Integrand size = 125, antiderivative size = 24 \[ \int \frac {\left (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx=\left (-\frac {1}{x}+x\right )^{1+e^{e^{15+(x-\log (2))^2}}} \]
Time = 0.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx=\left (-\frac {1}{x}+x\right )^{1+e^{e^{15+x^2+\log ^2(2)-x \log (4)}}} \]
Integrate[((-1 + x^2)*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2 )*(1 + x^2 + E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(1 + x^2 + E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(-2*x^2 + 2*x^4 + (2*x - 2*x^3)*Log[2])*Log[(-1 + x^2)/x])))/(x*(-x + x^3)),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.
| \(\displaystyle \int \frac {\left (x^2-1\right ) \left (\frac {x^2-1}{x}\right )^{e^{e^{x^2-2 x \log (2)+15+\log ^2(2)}}} \left (x^2+e^{e^{x^2-2 x \log (2)+15+\log ^2(2)}} \left (x^2+e^{x^2-2 x \log (2)+15+\log ^2(2)} \left (2 x^4+\left (2 x-2 x^3\right ) \log (2)-2 x^2\right ) \log \left (\frac {x^2-1}{x}\right )+1\right )+1\right )}{x \left (x^3-x\right )} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {\left (-\frac {1-x^2}{x}\right )^{e^{2^{-2 x} e^{x^2+15+\log ^2(2)}}} \left (x^2+e^{2^{-2 x} e^{x^2+15+\log ^2(2)}} \left (x^2-2^{1-2 x} e^{x^2+15+\log ^2(2)} \left (-x^4-\left (x-x^3\right ) \log (2)+x^2\right ) \log \left (-\frac {1-x^2}{x}\right )+1\right )+1\right )}{x^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (\frac {x^2-1}{x}\right )^{e^{2^{-2 x} e^{x^2+15+\log ^2(2)}}} \left (x^2+e^{2^{-2 x} e^{x^2+15+\log ^2(2)}} \left (x^2-2^{1-2 x} e^{x^2+15+\log ^2(2)} \left (-x^4-\left (x-x^3\right ) \log (2)+x^2\right ) \log \left (-\frac {1-x^2}{x}\right )+1\right )+1\right )}{x^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2^{1-2 x} (1-x) (x+1) (\log (2)-x) \log \left (\frac {x^2-1}{x}\right ) \left (\frac {x^2-1}{x}\right )^{e^{2^{-2 x} e^{x^2+15+\log ^2(2)}}} \exp \left (x^2+2^{-2 x} e^{x^2+15+\log ^2(2)}+15 \left (1+\frac {\log ^2(2)}{15}\right )\right )}{x}+\frac {\left (x^2+1\right ) \left (e^{4^{-x} e^{x^2+15+\log ^2(2)}}+1\right ) \left (\frac {x^2-1}{x}\right )^{e^{2^{-2 x} e^{x^2+15+\log ^2(2)}}}}{x^2}\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2^{1-2 x} (1-x) (x+1) (\log (2)-x) \log \left (\frac {x^2-1}{x}\right ) \left (\frac {x^2-1}{x}\right )^{e^{2^{-2 x} e^{x^2+15+\log ^2(2)}}} \exp \left (x^2+2^{-2 x} e^{x^2+15+\log ^2(2)}+15 \left (1+\frac {\log ^2(2)}{15}\right )\right )}{x}+\frac {\left (x^2+1\right ) \left (e^{4^{-x} e^{x^2+15+\log ^2(2)}}+1\right ) \left (\frac {x^2-1}{x}\right )^{e^{4^{-x} e^{x^2+15+\log ^2(2)}}}}{x^2}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {2^{1-2 x} (1-x) (x+1) (\log (2)-x) \log \left (\frac {x^2-1}{x}\right ) \left (\frac {x^2-1}{x}\right )^{e^{2^{-2 x} e^{x^2+15+\log ^2(2)}}} \exp \left (x^2+2^{-2 x} e^{x^2+15+\log ^2(2)}+15 \left (1+\frac {\log ^2(2)}{15}\right )\right )}{x}+\frac {\left (x^2+1\right ) \left (e^{4^{-x} e^{x^2+15+\log ^2(2)}}+1\right ) \left (\frac {x^2-1}{x}\right )^{e^{4^{-x} e^{x^2+15+\log ^2(2)}}}}{x^2}\right )dx\) |
Int[((-1 + x^2)*((-1 + x^2)/x)^E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(1 + x^2 + E^E^(15 + x^2 - 2*x*Log[2] + Log[2]^2)*(1 + x^2 + E^(15 + x^2 - 2*x *Log[2] + Log[2]^2)*(-2*x^2 + 2*x^4 + (2*x - 2*x^3)*Log[2])*Log[(-1 + x^2) /x])))/(x*(-x + x^3)),x]
3.29.16.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 5.58
\[\frac {\left (x^{2}-1\right ) x^{-{\mathrm e}^{4^{-x} {\mathrm e}^{\ln \left (2\right )^{2}+15+x^{2}}}} \left (x^{2}-1\right )^{{\mathrm e}^{4^{-x} {\mathrm e}^{\ln \left (2\right )^{2}+15+x^{2}}}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )+\operatorname {csgn}\left (i \left (x^{2}-1\right )\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left ({\mathrm e}^{\left (\frac {1}{4}\right )^{x} {\mathrm e}^{\ln \left (2\right )^{2}+15+x^{2}}}+1\right )}{2}}}{x}\]
