Integrand size = 99, antiderivative size = 27 \[ \int \frac {e^x \left (80 x^2+85 x^3+30 x^4+e^4 \left (75+75 x+25 x^2\right )+e^2 \left (15-165 x-165 x^2-55 x^3\right )\right )}{9 x^2+6 x^3+x^4+e^4 \left (9+6 x+x^2\right )+e^2 \left (-18 x-12 x^2-2 x^3\right )} \, dx=\frac {5 e^x \left (5-\frac {-1+x}{e^2-x}\right ) x}{3+x} \]
Time = 3.45 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^x \left (80 x^2+85 x^3+30 x^4+e^4 \left (75+75 x+25 x^2\right )+e^2 \left (15-165 x-165 x^2-55 x^3\right )\right )}{9 x^2+6 x^3+x^4+e^4 \left (9+6 x+x^2\right )+e^2 \left (-18 x-12 x^2-2 x^3\right )} \, dx=\frac {5 e^x \left (1+5 e^2-6 x\right ) x}{\left (e^2-x\right ) (3+x)} \]
Integrate[(E^x*(80*x^2 + 85*x^3 + 30*x^4 + E^4*(75 + 75*x + 25*x^2) + E^2* (15 - 165*x - 165*x^2 - 55*x^3)))/(9*x^2 + 6*x^3 + x^4 + E^4*(9 + 6*x + x^ 2) + E^2*(-18*x - 12*x^2 - 2*x^3)),x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.23 (sec) , antiderivative size = 618, normalized size of antiderivative = 22.89, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {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.
\(\displaystyle \int \frac {e^x \left (30 x^4+85 x^3+80 x^2+e^4 \left (25 x^2+75 x+75\right )+e^2 \left (-55 x^3-165 x^2-165 x+15\right )\right )}{x^4+6 x^3+9 x^2+e^4 \left (x^2+6 x+9\right )+e^2 \left (-2 x^3-12 x^2-18 x\right )} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {2 e^x \left (30 x^4+85 x^3+80 x^2+e^4 \left (25 x^2+75 x+75\right )+e^2 \left (-55 x^3-165 x^2-165 x+15\right )\right )}{\left (3+e^2\right )^3 \left (e^2-x\right )}+\frac {2 e^x \left (30 x^4+85 x^3+80 x^2+e^4 \left (25 x^2+75 x+75\right )+e^2 \left (-55 x^3-165 x^2-165 x+15\right )\right )}{\left (3+e^2\right )^3 (x+3)}+\frac {e^x \left (30 x^4+85 x^3+80 x^2+e^4 \left (25 x^2+75 x+75\right )+e^2 \left (-55 x^3-165 x^2-165 x+15\right )\right )}{\left (3+e^2\right )^2 \left (e^2-x\right )^2}+\frac {e^x \left (30 x^4+85 x^3+80 x^2+e^4 \left (25 x^2+75 x+75\right )+e^2 \left (-55 x^3-165 x^2-165 x+15\right )\right )}{\left (3+e^2\right )^2 (x+3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {15 \left (95+44 e^2+5 e^4\right ) \operatorname {ExpIntegralEi}(x+3)}{e^3 \left (3+e^2\right )^2}+\frac {15 \left (19+5 e^2\right ) \operatorname {ExpIntegralEi}(x+3)}{e^3 \left (3+e^2\right )}+\frac {30 \left (19+5 e^2\right ) \operatorname {ExpIntegralEi}(x+3)}{e^3 \left (3+e^2\right )^2}-\frac {5 e^{2+e^2} \left (1-e^4\right ) \operatorname {ExpIntegralEi}\left (x-e^2\right )}{\left (3+e^2\right )^2}+\frac {5 e^{2+e^2} \left (1-e^2\right ) \operatorname {ExpIntegralEi}\left (x-e^2\right )}{3+e^2}-\frac {10 e^{2+e^2} \left (1-e^2\right ) \operatorname {ExpIntegralEi}\left (x-e^2\right )}{\left (3+e^2\right )^2}-\frac {10 \left (1+11 e^2\right ) e^x x^2}{\left (3+e^2\right )^3}+\frac {60 e^x x^2}{\left (3+e^2\right )^2}-\frac {10 \left (17-5 e^2\right ) e^x x^2}{\left (3+e^2\right )^3}+\frac {10 \left (19+5 e^4\right ) e^x x}{\left (3+e^2\right )^3}-\frac {5 \left (19+11 e^2\right ) e^x x}{\left (3+e^2\right )^2}+\frac {20 \left (1+11 e^2\right ) e^x x}{\left (3+e^2\right )^3}+\frac {5 \left (17+e^2\right ) e^x x}{\left (3+e^2\right )^2}-\frac {120 e^x x}{\left (3+e^2\right )^2}-\frac {160 \left (1-e^2\right ) e^x x}{\left (3+e^2\right )^3}+\frac {20 \left (17-5 e^2\right ) e^x x}{\left (3+e^2\right )^3}+\frac {5 \left (1-e^2\right ) e^{x+2}}{\left (3+e^2\right ) \left (e^2-x\right )}-\frac {15 \left (19+5 e^2\right ) e^x}{\left (3+e^2\right ) (x+3)}+\frac {5 \left (76+33 e^2+5 e^4\right ) e^x}{\left (3+e^2\right )^2}-\frac {10 \left (19+5 e^4\right ) e^x}{\left (3+e^2\right )^3}+\frac {5 \left (16+e^2+e^4\right ) e^x}{\left (3+e^2\right )^2}+\frac {5 \left (19+11 e^2\right ) e^x}{\left (3+e^2\right )^2}-\frac {30 \left (19+11 e^2\right ) e^x}{\left (3+e^2\right )^3}-\frac {20 \left (1+11 e^2\right ) e^x}{\left (3+e^2\right )^3}-\frac {5 \left (17+e^2\right ) e^x}{\left (3+e^2\right )^2}+\frac {10 \left (17+e^2\right ) e^{x+2}}{\left (3+e^2\right )^3}+\frac {120 e^x}{\left (3+e^2\right )^2}+\frac {160 \left (1-e^2\right ) e^x}{\left (3+e^2\right )^3}-\frac {20 \left (17-5 e^2\right ) e^x}{\left (3+e^2\right )^3}\) |
Int[(E^x*(80*x^2 + 85*x^3 + 30*x^4 + E^4*(75 + 75*x + 25*x^2) + E^2*(15 - 165*x - 165*x^2 - 55*x^3)))/(9*x^2 + 6*x^3 + x^4 + E^4*(9 + 6*x + x^2) + E ^2*(-18*x - 12*x^2 - 2*x^3)),x]
(-20*E^x*(17 - 5*E^2))/(3 + E^2)^3 + (160*E^x*(1 - E^2))/(3 + E^2)^3 + (12 0*E^x)/(3 + E^2)^2 + (10*E^(2 + x)*(17 + E^2))/(3 + E^2)^3 - (5*E^x*(17 + E^2))/(3 + E^2)^2 - (20*E^x*(1 + 11*E^2))/(3 + E^2)^3 - (30*E^x*(19 + 11*E ^2))/(3 + E^2)^3 + (5*E^x*(19 + 11*E^2))/(3 + E^2)^2 + (5*E^x*(16 + E^2 + E^4))/(3 + E^2)^2 - (10*E^x*(19 + 5*E^4))/(3 + E^2)^3 + (5*E^x*(76 + 33*E^ 2 + 5*E^4))/(3 + E^2)^2 + (5*E^(2 + x)*(1 - E^2))/((3 + E^2)*(E^2 - x)) + (20*E^x*(17 - 5*E^2)*x)/(3 + E^2)^3 - (160*E^x*(1 - E^2)*x)/(3 + E^2)^3 - (120*E^x*x)/(3 + E^2)^2 + (5*E^x*(17 + E^2)*x)/(3 + E^2)^2 + (20*E^x*(1 + 11*E^2)*x)/(3 + E^2)^3 - (5*E^x*(19 + 11*E^2)*x)/(3 + E^2)^2 + (10*E^x*(19 + 5*E^4)*x)/(3 + E^2)^3 - (10*E^x*(17 - 5*E^2)*x^2)/(3 + E^2)^3 + (60*E^x *x^2)/(3 + E^2)^2 - (10*E^x*(1 + 11*E^2)*x^2)/(3 + E^2)^3 - (15*E^x*(19 + 5*E^2))/((3 + E^2)*(3 + x)) + (30*(19 + 5*E^2)*ExpIntegralEi[3 + x])/(E^3* (3 + E^2)^2) + (15*(19 + 5*E^2)*ExpIntegralEi[3 + x])/(E^3*(3 + E^2)) - (1 5*(95 + 44*E^2 + 5*E^4)*ExpIntegralEi[3 + x])/(E^3*(3 + E^2)^2) - (10*E^(2 + E^2)*(1 - E^2)*ExpIntegralEi[-E^2 + x])/(3 + E^2)^2 + (5*E^(2 + E^2)*(1 - E^2)*ExpIntegralEi[-E^2 + x])/(3 + E^2) - (5*E^(2 + E^2)*(1 - E^4)*ExpI ntegralEi[-E^2 + x])/(3 + E^2)^2
3.7.9.3.1 Defintions of rubi rules used
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 = 1.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22
method | result | size |
norman | \(\frac {\left (5+25 \,{\mathrm e}^{2}\right ) x \,{\mathrm e}^{x}-30 \,{\mathrm e}^{x} x^{2}}{\left (3+x \right ) \left ({\mathrm e}^{2}-x \right )}\) | \(33\) |
gosper | \(\frac {5 x \left (-6 x +5 \,{\mathrm e}^{2}+1\right ) {\mathrm e}^{x}}{{\mathrm e}^{2} x -x^{2}+3 \,{\mathrm e}^{2}-3 x}\) | \(34\) |
