Integrand size = 107, antiderivative size = 29 \[ \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8-16 x+8 x^2+\left (-2 x^4+2 x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (2 x^4+2 x^5\right ) \log (5)+e^{\frac {2}{x^2 \log (5)}} \left (-8 x+8 x^2+\left (-2 x^3+4 x^5\right ) \log (5)\right )\right )}{x^3 \log (5)} \, dx=e^{2 x} \left (e^{-\frac {2}{x^2 \log (5)}} (1-x)-x\right )^2 \]
Time = 5.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8-16 x+8 x^2+\left (-2 x^4+2 x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (2 x^4+2 x^5\right ) \log (5)+e^{\frac {2}{x^2 \log (5)}} \left (-8 x+8 x^2+\left (-2 x^3+4 x^5\right ) \log (5)\right )\right )}{x^3 \log (5)} \, dx=e^{2 x-\frac {4}{x^2 \log (5)}} \left (-1+x+e^{\frac {2}{x^2 \log (5)}} x\right )^2 \]
Integrate[(E^(2*x - 4/(x^2*Log[5]))*(8 - 16*x + 8*x^2 + (-2*x^4 + 2*x^5)*L og[5] + E^(4/(x^2*Log[5]))*(2*x^4 + 2*x^5)*Log[5] + E^(2/(x^2*Log[5]))*(-8 *x + 8*x^2 + (-2*x^3 + 4*x^5)*Log[5])))/(x^3*Log[5]),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.
\(\displaystyle \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8 x^2+\left (2 x^5-2 x^4\right ) \log (5)+\left (2 x^5+2 x^4\right ) \log (5) e^{\frac {4}{x^2 \log (5)}}+e^{\frac {2}{x^2 \log (5)}} \left (8 x^2+\left (4 x^5-2 x^3\right ) \log (5)-8 x\right )-16 x+8\right )}{x^3 \log (5)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 e^{2 x-\frac {4}{x^2 \log (5)}} \left (4 x^2-8 x-e^{\frac {2}{x^2 \log (5)}} \left (-4 x^2+4 x+\left (x^3-2 x^5\right ) \log (5)\right )-\left (x^4-x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (x^5+x^4\right ) \log (5)+4\right )}{x^3}dx}{\log (5)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (4 x^2-8 x-e^{\frac {2}{x^2 \log (5)}} \left (-4 x^2+4 x+\left (x^3-2 x^5\right ) \log (5)\right )-\left (x^4-x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (x^5+x^4\right ) \log (5)+4\right )}{x^3}dx}{\log (5)}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {2 \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (-e^{\frac {2}{x^2 \log (5)}} x-x+1\right ) \left (-e^{\frac {2}{x^2 \log (5)}} \log (5) x^4-\log (5) x^4-e^{\frac {2}{x^2 \log (5)}} \log (5) x^3-4 x+4\right )}{x^3}dx}{\log (5)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2 \int \left (e^{2 x} x \log (5) (x+1)+\frac {e^{2 x-\frac {4}{x^2 \log (5)}} (x-1) \left (\log (5) x^4+4 x-4\right )}{x^3}+\frac {e^{2 x-\frac {2}{x^2 \log (5)}} \left (\log (25) x^4-\log (5) x^2+4 x-4\right )}{x^2}\right )dx}{\log (5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\log (5) \int e^{2 x-\frac {2}{x^2 \log (5)}}dx-8 \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}}}{x^2}dx-4 \int \frac {e^{2 x-\frac {2}{x^2 \log (5)}}}{x^2}dx+4 \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}}}{x}dx+4 \int \frac {e^{2 x-\frac {2}{x^2 \log (5)}}}{x}dx-\log (5) \int e^{2 x-\frac {4}{x^2 \log (5)}} xdx+\log (5) \int e^{2 x-\frac {4}{x^2 \log (5)}} x^2dx+\log (25) \int e^{2 x-\frac {2}{x^2 \log (5)}} x^2dx+4 \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}}}{x^3}dx+\frac {1}{2} e^{2 x} x^2 \log (5)\right )}{\log (5)}\) |
Int[(E^(2*x - 4/(x^2*Log[5]))*(8 - 16*x + 8*x^2 + (-2*x^4 + 2*x^5)*Log[5] + E^(4/(x^2*Log[5]))*(2*x^4 + 2*x^5)*Log[5] + E^(2/(x^2*Log[5]))*(-8*x + 8 *x^2 + (-2*x^3 + 4*x^5)*Log[5])))/(x^3*Log[5]),x]
3.1.49.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(60\) vs. \(2(29)=58\).
