Integrand size = 112, antiderivative size = 31 \[ \int \frac {2+e^{1+e^{e^3} \left (-1+6 e^x-9 e^{2 x}\right )+x-6 e^x x+9 e^{2 x} x} \left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^2-18 x^3\right )+e^{2 x} \left (27 x^2+54 x^3\right )\right )}{3 x^2} \, dx=e^{1+\left (-1+3 e^x\right )^2 \left (-e^{e^3}+x\right )}-\frac {2}{3 x} \]
Time = 2.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {2+e^{1+e^{e^3} \left (-1+6 e^x-9 e^{2 x}\right )+x-6 e^x x+9 e^{2 x} x} \left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^2-18 x^3\right )+e^{2 x} \left (27 x^2+54 x^3\right )\right )}{3 x^2} \, dx=e^{1-e^{e^3}+6 e^x \left (e^{e^3}-x\right )-9 e^{2 x} \left (e^{e^3}-x\right )+x}-\frac {2}{3 x} \]
Integrate[(2 + E^(1 + E^E^3*(-1 + 6*E^x - 9*E^(2*x)) + x - 6*E^x*x + 9*E^( 2*x)*x)*(3*x^2 + E^E^3*(18*E^x*x^2 - 54*E^(2*x)*x^2) + E^x*(-18*x^2 - 18*x ^3) + E^(2*x)*(27*x^2 + 54*x^3)))/(3*x^2),x]
Time = 0.92 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {27, 2010, 2009}
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.
\(\displaystyle \int \frac {\left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^3-18 x^2\right )+e^{2 x} \left (54 x^3+27 x^2\right )\right ) \exp \left (e^{e^3} \left (6 e^x-9 e^{2 x}-1\right )-6 e^x x+9 e^{2 x} x+x+1\right )+2}{3 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {3 \exp \left (-e^{e^3} \left (1-6 e^x+9 e^{2 x}\right )-6 e^x x+9 e^{2 x} x+x+1\right ) \left (x^2+6 e^{e^3} \left (e^x x^2-3 e^{2 x} x^2\right )-6 e^x \left (x^3+x^2\right )+9 e^{2 x} \left (2 x^3+x^2\right )\right )+2}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{3} \int \left (3 \exp \left (-e^{e^3} \left (1-3 e^x\right )^2-6 e^x x+9 e^{2 x} x+x+1\right ) \left (1-3 e^x\right ) \left (-6 e^x x-3 e^x \left (1-2 e^{e^3}\right )+1\right )+\frac {2}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (3 \exp \left (-e^{e^3} \left (1-3 e^x\right )^2-6 e^x x+9 e^{2 x} x+x+1\right )-\frac {2}{x}\right )\) |
Int[(2 + E^(1 + E^E^3*(-1 + 6*E^x - 9*E^(2*x)) + x - 6*E^x*x + 9*E^(2*x)*x )*(3*x^2 + E^E^3*(18*E^x*x^2 - 54*E^(2*x)*x^2) + E^x*(-18*x^2 - 18*x^3) + E^(2*x)*(27*x^2 + 54*x^3)))/(3*x^2),x]
3.7.81.3.1 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]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 1.87 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32
method | result | size |
norman | \(\frac {-\frac {2}{3}+x \,{\mathrm e}^{\left (-9 \,{\mathrm e}^{2 x}+6 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{{\mathrm e}^{3}}+9 x \,{\mathrm e}^{2 x}-6 \,{\mathrm e}^{x} x +x +1}}{x}\) | \(41\) |
parallelrisch | \(\frac {3 x \,{\mathrm e}^{\left (-9 \,{\mathrm e}^{2 x}+6 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{{\mathrm e}^{3}}+9 x \,{\mathrm e}^{2 x}-6 \,{\mathrm e}^{x} x +x +1}-2}{3 x}\) | \(43\) |
risch | \(-\frac {2}{3 x}+{\mathrm e}^{6 \,{\mathrm e}^{{\mathrm e}^{3}+x}-6 \,{\mathrm e}^{x} x -9 \,{\mathrm e}^{2 x +{\mathrm e}^{3}}+9 x \,{\mathrm e}^{2 x}-{\mathrm e}^{{\mathrm e}^{3}}+x +1}\) | \(44\) |
int(1/3*(((-54*exp(x)^2*x^2+18*exp(x)*x^2)*exp(exp(3))+(54*x^3+27*x^2)*exp (x)^2+(-18*x^3-18*x^2)*exp(x)+3*x^2)*exp((-9*exp(x)^2+6*exp(x)-1)*exp(exp( 3))+9*x*exp(x)^2-6*exp(x)*x+x+1)+2)/x^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {2+e^{1+e^{e^3} \left (-1+6 e^x-9 e^{2 x}\right )+x-6 e^x x+9 e^{2 x} x} \left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^2-18 x^3\right )+e^{2 x} \left (27 x^2+54 x^3\right )\right )}{3 x^2} \, dx=\frac {3 \, x e^{\left (9 \, x e^{\left (2 \, x\right )} - 6 \, x e^{x} - {\left (9 \, e^{\left (2 \, x\right )} - 6 \, e^{x} + 1\right )} e^{\left (e^{3}\right )} + x + 1\right )} - 2}{3 \, x} \]
