Integrand size = 107, antiderivative size = 31 \[ \int \frac {e^{5 x} \left (e^{2 x} (4+20 x)+e^x \left (-28 x^2+6 e^3 x^2-12 x^3\right )\right )}{16 e^{2 x}+25 x^2+e^6 x^2+20 x^3+4 x^4+e^x \left (-40 x+8 e^3 x-16 x^2\right )+e^3 \left (-10 x^2-4 x^3\right )} \, dx=\frac {e^{5 x}}{\frac {4}{x}-e^{-x} \left (5-e^3+2 x\right )} \]
Time = 11.99 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{5 x} \left (e^{2 x} (4+20 x)+e^x \left (-28 x^2+6 e^3 x^2-12 x^3\right )\right )}{16 e^{2 x}+25 x^2+e^6 x^2+20 x^3+4 x^4+e^x \left (-40 x+8 e^3 x-16 x^2\right )+e^3 \left (-10 x^2-4 x^3\right )} \, dx=\frac {2 e^{6 x} x}{8 e^x+2 e^3 x-2 x (5+2 x)} \]
Integrate[(E^(5*x)*(E^(2*x)*(4 + 20*x) + E^x*(-28*x^2 + 6*E^3*x^2 - 12*x^3 )))/(16*E^(2*x) + 25*x^2 + E^6*x^2 + 20*x^3 + 4*x^4 + E^x*(-40*x + 8*E^3*x - 16*x^2) + E^3*(-10*x^2 - 4*x^3)),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.
| \(\displaystyle \int \frac {e^{5 x} \left (e^x \left (-12 x^3+6 e^3 x^2-28 x^2\right )+e^{2 x} (20 x+4)\right )}{4 x^4+20 x^3+e^6 x^2+25 x^2+e^x \left (-16 x^2+8 e^3 x-40 x\right )+e^3 \left (-4 x^3-10 x^2\right )+16 e^{2 x}} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{5 x} \left (e^x \left (-12 x^3+6 e^3 x^2-28 x^2\right )+e^{2 x} (20 x+4)\right )}{4 x^4+20 x^3+\left (25+e^6\right ) x^2+e^x \left (-16 x^2+8 e^3 x-40 x\right )+e^3 \left (-4 x^3-10 x^2\right )+16 e^{2 x}}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 e^{6 x} \left (-2 (3 x+7) x^2+3 e^3 x^2+2 e^x (5 x+1)\right )}{\left (-((2 x+5) x)+e^3 x+4 e^x\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {e^{6 x} \left (-2 (3 x+7) x^2+3 e^3 x^2+2 e^x (5 x+1)\right )}{\left (-((2 x+5) x)+e^3 x+4 e^x\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (\frac {e^{6 x} (5 x+1)}{2 \left (-2 x^2-5 \left (1-\frac {e^3}{5}\right ) x+4 e^x\right )}+\frac {e^{6 x} x \left (-2 x^2-\left (1-e^3\right ) x-e^3+5\right )}{2 \left (-2 x^2-5 \left (1-\frac {e^3}{5}\right ) x+4 e^x\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {1}{2} \left (5-e^3\right ) \int \frac {e^{6 x} x}{\left (-2 x^2-5 \left (1-\frac {e^3}{5}\right ) x+4 e^x\right )^2}dx-\frac {1}{2} \left (1-e^3\right ) \int \frac {e^{6 x} x^2}{\left (-2 x^2-5 \left (1-\frac {e^3}{5}\right ) x+4 e^x\right )^2}dx+\frac {1}{2} \int \frac {e^{6 x}}{-2 x^2-5 \left (1-\frac {e^3}{5}\right ) x+4 e^x}dx+\frac {5}{2} \int \frac {e^{6 x} x}{-2 x^2-5 \left (1-\frac {e^3}{5}\right ) x+4 e^x}dx-\int \frac {e^{6 x} x^3}{\left (-2 x^2-5 \left (1-\frac {e^3}{5}\right ) x+4 e^x\right )^2}dx\right )\) |
Int[(E^(5*x)*(E^(2*x)*(4 + 20*x) + E^x*(-28*x^2 + 6*E^3*x^2 - 12*x^3)))/(1 6*E^(2*x) + 25*x^2 + E^6*x^2 + 20*x^3 + 4*x^4 + E^x*(-40*x + 8*E^3*x - 16* x^2) + E^3*(-10*x^2 - 4*x^3)),x]
3.14.51.3.1 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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.66 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84
| method | result | size |
| risch | \(\frac {x \,{\mathrm e}^{6 x}}{x \,{\mathrm e}^{3}-2 x^{2}+4 \,{\mathrm e}^{x}-5 x}\) | \(26\) |
