E^2/x) + x + E^x*x + x^2)*Log[5]
Leaf count is larger than twice the leaf count of optimal. \(191\) vs. \(2(26)=52\).
Time = 1.29 (sec) , antiderivative size = 191, normalized size of antiderivative = 7.35, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2010, 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.
| \(\displaystyle \int \frac {e^{2 x} \left (2 x^4+2 x^3\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (2 x^3+2 x^2-2 e^2 x\right ) \log (5)+\left (4 x^3+2 x^2+e^2 \left (-2 x^2-2 x\right )-2 e^5\right ) \log (5)\right )+e^x \left (2 x^5+8 x^4+4 x^3+e^3 \left (2 x^3+2 x^2\right )\right ) \log (5)+\left (4 x^5+6 x^4+2 x^3+e^3 \left (4 x^3+2 x^2\right )\right ) \log (5)-2 e^{\frac {2 e^2}{x}+2} \log (5)}{x^2} \, dx\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \int \left (\frac {2 e^x \left (x^4+4 x^3+e^{\frac {e^2}{x}} x^2+2 \left (1+\frac {e^3}{2}\right ) x^2+e^{\frac {e^2}{x}} x+e^3 x-e^{\frac {e^2}{x}+2}\right ) \log (5)}{x}+\frac {2 \left (2 x^5+3 x^4+2 e^{\frac {e^2}{x}} x^3+\left (1+2 e^3\right ) x^3-\left (1-\frac {1}{e^2}\right ) e^{\frac {e^2}{x}+2} x^2+e^3 x^2-e^{\frac {e^2}{x}+2} x-e^{\frac {e^2}{x}+5}-e^{\frac {2 e^2}{x}+2}\right ) \log (5)}{x^2}+2 e^{2 x} x (x+1) \log (5)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 e^x x^3 \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+\left (x^2+x+e^3\right )^2 \log (5)-\frac {2 e^{x+\frac {e^2}{x}} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+2 e^{\frac {e^2}{x}} x \log (5)-4 e^x x \log (5)+2 \left (2+e^3\right ) e^x x \log (5)+2 e^{\frac {e^2}{x}+3} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+4 e^x \log (5)+2 e^{x+3} \log (5)-2 \left (2+e^3\right ) e^x \log (5)\) |
Input:
Int[(-2*E^(2 + (2*E^2)/x)*Log[5] + E^(2*x)*(2*x^3 + 2*x^4)*Log[5] + E^x*(4 *x^3 + 8*x^4 + 2*x^5 + E^3*(2*x^2 + 2*x^3))*Log[5] + (2*x^3 + 6*x^4 + 4*x^ 5 + E^3*(2*x^2 + 4*x^3))*Log[5] + E^(E^2/x)*(E^x*(-2*E^2*x + 2*x^2 + 2*x^3 )*Log[5] + (-2*E^5 + 2*x^2 + 4*x^3 + E^2*(-2*x - 2*x^2))*Log[5]))/x^2,x]
Output:
2*E^(3 + E^2/x)*Log[5] + E^((2*E^2)/x)*Log[5] + 4*E^x*Log[5] + 2*E^(3 + x) *Log[5] - 2*E^x*(2 + E^3)*Log[5] + 2*E^(E^2/x)*x*Log[5] - 4*E^x*x*Log[5] + 2*E^x*(2 + E^3)*x*Log[5] + 2*E^(E^2/x)*x^2*Log[5] + 2*E^x*x^2*Log[5] + E^ (2*x)*x^2*Log[5] + 2*E^x*x^3*Log[5] - (2*E^(E^2/x + x)*(E^2 - x^2)*Log[5]) /((1 - E^2/x^2)*x) + (E^3 + x + x^2)^2*Log[5]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Leaf count of result is larger than twice the leaf count of optimal. \(115\) vs. \(2(22)=44\).
