Integrand size = 66, antiderivative size = 33 \[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=-4-x+\frac {\left (-1+e^5\right ) \left (e^{2 x} (4-2 x)^2+x\right )}{3-x} \] Output:
(exp(5)-1)*(x+exp(x)^2*(4-2*x)^2)/(3-x)-4-x
Time = 2.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=\frac {3-3 e^5+4 e^{2 x} (-2+x)^2-4 e^{5+2 x} (-2+x)^2+3 x-x^2}{-3+x} \] Input:
Integrate[(-12 + 3*E^5 + 6*x - x^2 + E^(2*x)*(-64 + 104*x - 52*x^2 + 8*x^3 + E^5*(64 - 104*x + 52*x^2 - 8*x^3)))/(9 - 6*x + x^2),x]
Output:
(3 - 3*E^5 + 4*E^(2*x)*(-2 + x)^2 - 4*E^(5 + 2*x)*(-2 + x)^2 + 3*x - x^2)/ (-3 + x)
Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(33)=66\).
Time = 0.73 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {7277, 27, 7293, 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.
\(\displaystyle \int \frac {-x^2+e^{2 x} \left (8 x^3-52 x^2+e^5 \left (-8 x^3+52 x^2-104 x+64\right )+104 x-64\right )+6 x+3 e^5-12}{x^2-6 x+9} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {x^2-6 x+4 e^{2 x} \left (-2 x^3+13 x^2-26 x-e^5 \left (-2 x^3+13 x^2-26 x+16\right )+16\right )+3 \left (4-e^5\right )}{4 (3-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {x^2-6 x+4 e^{2 x} \left (-2 x^3+13 x^2-26 x-e^5 \left (-2 x^3+13 x^2-26 x+16\right )+16\right )+3 \left (4-e^5\right )}{(3-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {x^2-6 x+3 \left (4-e^5\right )}{(3-x)^2}+\frac {4 (-1+e) e^{2 x} \left (1+e+e^2+e^3+e^4\right ) (x-2) \left (2 x^2-9 x+8\right )}{(x-3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (1-e^5\right ) e^{2 x} x-x-\frac {4 \left (1-e^5\right ) e^{2 x}}{3-x}-\frac {3 \left (1-e^5\right )}{3-x}-4 \left (1-e^5\right ) e^{2 x}\) |
Input:
Int[(-12 + 3*E^5 + 6*x - x^2 + E^(2*x)*(-64 + 104*x - 52*x^2 + 8*x^3 + E^5 *(64 - 104*x + 52*x^2 - 8*x^3)))/(9 - 6*x + x^2),x]
Output:
-4*E^(2*x)*(1 - E^5) - (3*(1 - E^5))/(3 - x) - (4*E^(2*x)*(1 - E^5))/(3 - x) - x + 4*E^(2*x)*(1 - E^5)*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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67
method | result | size |
norman | \(\frac {\left (16-16 \,{\mathrm e}^{5}\right ) {\mathrm e}^{2 x}+\left (-16+16 \,{\mathrm e}^{5}\right ) x \,{\mathrm e}^{2 x}+\left (-4 \,{\mathrm e}^{5}+4\right ) x^{2} {\mathrm e}^{2 x}-x^{2}+12-3 \,{\mathrm e}^{5}}{-3+x}\) | \(55\) |
risch | \(-x +\frac {3}{-3+x}-\frac {3 \,{\mathrm e}^{5}}{-3+x}-\frac {4 \left (x^{2} {\mathrm e}^{5}-4 x \,{\mathrm e}^{5}-x^{2}+4 \,{\mathrm e}^{5}+4 x -4\right ) {\mathrm e}^{2 x}}{-3+x}\) | \(57\) |
parallelrisch | \(-\frac {4 \,{\mathrm e}^{2 x} {\mathrm e}^{5} x^{2}-16 \,{\mathrm e}^{2 x} {\mathrm e}^{5} x -4 \,{\mathrm e}^{2 x} x^{2}+16 \,{\mathrm e}^{2 x} {\mathrm e}^{5}+16 x \,{\mathrm e}^{2 x}+x^{2}-12-16 \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{5}}{-3+x}\) | \(67\) |
