3.85.74 $\int \frac{-16+16x+53x^2-10x^3-x^4}{1+10x+27x^2+10x^3+x^4}dx$

Optimal. Leaf size=25 $$\frac{1}{4}(-\frac{4(-4+x{)}^{2}x}{1+5x+x^2}+\log (5))$$

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Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, integrand size = 41, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.098, Rules used = {1680, 1814, 21, 8} $$\begin{array}{c}-\frac{4(-80x-13)}{21-4{(x+\frac{5}{2})}^2}-x\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-16 + 16*x + 53*x^2 - 10*x^3 - x^4)/(1 + 10*x + 27*x^2 + 10*x^3 + x^4),x]

[Out]

-x - (4*(-13 - 80*x))/(21 - 4*(5/2 + x)^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\mathrm{Subst}(\int \frac{6279-5984x+1448x^2-16x^4}{{(21-4x^2)}^2}dx,x,\frac{5}{2}+x)\\ & =-\frac{4(-13-80x)}{21-4{(\frac{5}{2}+x)}^2}-\frac{1}{42}\mathrm{Subst}(\int \frac{882-168x^2}{21-4x^2}dx,x,\frac{5}{2}+x)\\ & =-\frac{4(-13-80x)}{21-4{(\frac{5}{2}+x)}^2}-\mathrm{Subst}(\int 1dx,x,\frac{5}{2}+x)\\ & =-x-\frac{4(-13-80x)}{21-4{(\frac{5}{2}+x)}^2}\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.84 $$\begin{array}{c}-x-\frac{13+80x}{1+5x+x^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 16*x + 53*x^2 - 10*x^3 - x^4)/(1 + 10*x + 27*x^2 + 10*x^3 + x^4),x]

[Out]

-x - (13 + 80*x)/(1 + 5*x + x^2)

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fricas [A]  time = 0.87, size = 25, normalized size = 1.00 $$\begin{array}{c}-\frac{x^3+5x^2+81x+13}{x^2+5x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-10*x^3+53*x^2+16*x-16)/(x^4+10*x^3+27*x^2+10*x+1),x, algorithm="fricas")

[Out]

-(x^3 + 5*x^2 + 81*x + 13)/(x^2 + 5*x + 1)

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giac [A]  time = 0.14, size = 21, normalized size = 0.84 $$\begin{array}{c}-x-\frac{80x+13}{x^2+5x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-10*x^3+53*x^2+16*x-16)/(x^4+10*x^3+27*x^2+10*x+1),x, algorithm="giac")

[Out]

-x - (80*x + 13)/(x^2 + 5*x + 1)

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maple [A]  time = 0.02, size = 21, normalized size = 0.84

method	result	size
gosper	$-\frac{x^3+56x+8}{x^2+5x+1}$	$21$
default	$-x+\frac{-80x-13}{x^2+5x+1}$	$21$
risch	$-x+\frac{-80x-13}{x^2+5x+1}$	$21$
norman	$\frac{-x^3-56x-8}{x^2+5x+1}$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4-10*x^3+53*x^2+16*x-16)/(x^4+10*x^3+27*x^2+10*x+1),x,method=_RETURNVERBOSE)

[Out]

-(x^3+56*x+8)/(x^2+5*x+1)

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maxima [A]  time = 0.40, size = 21, normalized size = 0.84 $$\begin{array}{c}-x-\frac{80x+13}{x^2+5x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-10*x^3+53*x^2+16*x-16)/(x^4+10*x^3+27*x^2+10*x+1),x, algorithm="maxima")

[Out]

-x - (80*x + 13)/(x^2 + 5*x + 1)

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mupad [B]  time = 5.24, size = 21, normalized size = 0.84 $$\begin{array}{c}-x-\frac{80x+13}{x^2+5x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x^3 - 53*x^2 - 16*x + x^4 + 16)/(10*x + 27*x^2 + 10*x^3 + x^4 + 1),x)

[Out]

- x - (80*x + 13)/(5*x + x^2 + 1)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.60 $$\begin{array}{c}-x-\frac{80x+13}{x^2+5x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4-10*x**3+53*x**2+16*x-16)/(x**4+10*x**3+27*x**2+10*x+1),x)

[Out]

-x - (80*x + 13)/(x**2 + 5*x + 1)

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