| parallelrisch | \(\frac {{\mathrm e}^{5 x} x \,{\mathrm e}^{x}}{x \,{\mathrm e}^{3}-2 x^{2}+4 \,{\mathrm e}^{x}-5 x}\) | \(28\) |
int(((20*x+4)*exp(x)^2+(6*x^2*exp(3)-12*x^3-28*x^2)*exp(x))*exp(5*x)/(16*e xp(x)^2+(8*x*exp(3)-16*x^2-40*x)*exp(x)+x^2*exp(3)^2+(-4*x^3-10*x^2)*exp(3 )+4*x^4+20*x^3+25*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{5 x} \left (e^{2 x} (4+20 x)+e^x \left (-28 x^2+6 e^3 x^2-12 x^3\right )\right )}{16 e^{2 x}+25 x^2+e^6 x^2+20 x^3+4 x^4+e^x \left (-40 x+8 e^3 x-16 x^2\right )+e^3 \left (-10 x^2-4 x^3\right )} \, dx=-\frac {x e^{\left (6 \, x\right )}}{2 \, x^{2} - x e^{3} + 5 \, x - 4 \, e^{x}} \]
integrate(((20*x+4)*exp(x)^2+(6*x^2*exp(3)-12*x^3-28*x^2)*exp(x))*exp(5*x) /(16*exp(x)^2+(8*x*exp(3)-16*x^2-40*x)*exp(x)+x^2*exp(3)^2+(-4*x^3-10*x^2) *exp(3)+4*x^4+20*x^3+25*x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 646 vs. \(2 (19) = 38\).
Time = 0.48 (sec) , antiderivative size = 646, normalized size of antiderivative = 20.84 \[ \int \frac {e^{5 x} \left (e^{2 x} (4+20 x)+e^x \left (-28 x^2+6 e^3 x^2-12 x^3\right )\right )}{16 e^{2 x}+25 x^2+e^6 x^2+20 x^3+4 x^4+e^x \left (-40 x+8 e^3 x-16 x^2\right )+e^3 \left (-10 x^2-4 x^3\right )} \, dx =\text {Too large to display} \]
integrate(((20*x+4)*exp(x)**2+(6*x**2*exp(3)-12*x**3-28*x**2)*exp(x))*exp( 5*x)/(16*exp(x)**2+(8*x*exp(3)-16*x**2-40*x)*exp(x)+x**2*exp(3)**2+(-4*x** 3-10*x**2)*exp(3)+4*x**4+20*x**3+25*x**2),x)
x**11/128 + x**10*(25/256 - 5*exp(3)/256) + x**9*(-25*exp(3)/128 + 125/256 + 5*exp(6)/256) + x**8*(-5*exp(9)/512 - 375*exp(3)/512 + 625/512 + 75*exp (6)/512) + x**7*(-25*exp(9)/512 - 625*exp(3)/512 + 3125/2048 + 375*exp(6)/ 1024 + 5*exp(12)/2048) + x**6*(-exp(15)/4096 - 125*exp(9)/2048 - 3125*exp( 3)/4096 + 3125/4096 + 625*exp(6)/2048 + 25*exp(12)/4096) + x*exp(5*x)/4 + (134217728*x**3 - 67108864*x**2*exp(3) + 335544320*x**2)*exp(4*x)/10737418 24 + (67108864*x**5 - 67108864*x**4*exp(3) + 335544320*x**4 - 167772160*x* *3*exp(3) + 419430400*x**3 + 16777216*x**3*exp(6))*exp(3*x)/1073741824 + ( 33554432*x**7 - 50331648*x**6*exp(3) + 251658240*x**6 - 251658240*x**5*exp (3) + 629145600*x**5 + 25165824*x**5*exp(6) - 4194304*x**4*exp(9) - 314572 800*x**4*exp(3) + 524288000*x**4 + 62914560*x**4*exp(6))*exp(2*x)/10737418 24 + (16777216*x**9 - 33554432*x**8*exp(3) + 167772160*x**8 - 251658240*x* *7*exp(3) + 629145600*x**7 + 25165824*x**7*exp(6) - 8388608*x**6*exp(9) - 629145600*x**6*exp(3) + 1048576000*x**6 + 125829120*x**6*exp(6) - 20971520 *x**5*exp(9) - 524288000*x**5*exp(3) + 655360000*x**5 + 157286400*x**5*exp (6) + 1048576*x**5*exp(12))*exp(x)/1073741824 + (64*x**13 - 192*x**12*exp( 3) + 960*x**12 - 2400*x**11*exp(3) + 6000*x**11 + 240*x**11*exp(6) - 160*x **10*exp(9) - 12000*x**10*exp(3) + 20000*x**10 + 2400*x**10*exp(6) - 1200* x**9*exp(9) - 30000*x**9*exp(3) + 37500*x**9 + 9000*x**9*exp(6) + 60*x**9* exp(12) - 12*x**8*exp(15) - 3000*x**8*exp(9) - 37500*x**8*exp(3) + 3750...