int(((((-2*x^3+2*x)*ln(2)+2*x^4-2*x^2)*exp(ln(2)^2-2*x*ln(2)+x^2+15)*ln((x ^2-1)/x)+x^2+1)*exp(exp(ln(2)^2-2*x*ln(2)+x^2+15))+x^2+1)*exp(ln((x^2-1)/x )*exp(exp(ln(2)^2-2*x*ln(2)+x^2+15))+ln((x^2-1)/x))/(x^3-x),x)
(x^2-1)/x*x^(-exp(4^(-x)*exp(ln(2)^2+15+x^2)))*(x^2-1)^exp(4^(-x)*exp(ln(2 )^2+15+x^2))*exp(-1/2*I*Pi*csgn(I/x*(x^2-1))*(-csgn(I/x*(x^2-1))+csgn(I*(x ^2-1)))*(-csgn(I/x*(x^2-1))+csgn(I/x))*(exp((1/4)^x*exp(ln(2)^2+15+x^2))+1 ))
Timed out. \[ \int \frac {\left (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx=\text {Timed out} \]
integrate(((((-2*x^3+2*x)*log(2)+2*x^4-2*x^2)*exp(log(2)^2-2*x*log(2)+x^2+ 15)*log((x^2-1)/x)+x^2+1)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+x^2+1)*exp( log((x^2-1)/x)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+log((x^2-1)/x))/(x^3-x ),x, algorithm=\
Time = 4.89 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx=\frac {\left (x^{2} - 1\right ) e^{e^{e^{x^{2} - 2 x \log {\left (2 \right )} + \log {\left (2 \right )}^{2} + 15}} \log {\left (\frac {x^{2} - 1}{x} \right )}}}{x} \]
integrate(((((-2*x**3+2*x)*ln(2)+2*x**4-2*x**2)*exp(ln(2)**2-2*x*ln(2)+x** 2+15)*ln((x**2-1)/x)+x**2+1)*exp(exp(ln(2)**2-2*x*ln(2)+x**2+15))+x**2+1)* exp(ln((x**2-1)/x)*exp(exp(ln(2)**2-2*x*ln(2)+x**2+15))+ln((x**2-1)/x))/(x **3-x),x)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (22) = 44\).
Time = 0.52 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.04 \[ \int \frac {\left (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx=\frac {{\left (x^{2} - 1\right )} e^{\left (e^{\left (e^{\left (x^{2} - 2 \, x \log \left (2\right ) + \log \left (2\right )^{2} + 15\right )}\right )} \log \left (x + 1\right ) + e^{\left (e^{\left (x^{2} - 2 \, x \log \left (2\right ) + \log \left (2\right )^{2} + 15\right )}\right )} \log \left (x - 1\right ) - e^{\left (e^{\left (x^{2} - 2 \, x \log \left (2\right ) + \log \left (2\right )^{2} + 15\right )}\right )} \log \left (x\right )\right )}}{x} \]
integrate(((((-2*x^3+2*x)*log(2)+2*x^4-2*x^2)*exp(log(2)^2-2*x*log(2)+x^2+ 15)*log((x^2-1)/x)+x^2+1)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+x^2+1)*exp( log((x^2-1)/x)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+log((x^2-1)/x))/(x^3-x ),x, algorithm=\
(x^2 - 1)*e^(e^(e^(x^2 - 2*x*log(2) + log(2)^2 + 15))*log(x + 1) + e^(e^(x ^2 - 2*x*log(2) + log(2)^2 + 15))*log(x - 1) - e^(e^(x^2 - 2*x*log(2) + lo g(2)^2 + 15))*log(x))/x
Time = 1.75 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {\left (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx=e^{\left (e^{\left (e^{\left (x^{2} - 2 \, x \log \left (2\right ) + \log \left (2\right )^{2} + 15\right )}\right )} \log \left (x - \frac {1}{x}\right ) + \log \left (x - \frac {1}{x}\right )\right )} \]
integrate(((((-2*x^3+2*x)*log(2)+2*x^4-2*x^2)*exp(log(2)^2-2*x*log(2)+x^2+ 15)*log((x^2-1)/x)+x^2+1)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+x^2+1)*exp( log((x^2-1)/x)*exp(exp(log(2)^2-2*x*log(2)+x^2+15))+log((x^2-1)/x))/(x^3-x ),x, algorithm=\
Time = 9.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-1+x^2\right ) \left (\frac {-1+x^2}{x}\right )^{e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}}} \left (1+x^2+e^{e^{15+x^2-2 x \log (2)+\log ^2(2)}} \left (1+x^2+e^{15+x^2-2 x \log (2)+\log ^2(2)} \left (-2 x^2+2 x^4+\left (2 x-2 x^3\right ) \log (2)\right ) \log \left (\frac {-1+x^2}{x}\right )\right )\right )}{x \left (-x+x^3\right )} \, dx=\frac {{\left (x-\frac {1}{x}\right )}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\ln \left (2\right )}^2}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{15}}{2^{2\,x}}}}\,\left (x^2-1\right )}{x} \]