Integrand size = 80, antiderivative size = 35 \[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=\frac {1}{5} e^{x-\left (3-e^2\right ) \left (4+(-1+x) \left (8+\frac {3}{x}+x\right )\right )}-x \] Output:
1/5*exp(x-(-exp(2)+3)*((x+3/x+8)*(-1+x)+4))-x
Time = 0.61 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=\frac {1}{5} e^{3-e^2-\frac {3 \left (-3+e^2\right )}{x}+\left (-20+7 e^2\right ) x+\left (-3+e^2\right ) x^2}-x \] Input:
Integrate[(-5*x^2 + E^((9 + 3*x - 20*x^2 - 3*x^3 + E^2*(-3 - x + 7*x^2 + x ^3))/x)*(-9 - 20*x^2 - 6*x^3 + E^2*(3 + 7*x^2 + 2*x^3)))/(5*x^2),x]
Output:
E^(3 - E^2 - (3*(-3 + E^2))/x + (-20 + 7*E^2)*x + (-3 + E^2)*x^2)/5 - x
Time = 0.85 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {\left (-6 x^3-20 x^2+e^2 \left (2 x^3+7 x^2+3\right )-9\right ) \exp \left (\frac {-3 x^3-20 x^2+e^2 \left (x^3+7 x^2-x-3\right )+3 x+9}{x}\right )-5 x^2}{5 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {5 x^2+\exp \left (\frac {-3 x^3-20 x^2+3 x-e^2 \left (-x^3-7 x^2+x+3\right )+9}{x}\right ) \left (6 x^3+20 x^2-e^2 \left (2 x^3+7 x^2+3\right )+9\right )}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {5 x^2+\exp \left (\frac {-3 x^3-20 x^2+3 x-e^2 \left (-x^3-7 x^2+x+3\right )+9}{x}\right ) \left (6 x^3+20 x^2-e^2 \left (2 x^3+7 x^2+3\right )+9\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {\exp \left (-3 x^2-20 x+e^2 \left (x^2+7 x-1-\frac {3}{x}\right )+3+\frac {9}{x}\right ) \left (2 \left (3-e^2\right ) x^3+\left (20-7 e^2\right ) x^2+3 \left (3-e^2\right )\right )}{x^2}+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\exp \left (-3 x^2-e^2 \left (-x^2-7 x+\frac {3}{x}+1\right )-20 x+\frac {9}{x}+3\right )-5 x\right )\) |
Input:
Int[(-5*x^2 + E^((9 + 3*x - 20*x^2 - 3*x^3 + E^2*(-3 - x + 7*x^2 + x^3))/x )*(-9 - 20*x^2 - 6*x^3 + E^2*(3 + 7*x^2 + 2*x^3)))/(5*x^2),x]
Output:
(E^(3 + 9/x - 20*x - 3*x^2 - E^2*(1 + 3/x - 7*x - x^2)) - 5*x)/5
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 = 2.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23
method | result | size |
parallelrisch | \(-x +\frac {{\mathrm e}^{\frac {\left (x^{3}+7 x^{2}-x -3\right ) {\mathrm e}^{2}-3 x^{3}-20 x^{2}+3 x +9}{x}}}{5}\) | \(43\) |
parts | \(-x +\frac {{\mathrm e}^{\frac {\left (x^{3}+7 x^{2}-x -3\right ) {\mathrm e}^{2}-3 x^{3}-20 x^{2}+3 x +9}{x}}}{5}\) | \(43\) |
risch | \(-x +\frac {{\mathrm e}^{\frac {x^{3} {\mathrm e}^{2}+7 x^{2} {\mathrm e}^{2}-3 x^{3}-{\mathrm e}^{2} x -20 x^{2}-3 \,{\mathrm e}^{2}+3 x +9}{x}}}{5}\) | \(49\) |
norman | \(\frac {-x^{2}+\frac {x \,{\mathrm e}^{\frac {\left (x^{3}+7 x^{2}-x -3\right ) {\mathrm e}^{2}-3 x^{3}-20 x^{2}+3 x +9}{x}}}{5}}{x}\) | \(50\) |
Input:
int(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3)*exp( 2)-3*x^3-20*x^2+3*x+9)/x)-5*x^2)/x^2,x,method=_RETURNVERBOSE)
Output:
-x+1/5*exp(((x^3+7*x^2-x-3)*exp(2)-3*x^3-20*x^2+3*x+9)/x)
