\(\int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx\) [3242]

Optimal result

Integrand size = 27, antiderivative size = 25 \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=-x^2+\frac {x}{\log \left (e^{\frac {4 \left (9-\frac {x}{2}\right )}{x}}\right )} \]

[Out]

x/ln(exp(2*(9-1/2*x)/x)^2)-x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {27, 12, 1608, 785} \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=-x^2-\frac {x}{2}+\frac {162}{18-x} \]

[In]

Int[(-1260*x + 143*x^2 - 4*x^3)/(648 - 72*x + 2*x^2),x]

[Out]

162/(18 - x) - x/2 - x^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 785

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1260 x+143 x^2-4 x^3}{2 (-18+x)^2} \, dx \\ & = \frac {1}{2} \int \frac {-1260 x+143 x^2-4 x^3}{(-18+x)^2} \, dx \\ & = \frac {1}{2} \int \frac {x \left (-1260+143 x-4 x^2\right )}{(-18+x)^2} \, dx \\ & = \frac {1}{2} \int \left (-1+\frac {324}{(-18+x)^2}-4 x\right ) \, dx \\ & = \frac {162}{18-x}-\frac {x}{2}-x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=\frac {1}{2} \left (666-\frac {324}{-18+x}-x-2 x^2\right ) \]

[In]

Integrate[(-1260*x + 143*x^2 - 4*x^3)/(648 - 72*x + 2*x^2),x]

[Out]

(666 - 324/(-18 + x) - x - 2*x^2)/2

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64

method	result	size
gosper	\(-\frac {x^{2} \left (2 x -35\right )}{2 \left (-18+x \right )}\)	\(16\)
default	\(-x^{2}-\frac {x}{2}-\frac {162}{-18+x}\)	\(17\)
risch	\(-x^{2}-\frac {x}{2}-\frac {162}{-18+x}\)	\(17\)
norman	\(\frac {\frac {35}{2} x^{2}-x^{3}}{-18+x}\)	\(18\)
parallelrisch	\(-\frac {2 x^{3}-35 x^{2}}{2 \left (-18+x \right )}\)	\(19\)
meijerg	\(-\frac {9 x \left (-\frac {1}{162} x^{2}-\frac {1}{3} x +12\right )}{1-\frac {x}{18}}+\frac {143 x \left (-\frac {x}{6}+6\right )}{6 \left (1-\frac {x}{18}\right )}-\frac {35 x}{1-\frac {x}{18}}\)	\(47\)

[In]

int((-4*x^3+143*x^2-1260*x)/(2*x^2-72*x+648),x,method=_RETURNVERBOSE)

[Out]

-1/2*x^2*(2*x-35)/(-18+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=-\frac {2 \, x^{3} - 35 \, x^{2} - 18 \, x + 324}{2 \, {\left (x - 18\right )}} \]

[In]

integrate((-4*x^3+143*x^2-1260*x)/(2*x^2-72*x+648),x, algorithm="fricas")

[Out]

-1/2*(2*x^3 - 35*x^2 - 18*x + 324)/(x - 18)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.48 \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=- x^{2} - \frac {x}{2} - \frac {162}{x - 18} \]

[In]

integrate((-4*x**3+143*x**2-1260*x)/(2*x**2-72*x+648),x)

[Out]

-x**2 - x/2 - 162/(x - 18)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=-x^{2} - \frac {1}{2} \, x - \frac {162}{x - 18} \]

[In]

integrate((-4*x^3+143*x^2-1260*x)/(2*x^2-72*x+648),x, algorithm="maxima")

[Out]

-x^2 - 1/2*x - 162/(x - 18)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=-x^{2} - \frac {1}{2} \, x - \frac {162}{x - 18} \]

[In]

integrate((-4*x^3+143*x^2-1260*x)/(2*x^2-72*x+648),x, algorithm="giac")

[Out]

-x^2 - 1/2*x - 162/(x - 18)

Mupad [B] (verification not implemented)

Time = 8.53 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {-1260 x+143 x^2-4 x^3}{648-72 x+2 x^2} \, dx=-\frac {x}{2}-\frac {162}{x-18}-x^2 \]

[In]

int(-(1260*x - 143*x^2 + 4*x^3)/(2*x^2 - 72*x + 648),x)

[Out]

- x/2 - 162/(x - 18) - x^2