Integrand size = 142, antiderivative size = 23 \[ \int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 \left (-45 x+9 x^2\right )+e^3 \left (90 x^2-18 x^3\right )}} \left (30 x-8 x^2+e^3 (-10+4 x)\right )}{-225 x^5+90 x^6-9 x^7+e^9 \left (225 x^2-90 x^3+9 x^4\right )+e^6 \left (-675 x^3+270 x^4-27 x^5\right )+e^3 \left (675 x^4-270 x^5+27 x^6\right )} \, dx=e^{-\frac {2}{9 \left (e^3-x\right )^2 (-5+x) x}} \]
Time = 0.61 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 \left (-45 x+9 x^2\right )+e^3 \left (90 x^2-18 x^3\right )}} \left (30 x-8 x^2+e^3 (-10+4 x)\right )}{-225 x^5+90 x^6-9 x^7+e^9 \left (225 x^2-90 x^3+9 x^4\right )+e^6 \left (-675 x^3+270 x^4-27 x^5\right )+e^3 \left (675 x^4-270 x^5+27 x^6\right )} \, dx=e^{-\frac {2}{9 \left (e^3-x\right )^2 (-5+x) x}} \]
Integrate[(30*x - 8*x^2 + E^3*(-10 + 4*x))/(E^(2/(-45*x^3 + 9*x^4 + E^6*(- 45*x + 9*x^2) + E^3*(90*x^2 - 18*x^3)))*(-225*x^5 + 90*x^6 - 9*x^7 + E^9*( 225*x^2 - 90*x^3 + 9*x^4) + E^6*(-675*x^3 + 270*x^4 - 27*x^5) + E^3*(675*x ^4 - 270*x^5 + 27*x^6))),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.
\(\displaystyle \int \frac {\left (-8 x^2+30 x+e^3 (4 x-10)\right ) \exp \left (-\frac {2}{9 x^4-45 x^3+e^6 \left (9 x^2-45 x\right )+e^3 \left (90 x^2-18 x^3\right )}\right )}{-9 x^7+90 x^6-225 x^5+e^3 \left (27 x^6-270 x^5+675 x^4\right )+e^6 \left (-27 x^5+270 x^4-675 x^3\right )+e^9 \left (9 x^4-90 x^3+225 x^2\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-8 x^2+30 x+e^3 (4 x-10)\right ) \exp \left (-\frac {2}{9 x^4-45 x^3+e^6 \left (9 x^2-45 x\right )+e^3 \left (90 x^2-18 x^3\right )}\right )}{x^2 \left (-9 x^5+9 \left (10+3 e^3\right ) x^4-9 \left (25+30 e^3+3 e^6\right ) x^3+9 e^3 \left (75+30 e^3+e^6\right ) x^2-45 e^6 \left (15+2 e^3\right ) x+225 e^9\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-8 x^2+30 x+e^3 (4 x-10)\right ) \exp \left (-\frac {2}{9 x^4-45 x^3+e^6 \left (9 x^2-45 x\right )+e^3 \left (90 x^2-18 x^3\right )}\right )}{3 \left (e^3-5\right )^4 \left (e^3-x\right ) x^2}+\frac {\left (-8 x^2+30 x+e^3 (4 x-10)\right ) \exp \left (-\frac {2}{9 x^4-45 x^3+e^6 \left (9 x^2-45 x\right )+e^3 \left (90 x^2-18 x^3\right )}\right )}{3 \left (e^3-5\right )^4 (x-5) x^2}+\frac {2 \left (-8 x^2+30 x+e^3 (4 x-10)\right ) \exp \left (-\frac {2}{9 x^4-45 x^3+e^6 \left (9 x^2-45 x\right )+e^3 \left (90 x^2-18 x^3\right )}\right )}{9 \left (e^3-5\right )^3 \left (e^3-x\right )^2 x^2}+\frac {\left (-8 x^2+30 x+e^3 (4 x-10)\right ) \exp \left (-\frac {2}{9 x^4-45 x^3+e^6 \left (9 x^2-45 x\right )+e^3 \left (90 x^2-18 x^3\right )}\right )}{9 \left (e^3-5\right )^3 (x-5)^2 x^2}+\frac {\left (-8 x^2+30 x+e^3 (4 x-10)\right ) \exp \left (-\frac {2}{9 x^4-45 x^3+e^6 \left (9 x^2-45 x\right )+e^3 \left (90 x^2-18 x^3\right )}\right )}{9 \left (e^3-5\right )^2 \left (e^3-x\right )^3 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \int \frac {e^{-3-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{\left (e^3-x\right )^3}dx}{9 \left (5-e^3\right )}+\frac {2 \left (5+2 e^3\right ) \int \frac {e^{-6-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{\left (e^3-x\right )^2}dx}{9 \left (5-e^3\right )^2}-\frac {8 \int \frac {e^{-3-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{\left (e^3-x\right )^2}dx}{9 \left (5-e^3\right )^2}-\frac {4 \left (5+2 e^3\right ) \int \frac {e^{-6-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{e^3-x}dx}{9 \left (5-e^3\right )^3}+\frac {4 \int \frac {e^{-6-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{e^3-x}dx}{9 \left (5-e^3\right )^2}+\frac {4 \int \frac {e^{-3-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{e^3-x}dx}{3 \left (5-e^3\right )^3}+\frac {2 \int \frac {e^{-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{(x-5)^2}dx}{45 \left (5-e^3\right )^2}-\frac {10 \int \frac {e^{-6-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x^2}dx}{9 \left (5-e^3\right )^2}+\frac {20 \int \frac {e^{-3-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x^2}dx}{9 \left (5-e^3\right )^3}+\frac {2 \int \frac {e^{3-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x^2}dx}{45 \left (5-e^3\right )^3}+\frac {2 \int \frac {e^{3-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x^2}dx}{3 \left (5-e^3\right )^4}-\frac {10 \int \frac {e^{-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x^2}dx}{3 \left (5-e^3\right )^4}-\frac {4 \left (5+2 e^3\right ) \int \frac {e^{-6-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x}dx}{9 \left (5-e^3\right )^3}+\frac {4 \int \frac {e^{-6-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x}dx}{9 \left (5-e^3\right )^2}+\frac {4 \left (5+e^3\right ) \int \frac {e^{-3-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x}dx}{3 \left (5-e^3\right )^4}-\frac {2 \left (15+e^3\right ) \int \frac {e^{-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x}dx}{15 \left (5-e^3\right )^4}-\frac {2 \int \frac {e^{-\frac {2}{9 \left (e^3-x\right )^2 (x-5) x}}}{x}dx}{15 \left (5-e^3\right )^3}\) |
Int[(30*x - 8*x^2 + E^3*(-10 + 4*x))/(E^(2/(-45*x^3 + 9*x^4 + E^6*(-45*x + 9*x^2) + E^3*(90*x^2 - 18*x^3)))*(-225*x^5 + 90*x^6 - 9*x^7 + E^9*(225*x^ 2 - 90*x^3 + 9*x^4) + E^6*(-675*x^3 + 270*x^4 - 27*x^5) + E^3*(675*x^4 - 2 70*x^5 + 27*x^6))),x]
3.5.81.3.1 Defintions of rubi rules used
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] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 7.61 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26
method | result | size |
risch | \({\mathrm e}^{\frac {2}{9 x \left (-5+x \right ) \left (2 x \,{\mathrm e}^{3}-x^{2}-{\mathrm e}^{6}\right )}}\) | \(29\) |
gosper | \({\mathrm e}^{-\frac {2}{9 x \left (x \,{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{3}-5 \,{\mathrm e}^{6}+10 x \,{\mathrm e}^{3}-5 x^{2}\right )}}\) | \(42\) |
parallelrisch | \({\mathrm e}^{-\frac {2}{9 x \left (x \,{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{3}-5 \,{\mathrm e}^{6}+10 x \,{\mathrm e}^{3}-5 x^{2}\right )}}\) | \(42\) |