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=-x + \frac {1}{5} \, e^{\left (-\frac {3 \, x^{3} + 20 \, x^{2} - {\left (x^{3} + 7 \, x^{2} - x - 3\right )} e^{2} - 3 \, x - 9}{x}\right )} \] Input:
integrate(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3 )*exp(2)-3*x^3-20*x^2+3*x+9)/x)-5*x^2)/x^2,x, algorithm="fricas")
Output:
-x + 1/5*e^(-(3*x^3 + 20*x^2 - (x^3 + 7*x^2 - x - 3)*e^2 - 3*x - 9)/x)
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=- x + \frac {e^{\frac {- 3 x^{3} - 20 x^{2} + 3 x + \left (x^{3} + 7 x^{2} - x - 3\right ) e^{2} + 9}{x}}}{5} \] Input:
integrate(1/5*(((2*x**3+7*x**2+3)*exp(2)-6*x**3-20*x**2-9)*exp(((x**3+7*x* *2-x-3)*exp(2)-3*x**3-20*x**2+3*x+9)/x)-5*x**2)/x**2,x)
Output:
-x + exp((-3*x**3 - 20*x**2 + 3*x + (x**3 + 7*x**2 - x - 3)*exp(2) + 9)/x) /5
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=-x + \frac {1}{5} \, e^{\left (x^{2} e^{2} - 3 \, x^{2} + 7 \, x e^{2} - 20 \, x - \frac {3 \, e^{2}}{x} + \frac {9}{x} - e^{2} + 3\right )} \] Input:
integrate(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3 )*exp(2)-3*x^3-20*x^2+3*x+9)/x)-5*x^2)/x^2,x, algorithm="maxima")
Output:
-x + 1/5*e^(x^2*e^2 - 3*x^2 + 7*x*e^2 - 20*x - 3*e^2/x + 9/x - e^2 + 3)
Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=-x + \frac {1}{5} \, e^{\left (\frac {x^{3} e^{2} - 3 \, x^{3} + 7 \, x^{2} e^{2} - 20 \, x^{2} - x e^{2} + 3 \, x - 3 \, e^{2} + 9}{x}\right )} \] Input:
integrate(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3 )*exp(2)-3*x^3-20*x^2+3*x+9)/x)-5*x^2)/x^2,x, algorithm="giac")
Output:
-x + 1/5*e^((x^3*e^2 - 3*x^3 + 7*x^2*e^2 - 20*x^2 - x*e^2 + 3*x - 3*e^2 + 9)/x)
Time = 3.62 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.43 \[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=\frac {{\mathrm {e}}^{x^2\,{\mathrm {e}}^2}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^2}{x}}\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-20\,x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{9/x}\,{\mathrm {e}}^{7\,x\,{\mathrm {e}}^2}}{5}-x \] Input:
int(-((exp(-(exp(2)*(x - 7*x^2 - x^3 + 3) - 3*x + 20*x^2 + 3*x^3 - 9)/x)*( 20*x^2 - exp(2)*(7*x^2 + 2*x^3 + 3) + 6*x^3 + 9))/5 + x^2)/x^2,x)
Output:
(exp(x^2*exp(2))*exp(-(3*exp(2))/x)*exp(-exp(2))*exp(-20*x)*exp(3)*exp(-3* x^2)*exp(9/x)*exp(7*x*exp(2)))/5 - x
\[ \int \frac {-5 x^2+e^{\frac {9+3 x-20 x^2-3 x^3+e^2 \left (-3-x+7 x^2+x^3\right )}{x}} \left (-9-20 x^2-6 x^3+e^2 \left (3+7 x^2+2 x^3\right )\right )}{5 x^2} \, dx=\int \frac {\left (\left (2 x^{3}+7 x^{2}+3\right ) {\mathrm e}^{2}-6 x^{3}-20 x^{2}-9\right ) {\mathrm e}^{\frac {\left (x^{3}+7 x^{2}-x -3\right ) {\mathrm e}^{2}-3 x^{3}-20 x^{2}+3 x +9}{x}}-5 x^{2}}{5 x^{2}}d x \] Input:
int(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3)*exp( 2)-3*x^3-20*x^2+3*x+9)/x)-5*x^2)/x^2,x)
Output:
int(1/5*(((2*x^3+7*x^2+3)*exp(2)-6*x^3-20*x^2-9)*exp(((x^3+7*x^2-x-3)*exp( 2)-3*x^3-20*x^2+3*x+9)/x)-5*x^2)/x^2,x)