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

Optimal. Leaf size=10 \[ \frac {3}{(-16+x) x^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1594, 27, 74} \begin {gather*} -\frac {3}{(16-x) x^2} \end {gather*}

Antiderivative was successfully verified.

[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 74

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 1594

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \begin {gather*} \frac {3}{(-16+x) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3/((-16 + x)*x^2)

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fricas [A]  time = 0.46, size = 13, normalized size = 1.30 \begin {gather*} \frac {3}{x^{3} - 16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(x^3 - 16*x^2)

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giac [A]  time = 0.23, size = 16, normalized size = 1.60 \begin {gather*} \frac {3}{256 \, {\left (x - 16\right )}} - \frac {3 \, {\left (x + 16\right )}}{256 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 11, normalized size = 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	\(\frac {3}{x^{2} \left (x -16\right )}\)	\(11\)
default	\(\frac {3}{256 \left (x -16\right )}-\frac {3}{16 x^{2}}-\frac {3}{256 x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/x^2/(x-16)

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maxima [A]  time = 0.35, size = 13, normalized size = 1.30 \begin {gather*} \frac {3}{x^{3} - 16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(x^3 - 16*x^2)

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mupad [B]  time = 3.22, size = 15, normalized size = 1.50 \begin {gather*} -\frac {3}{16\,x^2-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/(16*x^2 - x^3)

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sympy [A]  time = 0.08, size = 8, normalized size = 0.80 \begin {gather*} \frac {3}{x^{3} - 16 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(x**3 - 16*x**2)

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