integrate(1/3*(((-54*exp(x)^2*x^2+18*exp(x)*x^2)*exp(exp(3))+(54*x^3+27*x^ 2)*exp(x)^2+(-18*x^3-18*x^2)*exp(x)+3*x^2)*exp((-9*exp(x)^2+6*exp(x)-1)*ex p(exp(3))+9*x*exp(x)^2-6*exp(x)*x+x+1)+2)/x^2,x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {2+e^{1+e^{e^3} \left (-1+6 e^x-9 e^{2 x}\right )+x-6 e^x x+9 e^{2 x} x} \left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^2-18 x^3\right )+e^{2 x} \left (27 x^2+54 x^3\right )\right )}{3 x^2} \, dx=e^{9 x e^{2 x} - 6 x e^{x} + x + \left (- 9 e^{2 x} + 6 e^{x} - 1\right ) e^{e^{3}} + 1} - \frac {2}{3 x} \]
integrate(1/3*(((-54*exp(x)**2*x**2+18*exp(x)*x**2)*exp(exp(3))+(54*x**3+2 7*x**2)*exp(x)**2+(-18*x**3-18*x**2)*exp(x)+3*x**2)*exp((-9*exp(x)**2+6*ex p(x)-1)*exp(exp(3))+9*x*exp(x)**2-6*exp(x)*x+x+1)+2)/x**2,x)
Time = 0.39 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {2+e^{1+e^{e^3} \left (-1+6 e^x-9 e^{2 x}\right )+x-6 e^x x+9 e^{2 x} x} \left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^2-18 x^3\right )+e^{2 x} \left (27 x^2+54 x^3\right )\right )}{3 x^2} \, dx=-\frac {2}{3 \, x} + e^{\left (9 \, x e^{\left (2 \, x\right )} - 6 \, x e^{x} + x - 9 \, e^{\left (2 \, x + e^{3}\right )} + 6 \, e^{\left (x + e^{3}\right )} - e^{\left (e^{3}\right )} + 1\right )} \]
integrate(1/3*(((-54*exp(x)^2*x^2+18*exp(x)*x^2)*exp(exp(3))+(54*x^3+27*x^ 2)*exp(x)^2+(-18*x^3-18*x^2)*exp(x)+3*x^2)*exp((-9*exp(x)^2+6*exp(x)-1)*ex p(exp(3))+9*x*exp(x)^2-6*exp(x)*x+x+1)+2)/x^2,x, algorithm=\
\[ \int \frac {2+e^{1+e^{e^3} \left (-1+6 e^x-9 e^{2 x}\right )+x-6 e^x x+9 e^{2 x} x} \left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^2-18 x^3\right )+e^{2 x} \left (27 x^2+54 x^3\right )\right )}{3 x^2} \, dx=\int { \frac {3 \, {\left (x^{2} + 9 \, {\left (2 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} - 6 \, {\left (x^{3} + x^{2}\right )} e^{x} - 6 \, {\left (3 \, x^{2} e^{\left (2 \, x\right )} - x^{2} e^{x}\right )} e^{\left (e^{3}\right )}\right )} e^{\left (9 \, x e^{\left (2 \, x\right )} - 6 \, x e^{x} - {\left (9 \, e^{\left (2 \, x\right )} - 6 \, e^{x} + 1\right )} e^{\left (e^{3}\right )} + x + 1\right )} + 2}{3 \, x^{2}} \,d x } \]
integrate(1/3*(((-54*exp(x)^2*x^2+18*exp(x)*x^2)*exp(exp(3))+(54*x^3+27*x^ 2)*exp(x)^2+(-18*x^3-18*x^2)*exp(x)+3*x^2)*exp((-9*exp(x)^2+6*exp(x)-1)*ex p(exp(3))+9*x*exp(x)^2-6*exp(x)*x+x+1)+2)/x^2,x, algorithm=\
integrate(1/3*(3*(x^2 + 9*(2*x^3 + x^2)*e^(2*x) - 6*(x^3 + x^2)*e^x - 6*(3 *x^2*e^(2*x) - x^2*e^x)*e^(e^3))*e^(9*x*e^(2*x) - 6*x*e^x - (9*e^(2*x) - 6 *e^x + 1)*e^(e^3) + x + 1) + 2)/x^2, x)
Time = 9.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {2+e^{1+e^{e^3} \left (-1+6 e^x-9 e^{2 x}\right )+x-6 e^x x+9 e^{2 x} x} \left (3 x^2+e^{e^3} \left (18 e^x x^2-54 e^{2 x} x^2\right )+e^x \left (-18 x^2-18 x^3\right )+e^{2 x} \left (27 x^2+54 x^3\right )\right )}{3 x^2} \, dx={\mathrm {e}}^{-6\,x\,{\mathrm {e}}^x}\,\mathrm {e}\,{\mathrm {e}}^{6\,{\mathrm {e}}^{{\mathrm {e}}^3}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{9\,x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-{\mathrm {e}}^{{\mathrm {e}}^3}}\,{\mathrm {e}}^{-9\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\mathrm {e}}^3}}\,{\mathrm {e}}^x-\frac {2}{3\,x} \]
int(((exp(x + 9*x*exp(2*x) - exp(exp(3))*(9*exp(2*x) - 6*exp(x) + 1) - 6*x *exp(x) + 1)*(exp(exp(3))*(18*x^2*exp(x) - 54*x^2*exp(2*x)) - exp(x)*(18*x ^2 + 18*x^3) + exp(2*x)*(27*x^2 + 54*x^3) + 3*x^2))/3 + 2/3)/x^2,x)