Integrand size = 149, antiderivative size = 29 \[ \int \frac {1985-2010 x+575 x^2-30 x^3-5 x^4+e^6 \left (125+5 x^2\right )+e^3 \left (1000-500 x+70 x^2\right )}{1-32 x+272 x^2+e^{12} x^2-258 x^3+96 x^4-16 x^5+x^6+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (-2 x+96 x^2-48 x^3+6 x^4\right )+e^3 \left (-16 x+260 x^2-192 x^3+48 x^4-4 x^5\right )} \, dx=\frac {25 \left (5-\frac {1}{5} x (3+x)\right )}{1-\left (4+e^3-x\right )^2 x} \]
Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {1985-2010 x+575 x^2-30 x^3-5 x^4+e^6 \left (125+5 x^2\right )+e^3 \left (1000-500 x+70 x^2\right )}{1-32 x+272 x^2+e^{12} x^2-258 x^3+96 x^4-16 x^5+x^6+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (-2 x+96 x^2-48 x^3+6 x^4\right )+e^3 \left (-16 x+260 x^2-192 x^3+48 x^4-4 x^5\right )} \, dx=\frac {5 \left (-25+3 x+x^2\right )}{-1+\left (4+e^3\right )^2 x-2 \left (4+e^3\right ) x^2+x^3} \]
Integrate[(1985 - 2010*x + 575*x^2 - 30*x^3 - 5*x^4 + E^6*(125 + 5*x^2) + E^3*(1000 - 500*x + 70*x^2))/(1 - 32*x + 272*x^2 + E^12*x^2 - 258*x^3 + 96 *x^4 - 16*x^5 + x^6 + E^9*(16*x^2 - 4*x^3) + E^6*(-2*x + 96*x^2 - 48*x^3 + 6*x^4) + E^3*(-16*x + 260*x^2 - 192*x^3 + 48*x^4 - 4*x^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.
| \(\displaystyle \int \frac {-5 x^4-30 x^3+575 x^2+e^6 \left (5 x^2+125\right )+e^3 \left (70 x^2-500 x+1000\right )-2010 x+1985}{x^6-16 x^5+96 x^4-258 x^3+e^{12} x^2+272 x^2+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (6 x^4-48 x^3+96 x^2-2 x\right )+e^3 \left (-4 x^5+48 x^4-192 x^3+260 x^2-16 x\right )-32 x+1} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {-5 x^4-30 x^3+575 x^2+e^6 \left (5 x^2+125\right )+e^3 \left (70 x^2-500 x+1000\right )-2010 x+1985}{x^6-16 x^5+96 x^4-258 x^3+\left (272+e^{12}\right ) x^2+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (6 x^4-48 x^3+96 x^2-2 x\right )+e^3 \left (-4 x^5+48 x^4-192 x^3+260 x^2-16 x\right )-32 x+1}dx\) |
| \(\Big \downarrow \) 2462 |
| \(\displaystyle \int \left (\frac {5 \left (x+2 e^3+14\right )}{-x^3+2 \left (4+e^3\right ) x^2-\left (4+e^3\right )^2 x+1}+\frac {5 \left (\left (19-22 e^3-2 e^6\right ) x^2-\left (179-44 e^3-30 e^6-2 e^9\right ) x+25 e^6+198 e^3+383\right )}{\left (-x^3+2 \left (4+e^3\right ) x^2-\left (4+e^3\right )^2 x+1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {5 \left (x+2 e^3+14\right )}{-x^3+2 \left (4+e^3\right ) x^2-\left (4+e^3\right )^2 x+1}+\frac {5 \left (\left (19-22 e^3-2 e^6\right ) x^2-\left (179-44 e^3-30 e^6-2 e^9\right ) x+25 e^6+198 e^3+383\right )}{\left (-x^3+2 \left (4+e^3\right ) x^2-\left (4+e^3\right )^2 x+1\right )^2}\right )dx\) |
Int[(1985 - 2010*x + 575*x^2 - 30*x^3 - 5*x^4 + E^6*(125 + 5*x^2) + E^3*(1 000 - 500*x + 70*x^2))/(1 - 32*x + 272*x^2 + E^12*x^2 - 258*x^3 + 96*x^4 - 16*x^5 + x^6 + E^9*(16*x^2 - 4*x^3) + E^6*(-2*x + 96*x^2 - 48*x^3 + 6*x^4 ) + E^3*(-16*x + 260*x^2 - 192*x^3 + 48*x^4 - 4*x^5)),x]
3.10.4.3.1 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[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48
