Integrand size = 102, antiderivative size = 30 \[ \int \frac {e^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\right )} \, dx=-4+3 e^{\frac {3}{5 \left (-e^x+5 x\right ) \left (2-x^2\right )}} \]
Time = 1.69 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\right )} \, dx=3 e^{\frac {3}{5 \left (e^x-5 x\right ) \left (-2+x^2\right )}} \]
Integrate[(E^(3/(50*x - 25*x^3 + E^x*(-10 + 5*x^2)))*(-90 + 135*x^2 + E^x* (18 - 18*x - 9*x^2)))/(500*x^2 - 500*x^4 + 125*x^6 + E^(2*x)*(20 - 20*x^2 + 5*x^4) + E^x*(-200*x + 200*x^3 - 50*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.
\(\displaystyle \int \frac {e^{\frac {3}{-25 x^3+e^x \left (5 x^2-10\right )+50 x}} \left (135 x^2+e^x \left (-9 x^2-18 x+18\right )-90\right )}{125 x^6-500 x^4+500 x^2+e^x \left (-50 x^5+200 x^3-200 x\right )+e^{2 x} \left (5 x^4-20 x^2+20\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {3}{5 \left (e^x-5 x\right ) \left (x^2-2\right )}} \left (135 x^2+e^x \left (-9 x^2-18 x+18\right )-90\right )}{5 \left (e^x-5 x\right )^2 \left (2-x^2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {9 e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}} \left (-15 x^2-e^x \left (-x^2-2 x+2\right )+10\right )}{\left (e^x-5 x\right )^2 \left (2-x^2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {9}{5} \int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}} \left (-15 x^2-e^x \left (-x^2-2 x+2\right )+10\right )}{\left (e^x-5 x\right )^2 \left (2-x^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {9}{5} \int \left (\frac {5 e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}} (x-1)}{\left (e^x-5 x\right )^2 \left (x^2-2\right )}+\frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}} \left (x^2+2 x-2\right )}{\left (e^x-5 x\right ) \left (x^2-2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {9}{5} \left (\frac {5 \int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}}}{\left (e^x-5 x\right )^2 \left (\sqrt {2}-x\right )}dx}{2 \sqrt {2}}-\frac {5}{2} \int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}}}{\left (e^x-5 x\right )^2 \left (\sqrt {2}-x\right )}dx-\frac {\int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}}}{\left (e^x-5 x\right ) \left (\sqrt {2}-x\right )}dx}{2 \sqrt {2}}+\frac {5 \int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}}}{\left (e^x-5 x\right )^2 \left (x+\sqrt {2}\right )}dx}{2 \sqrt {2}}+\frac {5}{2} \int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}}}{\left (e^x-5 x\right )^2 \left (x+\sqrt {2}\right )}dx-\frac {\int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}}}{\left (e^x-5 x\right ) \left (x+\sqrt {2}\right )}dx}{2 \sqrt {2}}+2 \int \frac {e^{-\frac {3}{5 \left (e^x-5 x\right ) \left (2-x^2\right )}} x}{\left (e^x-5 x\right ) \left (x^2-2\right )^2}dx\right )\) |
Int[(E^(3/(50*x - 25*x^3 + E^x*(-10 + 5*x^2)))*(-90 + 135*x^2 + E^x*(18 - 18*x - 9*x^2)))/(500*x^2 - 500*x^4 + 125*x^6 + E^(2*x)*(20 - 20*x^2 + 5*x^ 4) + E^x*(-200*x + 200*x^3 - 50*x^5)),x]