Time = 92.51 (sec) , antiderivative size = 116, normalized size of antiderivative = 4.46
| method | result | size |
| parts | \(2 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (5\right ) x^{3}+2 x \,{\mathrm e}^{x} \ln \left (5\right ) {\mathrm e}^{3}+2 x \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{x} \ln \left (5\right )+\ln \left (5\right ) \left ({\mathrm e}^{3}+x^{2}+x \right )^{2}+\ln \left (5\right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{2}}{x}}+2 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} x^{2} \ln \left (5\right )+2 \ln \left (5\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{3}+2 x \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} \ln \left (5\right )+\ln \left (5\right ) {\mathrm e}^{2 x} x^{2}\) | \(116\) |
| risch | \(x^{4} \ln \left (5\right )+2 \ln \left (5\right ) x^{2} {\mathrm e}^{3}+2 x^{3} \ln \left (5\right )+{\mathrm e}^{6} \ln \left (5\right )+2 x \,{\mathrm e}^{3} \ln \left (5\right )+x^{2} \ln \left (5\right )+\ln \left (5\right ) {\mathrm e}^{2 x} x^{2}+\left (2 x \,{\mathrm e}^{3} \ln \left (5\right )+2 x^{3} \ln \left (5\right )+2 x^{2} \ln \left (5\right )\right ) {\mathrm e}^{x}+\ln \left (5\right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{2}}{x}}+\left (2 \,{\mathrm e}^{3} \ln \left (5\right )+2 x^{2} \ln \left (5\right )+2 x \,{\mathrm e}^{x} \ln \left (5\right )+2 x \ln \left (5\right )\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{x}}\) | \(124\) |
| parallelrisch | \(x^{4} \ln \left (5\right )+2 \,{\mathrm e}^{x} \ln \left (5\right ) x^{3}+\ln \left (5\right ) {\mathrm e}^{2 x} x^{2}+2 \ln \left (5\right ) x^{2} {\mathrm e}^{3}+2 x \,{\mathrm e}^{x} \ln \left (5\right ) {\mathrm e}^{3}+2 x^{3} \ln \left (5\right )+2 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} x^{2} \ln \left (5\right )+2 x \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{x} \ln \left (5\right )+2 x \,{\mathrm e}^{3} \ln \left (5\right )+2 \ln \left (5\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} {\mathrm e}^{3}+x^{2} \ln \left (5\right )+2 x \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x}} \ln \left (5\right )+\ln \left (5\right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{2}}{x}}\) | \(139\) |
Input:
int((-2*exp(2)*ln(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*ln(5)*exp( x)+(-2*exp(2)*exp(3)+(-2*x^2-2*x)*exp(2)+4*x^3+2*x^2)*ln(5))*exp(exp(2)/x) +(2*x^4+2*x^3)*ln(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^4+4*x^3)*ln( 5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*ln(5))/x^2,x,method=_RE TURNVERBOSE)
Output:
2*x^2*ln(5)*exp(x)+2*exp(x)*ln(5)*x^3+2*x*exp(x)*ln(5)*exp(3)+2*x*exp(exp( 2)/x)*exp(x)*ln(5)+ln(5)*(exp(3)+x^2+x)^2+ln(5)*exp(exp(2)/x)^2+2*exp(exp( 2)/x)*x^2*ln(5)+2*ln(5)*exp(exp(2)/x)*exp(3)+2*x*exp(exp(2)/x)*ln(5)+x^2*l n(5)*exp(x)^2
Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (22) = 44\).
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.42 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=x^{2} e^{\left (2 \, x\right )} \log \left (5\right ) + 2 \, {\left (x^{3} + x^{2} + x e^{3}\right )} e^{x} \log \left (5\right ) + 2 \, {\left (x e^{x} \log \left (5\right ) + {\left (x^{2} + x + e^{3}\right )} \log \left (5\right )\right )} e^{\left (\frac {e^{2}}{x}\right )} + {\left (x^{4} + 2 \, x^{3} + x^{2} + 2 \, {\left (x^{2} + x\right )} e^{3}\right )} \log \left (5\right ) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \left (5\right ) \] Input:
integrate((-2*exp(2)*log(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*log (5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x^2-2*x)*exp(2)+4*x^3+2*x^2)*log(5))*exp( exp(2)/x)+(2*x^4+2*x^3)*log(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^4+ 4*x^3)*log(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*log(5))/x^2, x, algorithm="fricas")
Output:
x^2*e^(2*x)*log(5) + 2*(x^3 + x^2 + x*e^3)*e^x*log(5) + 2*(x*e^x*log(5) + (x^2 + x + e^3)*log(5))*e^(e^2/x) + (x^4 + 2*x^3 + x^2 + 2*(x^2 + x)*e^3)* log(5) + e^(2*e^2/x)*log(5)
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (22) = 44\).
