Integrand size = 68, antiderivative size = 32 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=-3+e^{4 \left (-1+\frac {4 \left (x+\frac {2}{\log (\log (5))}\right )}{-4+\frac {(1+x)^2}{x}}\right )} \]
Time = 1.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\frac {-4 \log (\log (5))+12 x^2 \log (\log (5))+8 x (4+\log (\log (5)))}{(-1+x)^2 \log (\log (5))}} \]
Integrate[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Lo g[Log[5]]))*(-32 - 32*x - 32*x*Log[Log[5]]))/((-1 + 3*x - 3*x^2 + x^3)*Log [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.
\(\displaystyle \int \frac {(-32 x-32 x \log (\log (5))-32) \exp \left (\frac {\left (12 x^2+8 x-4\right ) \log (\log (5))+32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}\right )}{\left (x^3-3 x^2+3 x-1\right ) \log (\log (5))} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {(x (-32-32 \log (\log (5)))-32) \exp \left (\frac {\left (12 x^2+8 x-4\right ) \log (\log (5))+32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}\right )}{\left (x^3-3 x^2+3 x-1\right ) \log (\log (5))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {32 e^{\frac {32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}} \log ^{-\frac {-12 x^2-8 x+4}{\left (x^2-2 x+1\right ) \log (\log (5))}}(5) ((1+\log (\log (5))) x+1)}{-x^3+3 x^2-3 x+1}dx}{\log (\log (5))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {32 \int \frac {e^{\frac {32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}} \log ^{-\frac {4 \left (-3 x^2-2 x+1\right )}{\left (x^2-2 x+1\right ) \log (\log (5))}}(5) ((1+\log (\log (5))) x+1)}{-x^3+3 x^2-3 x+1}dx}{\log (\log (5))}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \frac {32 \int \frac {e^{\frac {32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}} \log ^{-\frac {4 \left (-3 x^2-2 x+1\right )}{\left (x^2-2 x+1\right ) \log (\log (5))}}(5) ((1+\log (\log (5))) x+1)}{(1-x)^3}dx}{\log (\log (5))}\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {32 \int \frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right ) ((1+\log (\log (5))) x+1)}{(1-x)^3}dx}{\log (\log (5))}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {32 \int \left (\frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right ) (-1-\log (\log (5)))}{(x-1)^2}+\frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right ) (-2-\log (\log (5)))}{(x-1)^3}\right )dx}{\log (\log (5))}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {32 \left (-\left ((2+\log (\log (5))) \int \frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right )}{(x-1)^3}dx\right )-(1+\log (\log (5))) \int \frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right )}{(x-1)^2}dx\right )}{\log (\log (5))}\) |
Int[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Log[Log[ 5]]))*(-32 - 32*x - 32*x*Log[Log[5]]))/((-1 + 3*x - 3*x^2 + x^3)*Log[Log[5 ]]),x]
3.29.9.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}\) | \(36\) |
risch | \({\mathrm e}^{\frac {12 \ln \left (\ln \left (5\right )\right ) x^{2}+8 x \ln \left (\ln \left (5\right )\right )-4 \ln \left (\ln \left (5\right )\right )+32 x}{\ln \left (\ln \left (5\right )\right ) \left (-1+x \right )^{2}}}\) | \(37\) |
gosper | \({\mathrm e}^{\frac {12 \ln \left (\ln \left (5\right )\right ) x^{2}+8 x \ln \left (\ln \left (5\right )\right )-4 \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}\) | \(42\) |
norman | \(\frac {x^{2} {\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}-2 x \,{\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}+{\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}}{\left (-1+x \right )^{2}}\) | \(120\) |
default | \(\frac {-\frac {4 i {\mathrm e}^{12} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}+64 \,{\mathrm e}^{12} \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )-\frac {4 i {\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}+32 \,{\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\ln \left (\ln \left (5\right )\right )}\) | \(502\) |
derivativedivides | \(-\frac {32 \left (\frac {i {\mathrm e}^{12} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}-2 \,{\mathrm e}^{12} \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )+\frac {i {\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}-{\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )\right )}{\ln \left (\ln \left (5\right )\right )}\) | \(503\) |
int((-32*x*ln(ln(5))-32*x-32)*exp(((12*x^2+8*x-4)*ln(ln(5))+32*x)/(x^2-2*x +1)/ln(ln(5)))/(x^3-3*x^2+3*x-1)/ln(ln(5)),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {4 \, {\left ({\left (3 \, x^{2} + 2 \, x - 1\right )} \log \left (\log \left (5\right )\right ) + 8 \, x\right )}}{{\left (x^{2} - 2 \, x + 1\right )} \log \left (\log \left (5\right )\right )}\right )} \]
integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x )/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^2+3*x-1)/log(log(5)),x, algorithm=\
Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\frac {32 x + \left (12 x^{2} + 8 x - 4\right ) \log {\left (\log {\left (5 \right )} \right )}}{\left (x^{2} - 2 x + 1\right ) \log {\left (\log {\left (5 \right )} \right )}}} \]
integrate((-32*x*ln(ln(5))-32*x-32)*exp(((12*x**2+8*x-4)*ln(ln(5))+32*x)/( x**2-2*x+1)/ln(ln(5)))/(x**3-3*x**2+3*x-1)/ln(ln(5)),x)
Time = 0.44 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {32}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {16}{x^{2} - 2 \, x + 1} + \frac {32}{x \log \left (\log \left (5\right )\right ) - \log \left (\log \left (5\right )\right )} + \frac {32}{x - 1} + 12\right )} \]
integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x )/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^2+3*x-1)/log(log(5)),x, algorithm=\
e^(32/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 16/(x^2 - 2*x + 1) + 32/(x*log(log(5)) - log(log(5))) + 32/(x - 1) + 12)
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (29) = 58\).
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.12 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {12 \, x^{2} \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {8 \, x \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {32 \, x}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} - \frac {4 \, \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )}\right )} \]
integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x )/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^2+3*x-1)/log(log(5)),x, algorithm=\
e^(12*x^2*log(log(5))/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 8*x*log(log(5))/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 32*x/( x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) - 4*log(log(5))/(x^2*log( log(5)) - 2*x*log(log(5)) + log(log(5))))
Time = 13.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx={\mathrm {e}}^{\frac {32\,x}{\ln \left (\ln \left (5\right )\right )\,x^2-2\,\ln \left (\ln \left (5\right )\right )\,x+\ln \left (\ln \left (5\right )\right )}}\,{\ln \left (5\right )}^{\frac {12\,x^2+8\,x-4}{\ln \left ({\ln \left (5\right )}^{x^2-2\,x+1}\right )}} \]
int(-(exp((32*x + log(log(5))*(8*x + 12*x^2 - 4))/(log(log(5))*(x^2 - 2*x + 1)))*(32*x + 32*x*log(log(5)) + 32))/(log(log(5))*(3*x - 3*x^2 + x^3 - 1 )),x)