Integrand size = 117, antiderivative size = 32 \[ \int \frac {-75-150 e^{2 x}-150 x^2+e^{2 x/3} (-15+10 x)+e^x \left (75+5 e^{2 x/3}+300 x\right )}{100+4 e^{4 x/3}-400 x^2+400 e^{2 x} x^2+400 x^4+e^{2 x/3} \left (40-80 x^2\right )+e^x \left (400 x+80 e^{2 x/3} x-800 x^3\right )} \, dx=\frac {3}{4 \left (\frac {5+e^{2 x/3}}{5 \left (e^x-x\right )}+2 x\right )} \]
Time = 5.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-75-150 e^{2 x}-150 x^2+e^{2 x/3} (-15+10 x)+e^x \left (75+5 e^{2 x/3}+300 x\right )}{100+4 e^{4 x/3}-400 x^2+400 e^{2 x} x^2+400 x^4+e^{2 x/3} \left (40-80 x^2\right )+e^x \left (400 x+80 e^{2 x/3} x-800 x^3\right )} \, dx=\frac {15 \left (e^x-x\right )}{4 \left (5+e^{2 x/3}+10 e^x x-10 x^2\right )} \]
Integrate[(-75 - 150*E^(2*x) - 150*x^2 + E^((2*x)/3)*(-15 + 10*x) + E^x*(7 5 + 5*E^((2*x)/3) + 300*x))/(100 + 4*E^((4*x)/3) - 400*x^2 + 400*E^(2*x)*x ^2 + 400*x^4 + E^((2*x)/3)*(40 - 80*x^2) + E^x*(400*x + 80*E^((2*x)/3)*x - 800*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.
\(\displaystyle \int \frac {-150 x^2-150 e^{2 x}+e^{2 x/3} (10 x-15)+e^x \left (300 x+5 e^{2 x/3}+75\right )-75}{400 x^4+e^x \left (-800 x^3+80 e^{2 x/3} x+400 x\right )+400 e^{2 x} x^2-400 x^2+e^{2 x/3} \left (40-80 x^2\right )+4 e^{4 x/3}+100} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-150 x^2-150 e^{2 x}+e^{2 x/3} (10 x-15)+e^x \left (300 x+5 e^{2 x/3}+75\right )-75}{4 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {5 \left (30 x^2+30 e^{2 x}+e^{2 x/3} (3-2 x)-e^x \left (60 x+e^{2 x/3}+15\right )+15\right )}{\left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5}{4} \int \frac {30 x^2+30 e^{2 x}+e^{2 x/3} (3-2 x)-e^x \left (60 x+e^{2 x/3}+15\right )+15}{\left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {5}{4} \int \left (\frac {100 e^{2 x/3} x^3+1500 x^3-10 e^{x/3} x^2+600 e^{2 x/3} x^2+3000 x^2-30 e^{x/3} x+x+3}{1000 x^4 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )}-\frac {3000 e^{2 x/3} x^5+15000 x^5-100 e^{x/3} x^4-3000 e^{2 x/3} x^4-15000 x^4-300 e^{x/3} x^3-2000 e^{2 x/3} x^3-7490 x^3+50 e^{x/3} x^2-3000 e^{2 x/3} x^2-7470 x^2+150 e^{x/3} x-e^{2 x/3} x-5 x-3 e^{2 x/3}-15}{1000 x^4 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )^2}+\frac {3}{10 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{4} \left (15 \int \frac {1}{\left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+\frac {1}{10} \int \frac {e^{x/3}}{\left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+3 \int \frac {e^{2 x/3}}{\left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+\frac {3 \int \frac {e^{2 x/3}}{x^4 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx}{1000}-\frac {3}{20} \int \frac {e^{x/3}}{x^3 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+\frac {\int \frac {e^{2 x/3}}{x^3 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx}{1000}-\frac {1}{20} \int \frac {e^{x/3}}{x^2 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+3 \int \frac {e^{2 x/3}}{x^2 