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{5 x} \left (e^{2 x} (4+20 x)+e^x \left (-28 x^2+6 e^3 x^2-12 x^3\right )\right )}{16 e^{2 x}+25 x^2+e^6 x^2+20 x^3+4 x^4+e^x \left (-40 x+8 e^3 x-16 x^2\right )+e^3 \left (-10 x^2-4 x^3\right )} \, dx=-\frac {x e^{\left (6 \, x\right )}}{2 \, x^{2} - x {\left (e^{3} - 5\right )} - 4 \, e^{x}} \]
integrate(((20*x+4)*exp(x)^2+(6*x^2*exp(3)-12*x^3-28*x^2)*exp(x))*exp(5*x) /(16*exp(x)^2+(8*x*exp(3)-16*x^2-40*x)*exp(x)+x^2*exp(3)^2+(-4*x^3-10*x^2) *exp(3)+4*x^4+20*x^3+25*x^2),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{5 x} \left (e^{2 x} (4+20 x)+e^x \left (-28 x^2+6 e^3 x^2-12 x^3\right )\right )}{16 e^{2 x}+25 x^2+e^6 x^2+20 x^3+4 x^4+e^x \left (-40 x+8 e^3 x-16 x^2\right )+e^3 \left (-10 x^2-4 x^3\right )} \, dx=-\frac {x e^{\left (6 \, x\right )}}{2 \, x^{2} - x e^{3} + 5 \, x - 4 \, e^{x}} \]
integrate(((20*x+4)*exp(x)^2+(6*x^2*exp(3)-12*x^3-28*x^2)*exp(x))*exp(5*x) /(16*exp(x)^2+(8*x*exp(3)-16*x^2-40*x)*exp(x)+x^2*exp(3)^2+(-4*x^3-10*x^2) *exp(3)+4*x^4+20*x^3+25*x^2),x, algorithm=\
Timed out. \[ \int \frac {e^{5 x} \left (e^{2 x} (4+20 x)+e^x \left (-28 x^2+6 e^3 x^2-12 x^3\right )\right )}{16 e^{2 x}+25 x^2+e^6 x^2+20 x^3+4 x^4+e^x \left (-40 x+8 e^3 x-16 x^2\right )+e^3 \left (-10 x^2-4 x^3\right )} \, dx=-\int \frac {{\mathrm {e}}^{5\,x}\,\left ({\mathrm {e}}^x\,\left (28\,x^2-6\,x^2\,{\mathrm {e}}^3+12\,x^3\right )-{\mathrm {e}}^{2\,x}\,\left (20\,x+4\right )\right )}{16\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^3\,\left (4\,x^3+10\,x^2\right )+x^2\,{\mathrm {e}}^6+25\,x^2+20\,x^3+4\,x^4-{\mathrm {e}}^x\,\left (40\,x-8\,x\,{\mathrm {e}}^3+16\,x^2\right )} \,d x \]
int(-(exp(5*x)*(exp(x)*(28*x^2 - 6*x^2*exp(3) + 12*x^3) - exp(2*x)*(20*x + 4)))/(16*exp(2*x) - exp(3)*(10*x^2 + 4*x^3) + x^2*exp(6) + 25*x^2 + 20*x^ 3 + 4*x^4 - exp(x)*(40*x - 8*x*exp(3) + 16*x^2)),x)