parallelrisch | \(\frac {25 x \,{\mathrm e}^{2} {\mathrm e}^{x}-30 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} x}{{\mathrm e}^{2} x -x^{2}+3 \,{\mathrm e}^{2}-3 x}\) | \(41\) |
default | \(\text {Expression too large to display}\) | \(1338\) |
int(((25*x^2+75*x+75)*exp(2)^2+(-55*x^3-165*x^2-165*x+15)*exp(2)+30*x^4+85 *x^3+80*x^2)*exp(x)/((x^2+6*x+9)*exp(2)^2+(-2*x^3-12*x^2-18*x)*exp(2)+x^4+ 6*x^3+9*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^x \left (80 x^2+85 x^3+30 x^4+e^4 \left (75+75 x+25 x^2\right )+e^2 \left (15-165 x-165 x^2-55 x^3\right )\right )}{9 x^2+6 x^3+x^4+e^4 \left (9+6 x+x^2\right )+e^2 \left (-18 x-12 x^2-2 x^3\right )} \, dx=\frac {5 \, {\left (6 \, x^{2} - 5 \, x e^{2} - x\right )} e^{x}}{x^{2} - {\left (x + 3\right )} e^{2} + 3 \, x} \]
integrate(((25*x^2+75*x+75)*exp(2)^2+(-55*x^3-165*x^2-165*x+15)*exp(2)+30* x^4+85*x^3+80*x^2)*exp(x)/((x^2+6*x+9)*exp(2)^2+(-2*x^3-12*x^2-18*x)*exp(2 )+x^4+6*x^3+9*x^2),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^x \left (80 x^2+85 x^3+30 x^4+e^4 \left (75+75 x+25 x^2\right )+e^2 \left (15-165 x-165 x^2-55 x^3\right )\right )}{9 x^2+6 x^3+x^4+e^4 \left (9+6 x+x^2\right )+e^2 \left (-18 x-12 x^2-2 x^3\right )} \, dx=\frac {\left (30 x^{2} - 25 x e^{2} - 5 x\right ) e^{x}}{x^{2} - x e^{2} + 3 x - 3 e^{2}} \]
integrate(((25*x**2+75*x+75)*exp(2)**2+(-55*x**3-165*x**2-165*x+15)*exp(2) +30*x**4+85*x**3+80*x**2)*exp(x)/((x**2+6*x+9)*exp(2)**2+(-2*x**3-12*x**2- 18*x)*exp(2)+x**4+6*x**3+9*x**2),x)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^x \left (80 x^2+85 x^3+30 x^4+e^4 \left (75+75 x+25 x^2\right )+e^2 \left (15-165 x-165 x^2-55 x^3\right )\right )}{9 x^2+6 x^3+x^4+e^4 \left (9+6 x+x^2\right )+e^2 \left (-18 x-12 x^2-2 x^3\right )} \, dx=\frac {5 \, {\left (6 \, x^{2} - x {\left (5 \, e^{2} + 1\right )}\right )} e^{x}}{x^{2} - x {\left (e^{2} - 3\right )} - 3 \, e^{2}} \]
integrate(((25*x^2+75*x+75)*exp(2)^2+(-55*x^3-165*x^2-165*x+15)*exp(2)+30* x^4+85*x^3+80*x^2)*exp(x)/((x^2+6*x+9)*exp(2)^2+(-2*x^3-12*x^2-18*x)*exp(2 )+x^4+6*x^3+9*x^2),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {e^x \left (80 x^2+85 x^3+30 x^4+e^4 \left (75+75 x+25 x^2\right )+e^2 \left (15-165 x-165 x^2-55 x^3\right )\right )}{9 x^2+6 x^3+x^4+e^4 \left (9+6 x+x^2\right )+e^2 \left (-18 x-12 x^2-2 x^3\right )} \, dx=\frac {5 \, {\left (6 \, x^{2} e^{x} - 5 \, x e^{\left (x + 2\right )} - x e^{x}\right )}}{x^{2} - x e^{2} + 3 \, x - 3 \, e^{2}} \]
integrate(((25*x^2+75*x+75)*exp(2)^2+(-55*x^3-165*x^2-165*x+15)*exp(2)+30* x^4+85*x^3+80*x^2)*exp(x)/((x^2+6*x+9)*exp(2)^2+(-2*x^3-12*x^2-18*x)*exp(2 )+x^4+6*x^3+9*x^2),x, algorithm=\
Time = 9.63 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^x \left (80 x^2+85 x^3+30 x^4+e^4 \left (75+75 x+25 x^2\right )+e^2 \left (15-165 x-165 x^2-55 x^3\right )\right )}{9 x^2+6 x^3+x^4+e^4 \left (9+6 x+x^2\right )+e^2 \left (-18 x-12 x^2-2 x^3\right )} \, dx=-\frac {{\mathrm {e}}^x\,\left (30\,x^2-x\,\left (25\,{\mathrm {e}}^2+5\right )\right )}{-x^2+\left ({\mathrm {e}}^2-3\right )\,x+3\,{\mathrm {e}}^2} \]