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+\frac {3}{10} \int \frac {e^{x/3}}{x \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+2 \int \frac {e^{2 x/3}}{x \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx-3 \int \frac {e^{2 x/3} x}{\left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )^2}dx+\frac {3}{100} \int \frac {e^{x/3}}{x^3 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )}dx+\frac {1}{100} \int \frac {e^{x/3}}{x^2 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )}dx-\frac {3}{5} \int \frac {e^{2 x/3}}{x^2 \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )}dx-\frac {1}{10} \int \frac {e^{2 x/3}}{x \left (-10 x^2+10 e^x x+e^{2 x/3}+5\right )}dx+\frac {3}{200} \int \frac {1}{x^4 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )^2}dx+\frac {1}{200} \int \frac {1}{x^3 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )^2}dx+\frac {747}{100} \int \frac {1}{x^2 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )^2}dx+\frac {749}{100} \int \frac {1}{x \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )^2}dx-15 \int \frac {x}{\left (10 x^2-10 e^x x-e^{2 x/3}-5\right )^2}dx+\frac {3 \int \frac {1}{x^4 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )}dx}{1000}+\frac {\int \frac {1}{x^3 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )}dx}{1000}+3 \int \frac {1}{x^2 \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )}dx+\frac {3}{2} \int \frac {1}{x \left (10 x^2-10 e^x x-e^{2 x/3}-5\right )}dx-\frac {3}{10 x}\right )\) |
Int[(-75 - 150*E^(2*x) - 150*x^2 + E^((2*x)/3)*(-15 + 10*x) + E^x*(75 + 5* E^((2*x)/3) + 300*x))/(100 + 4*E^((4*x)/3) - 400*x^2 + 400*E^(2*x)*x^2 + 4 00*x^4 + E^((2*x)/3)*(40 - 80*x^2) + E^x*(400*x + 80*E^((2*x)/3)*x - 800*x ^3)),x]
3.4.68.3.1 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 = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-\frac {-15 x +15 \,{\mathrm e}^{x}}{4 \left (-10 \,{\mathrm e}^{x} x +10 x^{2}-{\mathrm e}^{\frac {2 x}{3}}-5\right )}\) | \(31\) |
risch | \(\frac {3}{8 x}+\frac {\frac {3 \,{\mathrm e}^{\frac {2 x}{3}}}{8}+\frac {15}{8}}{x \left (-10 \,{\mathrm e}^{x} x +10 x^{2}-{\mathrm e}^{\frac {2 x}{3}}-5\right )}\) | \(38\) |
int((-150*exp(x)^2+(5*exp(2/3*x)+300*x+75)*exp(x)+(10*x-15)*exp(2/3*x)-150 *x^2-75)/(400*exp(x)^2*x^2+(80*x*exp(2/3*x)-800*x^3+400*x)*exp(x)+4*exp(2/ 3*x)^2+(-80*x^2+40)*exp(2/3*x)+400*x^4-400*x^2+100),x,method=_RETURNVERBOS E)
Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-75-150 e^{2 x}-150 x^2+e^{2 x/3} (-15+10 x)+e^x \left (75+5 e^{2 x/3}+300 x\right )}{100+4 e^{4 x/3}-400 x^2+400 e^{2 x} x^2+400 x^4+e^{2 x/3} \left (40-80 x^2\right )+e^x \left (400 x+80 e^{2 x/3} x-800 x^3\right )} \, dx=\frac {15 \, {\left (x - e^{x}\right )}}{4 \, {\left (10 \, x^{2} - 10 \, x e^{x} - e^{\left (\frac {2}{3} \, x\right )} - 5\right )}} \]
