Integrand size = 131, antiderivative size = 32 \[ \int \frac {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx=1+\frac {x}{e^3 \left (7+\frac {x-x^2+\log \left (5-x^2\right )}{5+x}\right )} \]
Time = 0.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx=\frac {x (5+x)}{e^3 \left (35+8 x-x^2+\log \left (5-x^2\right )\right )} \]
Integrate[(-875 - 350*x + 100*x^2 + 68*x^3 + 13*x^4 + (-25 - 10*x + 5*x^2 + 2*x^3)*Log[5 - x^2])/(E^3*(-6125 - 2800*x + 1255*x^2 + 640*x^3 - 11*x^4 - 16*x^5 + x^6) + E^3*(-350 - 80*x + 80*x^2 + 16*x^3 - 2*x^4)*Log[5 - x^2] + E^3*(-5 + x^2)*Log[5 - x^2]^2),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.
\(\displaystyle \int \frac {13 x^4+68 x^3+100 x^2+\left (2 x^3+5 x^2-10 x-25\right ) \log \left (5-x^2\right )-350 x-875}{e^3 \left (x^2-5\right ) \log ^2\left (5-x^2\right )+e^3 \left (-2 x^4+16 x^3+80 x^2-80 x-350\right ) \log \left (5-x^2\right )+e^3 \left (x^6-16 x^5-11 x^4+640 x^3+1255 x^2-2800 x-6125\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-13 x^4-68 x^3-100 x^2-\left (2 x^3+5 x^2-10 x-25\right ) \log \left (5-x^2\right )+350 x+875}{e^3 \left (5-x^2\right ) \left (-x^2+\log \left (5-x^2\right )+8 x+35\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-13 x^4-68 x^3-100 x^2+350 x+\left (-2 x^3-5 x^2+10 x+25\right ) \log \left (5-x^2\right )+875}{\left (5-x^2\right ) \left (-x^2+8 x+\log \left (5-x^2\right )+35\right )^2}dx}{e^3}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {\int \left (\frac {-2 x-5}{x^2-8 x-\log \left (5-x^2\right )-35}+\frac {2 x \left (x^4+x^3-26 x^2-10 x+100\right )}{\left (x^2-5\right ) \left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}\right )dx}{e^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-10 \int \frac {1}{\left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx+5 \sqrt {5} \int \frac {1}{\left (\sqrt {5}-x\right ) \left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx+5 \int \frac {1}{\left (\sqrt {5}-x\right ) \left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx-42 \int \frac {x}{\left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx+2 \int \frac {x^2}{\left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx+5 \sqrt {5} \int \frac {1}{\left (x+\sqrt {5}\right ) \left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx-5 \int \frac {1}{\left (x+\sqrt {5}\right ) \left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx-5 \int \frac {1}{x^2-8 x-\log \left (5-x^2\right )-35}dx-2 \int \frac {x}{x^2-8 x-\log \left (5-x^2\right )-35}dx+2 \int \frac {x^3}{\left (x^2-8 x-\log \left (5-x^2\right )-35\right )^2}dx}{e^3}\) |
Int[(-875 - 350*x + 100*x^2 + 68*x^3 + 13*x^4 + (-25 - 10*x + 5*x^2 + 2*x^ 3)*Log[5 - x^2])/(E^3*(-6125 - 2800*x + 1255*x^2 + 640*x^3 - 11*x^4 - 16*x ^5 + x^6) + E^3*(-350 - 80*x + 80*x^2 + 16*x^3 - 2*x^4)*Log[5 - x^2] + E^3 *(-5 + x^2)*Log[5 - x^2]^2),x]
3.17.40.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.61 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {\left (5+x \right ) x \,{\mathrm e}^{-3}}{x^{2}-8 x -\ln \left (-x^{2}+5\right )-35}\) | \(29\) |
parallelrisch | \(\frac {\left (-x^{2}-5 x \right ) {\mathrm e}^{-3}}{x^{2}-8 x -\ln \left (-x^{2}+5\right )-35}\) | \(35\) |
norman | \(\frac {-13 \,{\mathrm e}^{-3} x -{\mathrm e}^{-3} \ln \left (-x^{2}+5\right )-35 \,{\mathrm e}^{-3}}{x^{2}-8 x -\ln \left (-x^{2}+5\right )-35}\) | \(50\) |
int(((2*x^3+5*x^2-10*x-25)*ln(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-875)/((x ^2-5)*exp(3)*ln(-x^2+5)^2+(-2*x^4+16*x^3+80*x^2-80*x-350)*exp(3)*ln(-x^2+5 )+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x,method=_RETUR NVERBOSE)
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx=-\frac {x^{2} + 5 \, x}{{\left (x^{2} - 8 \, x - 35\right )} e^{3} - e^{3} \log \left (-x^{2} + 5\right )} \]
integrate(((2*x^3+5*x^2-10*x-25)*log(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-8 75)/((x^2-5)*exp(3)*log(-x^2+5)^2+(-2*x^4+16*x^3+80*x^2-80*x-350)*exp(3)*l og(-x^2+5)+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x, alg orithm=\
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx=\frac {x^{2} + 5 x}{- x^{2} e^{3} + 8 x e^{3} + e^{3} \log {\left (5 - x^{2} \right )} + 35 e^{3}} \]
integrate(((2*x**3+5*x**2-10*x-25)*ln(-x**2+5)+13*x**4+68*x**3+100*x**2-35 0*x-875)/((x**2-5)*exp(3)*ln(-x**2+5)**2+(-2*x**4+16*x**3+80*x**2-80*x-350 )*exp(3)*ln(-x**2+5)+(x**6-16*x**5-11*x**4+640*x**3+1255*x**2-2800*x-6125) *exp(3)),x)
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx=-\frac {x^{2} + 5 \, x}{x^{2} e^{3} - 8 \, x e^{3} - e^{3} \log \left (-x^{2} + 5\right ) - 35 \, e^{3}} \]
integrate(((2*x^3+5*x^2-10*x-25)*log(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-8 75)/((x^2-5)*exp(3)*log(-x^2+5)^2+(-2*x^4+16*x^3+80*x^2-80*x-350)*exp(3)*l og(-x^2+5)+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x, alg orithm=\
Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx=-\frac {x^{2} + 5 \, x}{x^{2} e^{3} - 8 \, x e^{3} - e^{3} \log \left (-x^{2} + 5\right ) - 35 \, e^{3}} \]
integrate(((2*x^3+5*x^2-10*x-25)*log(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-8 75)/((x^2-5)*exp(3)*log(-x^2+5)^2+(-2*x^4+16*x^3+80*x^2-80*x-350)*exp(3)*l og(-x^2+5)+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x, alg orithm=\
Time = 10.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx=\frac {{\mathrm {e}}^{-3}\,\left (13\,x+\ln \left (5-x^2\right )+35\right )}{8\,x+\ln \left (5-x^2\right )-x^2+35} \]