3.10.4 \(\int \frac {x^{12}}{(1-x^4)^{3/2}} \, dx\) [904]

Optimal. Leaf size=61 \[ \frac {x^9}{2 \sqrt {1-x^4}}+\frac {15}{14} x \sqrt {1-x^4}+\frac {9}{14} x^5 \sqrt {1-x^4}-\frac {15}{14} F\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

-15/14*EllipticF(x,I)+1/2*x^9/(-x^4+1)^(1/2)+15/14*x*(-x^4+1)^(1/2)+9/14*x^5*(-x^4+1)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {294, 327, 227} \begin {gather*} -\frac {15}{14} F(\text {ArcSin}(x)|-1)+\frac {15}{14} \sqrt {1-x^4} x+\frac {x^9}{2 \sqrt {1-x^4}}+\frac {9}{14} \sqrt {1-x^4} x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^12/(1 - x^4)^(3/2),x]

[Out]

x^9/(2*Sqrt[1 - x^4]) + (15*x*Sqrt[1 - x^4])/14 + (9*x^5*Sqrt[1 - x^4])/14 - (15*EllipticF[ArcSin[x], -1])/14

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 294

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 327

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^{12}}{\left (1-x^4\right )^{3/2}} \, dx &=\frac {x^9}{2 \sqrt {1-x^4}}-\frac {9}{2} \int \frac {x^8}{\sqrt {1-x^4}} \, dx\\ &=\frac {x^9}{2 \sqrt {1-x^4}}+\frac {9}{14} x^5 \sqrt {1-x^4}-\frac {45}{14} \int \frac {x^4}{\sqrt {1-x^4}} \, dx\\ &=\frac {x^9}{2 \sqrt {1-x^4}}+\frac {15}{14} x \sqrt {1-x^4}+\frac {9}{14} x^5 \sqrt {1-x^4}-\frac {15}{14} \int \frac {1}{\sqrt {1-x^4}} \, dx\\ &=\frac {x^9}{2 \sqrt {1-x^4}}+\frac {15}{14} x \sqrt {1-x^4}+\frac {9}{14} x^5 \sqrt {1-x^4}-\frac {15}{14} F\left (\left .\sin ^{-1}(x)\right |-1\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.83, size = 54, normalized size = 0.89 \begin {gather*} -\frac {x \left (-15+6 x^4+2 x^8+15 \sqrt {1-x^4} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};x^4\right )\right )}{14 \sqrt {1-x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^12/(1 - x^4)^(3/2),x]

[Out]

-1/14*(x*(-15 + 6*x^4 + 2*x^8 + 15*Sqrt[1 - x^4]*Hypergeometric2F1[1/4, 1/2, 5/4, x^4]))/Sqrt[1 - x^4]

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Maple [A]
time = 0.16, size = 71, normalized size = 1.16

method	result	size
meijerg	\(\frac {x^{13} \hypergeom \left (\left [\frac {3}{2}, \frac {13}{4}\right ], \left [\frac {17}{4}\right ], x^{4}\right )}{13}\)	\(15\)
risch	\(-\frac {x \left (2 x^{8}+6 x^{4}-15\right )}{14 \sqrt {-x^{4}+1}}-\frac {15 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \EllipticF \left (x , i\right )}{14 \sqrt {-x^{4}+1}}\)	\(57\)
default	\(\frac {x}{2 \sqrt {-x^{4}+1}}+\frac {x^{5} \sqrt {-x^{4}+1}}{7}+\frac {4 x \sqrt {-x^{4}+1}}{7}-\frac {15 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \EllipticF \left (x , i\right )}{14 \sqrt {-x^{4}+1}}\)	\(71\)
elliptic	\(\frac {x}{2 \sqrt {-x^{4}+1}}+\frac {x^{5} \sqrt {-x^{4}+1}}{7}+\frac {4 x \sqrt {-x^{4}+1}}{7}-\frac {15 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \EllipticF \left (x , i\right )}{14 \sqrt {-x^{4}+1}}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(-x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*x/(-x^4+1)^(1/2)+1/7*x^5*(-x^4+1)^(1/2)+4/7*x*(-x^4+1)^(1/2)-15/14*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)^(
1/2)*EllipticF(x,I)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^12/(-x^4 + 1)^(3/2), x)

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Fricas [A]
time = 0.08, size = 32, normalized size = 0.52 \begin {gather*} \frac {{\left (2 \, x^{9} + 6 \, x^{5} - 15 \, x\right )} \sqrt {-x^{4} + 1}}{14 \, {\left (x^{4} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

1/14*(2*x^9 + 6*x^5 - 15*x)*sqrt(-x^4 + 1)/(x^4 - 1)

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Sympy [A]
time = 0.59, size = 31, normalized size = 0.51 \begin {gather*} \frac {x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {17}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**12/(-x**4+1)**(3/2),x)

[Out]

x**13*gamma(13/4)*hyper((3/2, 13/4), (17/4,), x**4*exp_polar(2*I*pi))/(4*gamma(17/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^12/(-x^4 + 1)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{12}}{{\left (1-x^4\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(1 - x^4)^(3/2),x)

[Out]

int(x^12/(1 - x^4)^(3/2), x)

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