\(\int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx\) [5203]

Optimal result

Integrand size = 22, antiderivative size = 10 \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=\frac {3}{(-16+x) x^2} \]

[Out]

3/x^2/(x-16)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1608, 27, 75} \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=-\frac {3}{(16-x) x^2} \]

[In]

Int[(96 - 9*x)/(256*x^3 - 32*x^4 + x^5),x]

[Out]

-3/((16 - x)*x^2)

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 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

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 {96-9 x}{x^3 \left (256-32 x+x^2\right )} \, dx \\ & = \int \frac {96-9 x}{(-16+x)^2 x^3} \, dx \\ & = -\frac {3}{(16-x) x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=\frac {3}{(-16+x) x^2} \]

[In]

Integrate[(96 - 9*x)/(256*x^3 - 32*x^4 + x^5),x]

[Out]

3/((-16 + x)*x^2)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
gosper	\(\frac {3}{x^{2} \left (x -16\right )}\)	\(11\)
norman	\(\frac {3}{x^{2} \left (x -16\right )}\)	\(11\)
risch	\(\frac {3}{x^{2} \left (x -16\right )}\)	\(11\)
parallelrisch	\(\frac {3}{x^{2} \left (x -16\right )}\)	\(11\)
default	\(\frac {3}{256 \left (x -16\right )}-\frac {3}{16 x^{2}}-\frac {3}{256 x}\)	\(19\)

[In]

int((96-9*x)/(x^5-32*x^4+256*x^3),x,method=_RETURNVERBOSE)

[Out]

3/x^2/(x-16)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.30 \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=\frac {3}{x^{3} - 16 \, x^{2}} \]

[In]

integrate((96-9*x)/(x^5-32*x^4+256*x^3),x, algorithm="fricas")

[Out]

3/(x^3 - 16*x^2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=\frac {3}{x^{3} - 16 x^{2}} \]

[In]

integrate((96-9*x)/(x**5-32*x**4+256*x**3),x)

[Out]

3/(x**3 - 16*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.30 \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=\frac {3}{x^{3} - 16 \, x^{2}} \]

[In]

integrate((96-9*x)/(x^5-32*x^4+256*x^3),x, algorithm="maxima")

[Out]

3/(x^3 - 16*x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.60 \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=\frac {3}{256 \, {\left (x - 16\right )}} - \frac {3 \, {\left (x + 16\right )}}{256 \, x^{2}} \]

[In]

integrate((96-9*x)/(x^5-32*x^4+256*x^3),x, algorithm="giac")

[Out]

3/256/(x - 16) - 3/256*(x + 16)/x^2

Mupad [B] (verification not implemented)

Time = 12.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.50 \[ \int \frac {96-9 x}{256 x^3-32 x^4+x^5} \, dx=-\frac {3}{16\,x^2-x^3} \]

[In]

int(-(9*x - 96)/(256*x^3 - 32*x^4 + x^5),x)

[Out]

-3/(16*x^2 - x^3)