Time = 3.42 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.10
method | result | size |
risch | \({\mathrm e}^{2 x} x^{2}+2 x \left (-1+x \right ) {\mathrm e}^{\frac {2 x^{3} \ln \left (5\right )-2}{x^{2} \ln \left (5\right )}}+\left (x^{2}-2 x +1\right ) {\mathrm e}^{\frac {2 x^{3} \ln \left (5\right )-4}{x^{2} \ln \left (5\right )}}\) | \(61\) |
parallelrisch | \(-\frac {\left (-2 \ln \left (5\right ) x^{4} {\mathrm e}^{2 x} {\mathrm e}^{\frac {4}{x^{2} \ln \left (5\right )}}-4 \ln \left (5\right ) x^{4} {\mathrm e}^{2 x} {\mathrm e}^{\frac {2}{x^{2} \ln \left (5\right )}}-2 \ln \left (5\right ) x^{4} {\mathrm e}^{2 x}+4 \ln \left (5\right ) x^{3} {\mathrm e}^{2 x} {\mathrm e}^{\frac {2}{x^{2} \ln \left (5\right )}}+4 \ln \left (5\right ) x^{3} {\mathrm e}^{2 x}-2 x^{2} \ln \left (5\right ) {\mathrm e}^{2 x}\right ) {\mathrm e}^{-\frac {4}{x^{2} \ln \left (5\right )}}}{2 \ln \left (5\right ) x^{2}}\) | \(121\) |
int(((2*x^5+2*x^4)*ln(5)*exp(2/x^2/ln(5))^2+((4*x^5-2*x^3)*ln(5)+8*x^2-8*x )*exp(2/x^2/ln(5))+(2*x^5-2*x^4)*ln(5)+8*x^2-16*x+8)*exp(x)^2/x^3/ln(5)/ex p(2/x^2/ln(5))^2,x,method=_RETURNVERBOSE)
exp(2*x)*x^2+2*x*(-1+x)*exp(2*(x^3*ln(5)-1)/x^2/ln(5))+(x^2-2*x+1)*exp(2*( x^3*ln(5)-2)/x^2/ln(5))
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.07 \[ \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8-16 x+8 x^2+\left (-2 x^4+2 x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (2 x^4+2 x^5\right ) \log (5)+e^{\frac {2}{x^2 \log (5)}} \left (-8 x+8 x^2+\left (-2 x^3+4 x^5\right ) \log (5)\right )\right )}{x^3 \log (5)} \, dx={\left (x^{2} e^{\left (\frac {4}{x^{2} \log \left (5\right )}\right )} + x^{2} + 2 \, {\left (x^{2} - x\right )} e^{\left (\frac {2}{x^{2} \log \left (5\right )}\right )} - 2 \, x + 1\right )} e^{\left (\frac {2 \, {\left (x^{3} \log \left (5\right ) - 2\right )}}{x^{2} \log \left (5\right )}\right )} \]
integrate(((2*x^5+2*x^4)*log(5)*exp(2/x^2/log(5))^2+((4*x^5-2*x^3)*log(5)+ 8*x^2-8*x)*exp(2/x^2/log(5))+(2*x^5-2*x^4)*log(5)+8*x^2-16*x+8)*exp(x)^2/x ^3/log(5)/exp(2/x^2/log(5))^2,x, algorithm=\
(x^2*e^(4/(x^2*log(5))) + x^2 + 2*(x^2 - x)*e^(2/(x^2*log(5))) - 2*x + 1)* e^(2*(x^3*log(5) - 2)/(x^2*log(5)))
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (20) = 40\).
Time = 1.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.34 \[ \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8-16 x+8 x^2+\left (-2 x^4+2 x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (2 x^4+2 x^5\right ) \log (5)+e^{\frac {2}{x^2 \log (5)}} \left (-8 x+8 x^2+\left (-2 x^3+4 x^5\right ) \log (5)\right )\right )}{x^3 \log (5)} \, dx=x^{2} e^{2 x} + \left (2 x^{2} e^{2 x} - 2 x e^{2 x}\right ) e^{- \frac {2}{x^{2} \log {\left (5 \right )}}} + \left (x^{2} e^{2 x} - 2 x e^{2 x} + e^{2 x}\right ) e^{- \frac {4}{x^{2} \log {\left (5 \right )}}} \]
integrate(((2*x**5+2*x**4)*ln(5)*exp(2/x**2/ln(5))**2+((4*x**5-2*x**3)*ln( 5)+8*x**2-8*x)*exp(2/x**2/ln(5))+(2*x**5-2*x**4)*ln(5)+8*x**2-16*x+8)*exp( x)**2/x**3/ln(5)/exp(2/x**2/ln(5))**2,x)
x**2*exp(2*x) + (2*x**2*exp(2*x) - 2*x*exp(2*x))*exp(-2/(x**2*log(5))) + ( x**2*exp(2*x) - 2*x*exp(2*x) + exp(2*x))*exp(-4/(x**2*log(5)))
Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (23) = 46\).