Time = 4.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 5.15 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=x^{4} \log {\left (5 \right )} + 2 x^{3} \log {\left (5 \right )} + x^{2} e^{2 x} \log {\left (5 \right )} + x^{2} \left (\log {\left (5 \right )} + 2 e^{3} \log {\left (5 \right )}\right ) + 2 x e^{3} \log {\left (5 \right )} + \left (2 x^{3} \log {\left (5 \right )} + 2 x^{2} \log {\left (5 \right )} + 2 x e^{3} \log {\left (5 \right )}\right ) e^{x} + \left (2 x^{2} \log {\left (5 \right )} + 2 x e^{x} \log {\left (5 \right )} + 2 x \log {\left (5 \right )} + 2 e^{3} \log {\left (5 \right )}\right ) e^{\frac {e^{2}}{x}} + e^{\frac {2 e^{2}}{x}} \log {\left (5 \right )} \] Input:
integrate((-2*exp(2)*ln(5)*exp(exp(2)/x)**2+((-2*exp(2)*x+2*x**3+2*x**2)*l n(5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x**2-2*x)*exp(2)+4*x**3+2*x**2)*ln(5))*e xp(exp(2)/x)+(2*x**4+2*x**3)*ln(5)*exp(x)**2+((2*x**3+2*x**2)*exp(3)+2*x** 5+8*x**4+4*x**3)*ln(5)*exp(x)+((4*x**3+2*x**2)*exp(3)+4*x**5+6*x**4+2*x**3 )*ln(5))/x**2,x)
Output:
x**4*log(5) + 2*x**3*log(5) + x**2*exp(2*x)*log(5) + x**2*(log(5) + 2*exp( 3)*log(5)) + 2*x*exp(3)*log(5) + (2*x**3*log(5) + 2*x**2*log(5) + 2*x*exp( 3)*log(5))*exp(x) + (2*x**2*log(5) + 2*x*exp(x)*log(5) + 2*x*log(5) + 2*ex p(3)*log(5))*exp(exp(2)/x) + exp(2*exp(2)/x)*log(5)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.20 (sec) , antiderivative size = 228, normalized size of antiderivative = 8.77 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=x^{4} \log \left (5\right ) + 2 \, x^{3} \log \left (5\right ) + 2 \, x^{2} e^{3} \log \left (5\right ) + x^{2} \log \left (5\right ) + 2 \, x e^{3} \log \left (5\right ) + 2 \, {\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{2} \log \left (5\right ) + \frac {1}{2} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} \log \left (5\right ) + \frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \left (5\right ) + 2 \, x e^{\left (x + \frac {e^{2}}{x}\right )} \log \left (5\right ) + 2 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} \log \left (5\right ) + 8 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} \log \left (5\right ) + 2 \, {\left (x e^{3} - e^{3}\right )} e^{x} \log \left (5\right ) + 4 \, {\left (x - 1\right )} e^{x} \log \left (5\right ) + 2 \, e^{4} \Gamma \left (-1, -\frac {e^{2}}{x}\right ) \log \left (5\right ) - 2 \, e^{2} \Gamma \left (-1, -\frac {e^{2}}{x}\right ) \log \left (5\right ) + 4 \, e^{4} \Gamma \left (-2, -\frac {e^{2}}{x}\right ) \log \left (5\right ) + 2 \, e^{\left (x + 3\right )} \log \left (5\right ) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \left (5\right ) + 2 \, e^{\left (\frac {e^{2}}{x} + 3\right )} \log \left (5\right ) \] Input:
