\(\int \frac {2-x^2}{(1-6 x+x^3)^5} \, dx\) [223]

Optimal result

Integrand size = 18, antiderivative size = 14 \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12 \left (1-6 x+x^3\right )^4} \]

[Out]

1/12/(x^3-6*x+1)^4

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1602} \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12 \left (x^3-6 x+1\right )^4} \]

[In]

Int[(2 - x^2)/(1 - 6*x + x^3)^5,x]

[Out]

1/(12*(1 - 6*x + x^3)^4)

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{12 \left (1-6 x+x^3\right )^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12 \left (1-6 x+x^3\right )^4} \]

[In]

Integrate[(2 - x^2)/(1 - 6*x + x^3)^5,x]

[Out]

1/(12*(1 - 6*x + x^3)^4)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
gosper	\(\frac {1}{12 \left (x^{3}-6 x +1\right )^{4}}\)	\(13\)
default	\(\frac {1}{12 \left (x^{3}-6 x +1\right )^{4}}\)	\(13\)
norman	\(\frac {1}{12 \left (x^{3}-6 x +1\right )^{4}}\)	\(13\)
risch	\(\frac {1}{12 \left (x^{3}-6 x +1\right )^{4}}\)	\(13\)
parallelrisch	\(\frac {1}{12 \left (x^{3}-6 x +1\right )^{4}}\)	\(13\)

[In]

int((-x^2+2)/(x^3-6*x+1)^5,x,method=_RETURNVERBOSE)

[Out]

1/12/(x^3-6*x+1)^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (12) = 24\).

Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 4.07 \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12 \, {\left (x^{12} - 24 \, x^{10} + 4 \, x^{9} + 216 \, x^{8} - 72 \, x^{7} - 858 \, x^{6} + 432 \, x^{5} + 1224 \, x^{4} - 860 \, x^{3} + 216 \, x^{2} - 24 \, x + 1\right )}} \]

[In]

integrate((-x^2+2)/(x^3-6*x+1)^5,x, algorithm="fricas")

[Out]

1/12/(x^12 - 24*x^10 + 4*x^9 + 216*x^8 - 72*x^7 - 858*x^6 + 432*x^5 + 1224*x^4 - 860*x^3 + 216*x^2 - 24*x + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (12) = 24\).

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 4.00 \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12 x^{12} - 288 x^{10} + 48 x^{9} + 2592 x^{8} - 864 x^{7} - 10296 x^{6} + 5184 x^{5} + 14688 x^{4} - 10320 x^{3} + 2592 x^{2} - 288 x + 12} \]

[In]

integrate((-x**2+2)/(x**3-6*x+1)**5,x)

[Out]

1/(12*x**12 - 288*x**10 + 48*x**9 + 2592*x**8 - 864*x**7 - 10296*x**6 + 5184*x**5 + 14688*x**4 - 10320*x**3 +
2592*x**2 - 288*x + 12)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12 \, {\left (x^{3} - 6 \, x + 1\right )}^{4}} \]

[In]

integrate((-x^2+2)/(x^3-6*x+1)^5,x, algorithm="maxima")

[Out]

1/12/(x^3 - 6*x + 1)^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12 \, {\left (x^{3} - 6 \, x + 1\right )}^{4}} \]

[In]

integrate((-x^2+2)/(x^3-6*x+1)^5,x, algorithm="giac")

[Out]

1/12/(x^3 - 6*x + 1)^4

Mupad [B] (verification not implemented)

Time = 9.37 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {2-x^2}{\left (1-6 x+x^3\right )^5} \, dx=\frac {1}{12\,{\left (x^3-6\,x+1\right )}^4} \]

[In]

int(-(x^2 - 2)/(x^3 - 6*x + 1)^5,x)

[Out]

1/(12*(x^3 - 6*x + 1)^4)