Integrand size = 72, antiderivative size = 18 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=-25+\frac {1}{64} e^{-\frac {10}{-3+x}}-x \] Output:
9/400*x^2/exp(5/(-3+x)+ln(6/5*x))^2-x-25
Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {1}{32} \left (\frac {1}{2} e^{\frac {10}{3-x}}-32 x\right ) \] Input:
Integrate[(9*x^2 + E^((2*(5 + (-3 + x)*Log[(6*x)/5]))/(-3 + x))*(-360 + 24 0*x - 40*x^2))/(E^((2*(5 + (-3 + x)*Log[(6*x)/5]))/(-3 + x))*(360 - 240*x + 40*x^2)),x]
Output:
(E^(10/(3 - x))/2 - 32*x)/32
Time = 0.69 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {7239, 27, 7281, 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.
| \(\displaystyle \int \frac {e^{-\frac {2 \left ((x-3) \log \left (\frac {6 x}{5}\right )+5\right )}{x-3}} \left (9 x^2+\left (-40 x^2+240 x-360\right ) e^{\frac {2 \left ((x-3) \log \left (\frac {6 x}{5}\right )+5\right )}{x-3}}\right )}{40 x^2-240 x+360} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {5 e^{-\frac {10}{x-3}}-32 (x-3)^2}{32 (3-x)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{32} \int \frac {5 e^{\frac {10}{3-x}}-32 (3-x)^2}{(3-x)^2}dx\) |
| \(\Big \downarrow \) 7281 |
| \(\displaystyle -\frac {1}{32} \int \frac {5 e^{\frac {10}{3-x}}-32 (3-x)^2}{(3-x)^2}d(3-x)\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle -\frac {1}{32} \int \left (\frac {5 e^{\frac {10}{3-x}}}{(3-x)^2}-32\right )d(3-x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{32} \left (32 (3-x)+\frac {1}{2} e^{\frac {10}{3-x}}\right )\) |
Input:
Int[(9*x^2 + E^((2*(5 + (-3 + x)*Log[(6*x)/5]))/(-3 + x))*(-360 + 240*x - 40*x^2))/(E^((2*(5 + (-3 + x)*Log[(6*x)/5]))/(-3 + x))*(360 - 240*x + 40*x ^2)),x]
Output:
(E^(10/(3 - x))/2 + 32*(3 - x))/32
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]]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Time = 0.88 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78
| method | result | size |
| risch | \(-x +\frac {9 x^{2} {\mathrm e}^{-\frac {2 \left (\ln \left (\frac {6 x}{5}\right ) x -3 \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{400}\) | \(32\) |
| default | \(-x +\frac {\left (54 x^{3}-162 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{-7200+2400 x}\) | \(42\) |
| parts | \(-x +\frac {\left (54 x^{3}-162 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{-7200+2400 x}\) | \(42\) |
| parallelrisch | \(\frac {\left (-10800 \,{\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}} x^{2}+243 x^{3}+10800 \,{\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}} x -729 x^{2}+64800 \,{\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{-32400+10800 x}\) | \(105\) |
| orering | \(\frac {\left (x +\frac {9}{10}\right ) \left (\left (-40 x^{2}+240 x -360\right ) {\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}}+9 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{40 x^{2}-240 x +360}-\frac {x \left (x -16\right ) \left (-3+x \right )^{2} \left (\frac {\left (\left (-80 x +240\right ) {\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}}+2 \left (-40 x^{2}+240 x -360\right ) {\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}} \left (\frac {\ln \left (\frac {6 x}{5}\right )+\frac {-3+x}{x}}{-3+x}-\frac {\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5}{\left (-3+x \right )^{2}}\right )+18 x \right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}}}{40 x^{2}-240 x +360}-\frac {\left (\left (-40 x^{2}+240 x -360\right ) {\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}}+9 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}} \left (80 x -240\right )}{\left (40 x^{2}-240 x +360\right )^{2}}-\frac {2 \left (\left (-40 x^{2}+240 x -360\right ) {\mathrm e}^{\frac {2 \left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+10}{-3+x}}+9 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{-3+x}} \left (\frac {\ln \left (\frac {6 x}{5}\right )+\frac {-3+x}{x}}{-3+x}-\frac {\left (-3+x \right ) \ln \left (\frac {6 x}{5}\right )+5}{\left (-3+x \right )^{2}}\right )}{40 x^{2}-240 x +360}\right )}{20 \left (-8+x \right )}\) | \(397\) |