parts | \(-x -\frac {-3+3 \,{\mathrm e}^{5}}{-3+x}+\frac {4 \,{\mathrm e}^{2 x}}{-3+x}+64 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{2 x}}{-3+x}-2 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )\right )-4 \,{\mathrm e}^{2 x}+4 x \,{\mathrm e}^{2 x}-104 \,{\mathrm e}^{5} \left (-7 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )-\frac {3 \,{\mathrm e}^{2 x}}{-3+x}\right )+52 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x}}{2}-24 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )-\frac {9 \,{\mathrm e}^{2 x}}{-3+x}\right )-8 \,{\mathrm e}^{5} \left (\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {11 \,{\mathrm e}^{2 x}}{4}-81 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )-\frac {27 \,{\mathrm e}^{2 x}}{-3+x}\right )\) | \(169\) |
default | \(\frac {3}{-3+x}-x -\frac {3 \,{\mathrm e}^{5}}{-3+x}+\frac {4 \,{\mathrm e}^{2 x}}{-3+x}+64 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{2 x}}{-3+x}-2 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )\right )-4 \,{\mathrm e}^{2 x}+4 x \,{\mathrm e}^{2 x}-104 \,{\mathrm e}^{5} \left (-7 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )-\frac {3 \,{\mathrm e}^{2 x}}{-3+x}\right )+52 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x}}{2}-24 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )-\frac {9 \,{\mathrm e}^{2 x}}{-3+x}\right )-8 \,{\mathrm e}^{5} \left (\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {11 \,{\mathrm e}^{2 x}}{4}-81 \,{\mathrm e}^{6} \operatorname {expIntegral}_{1}\left (6-2 x \right )-\frac {27 \,{\mathrm e}^{2 x}}{-3+x}\right )\) | \(172\) |
Input:
int((((-8*x^3+52*x^2-104*x+64)*exp(5)+8*x^3-52*x^2+104*x-64)*exp(x)^2+3*ex p(5)-x^2+6*x-12)/(x^2-6*x+9),x,method=_RETURNVERBOSE)
Output:
((16-16*exp(5))*exp(x)^2+(-16+16*exp(5))*x*exp(x)^2+(-4*exp(5)+4)*x^2*exp( x)^2-x^2+12-3*exp(5))/(-3+x)
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=-\frac {x^{2} - 4 \, {\left (x^{2} - {\left (x^{2} - 4 \, x + 4\right )} e^{5} - 4 \, x + 4\right )} e^{\left (2 \, x\right )} - 3 \, x + 3 \, e^{5} - 3}{x - 3} \] Input:
integrate((((-8*x^3+52*x^2-104*x+64)*exp(5)+8*x^3-52*x^2+104*x-64)*exp(x)^ 2+3*exp(5)-x^2+6*x-12)/(x^2-6*x+9),x, algorithm="fricas")
Output:
-(x^2 - 4*(x^2 - (x^2 - 4*x + 4)*e^5 - 4*x + 4)*e^(2*x) - 3*x + 3*e^5 - 3) /(x - 3)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=- x + \frac {\left (- 4 x^{2} e^{5} + 4 x^{2} - 16 x + 16 x e^{5} - 16 e^{5} + 16\right ) e^{2 x}}{x - 3} - \frac {-3 + 3 e^{5}}{x - 3} \] Input:
integrate((((-8*x**3+52*x**2-104*x+64)*exp(5)+8*x**3-52*x**2+104*x-64)*exp (x)**2+3*exp(5)-x**2+6*x-12)/(x**2-6*x+9),x)
Output:
-x + (-4*x**2*exp(5) + 4*x**2 - 16*x + 16*x*exp(5) - 16*exp(5) + 16)*exp(2 *x)/(x - 3) - (-3 + 3*exp(5))/(x - 3)
\[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=\int { -\frac {x^{2} - 4 \, {\left (2 \, x^{3} - 13 \, x^{2} - {\left (2 \, x^{3} - 13 \, x^{2} + 26 \, x - 16\right )} e^{5} + 26 \, x - 16\right )} e^{\left (2 \, x\right )} - 6 \, x - 3 \, e^{5} + 12}{x^{2} - 6 \, x + 9} \,d x } \] Input:
integrate((((-8*x^3+52*x^2-104*x+64)*exp(5)+8*x^3-52*x^2+104*x-64)*exp(x)^ 2+3*exp(5)-x^2+6*x-12)/(x^2-6*x+9),x, algorithm="maxima")
Output:
-x - 4*(x^3*(e^5 - 1) - 7*x^2*(e^5 - 1) + 16*x*(e^5 - 1))*e^(2*x)/(x^2 - 6 *x + 9) + 64*e^6*exp_integral_e(2, -2*x + 6)/(x - 3) - 3*e^5/(x - 3) + 3/( x - 3) + integrate(32*(x*(3*e^5 - 1) - 12*e^5 + 6)*e^(2*x)/(x^3 - 9*x^2 + 27*x - 27), x)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (29) = 58\).
