Integrand size = 81, antiderivative size = 31 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{5+2 x+x^2} \] Output:
ln(-x^2+1)/(x^2+2*x+5)*exp(3-exp(ln(5)^2))
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{3-e^{\log ^2(5)}} \log \left (1-x^2\right )}{5+2 x+x^2} \] Input:
Integrate[(E^(3 - E^Log[5]^2)*(10*x + 4*x^2 + 2*x^3 + (2 + 2*x - 2*x^2 - 2 *x^3)*Log[1 - x^2]))/(-25 - 20*x + 11*x^2 + 16*x^3 + 13*x^4 + 4*x^5 + x^6) ,x]
Output:
(E^(3 - E^Log[5]^2)*Log[1 - x^2])/(5 + 2*x + x^2)
Result contains complex when optimal does not.
Time = 3.21 (sec) , antiderivative size = 214, normalized size of antiderivative = 6.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {27, 27, 2463, 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.
\(\displaystyle \int \frac {e^{3-e^{\log ^2(5)}} \left (2 x^3+4 x^2+\left (-2 x^3-2 x^2+2 x+2\right ) \log \left (1-x^2\right )+10 x\right )}{x^6+4 x^5+13 x^4+16 x^3+11 x^2-20 x-25} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{3-e^{\log ^2(5)}} \int -\frac {2 \left (x^3+2 x^2+5 x+\left (-x^3-x^2+x+1\right ) \log \left (1-x^2\right )\right )}{-x^6-4 x^5-13 x^4-16 x^3-11 x^2+20 x+25}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -2 e^{3-e^{\log ^2(5)}} \int \frac {x^3+2 x^2+5 x+\left (-x^3-x^2+x+1\right ) \log \left (1-x^2\right )}{-x^6-4 x^5-13 x^4-16 x^3-11 x^2+20 x+25}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle -2 e^{3-e^{\log ^2(5)}} \int \left (-\frac {x^3+2 x^2+5 x+\left (-x^3-x^2+x+1\right ) \log \left (1-x^2\right )}{128 (x-1)}+\frac {x^3+2 x^2+5 x+\left (-x^3-x^2+x+1\right ) \log \left (1-x^2\right )}{32 (x+1)}+\frac {(-3 x-1) \left (x^3+2 x^2+5 x+\left (-x^3-x^2+x+1\right ) \log \left (1-x^2\right )\right )}{128 \left (x^2+2 x+5\right )}+\frac {(1-x) \left (x^3+2 x^2+5 x+\left (-x^3-x^2+x+1\right ) \log \left (1-x^2\right )\right )}{16 \left (x^2+2 x+5\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 e^{3-e^{\log ^2(5)}} \left (\frac {1}{16} \arctan \left (\frac {x+1}{2}\right )+\frac {\text {arctanh}(x)}{24}-\frac {1}{128} x \log \left (1-x^2\right )+\frac {1}{384} (x+1)^3 \log \left (1-x^2\right )+\frac {1}{128} \left (1-x^2\right ) \log \left (1-x^2\right )-\frac {i \log \left (1-x^2\right )}{4 (-2 x-(2-4 i))}-\frac {i \log \left (1-x^2\right )}{4 (2 x+(2+4 i))}+\frac {3}{32} \log \left (1-x^2\right )+\frac {3}{32} \log \left (x^2+2 x+5\right )-\frac {1}{384} x^3 \log \left (1-x^2\right )-\frac {1}{12} \log (1-x)-\frac {1}{8} \log (x+1)-\left (\frac {3}{32}-\frac {i}{32}\right ) \log (x+(1-2 i))-\left (\frac {3}{32}+\frac {i}{32}\right ) \log (x+(1+2 i))\right )\) |
Input:
Int[(E^(3 - E^Log[5]^2)*(10*x + 4*x^2 + 2*x^3 + (2 + 2*x - 2*x^2 - 2*x^3)* Log[1 - x^2]))/(-25 - 20*x + 11*x^2 + 16*x^3 + 13*x^4 + 4*x^5 + x^6),x]
Output:
-2*E^(3 - E^Log[5]^2)*(ArcTan[(1 + x)/2]/16 + ArcTanh[x]/24 - Log[1 - x]/1 2 - Log[1 + x]/8 - (3/32 - I/32)*Log[(1 - 2*I) + x] - (3/32 + I/32)*Log[(1 + 2*I) + x] + (3*Log[1 - x^2])/32 - ((I/4)*Log[1 - x^2])/((-2 + 4*I) - 2* x) - (x*Log[1 - x^2])/128 - (x^3*Log[1 - x^2])/384 + ((1 + x)^3*Log[1 - x^ 2])/384 - ((I/4)*Log[1 - x^2])/((2 + 4*I) + 2*x) + ((1 - x^2)*Log[1 - x^2] )/128 + (3*Log[5 + 2*x + x^2])/32)
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[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.70 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {\ln \left (-x^{2}+1\right ) {\mathrm e}^{3-{\mathrm e}^{\ln \left (5\right )^{2}}}}{x^{2}+2 x +5}\) | \(30\) |
parallelrisch | \(\frac {\ln \left (-x^{2}+1\right ) {\mathrm e}^{3-{\mathrm e}^{\ln \left (5\right )^{2}}}}{x^{2}+2 x +5}\) | \(30\) |
norman | \(\frac {{\mathrm e}^{-{\mathrm e}^{\ln \left (5\right )^{2}}} {\mathrm e}^{3} \ln \left (-x^{2}+1\right )}{x^{2}+2 x +5}\) | \(32\) |
orering | \(-\frac {\left (5 x^{4}+6 x^{3}+2 x^{2}-2 x +5\right ) \left (\left (-2 x^{3}-2 x^{2}+2 x +2\right ) \ln \left (-x^{2}+1\right )+2 x^{3}+4 x^{2}+10 x \right ) {\mathrm e}^{3-{\mathrm e}^{\ln \left (5\right )^{2}}}}{2 \left (1+x \right ) \left (2 x^{2}-x +1\right ) \left (x^{6}+4 x^{5}+13 x^{4}+16 x^{3}+11 x^{2}-20 x -25\right )}-\frac {\left (x^{2}+2 x +5\right ) x \left (-1+x \right ) \left (\frac {\left (\left (-6 x^{2}-4 x +2\right ) \ln \left (-x^{2}+1\right )-\frac {2 \left (-2 x^{3}-2 x^{2}+2 x +2\right ) x}{-x^{2}+1}+6 x^{2}+8 x +10\right ) {\mathrm e}^{3-{\mathrm e}^{\ln \left (5\right )^{2}}}}{x^{6}+4 x^{5}+13 x^{4}+16 x^{3}+11 x^{2}-20 x -25}-\frac {\left (\left (-2 x^{3}-2 x^{2}+2 x +2\right ) \ln \left (-x^{2}+1\right )+2 x^{3}+4 x^{2}+10 x \right ) {\mathrm e}^{3-{\mathrm e}^{\ln \left (5\right )^{2}}} \left (6 x^{5}+20 x^{4}+52 x^{3}+48 x^{2}+22 x -20\right )}{\left (x^{6}+4 x^{5}+13 x^{4}+16 x^{3}+11 x^{2}-20 x -25\right )^{2}}\right )}{2 \left (2 x^{2}-x +1\right )}\) | \(348\) |
Input:
int(((-2*x^3-2*x^2+2*x+2)*ln(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+13*x^4+1 6*x^3+11*x^2-20*x-25)/exp(exp(ln(5)^2)-3),x,method=_RETURNVERBOSE)
Output:
ln(-x^2+1)/(x^2+2*x+5)*exp(3-exp(ln(5)^2))
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{\left (-e^{\left (\log \left (5\right )^{2}\right )} + 3\right )} \log \left (-x^{2} + 1\right )}{x^{2} + 2 \, x + 5} \] Input:
integrate(((-2*x^3-2*x^2+2*x+2)*log(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+1 3*x^4+16*x^3+11*x^2-20*x-25)/exp(exp(log(5)^2)-3),x, algorithm="fricas")
Output:
e^(-e^(log(5)^2) + 3)*log(-x^2 + 1)/(x^2 + 2*x + 5)
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{3} \log {\left (1 - x^{2} \right )}}{x^{2} e^{e^{\log {\left (5 \right )}^{2}}} + 2 x e^{e^{\log {\left (5 \right )}^{2}}} + 5 e^{e^{\log {\left (5 \right )}^{2}}}} \] Input:
integrate(((-2*x**3-2*x**2+2*x+2)*ln(-x**2+1)+2*x**3+4*x**2+10*x)/(x**6+4* x**5+13*x**4+16*x**3+11*x**2-20*x-25)/exp(exp(ln(5)**2)-3),x)
Output:
exp(3)*log(1 - x**2)/(x**2*exp(exp(log(5)**2)) + 2*x*exp(exp(log(5)**2)) + 5*exp(exp(log(5)**2)))
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (29) = 58\).
