\(\int \frac {x^6}{(1-x^4)^{3/2}} \, dx\) [912]

Optimal result

Integrand size = 15, antiderivative size = 35 \[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=\frac {x^3}{2 \sqrt {1-x^4}}-\frac {3}{2} E(\arcsin (x)|-1)+\frac {3}{2} \operatorname {EllipticF}(\arcsin (x),-1) \]

[Out]

-3/2*EllipticE(x,I)+3/2*EllipticF(x,I)+1/2*x^3/(-x^4+1)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {294, 313, 227, 1195, 435} \[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=\frac {3}{2} \operatorname {EllipticF}(\arcsin (x),-1)-\frac {3}{2} E(\arcsin (x)|-1)+\frac {x^3}{2 \sqrt {1-x^4}} \]

[In]

Int[x^6/(1 - x^4)^(3/2),x]

[Out]

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

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 313

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[-q^(-1), Int[1/Sqrt[a + b*x^4]
, x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 435

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

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c], Int
[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {x^3}{2 \sqrt {1-x^4}}-\frac {3}{2} \int \frac {x^2}{\sqrt {1-x^4}} \, dx \\ & = \frac {x^3}{2 \sqrt {1-x^4}}+\frac {3}{2} \int \frac {1}{\sqrt {1-x^4}} \, dx-\frac {3}{2} \int \frac {1+x^2}{\sqrt {1-x^4}} \, dx \\ & = \frac {x^3}{2 \sqrt {1-x^4}}+\frac {3}{2} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {3}{2} \int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx \\ & = \frac {x^3}{2 \sqrt {1-x^4}}-\frac {3}{2} E\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {3}{2} F\left (\left .\sin ^{-1}(x)\right |-1\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=x^3 \left (-\frac {1}{\sqrt {1-x^4}}+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},x^4\right )\right ) \]

[In]

Integrate[x^6/(1 - x^4)^(3/2),x]

[Out]

x^3*(-(1/Sqrt[1 - x^4]) + Hypergeometric2F1[3/4, 3/2, 7/4, x^4])

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 4.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.43

method	result	size
meijerg	\(\frac {x^{7} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{2},\frac {7}{4};\frac {11}{4};x^{4}\right )}{7}\)	\(15\)
default	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(54\)
risch	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(54\)
elliptic	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(54\)

[In]

int(x^6/(-x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/7*x^7*hypergeom([3/2,7/4],[11/4],x^4)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (25) = 50\).

Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.83 \[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (-i \, x^{5} + i \, x\right )} E(\arcsin \left (\frac {1}{x}\right )\,|\,-1) + 3 \, {\left (i \, x^{5} - i \, x\right )} F(\arcsin \left (\frac {1}{x}\right )\,|\,-1) - {\left (2 \, x^{4} - 3\right )} \sqrt {-x^{4} + 1}}{2 \, {\left (x^{5} - x\right )}} \]

[In]

integrate(x^6/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

-1/2*(3*(-I*x^5 + I*x)*elliptic_e(arcsin(1/x), -1) + 3*(I*x^5 - I*x)*elliptic_f(arcsin(1/x), -1) - (2*x^4 - 3)
*sqrt(-x^4 + 1))/(x^5 - x)

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \]

[In]

integrate(x**6/(-x**4+1)**(3/2),x)

[Out]

x**7*gamma(7/4)*hyper((3/2, 7/4), (11/4,), x**4*exp_polar(2*I*pi))/(4*gamma(11/4))

Maxima [F]

\[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=\int { \frac {x^{6}}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^6/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^6/(-x^4 + 1)^(3/2), x)

Giac [F]

\[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=\int { \frac {x^{6}}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^6/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^6/(-x^4 + 1)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (1-x^4\right )^{3/2}} \, dx=\int \frac {x^6}{{\left (1-x^4\right )}^{3/2}} \,d x \]

[In]

int(x^6/(1 - x^4)^(3/2),x)

[Out]

int(x^6/(1 - x^4)^(3/2), x)