\(\int \frac {x^6}{\sqrt {-1+x^4}} \, dx\) [980]

Optimal result

Integrand size = 13, antiderivative size = 150 \[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=\frac {3 x \left (1+x^2\right )}{5 \sqrt {-1+x^4}}+\frac {1}{5} x^3 \sqrt {-1+x^4}-\frac {3 \sqrt {2} \sqrt {-1+x^2} \sqrt {1+x^2} E\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{5 \sqrt {-1+x^4}}+\frac {3 \sqrt {-1+x^2} \sqrt {1+x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right ),\frac {1}{2}\right )}{5 \sqrt {2} \sqrt {-1+x^4}} \]

[Out]

3/5*x*(x^2+1)/(x^4-1)^(1/2)+3/10*EllipticF(x*2^(1/2)/(x^2-1)^(1/2),1/2*2^(1/2))*(x^2-1)^(1/2)*(x^2+1)^(1/2)*2^
(1/2)/(x^4-1)^(1/2)-3/5*EllipticE(x*2^(1/2)/(x^2-1)^(1/2),1/2*2^(1/2))*2^(1/2)*(x^2-1)^(1/2)*(x^2+1)^(1/2)/(x^
4-1)^(1/2)+1/5*x^3*(x^4-1)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {327, 312, 228, 1199} \[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=\frac {3 \sqrt {x^2-1} \sqrt {x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right ),\frac {1}{2}\right )}{5 \sqrt {2} \sqrt {x^4-1}}-\frac {3 \sqrt {2} \sqrt {x^2-1} \sqrt {x^2+1} E\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right )|\frac {1}{2}\right )}{5 \sqrt {x^4-1}}+\frac {1}{5} \sqrt {x^4-1} x^3+\frac {3 \left (x^2+1\right ) x}{5 \sqrt {x^4-1}} \]

[In]

Int[x^6/Sqrt[-1 + x^4],x]

[Out]

(3*x*(1 + x^2))/(5*Sqrt[-1 + x^4]) + (x^3*Sqrt[-1 + x^4])/5 - (3*Sqrt[2]*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*Elliptic
E[ArcSin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(5*Sqrt[-1 + x^4]) + (3*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticF[Arc
Sin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(5*Sqrt[2]*Sqrt[-1 + x^4])

Rule 228

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*b, 2]}, Simp[Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2
)/q]/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]))*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rule 312

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[1/q, 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] && LtQ[a, 0] && GtQ[b, 0]

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]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[e*x*((q + c*x
^2)/(c*Sqrt[a + c*x^4])), x] - Simp[Sqrt[2]*e*q*Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2)/q]/(Sqrt[-a]*c*Sqrt[a + c*x
^4]))*EllipticE[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; EqQ[c*d + e*q, 0] && IntegerQ[q]] /; FreeQ[{a,
c, d, e}, x] && LtQ[a, 0] && GtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^3 \sqrt {-1+x^4}+\frac {3}{5} \int \frac {x^2}{\sqrt {-1+x^4}} \, dx \\ & = \frac {1}{5} x^3 \sqrt {-1+x^4}+\frac {3}{5} \int \frac {1}{\sqrt {-1+x^4}} \, dx-\frac {3}{5} \int \frac {1-x^2}{\sqrt {-1+x^4}} \, dx \\ & = \frac {3 x \left (1+x^2\right )}{5 \sqrt {-1+x^4}}+\frac {1}{5} x^3 \sqrt {-1+x^4}-\frac {3 \sqrt {2} \sqrt {-1+x^2} \sqrt {1+x^2} E\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{5 \sqrt {-1+x^4}}+\frac {3 \sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{5 \sqrt {2} \sqrt {-1+x^4}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.31 \[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=\frac {x^3 \left (-1+x^4+\sqrt {1-x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},x^4\right )\right )}{5 \sqrt {-1+x^4}} \]

[In]

Integrate[x^6/Sqrt[-1 + x^4],x]

[Out]

(x^3*(-1 + x^4 + Sqrt[1 - x^4]*Hypergeometric2F1[1/2, 3/4, 7/4, x^4]))/(5*Sqrt[-1 + x^4])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.61 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.22

method	result	size
meijerg	\(\frac {\sqrt {-\operatorname {signum}\left (x^{4}-1\right )}\, x^{7} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};x^{4}\right )}{7 \sqrt {\operatorname {signum}\left (x^{4}-1\right )}}\)	\(33\)
default	\(\frac {x^{3} \sqrt {x^{4}-1}}{5}-\frac {3 i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (F\left (i x , i\right )-E\left (i x , i\right )\right )}{5 \sqrt {x^{4}-1}}\)	\(57\)
risch	\(\frac {x^{3} \sqrt {x^{4}-1}}{5}-\frac {3 i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (F\left (i x , i\right )-E\left (i x , i\right )\right )}{5 \sqrt {x^{4}-1}}\)	\(57\)
elliptic	\(\frac {x^{3} \sqrt {x^{4}-1}}{5}-\frac {3 i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (F\left (i x , i\right )-E\left (i x , i\right )\right )}{5 \sqrt {x^{4}-1}}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.25 \[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=\frac {3 \, x E(\arcsin \left (\frac {1}{x}\right )\,|\,-1) - 3 \, x F(\arcsin \left (\frac {1}{x}\right )\,|\,-1) + {\left (x^{4} + 3\right )} \sqrt {x^{4} - 1}}{5 \, x} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.18 \[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=- \frac {i x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \]

[In]

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

[Out]

-I*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), x**4)/(4*gamma(11/4))

Maxima [F]

\[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {x^{4} - 1}} \,d x } \]

[In]

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

[Out]

integrate(x^6/sqrt(x^4 - 1), x)

Giac [F]

\[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {x^{4} - 1}} \,d x } \]

[In]

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

[Out]

integrate(x^6/sqrt(x^4 - 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\sqrt {-1+x^4}} \, dx=\int \frac {x^6}{\sqrt {x^4-1}} \,d x \]

[In]

int(x^6/(x^4 - 1)^(1/2),x)

[Out]

int(x^6/(x^4 - 1)^(1/2), x)