Time = 0.67 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.71 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=-\frac {1}{32} \, {\left (\frac {4 \, {\left (2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + {\left (x^{2} + 2 \, x - 3\right )} \log \left (-x + 1\right )\right )}}{x^{2} + 2 \, x + 5} + \frac {9 \, x + 35}{x^{2} + 2 \, x + 5} - \frac {2 \, {\left (7 \, x + 5\right )}}{x^{2} + 2 \, x + 5} + \frac {5 \, {\left (x - 5\right )}}{x^{2} + 2 \, x + 5} - 8 \, \log \left (x + 1\right ) - 4 \, \log \left (x - 1\right )\right )} e^{\left (-e^{\left (\log \left (5\right )^{2}\right )} + 3\right )} \] Input:
integrate(((-2*x^3-2*x^2+2*x+2)*log(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+1 3*x^4+16*x^3+11*x^2-20*x-25)/exp(exp(log(5)^2)-3),x, algorithm="maxima")
Output:
-1/32*(4*(2*(x^2 + 2*x + 1)*log(x + 1) + (x^2 + 2*x - 3)*log(-x + 1))/(x^2 + 2*x + 5) + (9*x + 35)/(x^2 + 2*x + 5) - 2*(7*x + 5)/(x^2 + 2*x + 5) + 5 *(x - 5)/(x^2 + 2*x + 5) - 8*log(x + 1) - 4*log(x - 1))*e^(-e^(log(5)^2) + 3)
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{\left (-e^{\left (\log \left (5\right )^{2}\right )} + 3\right )} \log \left (-x^{2} + 1\right )}{x^{2} + 2 \, x + 5} \] Input:
integrate(((-2*x^3-2*x^2+2*x+2)*log(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+1 3*x^4+16*x^3+11*x^2-20*x-25)/exp(exp(log(5)^2)-3),x, algorithm="giac")
Output:
e^(-e^(log(5)^2) + 3)*log(-x^2 + 1)/(x^2 + 2*x + 5)
Time = 2.67 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {\ln \left (1-x^2\right )\,{\mathrm {e}}^{3-{\mathrm {e}}^{{\ln \left (5\right )}^2}}}{x^2+2\,x+5} \] Input:
int((exp(3 - exp(log(5)^2))*(10*x + log(1 - x^2)*(2*x - 2*x^2 - 2*x^3 + 2) + 4*x^2 + 2*x^3))/(11*x^2 - 20*x + 16*x^3 + 13*x^4 + 4*x^5 + x^6 - 25),x)
Output:
(log(1 - x^2)*exp(3 - exp(log(5)^2)))/(2*x + x^2 + 5)
Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.97 \[ \int \frac {e^{3-e^{\log ^2(5)}} \left (10 x+4 x^2+2 x^3+\left (2+2 x-2 x^2-2 x^3\right ) \log \left (1-x^2\right )\right )}{-25-20 x+11 x^2+16 x^3+13 x^4+4 x^5+x^6} \, dx=\frac {e^{3} \left (-\mathrm {log}\left (-x^{2}+1\right ) x^{2}-2 \,\mathrm {log}\left (-x^{2}+1\right ) x +\mathrm {log}\left (x -1\right ) x^{2}+2 \,\mathrm {log}\left (x -1\right ) x +5 \,\mathrm {log}\left (x -1\right )+\mathrm {log}\left (x +1\right ) x^{2}+2 \,\mathrm {log}\left (x +1\right ) x +5 \,\mathrm {log}\left (x +1\right )\right )}{5 e^{e^{\mathrm {log}\left (5\right )^{2}}} \left (x^{2}+2 x +5\right )} \] Input:
int(((-2*x^3-2*x^2+2*x+2)*log(-x^2+1)+2*x^3+4*x^2+10*x)/(x^6+4*x^5+13*x^4+ 16*x^3+11*x^2-20*x-25)/exp(exp(log(5)^2)-3),x)
Output:
(e**3*( - log( - x**2 + 1)*x**2 - 2*log( - x**2 + 1)*x + log(x - 1)*x**2 + 2*log(x - 1)*x + 5*log(x - 1) + log(x + 1)*x**2 + 2*log(x + 1)*x + 5*log( x + 1)))/(5*e**(e**(log(5)**2))*(x**2 + 2*x + 5))