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

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

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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 {gather*} -\frac {4 (-80 x-13)}{21-4 \left (x+\frac {5}{2}\right )^2}-x \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {6279-5984 x+1448 x^2-16 x^4}{\left (21-4 x^2\right )^2} \, dx,x,\frac {5}{2}+x\right )\\ &=-\frac {4 (-13-80 x)}{21-4 \left (\frac {5}{2}+x\right )^2}-\frac {1}{42} \operatorname {Subst}\left (\int \frac {882-168 x^2}{21-4 x^2} \, dx,x,\frac {5}{2}+x\right )\\ &=-\frac {4 (-13-80 x)}{21-4 \left (\frac {5}{2}+x\right )^2}-\operatorname {Subst}\left (\int 1 \, dx,x,\frac {5}{2}+x\right )\\ &=-x-\frac {4 (-13-80 x)}{21-4 \left (\frac {5}{2}+x\right )^2}\\ \end {aligned} \end {gather*}

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

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 {gather*} -\frac {x^{3} + 5 \, x^{2} + 81 \, x + 13}{x^{2} + 5 \, x + 1} \end {gather*}

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 {gather*} -x - \frac {80 \, x + 13}{x^{2} + 5 \, x + 1} \end {gather*}

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	\(-\frac {x^{3}+56 x +8}{x^{2}+5 x +1}\)	\(21\)
default	\(-x +\frac {-80 x -13}{x^{2}+5 x +1}\)	\(21\)
risch	\(-x +\frac {-80 x -13}{x^{2}+5 x +1}\)	\(21\)
norman	\(\frac {-x^{3}-56 x -8}{x^{2}+5 x +1}\)	\(22\)

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 {gather*} -x - \frac {80 \, x + 13}{x^{2} + 5 \, x + 1} \end {gather*}

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 {gather*} -x-\frac {80\,x+13}{x^2+5\,x+1} \end {gather*}

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 {gather*} - x - \frac {80 x + 13}{x^{2} + 5 x + 1} \end {gather*}

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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