\(\int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx\) [375]

Optimal result

Integrand size = 30, antiderivative size = 17 \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=-4-3 x-\frac {x}{4+x}-\log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1608, 27, 1634} \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=-3 x+\frac {4}{x+4}-\log (x) \]

[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 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]

Rule 1634

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=-3 x+\frac {4}{4+x}-\log (x) \]

[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]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(-3 x -\ln \left (x \right )+\frac {4}{4+x}\)	\(16\)
risch	\(-3 x -\ln \left (x \right )+\frac {4}{4+x}\)	\(16\)
norman	\(\frac {-3 x^{2}+52}{4+x}-\ln \left (x \right )\)	\(19\)
parallelrisch	\(-\frac {x \ln \left (x \right )+3 x^{2}-52+4 \ln \left (x \right )}{4+x}\)	\(23\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=-\frac {3 \, x^{2} + {\left (x + 4\right )} \log \left (x\right ) + 12 \, x - 4}{x + 4} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=- 3 x - \log {\left (x \right )} + \frac {4}{x + 4} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=-3 \, x + \frac {4}{x + 4} - \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=-3 \, x + \frac {4}{x + 4} - \log \left ({\left | x \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-16-60 x-25 x^2-3 x^3}{16 x+8 x^2+x^3} \, dx=\frac {4}{x+4}-\ln \left (x\right )-3\,x \]

[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