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12 \[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=\frac {4 \, x^{2} e^{\left (2 \, x\right )} - 4 \, x^{2} e^{\left (2 \, x + 5\right )} - x^{2} - 16 \, x e^{\left (2 \, x\right )} + 16 \, x e^{\left (2 \, x + 5\right )} + 3 \, x - 3 \, e^{5} + 16 \, e^{\left (2 \, x\right )} - 16 \, e^{\left (2 \, x + 5\right )} + 3}{x - 3} \] Input:
integrate((((-8*x^3+52*x^2-104*x+64)*exp(5)+8*x^3-52*x^2+104*x-64)*exp(x)^ 2+3*exp(5)-x^2+6*x-12)/(x^2-6*x+9),x, algorithm="giac")
Output:
(4*x^2*e^(2*x) - 4*x^2*e^(2*x + 5) - x^2 - 16*x*e^(2*x) + 16*x*e^(2*x + 5) + 3*x - 3*e^5 + 16*e^(2*x) - 16*e^(2*x + 5) + 3)/(x - 3)
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=-\frac {x\,\left ({\mathrm {e}}^5-4\right )+{\mathrm {e}}^{2\,x}\,\left (16\,{\mathrm {e}}^5-16\right )+x^2-x\,{\mathrm {e}}^{2\,x}\,\left (16\,{\mathrm {e}}^5-16\right )+x^2\,{\mathrm {e}}^{2\,x}\,\left (4\,{\mathrm {e}}^5-4\right )}{x-3} \] Input:
int(-(exp(2*x)*(exp(5)*(104*x - 52*x^2 + 8*x^3 - 64) - 104*x + 52*x^2 - 8* x^3 + 64) - 3*exp(5) - 6*x + x^2 + 12)/(x^2 - 6*x + 9),x)
Output:
-(x*(exp(5) - 4) + exp(2*x)*(16*exp(5) - 16) + x^2 - x*exp(2*x)*(16*exp(5) - 16) + x^2*exp(2*x)*(4*exp(5) - 4))/(x - 3)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.42 \[ \int \frac {-12+3 e^5+6 x-x^2+e^{2 x} \left (-64+104 x-52 x^2+8 x^3+e^5 \left (64-104 x+52 x^2-8 x^3\right )\right )}{9-6 x+x^2} \, dx=\frac {-4 e^{2 x} e^{5} x^{2}+16 e^{2 x} e^{5} x -16 e^{2 x} e^{5}+4 e^{2 x} x^{2}-16 e^{2 x} x +16 e^{2 x}-e^{5} x -x^{2}+4 x}{x -3} \] Input:
int((((-8*x^3+52*x^2-104*x+64)*exp(5)+8*x^3-52*x^2+104*x-64)*exp(x)^2+3*ex p(5)-x^2+6*x-12)/(x^2-6*x+9),x)
Output:
( - 4*e**(2*x)*e**5*x**2 + 16*e**(2*x)*e**5*x - 16*e**(2*x)*e**5 + 4*e**(2 *x)*x**2 - 16*e**(2*x)*x + 16*e**(2*x) - e**5*x - x**2 + 4*x)/(x - 3)