norman | \(\frac {x^{4} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}+\left ({\mathrm e}^{6}+10 \,{\mathrm e}^{3}\right ) x^{2} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}+\left (-2 \,{\mathrm e}^{3}-5\right ) x^{3} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}-5 x \,{\mathrm e}^{6} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}}{x \left (-5+x \right ) \left (-x +{\mathrm e}^{3}\right )^{2}}\) | \(229\) |
int(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18*x^3+90 *x^2)*exp(3)+9*x^4-45*x^3))/((9*x^4-90*x^3+225*x^2)*exp(3)^3+(-27*x^5+270* x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6-225*x^5 ),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 \left (-45 x+9 x^2\right )+e^3 \left (90 x^2-18 x^3\right )}} \left (30 x-8 x^2+e^3 (-10+4 x)\right )}{-225 x^5+90 x^6-9 x^7+e^9 \left (225 x^2-90 x^3+9 x^4\right )+e^6 \left (-675 x^3+270 x^4-27 x^5\right )+e^3 \left (675 x^4-270 x^5+27 x^6\right )} \, dx=e^{\left (-\frac {2}{9 \, {\left (x^{4} - 5 \, x^{3} + {\left (x^{2} - 5 \, x\right )} e^{6} - 2 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3}\right )}}\right )} \]
integrate(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18* x^3+90*x^2)*exp(3)+9*x^4-45*x^3))/((9*x^4-90*x^3+225*x^2)*exp(3)^3+(-27*x^ 5+270*x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6-2 25*x^5),x, algorithm=\
Time = 0.99 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 \left (-45 x+9 x^2\right )+e^3 \left (90 x^2-18 x^3\right )}} \left (30 x-8 x^2+e^3 (-10+4 x)\right )}{-225 x^5+90 x^6-9 x^7+e^9 \left (225 x^2-90 x^3+9 x^4\right )+e^6 \left (-675 x^3+270 x^4-27 x^5\right )+e^3 \left (675 x^4-270 x^5+27 x^6\right )} \, dx=e^{- \frac {2}{9 x^{4} - 45 x^{3} + \left (9 x^{2} - 45 x\right ) e^{6} + \left (- 18 x^{3} + 90 x^{2}\right ) e^{3}}} \]
integrate(((4*x-10)*exp(3)-8*x**2+30*x)*exp(-2/((9*x**2-45*x)*exp(3)**2+(- 18*x**3+90*x**2)*exp(3)+9*x**4-45*x**3))/((9*x**4-90*x**3+225*x**2)*exp(3) **3+(-27*x**5+270*x**4-675*x**3)*exp(3)**2+(27*x**6-270*x**5+675*x**4)*exp (3)-9*x**7+90*x**6-225*x**5),x)
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (19) = 38\).
Time = 0.50 (sec) , antiderivative size = 125, normalized size of antiderivative = 5.43 \[ \int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 \left (-45 x+9 x^2\right )+e^3 \left (90 x^2-18 x^3\right )}} \left (30 x-8 x^2+e^3 (-10+4 x)\right )}{-225 x^5+90 x^6-9 x^7+e^9 \left (225 x^2-90 x^3+9 x^4\right )+e^6 \left (-675 x^3+270 x^4-27 x^5\right )+e^3 \left (675 x^4-270 x^5+27 x^6\right )} \, dx=e^{\left (\frac {2 \, e^{\left (-6\right )}}{45 \, x} - \frac {2}{9 \, {\left (x^{2} {\left (e^{6} - 5 \, e^{3}\right )} - 2 \, x {\left (e^{9} - 5 \, e^{6}\right )} + e^{12} - 5 \, e^{9}\right )}} - \frac {10}{9 \, {\left (x {\left (e^{12} - 10 \, e^{9} + 25 \, e^{6}\right )} - e^{15} + 10 \, e^{12} - 25 \, e^{9}\right )}} + \frac {4}{9 \, {\left (x {\left (e^{9} - 10 \, e^{6} + 25 \, e^{3}\right )} - e^{12} + 10 \, e^{9} - 25 \, e^{6}\right )}} - \frac {2}{45 \, {\left (x {\left (e^{6} - 10 \, e^{3} + 25\right )} - 5 \, e^{6} + 50 \, e^{3} - 125\right )}}\right )} \]