integrate((-150*exp(x)^2+(5*exp(2/3*x)+300*x+75)*exp(x)+(10*x-15)*exp(2/3* x)-150*x^2-75)/(400*exp(x)^2*x^2+(80*x*exp(2/3*x)-800*x^3+400*x)*exp(x)+4* exp(2/3*x)^2+(-80*x^2+40)*exp(2/3*x)+400*x^4-400*x^2+100),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {-75-150 e^{2 x}-150 x^2+e^{2 x/3} (-15+10 x)+e^x \left (75+5 e^{2 x/3}+300 x\right )}{100+4 e^{4 x/3}-400 x^2+400 e^{2 x} x^2+400 x^4+e^{2 x/3} \left (40-80 x^2\right )+e^x \left (400 x+80 e^{2 x/3} x-800 x^3\right )} \, dx=\frac {- 3 e^{\frac {2 x}{3}} - 15}{- 80 x^{3} + 80 x^{2} e^{x} + 8 x e^{\frac {2 x}{3}} + 40 x} + \frac {3}{8 x} \]
integrate((-150*exp(x)**2+(5*exp(2/3*x)+300*x+75)*exp(x)+(10*x-15)*exp(2/3 *x)-150*x**2-75)/(400*exp(x)**2*x**2+(80*x*exp(2/3*x)-800*x**3+400*x)*exp( x)+4*exp(2/3*x)**2+(-80*x**2+40)*exp(2/3*x)+400*x**4-400*x**2+100),x)
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-75-150 e^{2 x}-150 x^2+e^{2 x/3} (-15+10 x)+e^x \left (75+5 e^{2 x/3}+300 x\right )}{100+4 e^{4 x/3}-400 x^2+400 e^{2 x} x^2+400 x^4+e^{2 x/3} \left (40-80 x^2\right )+e^x \left (400 x+80 e^{2 x/3} x-800 x^3\right )} \, dx=\frac {15 \, {\left (x - e^{x}\right )}}{4 \, {\left (10 \, x^{2} - 10 \, x e^{x} - e^{\left (\frac {2}{3} \, x\right )} - 5\right )}} \]
integrate((-150*exp(x)^2+(5*exp(2/3*x)+300*x+75)*exp(x)+(10*x-15)*exp(2/3* x)-150*x^2-75)/(400*exp(x)^2*x^2+(80*x*exp(2/3*x)-800*x^3+400*x)*exp(x)+4* exp(2/3*x)^2+(-80*x^2+40)*exp(2/3*x)+400*x^4-400*x^2+100),x, algorithm=\
Time = 0.90 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {-75-150 e^{2 x}-150 x^2+e^{2 x/3} (-15+10 x)+e^x \left (75+5 e^{2 x/3}+300 x\right )}{100+4 e^{4 x/3}-400 x^2+400 e^{2 x} x^2+400 x^4+e^{2 x/3} \left (40-80 x^2\right )+e^x \left (400 x+80 e^{2 x/3} x-800 x^3\right )} \, dx=\frac {3 \, {\left (10 \, x^{2} - 10 \, x e^{x} + e^{\left (\frac {2}{3} \, x\right )} + 5\right )}}{8 \, {\left (10 \, x^{3} - 10 \, x^{2} e^{x} - x e^{\left (\frac {2}{3} \, x\right )} - 5 \, x\right )}} \]
integrate((-150*exp(x)^2+(5*exp(2/3*x)+300*x+75)*exp(x)+(10*x-15)*exp(2/3* x)-150*x^2-75)/(400*exp(x)^2*x^2+(80*x*exp(2/3*x)-800*x^3+400*x)*exp(x)+4* exp(2/3*x)^2+(-80*x^2+40)*exp(2/3*x)+400*x^4-400*x^2+100),x, algorithm=\
Timed out. \[ \int \frac {-75-150 e^{2 x}-150 x^2+e^{2 x/3} (-15+10 x)+e^x \left (75+5 e^{2 x/3}+300 x\right )}{100+4 e^{4 x/3}-400 x^2+400 e^{2 x} x^2+400 x^4+e^{2 x/3} \left (40-80 x^2\right )+e^x \left (400 x+80 e^{2 x/3} x-800 x^3\right )} \, dx=\int -\frac {150\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (300\,x+5\,{\mathrm {e}}^{\frac {2\,x}{3}}+75\right )-{\mathrm {e}}^{\frac {2\,x}{3}}\,\left (10\,x-15\right )+150\,x^2+75}{4\,{\mathrm {e}}^{\frac {4\,x}{3}}-{\mathrm {e}}^{\frac {2\,x}{3}}\,\left (80\,x^2-40\right )+400\,x^2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (400\,x+80\,x\,{\mathrm {e}}^{\frac {2\,x}{3}}-800\,x^3\right )-400\,x^2+400\,x^4+100} \,d x \]
int(-(150*exp(2*x) - exp(x)*(300*x + 5*exp((2*x)/3) + 75) - exp((2*x)/3)*( 10*x - 15) + 150*x^2 + 75)/(4*exp((4*x)/3) - exp((2*x)/3)*(80*x^2 - 40) + 400*x^2*exp(2*x) + exp(x)*(400*x + 80*x*exp((2*x)/3) - 800*x^3) - 400*x^2 + 400*x^4 + 100),x)