Integrand size = 106, antiderivative size = 30 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {e^4}{x \left (1-e^6+5 x+\frac {16}{-\frac {4}{3}+2 x}\right )} \] Output:
exp(4)/(5*x-exp(3)^2+16/(2*x-4/3)+1)/x
Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {e^4 (-2+3 x)}{x \left (22+e^6 (2-3 x)-7 x+15 x^2\right )} \] Input:
Integrate[(E^4*(44 - 28*x + 111*x^2 - 90*x^3 + E^6*(4 - 12*x + 9*x^2)))/(4 84*x^2 - 308*x^3 + 709*x^4 - 210*x^5 + 225*x^6 + E^12*(4*x^2 - 12*x^3 + 9* x^4) + E^6*(88*x^2 - 160*x^3 + 102*x^4 - 90*x^5)),x]
Output:
(E^4*(-2 + 3*x))/(x*(22 + E^6*(2 - 3*x) - 7*x + 15*x^2))
Time = 0.58 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {27, 2026, 2462, 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.
\(\displaystyle \int \frac {e^4 \left (-90 x^3+111 x^2+e^6 \left (9 x^2-12 x+4\right )-28 x+44\right )}{225 x^6-210 x^5+709 x^4-308 x^3+484 x^2+e^{12} \left (9 x^4-12 x^3+4 x^2\right )+e^6 \left (-90 x^5+102 x^4-160 x^3+88 x^2\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^4 \int \frac {-90 x^3+111 x^2-28 x+e^6 \left (9 x^2-12 x+4\right )+44}{225 x^6-210 x^5+709 x^4-308 x^3+484 x^2+e^{12} \left (9 x^4-12 x^3+4 x^2\right )+2 e^6 \left (-45 x^5+51 x^4-80 x^3+44 x^2\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle e^4 \int \frac {-90 x^3+111 x^2-28 x+e^6 \left (9 x^2-12 x+4\right )+44}{x^2 \left (225 x^4-30 \left (7+3 e^6\right ) x^3+\left (709+102 e^6+9 e^{12}\right ) x^2-4 \left (77+40 e^6+3 e^{12}\right ) x+4 \left (11+e^6\right )^2\right )}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle e^4 \int \left (\frac {2 \left (421+69 e^6\right )-15 \left (59+3 e^6\right ) x}{\left (11+e^6\right ) \left (15 x^2-\left (7+3 e^6\right ) x+2 \left (11+e^6\right )\right )^2}+\frac {15}{\left (-11-e^6\right ) \left (15 x^2-\left (7+3 e^6\right ) x+2 \left (11+e^6\right )\right )}+\frac {1}{\left (11+e^6\right ) x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^4 \left (\frac {15 x+26}{\left (11+e^6\right ) \left (15 x^2-\left (7+3 e^6\right ) x+2 \left (11+e^6\right )\right )}-\frac {1}{\left (11+e^6\right ) x}\right )\) |
Input:
Int[(E^4*(44 - 28*x + 111*x^2 - 90*x^3 + E^6*(4 - 12*x + 9*x^2)))/(484*x^2 - 308*x^3 + 709*x^4 - 210*x^5 + 225*x^6 + E^12*(4*x^2 - 12*x^3 + 9*x^4) + E^6*(88*x^2 - 160*x^3 + 102*x^4 - 90*x^5)),x]
Output:
E^4*(-(1/((11 + E^6)*x)) + (26 + 15*x)/((11 + E^6)*(2*(11 + E^6) - (7 + 3* E^6)*x + 15*x^2)))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, 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 = 0.51 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {3 \,{\mathrm e}^{4} \left (\frac {2}{3}-x \right )}{\left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right ) x}\) | \(34\) |
gosper | \(-\frac {\left (-2+3 x \right ) {\mathrm e}^{4}}{x \left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right )}\) | \(38\) |
parallelrisch | \(-\frac {{\mathrm e}^{4} \left (-30+45 x \right )}{15 x \left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right )}\) | \(38\) |
norman | \(\frac {-3 x \,{\mathrm e}^{4}+2 \,{\mathrm e}^{4}}{x \left (3 x \,{\mathrm e}^{6}-2 \,{\mathrm e}^{6}-15 x^{2}+7 x -22\right )}\) | \(40\) |
default | \({\mathrm e}^{4} \left (-\frac {{\mathrm e}^{18}+33 \,{\mathrm e}^{12}+363 \,{\mathrm e}^{6}+1331}{\left ({\mathrm e}^{12}+22 \,{\mathrm e}^{6}+121\right )^{2} x}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (225 \textit {\_Z}^{4}+\left (-90 \,{\mathrm e}^{6}-210\right ) \textit {\_Z}^{3}+\left (9 \,{\mathrm e}^{12}+102 \,{\mathrm e}^{6}+709\right ) \textit {\_Z}^{2}+\left (-12 \,{\mathrm e}^{12}-160 \,{\mathrm e}^{6}-308\right ) \textit {\_Z} +4 \,{\mathrm e}^{12}+88 \,{\mathrm e}^{6}+484\right )}{\sum }\frac {\left (681472+225 \left (-{\mathrm e}^{18}-33 \,{\mathrm e}^{12}-363 \,{\mathrm e}^{6}-1331\right ) \textit {\_R}^{2}+780 \left (-{\mathrm e}^{18}-33 \,{\mathrm e}^{12}-363 \,{\mathrm e}^{6}-1331\right ) \textit {\_R} +56100 \,{\mathrm e}^{12}+4076 \,{\mathrm e}^{18}+108 \,{\mathrm e}^{24}+329604 \,{\mathrm e}^{6}\right ) \ln \left (x -\textit {\_R} \right )}{-154+9 \textit {\_R} \,{\mathrm e}^{12}-6 \,{\mathrm e}^{12}-135 \textit {\_R}^{2} {\mathrm e}^{6}+102 \textit {\_R} \,{\mathrm e}^{6}+450 \textit {\_R}^{3}-80 \,{\mathrm e}^{6}-315 \textit {\_R}^{2}+709 \textit {\_R}}}{2 \left ({\mathrm e}^{12}+22 \,{\mathrm e}^{6}+121\right )^{2}}\right )\) | \(202\) |
