Integrand size = 115, antiderivative size = 32 \[ \int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx=-x+\frac {\frac {5+x}{2+e^{e+x^2}}-\log (x)}{3+e^3} \]
Time = 2.83 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx=-\frac {\frac {-5-x}{2+e^{e+x^2}}+\left (3+e^3\right ) x+\log (x)}{3+e^3} \]
Integrate[(-4 - 10*x - 4*E^3*x + E^(2*E + 2*x^2)*(-1 - 3*x - E^3*x) + E^(E + x^2)*(-4 - 11*x - 4*E^3*x - 10*x^2 - 2*x^3))/(12*x + 4*E^3*x + E^(2*E + 2*x^2)*(3*x + E^3*x) + E^(E + x^2)*(12*x + 4*E^3*x)),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.
\(\displaystyle \int \frac {e^{2 x^2+2 e} \left (-e^3 x-3 x-1\right )+e^{x^2+e} \left (-2 x^3-10 x^2-4 e^3 x-11 x-4\right )-4 e^3 x-10 x-4}{e^{2 x^2+2 e} \left (e^3 x+3 x\right )+e^{x^2+e} \left (4 e^3 x+12 x\right )+4 e^3 x+12 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{2 x^2+2 e} \left (-e^3 x-3 x-1\right )+e^{x^2+e} \left (-2 x^3-10 x^2-4 e^3 x-11 x-4\right )-4 e^3 x-10 x-4}{e^{2 x^2+2 e} \left (e^3 x+3 x\right )+e^{x^2+e} \left (4 e^3 x+12 x\right )+\left (12+4 e^3\right ) x}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{2 x^2+2 e} \left (-e^3 x-3 x-1\right )+e^{x^2+e} \left (-2 x^3-10 x^2-4 e^3 x-11 x-4\right )+\left (-10-4 e^3\right ) x-4}{e^{2 x^2+2 e} \left (e^3 x+3 x\right )+e^{x^2+e} \left (4 e^3 x+12 x\right )+\left (12+4 e^3\right ) x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x^2+2 e} \left (-e^3 x-3 x-1\right )+e^{x^2+e} \left (-2 x^3-10 x^2-4 e^3 x-11 x-4\right )+\left (-10-4 e^3\right ) x-4}{\left (3+e^3\right ) \left (e^{x^2+e}+2\right )^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {2 \left (5+2 e^3\right ) x+e^{2 x^2+2 e} \left (e^3 x+3 x+1\right )+e^{x^2+e} \left (2 x^3+10 x^2+4 e^3 x+11 x+4\right )+4}{\left (2+e^{x^2+e}\right )^2 x}dx}{3+e^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {2 \left (5+2 e^3\right ) x+e^{2 x^2+2 e} \left (e^3 x+3 x+1\right )+e^{x^2+e} \left (2 x^3+10 x^2+4 e^3 x+11 x+4\right )+4}{\left (2+e^{x^2+e}\right )^2 x}dx}{3+e^3}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (-\frac {4 x (x+5)}{\left (2+e^{x^2+e}\right )^2}+\frac {\left (3+e^3\right ) x+1}{x}+\frac {2 x^2+10 x-1}{2+e^{x^2+e}}\right )dx}{3+e^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\int \frac {1}{2+e^{x^2+e}}dx-4 \int \frac {x^2}{\left (2+e^{x^2+e}\right )^2}dx+2 \int \frac {x^2}{2+e^{x^2+e}}dx-\frac {5}{e^{x^2+e}+2}+\left (3+e^3\right ) x+\log (x)}{3+e^3}\) |
Int[(-4 - 10*x - 4*E^3*x + E^(2*E + 2*x^2)*(-1 - 3*x - E^3*x) + E^(E + x^2 )*(-4 - 11*x - 4*E^3*x - 10*x^2 - 2*x^3))/(12*x + 4*E^3*x + E^(2*E + 2*x^2 )*(3*x + E^3*x) + E^(E + x^2)*(12*x + 4*E^3*x)),x]
3.16.68.3.1 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]]
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12
method | result | size |
risch | \(-x -\frac {\ln \left (x \right )}{{\mathrm e}^{3}+3}+\frac {5+x}{\left ({\mathrm e}^{3}+3\right ) \left (2+{\mathrm e}^{{\mathrm e}+x^{2}}\right )}\) | \(36\) |
norman | \(\frac {-x \,{\mathrm e}^{{\mathrm e}+x^{2}}-\frac {\left (5+2 \,{\mathrm e}^{3}\right ) x}{{\mathrm e}^{3}+3}+\frac {5}{{\mathrm e}^{3}+3}}{2+{\mathrm e}^{{\mathrm e}+x^{2}}}-\frac {\ln \left (x \right )}{{\mathrm e}^{3}+3}\) | \(58\) |
parallelrisch | \(-\frac {{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}+x^{2}} x +\ln \left (x \right ) {\mathrm e}^{{\mathrm e}+x^{2}}+2 x \,{\mathrm e}^{3}+3 x \,{\mathrm e}^{{\mathrm e}+x^{2}}+2 \ln \left (x \right )+5 x -5}{\left ({\mathrm e}^{3}+3\right ) \left (2+{\mathrm e}^{{\mathrm e}+x^{2}}\right )}\) | \(65\) |
