Integrand size = 120, antiderivative size = 27 \[ \int \frac {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx=\frac {x}{-3+e^{-2 \left (\frac {e-e^4+x}{x}-\log (2)\right )}} \]
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx=\frac {e^{2+\frac {2 e}{x}} x}{-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}} \]
Integrate[(-3*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*x + E^((2*E - 2*E ^4 + 2*x - 2*x*Log[2])/x)*(-2*E + 2*E^4 + x))/(x - 6*E^((2*E - 2*E^4 + 2*x - 2*x*Log[2])/x)*x + 9*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*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.
\(\displaystyle \int \frac {\left (x+2 e^4-2 e\right ) e^{\frac {2 x-2 x \log (2)-2 e^4+2 e}{x}}-3 x e^{\frac {2 \left (2 x-2 x \log (2)-2 e^4+2 e\right )}{x}}}{x-6 x e^{\frac {2 x-2 x \log (2)-2 e^4+2 e}{x}}+9 x e^{\frac {2 \left (2 x-2 x \log (2)-2 e^4+2 e\right )}{x}}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {16 e^{\frac {4 e^4}{x}} \left (\left (x+2 e^4-2 e\right ) e^{\frac {2 x-2 x \log (2)-2 e^4+2 e}{x}}-3 x e^{\frac {2 \left (2 x-2 x \log (2)-2 e^4+2 e\right )}{x}}\right )}{\left (3 e^{\frac {2 e}{x}+2}-4 e^{\frac {2 e^4}{x}}\right )^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 16 \int -\frac {e^{\frac {4 e^4}{x}} \left (4 e^{\frac {2 \left (x+e \left (1-e^3\right )\right )}{x}} \left (2 e \left (1-e^3\right )-x\right )+3 e^{\frac {4 \left (x+e \left (1-e^3\right )\right )}{x}} x\right )}{16 \left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {e^{\frac {4 e^4}{x}} \left (4 e^{\frac {2 \left (x+e \left (1-e^3\right )\right )}{x}} \left (2 e \left (1-e^3\right )-x\right )+3 e^{\frac {4 \left (x+e \left (1-e^3\right )\right )}{x}} x\right )}{\left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {4 e^{2+\frac {2 e-2 e^4}{x}+\frac {4 e^4}{x}} \left (-x-2 e^4+2 e\right )}{\left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2 x}+\frac {3 e^{4+\frac {4 e-4 e^4}{x}+\frac {4 e^4}{x}}}{\left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{3} \text {Subst}\left (\int \frac {1}{\left (4-3 e^{2 e \left (1-e^3\right ) x+2}\right ) x^2}dx,x,\frac {1}{x}\right )-3 \int \frac {e^{4+\frac {4 e}{x}}}{\left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2}dx+4 \int \frac {e^{2+\frac {2 \left (e+e^4\right )}{x}}}{\left (-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}\right )^2}dx+\frac {4 x}{3 \left (4-3 e^{\frac {2 \left (x+e \left (1-e^3\right )\right )}{x}}\right )}\) |
Int[(-3*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*x + E^((2*E - 2*E^4 + 2 *x - 2*x*Log[2])/x)*(-2*E + 2*E^4 + x))/(x - 6*E^((2*E - 2*E^4 + 2*x - 2*x *Log[2])/x)*x + 9*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*x),x]
3.23.2.3.1 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 = 0.59 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {x}{3}-\frac {x}{3 \left (\frac {3 \,{\mathrm e}^{\frac {2 \,{\mathrm e}+2 x -2 \,{\mathrm e}^{4}}{x}}}{4}-1\right )}\) | \(28\) |
parallelrisch | \(-\frac {x \,{\mathrm e}^{-\frac {2 \left (x \ln \left (2\right )-{\mathrm e}+{\mathrm e}^{4}-x \right )}{x}}}{-1+3 \,{\mathrm e}^{-\frac {2 \left (x \ln \left (2\right )-{\mathrm e}+{\mathrm e}^{4}-x \right )}{x}}}\) | \(50\) |
norman | \(-\frac {x \,{\mathrm e}^{\frac {-2 x \ln \left (2\right )-2 \,{\mathrm e}^{4}+2 \,{\mathrm e}+2 x}{x}}}{3 \,{\mathrm e}^{\frac {-2 x \ln \left (2\right )-2 \,{\mathrm e}^{4}+2 \,{\mathrm e}+2 x}{x}}-1}\) | \(54\) |
int((-3*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*exp(1)+x )*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x))/(9*x*exp((-2*x*ln(2)-2*exp(4) +2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)+x),x,met hod=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx=-\frac {x e^{\left (-\frac {2 \, {\left (x \log \left (2\right ) - x + e^{4} - e\right )}}{x}\right )}}{3 \, e^{\left (-\frac {2 \, {\left (x \log \left (2\right ) - x + e^{4} - e\right )}}{x}\right )} - 1} \]
integrate((-3*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*e xp(1)+x)*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x))/(9*x*exp((-2*x*log(2) -2*exp(4)+2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x )+x),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx=- \frac {x}{3} - \frac {x}{9 e^{\frac {- 2 x \log {\left (2 \right )} + 2 x - 2 e^{4} + 2 e}{x}} - 3} \]
integrate((-3*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)**2+(2*exp(4)-2*e xp(1)+x)*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x))/(9*x*exp((-2*x*ln(2)-2 *exp(4)+2*exp(1)+2*x)/x)**2-6*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)+ x),x)
Time = 0.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx=\frac {x e^{\left (\frac {2 \, e}{x} + 2\right )}}{4 \, e^{\left (\frac {2 \, e^{4}}{x}\right )} - 3 \, e^{\left (\frac {2 \, e}{x} + 2\right )}} \]
integrate((-3*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*e xp(1)+x)*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x))/(9*x*exp((-2*x*log(2) -2*exp(4)+2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x )+x),x, algorithm=\
Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx=-\frac {x e^{\left (\frac {2 \, {\left (x - e^{4} + e\right )}}{x}\right )}}{3 \, e^{\left (\frac {2 \, {\left (x - e^{4} + e\right )}}{x}\right )} - 4} \]
integrate((-3*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*e xp(1)+x)*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x))/(9*x*exp((-2*x*log(2) -2*exp(4)+2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x )+x),x, algorithm=\
Time = 14.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx=-\frac {x}{3}-\frac {x}{3\,\left (\frac {3\,{\mathrm {e}}^{\frac {2\,\mathrm {e}}{x}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^4}{x}}\,{\mathrm {e}}^2}{4}-1\right )} \]