3.899 \(\int \frac {x^9}{(1-x^4)^{3/2}} \, dx\)

Optimal. Leaf size=45 \[ -\frac {3}{4} \sin ^{-1}\left (x^2\right )+\frac {x^6}{2 \sqrt {1-x^4}}+\frac {3}{4} \sqrt {1-x^4} x^2 \]

[Out]

-3/4*arcsin(x^2)+1/2*x^6/(-x^4+1)^(1/2)+3/4*x^2*(-x^4+1)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {275, 288, 321, 216} \[ \frac {x^6}{2 \sqrt {1-x^4}}+\frac {3}{4} \sqrt {1-x^4} x^2-\frac {3}{4} \sin ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^9/(1 - x^4)^(3/2),x]

[Out]

x^6/(2*Sqrt[1 - x^4]) + (3*x^2*Sqrt[1 - x^4])/4 - (3*ArcSin[x^2])/4

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^9}{\left (1-x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {x^6}{2 \sqrt {1-x^4}}-\frac {3}{2} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2}} \, dx,x,x^2\right )\\ &=\frac {x^6}{2 \sqrt {1-x^4}}+\frac {3}{4} x^2 \sqrt {1-x^4}-\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,x^2\right )\\ &=\frac {x^6}{2 \sqrt {1-x^4}}+\frac {3}{4} x^2 \sqrt {1-x^4}-\frac {3}{4} \sin ^{-1}\left (x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.91 \[ -\frac {x^6-3 x^2+3 \sqrt {1-x^4} \sin ^{-1}\left (x^2\right )}{4 \sqrt {1-x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(1 - x^4)^(3/2),x]

[Out]

-1/4*(-3*x^2 + x^6 + 3*Sqrt[1 - x^4]*ArcSin[x^2])/Sqrt[1 - x^4]

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fricas [A]  time = 0.86, size = 52, normalized size = 1.16 \[ \frac {6 \, {\left (x^{4} - 1\right )} \arctan \left (\frac {\sqrt {-x^{4} + 1} - 1}{x^{2}}\right ) + {\left (x^{6} - 3 \, x^{2}\right )} \sqrt {-x^{4} + 1}}{4 \, {\left (x^{4} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

1/4*(6*(x^4 - 1)*arctan((sqrt(-x^4 + 1) - 1)/x^2) + (x^6 - 3*x^2)*sqrt(-x^4 + 1))/(x^4 - 1)

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giac [A]  time = 0.19, size = 33, normalized size = 0.73 \[ \frac {{\left (x^{4} - 3\right )} \sqrt {-x^{4} + 1} x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {3}{4} \, \arcsin \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

1/4*(x^4 - 3)*sqrt(-x^4 + 1)*x^2/(x^4 - 1) - 3/4*arcsin(x^2)

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maple [B]  time = 0.02, size = 76, normalized size = 1.69 \[ \frac {\sqrt {-x^{4}+1}\, x^{2}}{4}-\frac {3 \arcsin \left (x^{2}\right )}{4}-\frac {\sqrt {-2 x^{2}-\left (x^{2}-1\right )^{2}+2}}{4 \left (x^{2}-1\right )}-\frac {\sqrt {2 x^{2}-\left (x^{2}+1\right )^{2}+2}}{4 \left (x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(-x^4+1)^(3/2),x)

[Out]

1/4*(-x^4+1)^(1/2)*x^2-3/4*arcsin(x^2)-1/4/(x^2-1)*(-2*x^2-(x^2-1)^2+2)^(1/2)-1/4/(x^2+1)*(2*x^2-(x^2+1)^2+2)^
(1/2)

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maxima [A]  time = 2.95, size = 60, normalized size = 1.33 \[ -\frac {\frac {3 \, {\left (x^{4} - 1\right )}}{x^{4}} - 2}{4 \, {\left (\frac {\sqrt {-x^{4} + 1}}{x^{2}} + \frac {{\left (-x^{4} + 1\right )}^{\frac {3}{2}}}{x^{6}}\right )}} + \frac {3}{4} \, \arctan \left (\frac {\sqrt {-x^{4} + 1}}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(3*(x^4 - 1)/x^4 - 2)/(sqrt(-x^4 + 1)/x^2 + (-x^4 + 1)^(3/2)/x^6) + 3/4*arctan(sqrt(-x^4 + 1)/x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^9}{{\left (1-x^4\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(1 - x^4)^(3/2),x)

[Out]

int(x^9/(1 - x^4)^(3/2), x)

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sympy [A]  time = 4.16, size = 82, normalized size = 1.82 \[ \begin {cases} \frac {i x^{6}}{4 \sqrt {x^{4} - 1}} - \frac {3 i x^{2}}{4 \sqrt {x^{4} - 1}} + \frac {3 i \operatorname {acosh}{\left (x^{2} \right )}}{4} & \text {for}\: \left |{x^{4}}\right | > 1 \\- \frac {x^{6}}{4 \sqrt {1 - x^{4}}} + \frac {3 x^{2}}{4 \sqrt {1 - x^{4}}} - \frac {3 \operatorname {asin}{\left (x^{2} \right )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(-x**4+1)**(3/2),x)

[Out]

Piecewise((I*x**6/(4*sqrt(x**4 - 1)) - 3*I*x**2/(4*sqrt(x**4 - 1)) + 3*I*acosh(x**2)/4, Abs(x**4) > 1), (-x**6
/(4*sqrt(1 - x**4)) + 3*x**2/(4*sqrt(1 - x**4)) - 3*asin(x**2)/4, True))

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