| method | result | size |
| risch | \(\frac {5 x^{2}+15 x -125}{x \,{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{3}+8 x \,{\mathrm e}^{3}-8 x^{2}+16 x -1}\) | \(43\) |
| gosper | \(\frac {5 x^{2}+15 x -125}{x \,{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{3}+8 x \,{\mathrm e}^{3}-8 x^{2}+16 x -1}\) | \(44\) |
| norman | \(\frac {5 x^{2}+15 x -125}{x \,{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{3}+8 x \,{\mathrm e}^{3}-8 x^{2}+16 x -1}\) | \(45\) |
| parallelrisch | \(\frac {5 x^{2}+15 x -125}{x \,{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{3}+8 x \,{\mathrm e}^{3}-8 x^{2}+16 x -1}\) | \(45\) |
| default | \(-\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\textit {\_Z}^{6}+\left (-4 \,{\mathrm e}^{3}-16\right ) \textit {\_Z}^{5}+\left (48 \,{\mathrm e}^{3}+6 \,{\mathrm e}^{6}+96\right ) \textit {\_Z}^{4}+\left (-192 \,{\mathrm e}^{3}-4 \,{\mathrm e}^{9}-48 \,{\mathrm e}^{6}-258\right ) \textit {\_Z}^{3}+\left (260 \,{\mathrm e}^{3}+{\mathrm e}^{12}+16 \,{\mathrm e}^{9}+96 \,{\mathrm e}^{6}+272\right ) \textit {\_Z}^{2}+\left (-16 \,{\mathrm e}^{3}-2 \,{\mathrm e}^{6}-32\right ) \textit {\_Z} \right )}{\sum }\frac {\left (397-\textit {\_R}^{4}-6 \textit {\_R}^{3}+\left (14 \,{\mathrm e}^{3}+{\mathrm e}^{6}+115\right ) \textit {\_R}^{2}+2 \left (-201-50 \,{\mathrm e}^{3}\right ) \textit {\_R} +25 \,{\mathrm e}^{6}+200 \,{\mathrm e}^{3}\right ) \ln \left (x -\textit {\_R} \right )}{16-\textit {\_R} \,{\mathrm e}^{12}+6 \textit {\_R}^{2} {\mathrm e}^{9}-12 \textit {\_R}^{3} {\mathrm e}^{6}+10 \textit {\_R}^{4} {\mathrm e}^{3}-3 \textit {\_R}^{5}-16 \textit {\_R} \,{\mathrm e}^{9}+72 \textit {\_R}^{2} {\mathrm e}^{6}-96 \textit {\_R}^{3} {\mathrm e}^{3}+40 \textit {\_R}^{4}-96 \textit {\_R} \,{\mathrm e}^{6}+288 \textit {\_R}^{2} {\mathrm e}^{3}-192 \textit {\_R}^{3}+{\mathrm e}^{6}-260 \textit {\_R} \,{\mathrm e}^{3}+387 \textit {\_R}^{2}+8 \,{\mathrm e}^{3}-272 \textit {\_R}}\right )}{2}\) | \(229\) |
int(((5*x^2+125)*exp(3)^2+(70*x^2-500*x+1000)*exp(3)-5*x^4-30*x^3+575*x^2- 2010*x+1985)/(x^2*exp(3)^4+(-4*x^3+16*x^2)*exp(3)^3+(6*x^4-48*x^3+96*x^2-2 *x)*exp(3)^2+(-4*x^5+48*x^4-192*x^3+260*x^2-16*x)*exp(3)+x^6-16*x^5+96*x^4 -258*x^3+272*x^2-32*x+1),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {1985-2010 x+575 x^2-30 x^3-5 x^4+e^6 \left (125+5 x^2\right )+e^3 \left (1000-500 x+70 x^2\right )}{1-32 x+272 x^2+e^{12} x^2-258 x^3+96 x^4-16 x^5+x^6+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (-2 x+96 x^2-48 x^3+6 x^4\right )+e^3 \left (-16 x+260 x^2-192 x^3+48 x^4-4 x^5\right )} \, dx=\frac {5 \, {\left (x^{2} + 3 \, x - 25\right )}}{x^{3} - 8 \, x^{2} + x e^{6} - 2 \, {\left (x^{2} - 4 \, x\right )} e^{3} + 16 \, x - 1} \]
integrate(((5*x^2+125)*exp(3)^2+(70*x^2-500*x+1000)*exp(3)-5*x^4-30*x^3+57 5*x^2-2010*x+1985)/(x^2*exp(3)^4+(-4*x^3+16*x^2)*exp(3)^3+(6*x^4-48*x^3+96 *x^2-2*x)*exp(3)^2+(-4*x^5+48*x^4-192*x^3+260*x^2-16*x)*exp(3)+x^6-16*x^5+ 96*x^4-258*x^3+272*x^2-32*x+1),x, algorithm=\
