3.4.75 \(\int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx\)

Optimal. Leaf size=17 \[ -4-3 x-\frac {x}{4+x}-\log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1594, 27, 1620} \begin {gather*} -3 x+\frac {4}{x+4}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - 60*x - 25*x^2 - 3*x^3)/(16*x + 8*x^2 + x^3),x]

[Out]

-3*x + 4/(4 + x) - Log[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 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]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16-60 x-25 x^2-3 x^3}{x \left (16+8 x+x^2\right )} \, dx\\ &=\int \frac {-16-60 x-25 x^2-3 x^3}{x (4+x)^2} \, dx\\ &=\int \left (-3-\frac {1}{x}-\frac {4}{(4+x)^2}\right ) \, dx\\ &=-3 x+\frac {4}{4+x}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.88 \begin {gather*} -3 x+\frac {4}{4+x}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 60*x - 25*x^2 - 3*x^3)/(16*x + 8*x^2 + x^3),x]

[Out]

-3*x + 4/(4 + x) - Log[x]

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fricas [A]  time = 0.63, size = 23, normalized size = 1.35 \begin {gather*} -\frac {3 \, x^{2} + {\left (x + 4\right )} \log \relax (x) + 12 \, x - 4}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^3-25*x^2-60*x-16)/(x^3+8*x^2+16*x),x, algorithm="fricas")

[Out]

-(3*x^2 + (x + 4)*log(x) + 12*x - 4)/(x + 4)

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giac [A]  time = 0.59, size = 16, normalized size = 0.94 \begin {gather*} -3 \, x + \frac {4}{x + 4} - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^3-25*x^2-60*x-16)/(x^3+8*x^2+16*x),x, algorithm="giac")

[Out]

-3*x + 4/(x + 4) - log(abs(x))

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maple [A]  time = 0.03, size = 16, normalized size = 0.94

method	result	size
default	\(-3 x -\ln \relax (x )+\frac {4}{4+x}\)	\(16\)
risch	\(-3 x -\ln \relax (x )+\frac {4}{4+x}\)	\(16\)
norman	\(\frac {-3 x^{2}+52}{4+x}-\ln \relax (x )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^3-25*x^2-60*x-16)/(x^3+8*x^2+16*x),x,method=_RETURNVERBOSE)

[Out]

-3*x-ln(x)+4/(4+x)

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maxima [A]  time = 0.65, size = 15, normalized size = 0.88 \begin {gather*} -3 \, x + \frac {4}{x + 4} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^3-25*x^2-60*x-16)/(x^3+8*x^2+16*x),x, algorithm="maxima")

[Out]

-3*x + 4/(x + 4) - log(x)

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mupad [B]  time = 0.39, size = 15, normalized size = 0.88 \begin {gather*} \frac {4}{x+4}-\ln \relax (x)-3\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(60*x + 25*x^2 + 3*x^3 + 16)/(16*x + 8*x^2 + x^3),x)

[Out]

4/(x + 4) - log(x) - 3*x

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sympy [A]  time = 0.08, size = 10, normalized size = 0.59 \begin {gather*} - 3 x - \log {\relax (x )} + \frac {4}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**3-25*x**2-60*x-16)/(x**3+8*x**2+16*x),x)

[Out]

-3*x - log(x) + 4/(x + 4)

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