integrate((-2*exp(2)*log(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*log (5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x^2-2*x)*exp(2)+4*x^3+2*x^2)*log(5))*exp( exp(2)/x)+(2*x^4+2*x^3)*log(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^4+ 4*x^3)*log(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*log(5))/x^2, x, algorithm="maxima")
Output:
x^4*log(5) + 2*x^3*log(5) + 2*x^2*e^3*log(5) + x^2*log(5) + 2*x*e^3*log(5) + 2*Ei(e^2/x)*e^2*log(5) + 1/2*(2*x^2 - 2*x + 1)*e^(2*x)*log(5) + 1/2*(2* x - 1)*e^(2*x)*log(5) + 2*x*e^(x + e^2/x)*log(5) + 2*(x^3 - 3*x^2 + 6*x - 6)*e^x*log(5) + 8*(x^2 - 2*x + 2)*e^x*log(5) + 2*(x*e^3 - e^3)*e^x*log(5) + 4*(x - 1)*e^x*log(5) + 2*e^4*gamma(-1, -e^2/x)*log(5) - 2*e^2*gamma(-1, -e^2/x)*log(5) + 4*e^4*gamma(-2, -e^2/x)*log(5) + 2*e^(x + 3)*log(5) + e^( 2*e^2/x)*log(5) + 2*e^(e^2/x + 3)*log(5)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.19 (sec) , antiderivative size = 238, normalized size of antiderivative = 9.15 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=x^{4} \log \left (5\right ) - 2 \, x^{2} {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{8}}{x^{2}} - \frac {e^{\left (\frac {e^{2}}{x} + 6\right )}}{x} - e^{\left (\frac {e^{2}}{x} + 4\right )}\right )} e^{\left (-4\right )} \log \left (5\right ) + 2 \, x^{3} e^{x} \log \left (5\right ) + 2 \, x^{3} \log \left (5\right ) + 2 \, x^{2} e^{3} \log \left (5\right ) + 2 \, x {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{6}}{x} - e^{\left (\frac {e^{2}}{x} + 4\right )}\right )} e^{\left (-2\right )} \log \left (5\right ) - 2 \, x {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{6}}{x} - e^{\left (\frac {e^{2}}{x} + 4\right )}\right )} e^{\left (-4\right )} \log \left (5\right ) + x^{2} e^{\left (2 \, x\right )} \log \left (5\right ) + 2 \, x^{2} e^{x} \log \left (5\right ) + x^{2} \log \left (5\right ) + 2 \, x e^{3} \log \left (5\right ) + 2 \, {\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{2} \log \left (5\right ) + 2 \, x e^{\left (x + 3\right )} \log \left (5\right ) + 2 \, x e^{\left (\frac {x^{2} + e^{2}}{x}\right )} \log \left (5\right ) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \left (5\right ) + 2 \, e^{\left (\frac {e^{2}}{x} + 3\right )} \log \left (5\right ) \] Input:
integrate((-2*exp(2)*log(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*log (5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x^2-2*x)*exp(2)+4*x^3+2*x^2)*log(5))*exp( exp(2)/x)+(2*x^4+2*x^3)*log(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^4+ 4*x^3)*log(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*log(5))/x^2, x, algorithm="giac")
Output:
x^4*log(5) - 2*x^2*(Ei(e^2/x)*e^8/x^2 - e^(e^2/x + 6)/x - e^(e^2/x + 4))*e ^(-4)*log(5) + 2*x^3*e^x*log(5) + 2*x^3*log(5) + 2*x^2*e^3*log(5) + 2*x*(E i(e^2/x)*e^6/x - e^(e^2/x + 4))*e^(-2)*log(5) - 2*x*(Ei(e^2/x)*e^6/x - e^( e^2/x + 4))*e^(-4)*log(5) + x^2*e^(2*x)*log(5) + 2*x^2*e^x*log(5) + x^2*lo g(5) + 2*x*e^3*log(5) + 2*Ei(e^2/x)*e^2*log(5) + 2*x*e^(x + 3)*log(5) + 2* x*e^((x^2 + e^2)/x)*log(5) + e^(2*e^2/x)*log(5) + 2*e^(e^2/x + 3)*log(5)