int(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-11*x-4) *exp(exp(1)+x^2)-4*x*exp(3)-10*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x^2)^2+(4*x *exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.88 \[ \int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx=-\frac {2 \, x e^{3} + {\left (x e^{3} + 3 \, x\right )} e^{\left (x^{2} + e\right )} + {\left (e^{\left (x^{2} + e\right )} + 2\right )} \log \left (x\right ) + 5 \, x - 5}{{\left (e^{3} + 3\right )} e^{\left (x^{2} + e\right )} + 2 \, e^{3} + 6} \]
integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-1 1*x-4)*exp(exp(1)+x^2)-4*x*exp(3)-10*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x^2)^ 2+(4*x*exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x, algorithm=\
-(2*x*e^3 + (x*e^3 + 3*x)*e^(x^2 + e) + (e^(x^2 + e) + 2)*log(x) + 5*x - 5 )/((e^3 + 3)*e^(x^2 + e) + 2*e^3 + 6)
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx=\frac {x + 5}{\left (3 + e^{3}\right ) e^{x^{2} + e} + 6 + 2 e^{3}} + \frac {- x \left (3 + e^{3}\right ) - \log {\left (x \right )}}{3 + e^{3}} \]
integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x**2)**2+(-4*x*exp(3)-2*x**3-10*x* *2-11*x-4)*exp(exp(1)+x**2)-4*x*exp(3)-10*x-4)/((x*exp(3)+3*x)*exp(exp(1)+ x**2)**2+(4*x*exp(3)+12*x)*exp(exp(1)+x**2)+4*x*exp(3)+12*x),x)
(x + 5)/((3 + exp(3))*exp(x**2 + E) + 6 + 2*exp(3)) + (-x*(3 + exp(3)) - l og(x))/(3 + exp(3))
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (31) = 62\).
Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00 \[ \int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx=-\frac {x {\left (e^{\left (e + 3\right )} + 3 \, e^{e}\right )} e^{\left (x^{2}\right )} + x {\left (2 \, e^{3} + 5\right )} - 5}{{\left (e^{\left (e + 3\right )} + 3 \, e^{e}\right )} e^{\left (x^{2}\right )} + 2 \, e^{3} + 6} - \frac {\log \left (x\right )}{e^{3} + 3} \]
integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-1 1*x-4)*exp(exp(1)+x^2)-4*x*exp(3)-10*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x^2)^ 2+(4*x*exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x, algorithm=\
-(x*(e^(e + 3) + 3*e^e)*e^(x^2) + x*(2*e^3 + 5) - 5)/((e^(e + 3) + 3*e^e)* e^(x^2) + 2*e^3 + 6) - log(x)/(e^3 + 3)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (31) = 62\).
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.22 \[ \int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx=-\frac {2 \, x e^{3} + x e^{\left (x^{2} + e + 3\right )} + 3 \, x e^{\left (x^{2} + e\right )} + e^{\left (x^{2} + e\right )} \log \left (x\right ) + 5 \, x + 2 \, \log \left (x\right ) - 5}{2 \, e^{3} + e^{\left (x^{2} + e + 3\right )} + 3 \, e^{\left (x^{2} + e\right )} + 6} \]
integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-1 1*x-4)*exp(exp(1)+x^2)-4*x*exp(3)-10*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x^2)^ 2+(4*x*exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x, algorithm=\
-(2*x*e^3 + x*e^(x^2 + e + 3) + 3*x*e^(x^2 + e) + e^(x^2 + e)*log(x) + 5*x + 2*log(x) - 5)/(2*e^3 + e^(x^2 + e + 3) + 3*e^(x^2 + e) + 6)
Time = 9.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.38 \[ \int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx=-\frac {3\,x+\ln \left (x\right )+x\,{\mathrm {e}}^3}{{\mathrm {e}}^3+3}-\frac {5\,x+2\,\ln \left (x\right )+2\,x\,{\mathrm {e}}^3-\frac {\left (2\,{\mathrm {e}}^3+6\right )\,\left (3\,x+\ln \left (x\right )+x\,{\mathrm {e}}^3+\frac {5}{2}\right )}{{\mathrm {e}}^3+3}}{\left ({\mathrm {e}}^{x^2+\mathrm {e}}+2\right )\,\left ({\mathrm {e}}^3+3\right )} \]
int(-(10*x + 4*x*exp(3) + exp(exp(1) + x^2)*(11*x + 4*x*exp(3) + 10*x^2 + 2*x^3 + 4) + exp(2*exp(1) + 2*x^2)*(3*x + x*exp(3) + 1) + 4)/(12*x + 4*x*e xp(3) + exp(2*exp(1) + 2*x^2)*(3*x + x*exp(3)) + exp(exp(1) + x^2)*(12*x + 4*x*exp(3))),x)