Input:
int(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4-12*x^3 +4*x^2)*exp(3)^4+(-90*x^5+102*x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-210*x^5 +709*x^4-308*x^3+484*x^2),x,method=_RETURNVERBOSE)
Output:
3*exp(4)*(2/3-x)/(3*x*exp(6)-2*exp(6)-15*x^2+7*x-22)/x
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {{\left (3 \, x - 2\right )} e^{4}}{15 \, x^{3} - 7 \, x^{2} - {\left (3 \, x^{2} - 2 \, x\right )} e^{6} + 22 \, x} \] Input:
integrate(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4- 12*x^3+4*x^2)*exp(3)^4+(-90*x^5+102*x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-2 10*x^5+709*x^4-308*x^3+484*x^2),x, algorithm="fricas")
Output:
(3*x - 2)*e^4/(15*x^3 - 7*x^2 - (3*x^2 - 2*x)*e^6 + 22*x)
Time = 1.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=- \frac {- 3 x e^{4} + 2 e^{4}}{15 x^{3} + x^{2} \left (- 3 e^{6} - 7\right ) + x \left (22 + 2 e^{6}\right )} \] Input:
integrate(((9*x**2-12*x+4)*exp(3)**2-90*x**3+111*x**2-28*x+44)*exp(4)/((9* x**4-12*x**3+4*x**2)*exp(3)**4+(-90*x**5+102*x**4-160*x**3+88*x**2)*exp(3) **2+225*x**6-210*x**5+709*x**4-308*x**3+484*x**2),x)
Output:
-(-3*x*exp(4) + 2*exp(4))/(15*x**3 + x**2*(-3*exp(6) - 7) + x*(22 + 2*exp( 6)))
Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {{\left (3 \, x - 2\right )} e^{4}}{15 \, x^{3} - x^{2} {\left (3 \, e^{6} + 7\right )} + 2 \, x {\left (e^{6} + 11\right )}} \] Input:
integrate(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4- 12*x^3+4*x^2)*exp(3)^4+(-90*x^5+102*x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-2 10*x^5+709*x^4-308*x^3+484*x^2),x, algorithm="maxima")
Output:
(3*x - 2)*e^4/(15*x^3 - x^2*(3*e^6 + 7) + 2*x*(e^6 + 11))
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {{\left (3 \, x - 2\right )} e^{4}}{15 \, x^{3} - 3 \, x^{2} e^{6} - 7 \, x^{2} + 2 \, x e^{6} + 22 \, x} \] Input:
integrate(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4- 12*x^3+4*x^2)*exp(3)^4+(-90*x^5+102*x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-2 10*x^5+709*x^4-308*x^3+484*x^2),x, algorithm="giac")
Output:
(3*x - 2)*e^4/(15*x^3 - 3*x^2*e^6 - 7*x^2 + 2*x*e^6 + 22*x)
Time = 2.81 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=-\frac {2\,{\mathrm {e}}^4-3\,x\,{\mathrm {e}}^4}{15\,x^3+\left (-3\,{\mathrm {e}}^6-7\right )\,x^2+\left (2\,{\mathrm {e}}^6+22\right )\,x} \] Input:
int((exp(4)*(exp(6)*(9*x^2 - 12*x + 4) - 28*x + 111*x^2 - 90*x^3 + 44))/(e xp(12)*(4*x^2 - 12*x^3 + 9*x^4) + 484*x^2 - 308*x^3 + 709*x^4 - 210*x^5 + 225*x^6 + exp(6)*(88*x^2 - 160*x^3 + 102*x^4 - 90*x^5)),x)
Output:
-(2*exp(4) - 3*x*exp(4))/(15*x^3 - x^2*(3*exp(6) + 7) + x*(2*exp(6) + 22))
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {e^4 \left (44-28 x+111 x^2-90 x^3+e^6 \left (4-12 x+9 x^2\right )\right )}{484 x^2-308 x^3+709 x^4-210 x^5+225 x^6+e^{12} \left (4 x^2-12 x^3+9 x^4\right )+e^6 \left (88 x^2-160 x^3+102 x^4-90 x^5\right )} \, dx=\frac {e^{4} \left (-3 x +2\right )}{x \left (3 e^{6} x -2 e^{6}-15 x^{2}+7 x -22\right )} \] Input:
int(((9*x^2-12*x+4)*exp(3)^2-90*x^3+111*x^2-28*x+44)*exp(4)/((9*x^4-12*x^3 +4*x^2)*exp(3)^4+(-90*x^5+102*x^4-160*x^3+88*x^2)*exp(3)^2+225*x^6-210*x^5 +709*x^4-308*x^3+484*x^2),x)
Output:
(e**4*( - 3*x + 2))/(x*(3*e**6*x - 2*e**6 - 15*x**2 + 7*x - 22))