Time = 0.36 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.24 \[ \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8-16 x+8 x^2+\left (-2 x^4+2 x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (2 x^4+2 x^5\right ) \log (5)+e^{\frac {2}{x^2 \log (5)}} \left (-8 x+8 x^2+\left (-2 x^3+4 x^5\right ) \log (5)\right )\right )}{x^3 \log (5)} \, dx=\frac {{\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} \log \left (5\right ) + {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \left (5\right ) + 4 \, {\left (x^{2} \log \left (5\right ) - x \log \left (5\right )\right )} e^{\left (2 \, x - \frac {2}{x^{2} \log \left (5\right )}\right )} + 2 \, {\left (x^{2} \log \left (5\right ) - 2 \, x \log \left (5\right ) + \log \left (5\right )\right )} e^{\left (2 \, x - \frac {4}{x^{2} \log \left (5\right )}\right )}}{2 \, \log \left (5\right )} \]
integrate(((2*x^5+2*x^4)*log(5)*exp(2/x^2/log(5))^2+((4*x^5-2*x^3)*log(5)+ 8*x^2-8*x)*exp(2/x^2/log(5))+(2*x^5-2*x^4)*log(5)+8*x^2-16*x+8)*exp(x)^2/x ^3/log(5)/exp(2/x^2/log(5))^2,x, algorithm=\
1/2*((2*x^2 - 2*x + 1)*e^(2*x)*log(5) + (2*x - 1)*e^(2*x)*log(5) + 4*(x^2* log(5) - x*log(5))*e^(2*x - 2/(x^2*log(5))) + 2*(x^2*log(5) - 2*x*log(5) + log(5))*e^(2*x - 4/(x^2*log(5))))/log(5)
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 4.55 \[ \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8-16 x+8 x^2+\left (-2 x^4+2 x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (2 x^4+2 x^5\right ) \log (5)+e^{\frac {2}{x^2 \log (5)}} \left (-8 x+8 x^2+\left (-2 x^3+4 x^5\right ) \log (5)\right )\right )}{x^3 \log (5)} \, dx=\frac {x^{2} e^{\left (2 \, x\right )} \log \left (5\right ) + 2 \, x^{2} e^{\left (\frac {2 \, {\left (x^{3} \log \left (5\right ) - 1\right )}}{x^{2} \log \left (5\right )}\right )} \log \left (5\right ) + x^{2} e^{\left (\frac {2 \, {\left (x^{3} \log \left (5\right ) - 2\right )}}{x^{2} \log \left (5\right )}\right )} \log \left (5\right ) - 2 \, x e^{\left (\frac {2 \, {\left (x^{3} \log \left (5\right ) - 1\right )}}{x^{2} \log \left (5\right )}\right )} \log \left (5\right ) - 2 \, x e^{\left (\frac {2 \, {\left (x^{3} \log \left (5\right ) - 2\right )}}{x^{2} \log \left (5\right )}\right )} \log \left (5\right ) + e^{\left (\frac {2 \, {\left (x^{3} \log \left (5\right ) - 2\right )}}{x^{2} \log \left (5\right )}\right )} \log \left (5\right )}{\log \left (5\right )} \]
integrate(((2*x^5+2*x^4)*log(5)*exp(2/x^2/log(5))^2+((4*x^5-2*x^3)*log(5)+ 8*x^2-8*x)*exp(2/x^2/log(5))+(2*x^5-2*x^4)*log(5)+8*x^2-16*x+8)*exp(x)^2/x ^3/log(5)/exp(2/x^2/log(5))^2,x, algorithm=\
(x^2*e^(2*x)*log(5) + 2*x^2*e^(2*(x^3*log(5) - 1)/(x^2*log(5)))*log(5) + x ^2*e^(2*(x^3*log(5) - 2)/(x^2*log(5)))*log(5) - 2*x*e^(2*(x^3*log(5) - 1)/ (x^2*log(5)))*log(5) - 2*x*e^(2*(x^3*log(5) - 2)/(x^2*log(5)))*log(5) + e^ (2*(x^3*log(5) - 2)/(x^2*log(5)))*log(5))/log(5)
Time = 10.57 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{2 x-\frac {4}{x^2 \log (5)}} \left (8-16 x+8 x^2+\left (-2 x^4+2 x^5\right ) \log (5)+e^{\frac {4}{x^2 \log (5)}} \left (2 x^4+2 x^5\right ) \log (5)+e^{\frac {2}{x^2 \log (5)}} \left (-8 x+8 x^2+\left (-2 x^3+4 x^5\right ) \log (5)\right )\right )}{x^3 \log (5)} \, dx={\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {4}{x^2\,\ln \left (5\right )}}\,{\left (x+x\,{\mathrm {e}}^{\frac {2}{x^2\,\ln \left (5\right )}}-1\right )}^2 \]