Integrand size = 71, antiderivative size = 28 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=\frac {e^{16} \left (-e^{3 x}+\frac {3}{x}+\frac {5}{e^3 x}\right )}{x} \] Output:
exp(16+ln((3/x-exp(3*x)+5/x/exp(3))/x))
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=-\frac {e^{13} \left (-5-3 e^3+e^{3+3 x} x\right )}{x^2} \] Input:
Integrate[(E^13*(5 + 3*E^3 - E^(3 + 3*x)*x)*(10 + 6*E^3 + E^(3 + 3*x)*(-x + 3*x^2)))/(x^2*(-5*x - 3*E^3*x + E^(3 + 3*x)*x^2)),x]
Output:
-((E^13*(-5 - 3*E^3 + E^(3 + 3*x)*x))/x^2)
Time = 0.64 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {6, 27, 25, 7239, 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.
\(\displaystyle \int \frac {e^{13} \left (-e^{3 x+3} x+3 e^3+5\right ) \left (e^{3 x+3} \left (3 x^2-x\right )+6 e^3+10\right )}{x^2 \left (e^{3 x+3} x^2-3 e^3 x-5 x\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{13} \left (-e^{3 x+3} x+3 e^3+5\right ) \left (e^{3 x+3} \left (3 x^2-x\right )+6 e^3+10\right )}{x^2 \left (e^{3 x+3} x^2+\left (-5-3 e^3\right ) x\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{13} \int -\frac {\left (-e^{3 x+3} x+3 e^3+5\right ) \left (2 \left (5+3 e^3\right )-e^{3 x+3} \left (x-3 x^2\right )\right )}{x^2 \left (\left (5+3 e^3\right ) x-e^{3 x+3} x^2\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e^{13} \int \frac {\left (-e^{3 x+3} x+3 e^3+5\right ) \left (2 \left (5+3 e^3\right )-e^{3 x+3} \left (x-3 x^2\right )\right )}{x^2 \left (\left (5+3 e^3\right ) x-e^{3 x+3} x^2\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle -e^{13} \int \frac {e^{3 x+3} x (3 x-1)+10 \left (1+\frac {3 e^3}{5}\right )}{x^3}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -e^{13} \int \left (\frac {e^{3 x+3} (3 x-1)}{x^2}+\frac {2 \left (5+3 e^3\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -e^{13} \left (\frac {e^{3 x+3}}{x}-\frac {5+3 e^3}{x^2}\right )\) |
Input:
Int[(E^13*(5 + 3*E^3 - E^(3 + 3*x)*x)*(10 + 6*E^3 + E^(3 + 3*x)*(-x + 3*x^ 2)))/(x^2*(-5*x - 3*E^3*x + E^(3 + 3*x)*x^2)),x]
Output:
-(E^13*(-((5 + 3*E^3)/x^2) + E^(3 + 3*x)/x))
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]]
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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 4.74 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {{\mathrm e}^{-3} {\mathrm e}^{16} \left (3 \,{\mathrm e}^{3}+5\right )-x \,{\mathrm e}^{16} {\mathrm e}^{3 x}}{x^{2}}\) | \(28\) |
parallelrisch | \(-\frac {\left (-30 \,{\mathrm e}^{9} {\mathrm e}^{\ln \left (-\frac {\left (x \,{\mathrm e}^{3} {\mathrm e}^{3 x}-3 \,{\mathrm e}^{3}-5\right ) {\mathrm e}^{-3}}{x^{2}}\right )+16}-25 \,{\mathrm e}^{\ln \left (-\frac {\left (x \,{\mathrm e}^{3} {\mathrm e}^{3 x}-3 \,{\mathrm e}^{3}-5\right ) {\mathrm e}^{-3}}{x^{2}}\right )+16} {\mathrm e}^{6}-9 \,{\mathrm e}^{12} {\mathrm e}^{\ln \left (-\frac {\left (x \,{\mathrm e}^{3} {\mathrm e}^{3 x}-3 \,{\mathrm e}^{3}-5\right ) {\mathrm e}^{-3}}{x^{2}}\right )+16}-{\mathrm e}^{12} x^{2} {\mathrm e}^{6 x} {\mathrm e}^{\ln \left (-\frac {\left (x \,{\mathrm e}^{3} {\mathrm e}^{3 x}-3 \,{\mathrm e}^{3}-5\right ) {\mathrm e}^{-3}}{x^{2}}\right )+16}+6 \,{\mathrm e}^{12} {\mathrm e}^{\ln \left (-\frac {\left (x \,{\mathrm e}^{3} {\mathrm e}^{3 x}-3 \,{\mathrm e}^{3}-5\right ) {\mathrm e}^{-3}}{x^{2}}\right )+16} x \,{\mathrm e}^{3 x}+10 \,{\mathrm e}^{9} {\mathrm e}^{\ln \left (-\frac {\left (x \,{\mathrm e}^{3} {\mathrm e}^{3 x}-3 \,{\mathrm e}^{3}-5\right ) {\mathrm e}^{-3}}{x^{2}}\right )+16} x \,{\mathrm e}^{3 x}\right ) {\mathrm e}^{-6}}{\left (x \,{\mathrm e}^{3} {\mathrm e}^{3 x}-3 \,{\mathrm e}^{3}-5\right )^{2}}\) | \(241\) |