Input:
int(((-40*x^2+240*x-360)*exp(((-3+x)*ln(6/5*x)+5)/(-3+x))^2+9*x^2)/(40*x^2 -240*x+360)/exp(((-3+x)*ln(6/5*x)+5)/(-3+x))^2,x,method=_RETURNVERBOSE)
Output:
-x+9/400*x^2*exp(-2*(ln(6/5*x)*x-3*ln(6/5*x)+5)/(-3+x))
Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.61 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {1}{400} \, {\left (9 \, x^{2} - 400 \, x e^{\left (\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}\right )} e^{\left (-\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )} \] Input:
integrate(((-40*x^2+240*x-360)*exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2+9*x^2)/ (40*x^2-240*x+360)/exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2,x, algorithm="frica s")
Output:
1/400*(9*x^2 - 400*x*e^(2*((x - 3)*log(6/5*x) + 5)/(x - 3)))*e^(-2*((x - 3 )*log(6/5*x) + 5)/(x - 3))
Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {9 x^{2} e^{- \frac {2 \left (\left (x - 3\right ) \log {\left (\frac {6 x}{5} \right )} + 5\right )}{x - 3}}}{400} - x \] Input:
integrate(((-40*x**2+240*x-360)*exp(((-3+x)*ln(6/5*x)+5)/(-3+x))**2+9*x**2 )/(40*x**2-240*x+360)/exp(((-3+x)*ln(6/5*x)+5)/(-3+x))**2,x)
Output:
9*x**2*exp(-2*((x - 3)*log(6*x/5) + 5)/(x - 3))/400 - x
Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=-x + \frac {1}{64} \, e^{\left (-\frac {10}{x - 3}\right )} \] Input:
integrate(((-40*x^2+240*x-360)*exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2+9*x^2)/ (40*x^2-240*x+360)/exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2,x, algorithm="maxim a")
Output:
-x + 1/64*e^(-10/(x - 3))
\[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\int { \frac {{\left (9 \, x^{2} - 40 \, {\left (x^{2} - 6 \, x + 9\right )} e^{\left (\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}\right )} e^{\left (-\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}}{40 \, {\left (x^{2} - 6 \, x + 9\right )}} \,d x } \] Input:
integrate(((-40*x^2+240*x-360)*exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2+9*x^2)/ (40*x^2-240*x+360)/exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2,x, algorithm="giac" )
Output:
integrate(1/40*(9*x^2 - 40*(x^2 - 6*x + 9)*e^(2*((x - 3)*log(6/5*x) + 5)/( x - 3)))*e^(-2*((x - 3)*log(6/5*x) + 5)/(x - 3))/(x^2 - 6*x + 9), x)
Timed out. \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=-\int \frac {{\mathrm {e}}^{-\frac {2\,\left (\ln \left (\frac {6\,x}{5}\right )\,\left (x-3\right )+5\right )}{x-3}}\,\left ({\mathrm {e}}^{\frac {2\,\left (\ln \left (\frac {6\,x}{5}\right )\,\left (x-3\right )+5\right )}{x-3}}\,\left (40\,x^2-240\,x+360\right )-9\,x^2\right )}{40\,x^2-240\,x+360} \,d x \] Input:
int(-(exp(-(2*(log((6*x)/5)*(x - 3) + 5))/(x - 3))*(exp((2*(log((6*x)/5)*( x - 3) + 5))/(x - 3))*(40*x^2 - 240*x + 360) - 9*x^2))/(40*x^2 - 240*x + 3 60),x)
Output:
-int((exp(-(2*(log((6*x)/5)*(x - 3) + 5))/(x - 3))*(exp((2*(log((6*x)/5)*( x - 3) + 5))/(x - 3))*(40*x^2 - 240*x + 360) - 9*x^2))/(40*x^2 - 240*x + 3 60), x)
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{360-240 x+40 x^2} \, dx=\frac {-64 e^{\frac {10}{x -3}} x +1}{64 e^{\frac {10}{x -3}}} \] Input:
int(((-40*x^2+240*x-360)*exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2+9*x^2)/(40*x^ 2-240*x+360)/exp(((-3+x)*log(6/5*x)+5)/(-3+x))^2,x)
Output:
( - 64*e**(10/(x - 3))*x + 1)/(64*e**(10/(x - 3)))