integrate(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18* x^3+90*x^2)*exp(3)+9*x^4-45*x^3))/((9*x^4-90*x^3+225*x^2)*exp(3)^3+(-27*x^ 5+270*x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6-2 25*x^5),x, algorithm=\
e^(2/45*e^(-6)/x - 2/9/(x^2*(e^6 - 5*e^3) - 2*x*(e^9 - 5*e^6) + e^12 - 5*e ^9) - 10/9/(x*(e^12 - 10*e^9 + 25*e^6) - e^15 + 10*e^12 - 25*e^9) + 4/9/(x *(e^9 - 10*e^6 + 25*e^3) - e^12 + 10*e^9 - 25*e^6) - 2/45/(x*(e^6 - 10*e^3 + 25) - 5*e^6 + 50*e^3 - 125))
\[ \int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 \left (-45 x+9 x^2\right )+e^3 \left (90 x^2-18 x^3\right )}} \left (30 x-8 x^2+e^3 (-10+4 x)\right )}{-225 x^5+90 x^6-9 x^7+e^9 \left (225 x^2-90 x^3+9 x^4\right )+e^6 \left (-675 x^3+270 x^4-27 x^5\right )+e^3 \left (675 x^4-270 x^5+27 x^6\right )} \, dx=\int { \frac {2 \, {\left (4 \, x^{2} - {\left (2 \, x - 5\right )} e^{3} - 15 \, x\right )} e^{\left (-\frac {2}{9 \, {\left (x^{4} - 5 \, x^{3} + {\left (x^{2} - 5 \, x\right )} e^{6} - 2 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3}\right )}}\right )}}{9 \, {\left (x^{7} - 10 \, x^{6} + 25 \, x^{5} - {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} e^{9} + 3 \, {\left (x^{5} - 10 \, x^{4} + 25 \, x^{3}\right )} e^{6} - 3 \, {\left (x^{6} - 10 \, x^{5} + 25 \, x^{4}\right )} e^{3}\right )}} \,d x } \]
integrate(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18* x^3+90*x^2)*exp(3)+9*x^4-45*x^3))/((9*x^4-90*x^3+225*x^2)*exp(3)^3+(-27*x^ 5+270*x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6-2 25*x^5),x, algorithm=\
integrate(2/9*(4*x^2 - (2*x - 5)*e^3 - 15*x)*e^(-2/9/(x^4 - 5*x^3 + (x^2 - 5*x)*e^6 - 2*(x^3 - 5*x^2)*e^3))/(x^7 - 10*x^6 + 25*x^5 - (x^4 - 10*x^3 + 25*x^2)*e^9 + 3*(x^5 - 10*x^4 + 25*x^3)*e^6 - 3*(x^6 - 10*x^5 + 25*x^4)*e ^3), x)
Time = 11.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.70 \[ \int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 \left (-45 x+9 x^2\right )+e^3 \left (90 x^2-18 x^3\right )}} \left (30 x-8 x^2+e^3 (-10+4 x)\right )}{-225 x^5+90 x^6-9 x^7+e^9 \left (225 x^2-90 x^3+9 x^4\right )+e^6 \left (-675 x^3+270 x^4-27 x^5\right )+e^3 \left (675 x^4-270 x^5+27 x^6\right )} \, dx={\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-6}}{45\,x}-\frac {2}{45\,{\left ({\mathrm {e}}^3-5\right )}^2\,\left (x-5\right )}-\frac {2\,{\mathrm {e}}^{-6}\,\left (5\,x-10\,{\mathrm {e}}^3+3\,{\mathrm {e}}^6-2\,x\,{\mathrm {e}}^3\right )}{9\,{\left ({\mathrm {e}}^3-5\right )}^2\,\left (x^2-2\,{\mathrm {e}}^3\,x+{\mathrm {e}}^6\right )}} \]
int((exp(2/(exp(6)*(45*x - 9*x^2) - exp(3)*(90*x^2 - 18*x^3) + 45*x^3 - 9* x^4))*(30*x - 8*x^2 + exp(3)*(4*x - 10)))/(exp(9)*(225*x^2 - 90*x^3 + 9*x^ 4) + exp(3)*(675*x^4 - 270*x^5 + 27*x^6) - exp(6)*(675*x^3 - 270*x^4 + 27* x^5) - 225*x^5 + 90*x^6 - 9*x^7),x)