Time = 4.93 (sec) , antiderivative size = 91, normalized size of antiderivative = 3.50 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=x^2\,\left (\ln \left (5\right )+2\,{\mathrm {e}}^3\,\ln \left (5\right )\right )+2\,x^3\,\ln \left (5\right )+x^4\,\ln \left (5\right )+{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^2}{x}}\,\ln \left (5\right )+2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x}}\,\ln \left (5\right )\,\left (x+{\mathrm {e}}^3+x\,{\mathrm {e}}^x+x^2\right )+2\,x\,{\mathrm {e}}^3\,\ln \left (5\right )+x^2\,{\mathrm {e}}^{2\,x}\,\ln \left (5\right )+2\,x\,{\mathrm {e}}^x\,\ln \left (5\right )\,\left (x^2+x+{\mathrm {e}}^3\right ) \] Input:
int((log(5)*(exp(3)*(2*x^2 + 4*x^3) + 2*x^3 + 6*x^4 + 4*x^5) - exp(exp(2)/ x)*(log(5)*(2*exp(5) + exp(2)*(2*x + 2*x^2) - 2*x^2 - 4*x^3) - exp(x)*log( 5)*(2*x^2 - 2*x*exp(2) + 2*x^3)) - 2*exp((2*exp(2))/x)*exp(2)*log(5) + exp (x)*log(5)*(exp(3)*(2*x^2 + 2*x^3) + 4*x^3 + 8*x^4 + 2*x^5) + exp(2*x)*log (5)*(2*x^3 + 2*x^4))/x^2,x)
Output:
x^2*(log(5) + 2*exp(3)*log(5)) + 2*x^3*log(5) + x^4*log(5) + exp((2*exp(2) )/x)*log(5) + 2*exp(exp(2)/x)*log(5)*(x + exp(3) + x*exp(x) + x^2) + 2*x*e xp(3)*log(5) + x^2*exp(2*x)*log(5) + 2*x*exp(x)*log(5)*(x + exp(3) + x^2)
Time = 3.82 (sec) , antiderivative size = 129, normalized size of antiderivative = 4.96 \[ \int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} \left (2 x^3+2 x^4\right ) \log (5)+e^x \left (4 x^3+8 x^4+2 x^5+e^3 \left (2 x^2+2 x^3\right )\right ) \log (5)+\left (2 x^3+6 x^4+4 x^5+e^3 \left (2 x^2+4 x^3\right )\right ) \log (5)+e^{\frac {e^2}{x}} \left (e^x \left (-2 e^2 x+2 x^2+2 x^3\right ) \log (5)+\left (-2 e^5+2 x^2+4 x^3+e^2 \left (-2 x-2 x^2\right )\right ) \log (5)\right )}{x^2} \, dx=\mathrm {log}\left (5\right ) \left (e^{\frac {2 e^{2}}{x}}+2 e^{\frac {e^{2}+x^{2}}{x}} x +2 e^{\frac {e^{2}}{x}} e^{3}+2 e^{\frac {e^{2}}{x}} x^{2}+2 e^{\frac {e^{2}}{x}} x +e^{2 x} x^{2}+2 e^{x} e^{3} x +2 e^{x} x^{3}+2 e^{x} x^{2}+2 e^{3} x^{2}+2 e^{3} x +x^{4}+2 x^{3}+x^{2}\right ) \] Input:
int((-2*exp(2)*log(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*log(5)*ex p(x)+(-2*exp(2)*exp(3)+(-2*x^2-2*x)*exp(2)+4*x^3+2*x^2)*log(5))*exp(exp(2) /x)+(2*x^4+2*x^3)*log(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^4+4*x^3) *log(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*log(5))/x^2,x)
Output:
log(5)*(e**((2*e**2)/x) + 2*e**((e**2 + x**2)/x)*x + 2*e**(e**2/x)*e**3 + 2*e**(e**2/x)*x**2 + 2*e**(e**2/x)*x + e**(2*x)*x**2 + 2*e**x*e**3*x + 2*e **x*x**3 + 2*e**x*x**2 + 2*e**3*x**2 + 2*e**3*x + x**4 + 2*x**3 + x**2)