3.7.36.3.1 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]]
Time = 4.73 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70
method | result | size |
risch | \(3 \,{\mathrm e}^{\frac {3}{5 \left (x^{2}-2\right ) \left ({\mathrm e}^{x}-5 x \right )}}\) | \(21\) |
parallelrisch | \(3 \,{\mathrm e}^{\frac {3}{5 \left ({\mathrm e}^{x} x^{2}-5 x^{3}-2 \,{\mathrm e}^{x}+10 x \right )}}\) | \(27\) |
int(((-9*x^2-18*x+18)*exp(x)+135*x^2-90)*exp(3/((5*x^2-10)*exp(x)-25*x^3+5 0*x))/((5*x^4-20*x^2+20)*exp(x)^2+(-50*x^5+200*x^3-200*x)*exp(x)+125*x^6-5 00*x^4+500*x^2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\right )} \, dx=3 \, e^{\left (-\frac {3}{5 \, {\left (5 \, x^{3} - {\left (x^{2} - 2\right )} e^{x} - 10 \, x\right )}}\right )} \]
integrate(((-9*x^2-18*x+18)*exp(x)+135*x^2-90)*exp(3/((5*x^2-10)*exp(x)-25 *x^3+50*x))/((5*x^4-20*x^2+20)*exp(x)^2+(-50*x^5+200*x^3-200*x)*exp(x)+125 *x^6-500*x^4+500*x^2),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\right )} \, dx=3 e^{\frac {3}{- 25 x^{3} + 50 x + \left (5 x^{2} - 10\right ) e^{x}}} \]
integrate(((-9*x**2-18*x+18)*exp(x)+135*x**2-90)*exp(3/((5*x**2-10)*exp(x) -25*x**3+50*x))/((5*x**4-20*x**2+20)*exp(x)**2+(-50*x**5+200*x**3-200*x)*e xp(x)+125*x**6-500*x**4+500*x**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (24) = 48\).
Time = 0.45 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.53 \[ \int \frac {e^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\right )} \, dx=3 \, e^{\left (-\frac {3 \, x}{50 \, x^{2} - {\left (x^{2} - 2\right )} e^{\left (2 \, x\right )} - 100} - \frac {3 \, e^{x}}{5 \, {\left (50 \, x^{2} - {\left (x^{2} - 2\right )} e^{\left (2 \, x\right )} - 100\right )}} - \frac {15}{5 \, x e^{\left (2 \, x\right )} - 250 \, x - e^{\left (3 \, x\right )} + 50 \, e^{x}}\right )} \]
integrate(((-9*x^2-18*x+18)*exp(x)+135*x^2-90)*exp(3/((5*x^2-10)*exp(x)-25 *x^3+50*x))/((5*x^4-20*x^2+20)*exp(x)^2+(-50*x^5+200*x^3-200*x)*exp(x)+125 *x^6-500*x^4+500*x^2),x, algorithm=\
3*e^(-3*x/(50*x^2 - (x^2 - 2)*e^(2*x) - 100) - 3/5*e^x/(50*x^2 - (x^2 - 2) *e^(2*x) - 100) - 15/(5*x*e^(2*x) - 250*x - e^(3*x) + 50*e^x))
Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\right )} \, dx=3 \, e^{\left (-\frac {3}{5 \, {\left (5 \, x^{3} - x^{2} e^{x} - 10 \, x + 2 \, e^{x}\right )}}\right )} \]
integrate(((-9*x^2-18*x+18)*exp(x)+135*x^2-90)*exp(3/((5*x^2-10)*exp(x)-25 *x^3+50*x))/((5*x^4-20*x^2+20)*exp(x)^2+(-50*x^5+200*x^3-200*x)*exp(x)+125 *x^6-500*x^4+500*x^2),x, algorithm=\
Time = 9.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {3}{50 x-25 x^3+e^x \left (-10+5 x^2\right )}} \left (-90+135 x^2+e^x \left (18-18 x-9 x^2\right )\right )}{500 x^2-500 x^4+125 x^6+e^{2 x} \left (20-20 x^2+5 x^4\right )+e^x \left (-200 x+200 x^3-50 x^5\right )} \, dx=3\,{\mathrm {e}}^{\frac {3}{5\,\left (10\,x-2\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^x-5\,x^3\right )}} \]