Integrand size = 124, antiderivative size = 23 \[ \int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx=(2+x) \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(23)=46\).
Time = 5.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx=-2 \log (x)+x \log \left (-1-e^4+e^x+\frac {2}{5 x}+x\right )+2 \log \left (2-5 x-5 e^4 x+5 e^x x+5 x^2\right ) \]
Integrate[(-4 - 2*x + 10*x^2 + 5*x^3 + E^x*(10*x^2 + 5*x^3) + (2*x - 5*x^2 - 5*E^4*x^2 + 5*E^x*x^2 + 5*x^3)*Log[(2 - 5*x - 5*E^4*x + 5*E^x*x + 5*x^2 )/(5*x)])/(2*x - 5*x^2 - 5*E^4*x^2 + 5*E^x*x^2 + 5*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.
\(\displaystyle \int \frac {5 x^3+10 x^2+e^x \left (5 x^3+10 x^2\right )+\left (5 x^3+5 e^x x^2-5 e^4 x^2-5 x^2+2 x\right ) \log \left (\frac {5 x^2+5 e^x x-5 e^4 x-5 x+2}{5 x}\right )-2 x-4}{5 x^3+5 e^x x^2-5 e^4 x^2-5 x^2+2 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {5 x^3+10 x^2+e^x \left (5 x^3+10 x^2\right )+\left (5 x^3+5 e^x x^2-5 e^4 x^2-5 x^2+2 x\right ) \log \left (\frac {5 x^2+5 e^x x-5 e^4 x-5 x+2}{5 x}\right )-2 x-4}{5 x^3+5 e^x x^2+\left (-5-5 e^4\right ) x^2+2 x}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {(x+2) \left (5 \left (e^x+1\right ) x^2-2\right )}{x \left (5 x^2+5 \left (e^x-1-e^4\right ) x+2\right )}+\log \left (x+e^x+\frac {2}{5 x}-e^4-1\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {1}{-5 x^2-5 e^x x+5 \left (1+e^4\right ) x-2}dx+4 \int \frac {1}{x \left (-5 x^2-5 e^x x+5 \left (1+e^4\right ) x-2\right )}dx-2 \int \frac {x}{-5 x^2-5 e^x x+5 \left (1+e^4\right ) x-2}dx+2 \left (9+5 e^4\right ) \int \frac {x}{5 x^2+5 e^x x-5 \left (1+e^4\right ) x+2}dx-5 \left (2+e^4\right ) \int \frac {x^2}{5 x^2+5 e^x x-5 \left (1+e^4\right ) x+2}dx+5 e^4 \int \frac {x^2}{5 x^2+5 e^x x-5 \left (1+e^4\right ) x+2}dx+2 x+x \log \left (x+e^x+\frac {2}{5 x}-e^4-1\right )\) |
Int[(-4 - 2*x + 10*x^2 + 5*x^3 + E^x*(10*x^2 + 5*x^3) + (2*x - 5*x^2 - 5*E ^4*x^2 + 5*E^x*x^2 + 5*x^3)*Log[(2 - 5*x - 5*E^4*x + 5*E^x*x + 5*x^2)/(5*x )])/(2*x - 5*x^2 - 5*E^4*x^2 + 5*E^x*x^2 + 5*x^3),x]
3.19.97.3.1 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[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(57\) vs. \(2(19)=38\).
Time = 0.89 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.52
method | result | size |
norman | \(2 \ln \left (\frac {5 \,{\mathrm e}^{x} x -5 x \,{\mathrm e}^{4}+5 x^{2}-5 x +2}{5 x}\right )+\ln \left (\frac {5 \,{\mathrm e}^{x} x -5 x \,{\mathrm e}^{4}+5 x^{2}-5 x +2}{5 x}\right ) x\) | \(58\) |
parallelrisch | \(\ln \left (-\frac {5 x \,{\mathrm e}^{4}-5 \,{\mathrm e}^{x} x -5 x^{2}+5 x -2}{5 x}\right ) x +2 \ln \left (-\frac {5 x \,{\mathrm e}^{4}-5 \,{\mathrm e}^{x} x -5 x^{2}+5 x -2}{5 x}\right )\) | \(58\) |
risch | \(x \ln \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )-x \ln \left (x \right )-i \pi x {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )}^{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )\right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )}^{3}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4}-\frac {2}{5}-x^{2}+\left (1-{\mathrm e}^{x}\right ) x \right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+i \pi x +2 \ln \left ({\mathrm e}^{x}-\frac {5 x \,{\mathrm e}^{4}-5 x^{2}+5 x -2}{5 x}\right )\) | \(283\) |
int(((5*exp(x)*x^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x)*ln(1/5*(5*exp(x)*x-5*x*ex p(4)+5*x^2-5*x+2)/x)+(5*x^3+10*x^2)*exp(x)+5*x^3+10*x^2-2*x-4)/(5*exp(x)*x ^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x),x,method=_RETURNVERBOSE)
2*ln(1/5*(5*exp(x)*x-5*x*exp(4)+5*x^2-5*x+2)/x)+ln(1/5*(5*exp(x)*x-5*x*exp (4)+5*x^2-5*x+2)/x)*x
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx={\left (x + 2\right )} \log \left (\frac {5 \, x^{2} - 5 \, x e^{4} + 5 \, x e^{x} - 5 \, x + 2}{5 \, x}\right ) \]
integrate(((5*exp(x)*x^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x)*log(1/5*(5*exp(x)*x -5*x*exp(4)+5*x^2-5*x+2)/x)+(5*x^3+10*x^2)*exp(x)+5*x^3+10*x^2-2*x-4)/(5*e xp(x)*x^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).
Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13 \[ \int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx=x \log {\left (\frac {x^{2} + x e^{x} - x e^{4} - x + \frac {2}{5}}{x} \right )} + 2 \log {\left (e^{x} + \frac {5 x^{2} - 5 x e^{4} - 5 x + 2}{5 x} \right )} \]
integrate(((5*exp(x)*x**2-5*x**2*exp(4)+5*x**3-5*x**2+2*x)*ln(1/5*(5*exp(x )*x-5*x*exp(4)+5*x**2-5*x+2)/x)+(5*x**3+10*x**2)*exp(x)+5*x**3+10*x**2-2*x -4)/(5*exp(x)*x**2-5*x**2*exp(4)+5*x**3-5*x**2+2*x),x)
x*log((x**2 + x*exp(x) - x*exp(4) - x + 2/5)/x) + 2*log(exp(x) + (5*x**2 - 5*x*exp(4) - 5*x + 2)/(5*x))
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (19) = 38\).
Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.61 \[ \int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx=-x \log \left (5\right ) + x \log \left (5 \, x^{2} - 5 \, x {\left (e^{4} + 1\right )} + 5 \, x e^{x} + 2\right ) - x \log \left (x\right ) + 2 \, \log \left (\frac {5 \, x^{2} - 5 \, x {\left (e^{4} + 1\right )} + 5 \, x e^{x} + 2}{5 \, x}\right ) \]
integrate(((5*exp(x)*x^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x)*log(1/5*(5*exp(x)*x -5*x*exp(4)+5*x^2-5*x+2)/x)+(5*x^3+10*x^2)*exp(x)+5*x^3+10*x^2-2*x-4)/(5*e xp(x)*x^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x),x, algorithm=\
-x*log(5) + x*log(5*x^2 - 5*x*(e^4 + 1) + 5*x*e^x + 2) - x*log(x) + 2*log( 1/5*(5*x^2 - 5*x*(e^4 + 1) + 5*x*e^x + 2)/x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (19) = 38\).
Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.43 \[ \int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx=x \log \left (\frac {5 \, x^{2} - 5 \, x e^{4} + 5 \, x e^{x} - 5 \, x + 2}{5 \, x}\right ) + 2 \, \log \left (-5 \, x^{2} + 5 \, x e^{4} - 5 \, x e^{x} + 5 \, x - 2\right ) - 2 \, \log \left (x\right ) \]
integrate(((5*exp(x)*x^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x)*log(1/5*(5*exp(x)*x -5*x*exp(4)+5*x^2-5*x+2)/x)+(5*x^3+10*x^2)*exp(x)+5*x^3+10*x^2-2*x-4)/(5*e xp(x)*x^2-5*x^2*exp(4)+5*x^3-5*x^2+2*x),x, algorithm=\
x*log(1/5*(5*x^2 - 5*x*e^4 + 5*x*e^x - 5*x + 2)/x) + 2*log(-5*x^2 + 5*x*e^ 4 - 5*x*e^x + 5*x - 2) - 2*log(x)
Timed out. \[ \int \frac {-4-2 x+10 x^2+5 x^3+e^x \left (10 x^2+5 x^3\right )+\left (2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3\right ) \log \left (\frac {2-5 x-5 e^4 x+5 e^x x+5 x^2}{5 x}\right )}{2 x-5 x^2-5 e^4 x^2+5 e^x x^2+5 x^3} \, dx=\int \frac {{\mathrm {e}}^x\,\left (5\,x^3+10\,x^2\right )-2\,x+\ln \left (\frac {x\,{\mathrm {e}}^x-x\,{\mathrm {e}}^4-x+x^2+\frac {2}{5}}{x}\right )\,\left (2\,x+5\,x^2\,{\mathrm {e}}^x-5\,x^2\,{\mathrm {e}}^4-5\,x^2+5\,x^3\right )+10\,x^2+5\,x^3-4}{2\,x+5\,x^2\,{\mathrm {e}}^x-5\,x^2\,{\mathrm {e}}^4-5\,x^2+5\,x^3} \,d x \]
int((exp(x)*(10*x^2 + 5*x^3) - 2*x + log((x*exp(x) - x*exp(4) - x + x^2 + 2/5)/x)*(2*x + 5*x^2*exp(x) - 5*x^2*exp(4) - 5*x^2 + 5*x^3) + 10*x^2 + 5*x ^3 - 4)/(2*x + 5*x^2*exp(x) - 5*x^2*exp(4) - 5*x^2 + 5*x^3),x)