3.85.66 $\int \frac{-121-50x-22x^2-x^4}{121+22x^2+x^4}dx$

Optimal. Leaf size=27 $$-22-x+\frac{5}{3-\frac{1}{5}(\frac{4}{x}-x)x}+\log (2)$$

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, integrand size = 28, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.143, Rules used = {28, 1814, 21, 8} $$\begin{array}{c}\frac{25}{x^2+11}-x\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-121 - 50*x - 22*x^2 - x^4)/(121 + 22*x^2 + x^4),x]

[Out]

-x + 25/(11 + 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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-121-50x-22x^2-x^4}{{(11+x^2)}^2}dx\\ & =\frac{25}{11+x^2}-\frac{1}{22}\int \frac{242+22x^2}{11+x^2}dx\\ & =\frac{25}{11+x^2}-\int 1dx\\ & =-x+\frac{25}{11+x^2}\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.48 $$\begin{array}{c}-x+\frac{25}{11+x^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-121 - 50*x - 22*x^2 - x^4)/(121 + 22*x^2 + x^4),x]

[Out]

-x + 25/(11 + x^2)

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fricas [A]  time = 0.60, size = 17, normalized size = 0.63 $$\begin{array}{c}-\frac{x^3+11x-25}{x^2+11}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-22*x^2-50*x-121)/(x^4+22*x^2+121),x, algorithm="fricas")

[Out]

-(x^3 + 11*x - 25)/(x^2 + 11)

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giac [A]  time = 0.24, size = 13, normalized size = 0.48 $$\begin{array}{c}-x+\frac{25}{x^2+11}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-22*x^2-50*x-121)/(x^4+22*x^2+121),x, algorithm="giac")

[Out]

-x + 25/(x^2 + 11)

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maple [A]  time = 0.02, size = 14, normalized size = 0.52

method	result	size
default	$-x+\frac{25}{x^2+11}$	$14$
risch	$-x+\frac{25}{x^2+11}$	$14$
gosper	$-\frac{x^3+11x-25}{x^2+11}$	$18$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4-22*x^2-50*x-121)/(x^4+22*x^2+121),x,method=_RETURNVERBOSE)

[Out]

-x+25/(x^2+11)

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maxima [A]  time = 0.36, size = 13, normalized size = 0.48 $$\begin{array}{c}-x+\frac{25}{x^2+11}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-22*x^2-50*x-121)/(x^4+22*x^2+121),x, algorithm="maxima")

[Out]

-x + 25/(x^2 + 11)

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mupad [B]  time = 5.15, size = 13, normalized size = 0.48 $$\begin{array}{c}\frac{25}{x^2+11}-x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*x + 22*x^2 + x^4 + 121)/(22*x^2 + x^4 + 121),x)

[Out]

25/(x^2 + 11) - x

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sympy [A]  time = 0.08, size = 7, normalized size = 0.26 $$\begin{array}{c}-x+\frac{25}{x^2+11}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4-22*x**2-50*x-121)/(x**4+22*x**2+121),x)

[Out]

-x + 25/(x**2 + 11)

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