Time = 1.36 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {1985-2010 x+575 x^2-30 x^3-5 x^4+e^6 \left (125+5 x^2\right )+e^3 \left (1000-500 x+70 x^2\right )}{1-32 x+272 x^2+e^{12} x^2-258 x^3+96 x^4-16 x^5+x^6+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (-2 x+96 x^2-48 x^3+6 x^4\right )+e^3 \left (-16 x+260 x^2-192 x^3+48 x^4-4 x^5\right )} \, dx=- \frac {- 5 x^{2} - 15 x + 125}{x^{3} + x^{2} \left (- 2 e^{3} - 8\right ) + x \left (16 + 8 e^{3} + e^{6}\right ) - 1} \]
integrate(((5*x**2+125)*exp(3)**2+(70*x**2-500*x+1000)*exp(3)-5*x**4-30*x* *3+575*x**2-2010*x+1985)/(x**2*exp(3)**4+(-4*x**3+16*x**2)*exp(3)**3+(6*x* *4-48*x**3+96*x**2-2*x)*exp(3)**2+(-4*x**5+48*x**4-192*x**3+260*x**2-16*x) *exp(3)+x**6-16*x**5+96*x**4-258*x**3+272*x**2-32*x+1),x)
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {1985-2010 x+575 x^2-30 x^3-5 x^4+e^6 \left (125+5 x^2\right )+e^3 \left (1000-500 x+70 x^2\right )}{1-32 x+272 x^2+e^{12} x^2-258 x^3+96 x^4-16 x^5+x^6+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (-2 x+96 x^2-48 x^3+6 x^4\right )+e^3 \left (-16 x+260 x^2-192 x^3+48 x^4-4 x^5\right )} \, dx=\frac {5 \, {\left (x^{2} + 3 \, x - 25\right )}}{x^{3} - 2 \, x^{2} {\left (e^{3} + 4\right )} + x {\left (e^{6} + 8 \, e^{3} + 16\right )} - 1} \]
integrate(((5*x^2+125)*exp(3)^2+(70*x^2-500*x+1000)*exp(3)-5*x^4-30*x^3+57 5*x^2-2010*x+1985)/(x^2*exp(3)^4+(-4*x^3+16*x^2)*exp(3)^3+(6*x^4-48*x^3+96 *x^2-2*x)*exp(3)^2+(-4*x^5+48*x^4-192*x^3+260*x^2-16*x)*exp(3)+x^6-16*x^5+ 96*x^4-258*x^3+272*x^2-32*x+1),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {1985-2010 x+575 x^2-30 x^3-5 x^4+e^6 \left (125+5 x^2\right )+e^3 \left (1000-500 x+70 x^2\right )}{1-32 x+272 x^2+e^{12} x^2-258 x^3+96 x^4-16 x^5+x^6+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (-2 x+96 x^2-48 x^3+6 x^4\right )+e^3 \left (-16 x+260 x^2-192 x^3+48 x^4-4 x^5\right )} \, dx=\frac {5 \, {\left (x^{2} + 3 \, x - 25\right )}}{x^{3} - 2 \, x^{2} e^{3} - 8 \, x^{2} + x e^{6} + 8 \, x e^{3} + 16 \, x - 1} \]
integrate(((5*x^2+125)*exp(3)^2+(70*x^2-500*x+1000)*exp(3)-5*x^4-30*x^3+57 5*x^2-2010*x+1985)/(x^2*exp(3)^4+(-4*x^3+16*x^2)*exp(3)^3+(6*x^4-48*x^3+96 *x^2-2*x)*exp(3)^2+(-4*x^5+48*x^4-192*x^3+260*x^2-16*x)*exp(3)+x^6-16*x^5+ 96*x^4-258*x^3+272*x^2-32*x+1),x, algorithm=\
Time = 8.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {1985-2010 x+575 x^2-30 x^3-5 x^4+e^6 \left (125+5 x^2\right )+e^3 \left (1000-500 x+70 x^2\right )}{1-32 x+272 x^2+e^{12} x^2-258 x^3+96 x^4-16 x^5+x^6+e^9 \left (16 x^2-4 x^3\right )+e^6 \left (-2 x+96 x^2-48 x^3+6 x^4\right )+e^3 \left (-16 x+260 x^2-192 x^3+48 x^4-4 x^5\right )} \, dx=-\frac {5\,x^2+15\,x-125}{-x^3+\left (2\,{\mathrm {e}}^3+8\right )\,x^2+\left (-8\,{\mathrm {e}}^3-{\mathrm {e}}^6-16\right )\,x+1} \]
int((exp(3)*(70*x^2 - 500*x + 1000) - 2010*x + exp(6)*(5*x^2 + 125) + 575* x^2 - 30*x^3 - 5*x^4 + 1985)/(exp(9)*(16*x^2 - 4*x^3) - 32*x + x^2*exp(12) - exp(6)*(2*x - 96*x^2 + 48*x^3 - 6*x^4) - exp(3)*(16*x - 260*x^2 + 192*x ^3 - 48*x^4 + 4*x^5) + 272*x^2 - 258*x^3 + 96*x^4 - 16*x^5 + x^6 + 1),x)