Integrand size = 70, antiderivative size = 38 \[ \int \frac {e^{-x} \left (-8 x+2 x^2+4 x^3-x^4-x^5+\left (4-4 x-4 x^2+4 x^3+x^4-x^5\right ) \log ^2(5)\right )}{4-4 x^2+x^4} \, dx=-3-e^{-x} \left (-3-x+\frac {4-x^2}{2-x^2}-x \log ^2(5)\right ) \]
Time = 5.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {e^{-x} \left (-8 x+2 x^2+4 x^3-x^4-x^5+\left (4-4 x-4 x^2+4 x^3+x^4-x^5\right ) \log ^2(5)\right )}{4-4 x^2+x^4} \, dx=e^{-x} \left (2+\frac {2}{-2+x^2}+x \left (1+\log ^2(5)\right )\right ) \]
Integrate[(-8*x + 2*x^2 + 4*x^3 - x^4 - x^5 + (4 - 4*x - 4*x^2 + 4*x^3 + x ^4 - x^5)*Log[5]^2)/(E^x*(4 - 4*x^2 + x^4)),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.
\(\displaystyle \int \frac {e^{-x} \left (-x^5-x^4+4 x^3+2 x^2+\left (-x^5+x^4+4 x^3-4 x^2-4 x+4\right ) \log ^2(5)-8 x\right )}{x^4-4 x^2+4} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int -\frac {e^{-x} \left (x^5+x^4-4 x^3-2 x^2-\left (-x^5+x^4+4 x^3-4 x^2-4 x+4\right ) \log ^2(5)+8 x\right )}{\left (2-x^2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {e^{-x} \left (x^5+x^4-4 x^3-2 x^2+8 x-\left (-x^5+x^4+4 x^3-4 x^2-4 x+4\right ) \log ^2(5)\right )}{\left (2-x^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {4 e^{-x} x}{\left (x^2-2\right )^2}+e^{-x} \left (1+\log ^2(5)\right ) x+\frac {2 e^{-x}}{x^2-2}+e^{-x} \left (1-\log ^2(5)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {e^{-x} x}{\left (x^2-2\right )^2}dx+\frac {e^{\sqrt {2}} \operatorname {ExpIntegralEi}\left (-x-\sqrt {2}\right )}{\sqrt {2}}-\frac {e^{-\sqrt {2}} \operatorname {ExpIntegralEi}\left (\sqrt {2}-x\right )}{\sqrt {2}}+e^{-x} x \left (1+\log ^2(5)\right )+e^{-x} \left (1+\log ^2(5)\right )+e^{-x} \left (1-\log ^2(5)\right )\) |
Int[(-8*x + 2*x^2 + 4*x^3 - x^4 - x^5 + (4 - 4*x - 4*x^2 + 4*x^3 + x^4 - x ^5)*Log[5]^2)/(E^x*(4 - 4*x^2 + x^4)),x]
3.25.59.3.1 Defintions of rubi rules used
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {\left (-2+\left (-2 \ln \left (5\right )^{2}-2\right ) x +\left (\ln \left (5\right )^{2}+1\right ) x^{3}+2 x^{2}\right ) {\mathrm e}^{-x}}{x^{2}-2}\) | \(40\) |
gosper | \(\frac {\left (x^{3} \ln \left (5\right )^{2}-2 x \ln \left (5\right )^{2}+x^{3}+2 x^{2}-2 x -2\right ) {\mathrm e}^{-x}}{x^{2}-2}\) | \(41\) |
risch | \(\frac {\left (x^{3} \ln \left (5\right )^{2}-2 x \ln \left (5\right )^{2}+x^{3}+2 x^{2}-2 x -2\right ) {\mathrm e}^{-x}}{x^{2}-2}\) | \(41\) |
parallelrisch | \(\frac {\left (x^{3} \ln \left (5\right )^{2}-2 x \ln \left (5\right )^{2}+x^{3}+2 x^{2}-2 x -2\right ) {\mathrm e}^{-x}}{x^{2}-2}\) | \(41\) |
default | \(\ln \left (5\right )^{2} \left (-{\mathrm e}^{-x}-\frac {{\mathrm e}^{-x} x}{x^{2}-2}+\frac {{\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{2}-\frac {3 \sqrt {2}\, {\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{4}+\frac {{\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{2}+\frac {3 \sqrt {2}\, {\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{4}\right )+{\mathrm e}^{-x}+\left (1+x \right ) {\mathrm e}^{-x}+\frac {2 \,{\mathrm e}^{-x}}{x^{2}-2}+4 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-x} x}{4 \left (x^{2}-2\right )}+\frac {{\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{8}+\frac {\sqrt {2}\, {\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{16}+\frac {{\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{8}-\frac {\sqrt {2}\, {\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{16}\right )-4 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-x}}{2 \left (x^{2}-2\right )}+\frac {\sqrt {2}\, {\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{8}-\frac {\sqrt {2}\, {\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{8}\right )-4 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-x} x}{2 \left (x^{2}-2\right )}+\frac {{\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{4}-\frac {\sqrt {2}\, {\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{8}+\frac {{\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{4}+\frac {\sqrt {2}\, {\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{8}\right )+4 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-x}}{x^{2}-2}-\frac {{\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{2}+\frac {\sqrt {2}\, {\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{4}-\frac {{\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{2}-\frac {\sqrt {2}\, {\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{4}\right )-\ln \left (5\right )^{2} \left (-\left (1+x \right ) {\mathrm e}^{-x}-\frac {2 \,{\mathrm e}^{-x}}{x^{2}-2}-2 \,{\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )+\frac {\sqrt {2}\, {\mathrm e}^{-\sqrt {2}} \operatorname {Ei}_{1}\left (-\sqrt {2}+x \right )}{2}-2 \,{\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )-\frac {\sqrt {2}\, {\mathrm e}^{\sqrt {2}} \operatorname {Ei}_{1}\left (\sqrt {2}+x \right )}{2}\right )\) | \(530\) |
int(((-x^5+x^4+4*x^3-4*x^2-4*x+4)*ln(5)^2-x^5-x^4+4*x^3+2*x^2-8*x)/(x^4-4* x^2+4)/exp(x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x} \left (-8 x+2 x^2+4 x^3-x^4-x^5+\left (4-4 x-4 x^2+4 x^3+x^4-x^5\right ) \log ^2(5)\right )}{4-4 x^2+x^4} \, dx=\frac {{\left (x^{3} + {\left (x^{3} - 2 \, x\right )} \log \left (5\right )^{2} + 2 \, x^{2} - 2 \, x - 2\right )} e^{\left (-x\right )}}{x^{2} - 2} \]
integrate(((-x^5+x^4+4*x^3-4*x^2-4*x+4)*log(5)^2-x^5-x^4+4*x^3+2*x^2-8*x)/ (x^4-4*x^2+4)/exp(x),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x} \left (-8 x+2 x^2+4 x^3-x^4-x^5+\left (4-4 x-4 x^2+4 x^3+x^4-x^5\right ) \log ^2(5)\right )}{4-4 x^2+x^4} \, dx=\frac {\left (x^{3} + x^{3} \log {\left (5 \right )}^{2} + 2 x^{2} - 2 x \log {\left (5 \right )}^{2} - 2 x - 2\right ) e^{- x}}{x^{2} - 2} \]
integrate(((-x**5+x**4+4*x**3-4*x**2-4*x+4)*ln(5)**2-x**5-x**4+4*x**3+2*x* *2-8*x)/(x**4-4*x**2+4)/exp(x),x)
Time = 0.36 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-8 x+2 x^2+4 x^3-x^4-x^5+\left (4-4 x-4 x^2+4 x^3+x^4-x^5\right ) \log ^2(5)\right )}{4-4 x^2+x^4} \, dx=\frac {{\left ({\left (\log \left (5\right )^{2} + 1\right )} x^{3} - 2 \, {\left (\log \left (5\right )^{2} + 1\right )} x + 2 \, x^{2} - 2\right )} e^{\left (-x\right )}}{x^{2} - 2} \]
integrate(((-x^5+x^4+4*x^3-4*x^2-4*x+4)*log(5)^2-x^5-x^4+4*x^3+2*x^2-8*x)/ (x^4-4*x^2+4)/exp(x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (30) = 60\).
Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.63 \[ \int \frac {e^{-x} \left (-8 x+2 x^2+4 x^3-x^4-x^5+\left (4-4 x-4 x^2+4 x^3+x^4-x^5\right ) \log ^2(5)\right )}{4-4 x^2+x^4} \, dx=\frac {x^{3} e^{\left (-x\right )} \log \left (5\right )^{2} + x^{3} e^{\left (-x\right )} - 2 \, x e^{\left (-x\right )} \log \left (5\right )^{2} + 2 \, x^{2} e^{\left (-x\right )} - 2 \, x e^{\left (-x\right )} - 2 \, e^{\left (-x\right )}}{x^{2} - 2} \]
integrate(((-x^5+x^4+4*x^3-4*x^2-4*x+4)*log(5)^2-x^5-x^4+4*x^3+2*x^2-8*x)/ (x^4-4*x^2+4)/exp(x),x, algorithm=\
(x^3*e^(-x)*log(5)^2 + x^3*e^(-x) - 2*x*e^(-x)*log(5)^2 + 2*x^2*e^(-x) - 2 *x*e^(-x) - 2*e^(-x))/(x^2 - 2)
Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-x} \left (-8 x+2 x^2+4 x^3-x^4-x^5+\left (4-4 x-4 x^2+4 x^3+x^4-x^5\right ) \log ^2(5)\right )}{4-4 x^2+x^4} \, dx={\mathrm {e}}^{-x}\,\left (x+x\,{\ln \left (5\right )}^2+2\right )+\frac {2\,{\mathrm {e}}^{-x}}{x^2-2} \]