risch | \(-\frac {\left ({\mathrm e}^{3 x +3} x -3 \,{\mathrm e}^{3}-5\right ) {\mathrm e}^{13+\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3 x +3} x +3 \,{\mathrm e}^{3}+5\right )}{x^{2}}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3 x +3} x +3 \,{\mathrm e}^{3}+5\right )}{x^{2}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{2}-\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3 x +3} x +3 \,{\mathrm e}^{3}+5\right )}{x^{2}}\right )}^{2} \operatorname {csgn}\left (i \left (-{\mathrm e}^{3 x +3} x +3 \,{\mathrm e}^{3}+5\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3 x +3} x +3 \,{\mathrm e}^{3}+5\right )}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{3 x +3} x +3 \,{\mathrm e}^{3}+5\right )\right )}{2}-i {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3 x +3} x +3 \,{\mathrm e}^{3}+5\right )}{x^{2}}\right )}^{2} \pi }}{x^{2}}\) | \(262\) |
Input:
int(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(ln((-x*exp(3)*exp(3*x)+3*e xp(3)+5)/x^2/exp(3))+16)/(x^2*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x,method=_RE TURNVERBOSE)
Output:
(1/exp(3)*exp(16)*(3*exp(3)+5)-x*exp(16)*exp(3*x))/x^2
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=-\frac {x e^{\left (3 \, x + 16\right )} - 3 \, e^{16} - 5 \, e^{13}}{x^{2}} \] Input:
integrate(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(log((-x*exp(3)*exp(3 *x)+3*exp(3)+5)/x^2/exp(3))+16)/(x^2*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x, al gorithm="fricas")
Output:
-(x*e^(3*x + 16) - 3*e^16 - 5*e^13)/x^2
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=- \frac {e^{16} e^{3 x}}{x} - \frac {- 6 e^{16} - 10 e^{13}}{2 x^{2}} \] Input:
integrate(((3*x**2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(ln((-x*exp(3)*exp(3 *x)+3*exp(3)+5)/x**2/exp(3))+16)/(x**2*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x)
Output:
-exp(16)*exp(3*x)/x - (-6*exp(16) - 10*exp(13))/(2*x**2)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=-\frac {{\left (x e^{\left (3 \, x + 3\right )} - 3 \, e^{3} - 5\right )} e^{13}}{x^{2}} \] Input:
integrate(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(log((-x*exp(3)*exp(3 *x)+3*exp(3)+5)/x^2/exp(3))+16)/(x^2*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x, al gorithm="maxima")
Output:
-(x*e^(3*x + 3) - 3*e^3 - 5)*e^13/x^2
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=-\frac {x e^{\left (3 \, x + 16\right )} - 3 \, e^{16} - 5 \, e^{13}}{x^{2}} \] Input:
integrate(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(log((-x*exp(3)*exp(3 *x)+3*exp(3)+5)/x^2/exp(3))+16)/(x^2*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x, al gorithm="giac")
Output:
-(x*e^(3*x + 16) - 3*e^16 - 5*e^13)/x^2
Time = 6.84 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=-\frac {x\,{\mathrm {e}}^{3\,x+16}-{\mathrm {e}}^{13}\,\left (3\,{\mathrm {e}}^3+5\right )}{x^2} \] Input:
int(-(exp(log((exp(-3)*(3*exp(3) - x*exp(3*x)*exp(3) + 5))/x^2) + 16)*(6*e xp(3) - exp(3*x)*exp(3)*(x - 3*x^2) + 10))/(5*x + 3*x*exp(3) - x^2*exp(3*x )*exp(3)),x)
Output:
-(x*exp(3*x + 16) - exp(13)*(3*exp(3) + 5))/x^2
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (-5 x-3 e^3 x+e^{3+3 x} x^2\right )} \, dx=\frac {e^{13} \left (-e^{3 x} e^{3} x +3 e^{3}+5\right )}{x^{2}} \] Input:
int(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(log((-x*exp(3)*exp(3*x)+3* exp(3)+5)/x^2/exp(3))+16)/(x^2*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x)
Output:
(e**13*( - e**(3*x)*e**3*x + 3*e**3 + 5))/x**2