\(\int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx\) [114]

Optimal result

Integrand size = 18, antiderivative size = 16 \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=\frac {\left (-1+x^4\right )^{3/2}}{3 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1493, 460} \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=\frac {\left (x^4-1\right )^{3/2}}{3 x^3} \]

[In]

Int[(-1 + x^8)/(x^4*Sqrt[-1 + x^4]),x]

[Out]

(-1 + x^4)^(3/2)/(3*x^3)

Rule 460

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

Rule 1493

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(f*x)^
m*(d + e*x^n)^(q + p)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, f, q, m, n, q}, x] && EqQ[n2, 2*n] && EqQ[
c*d^2 + a*e^2, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=\frac {\left (-1+x^4\right )^{3/2}}{3 x^3} \]

[In]

Integrate[(-1 + x^8)/(x^4*Sqrt[-1 + x^4]),x]

[Out]

(-1 + x^4)^(3/2)/(3*x^3)

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
trager	\(\frac {\left (x^{4}-1\right )^{\frac {3}{2}}}{3 x^{3}}\)	\(13\)
elliptic	\(\frac {\left (x^{4}-1\right )^{\frac {3}{2}}}{3 x^{3}}\)	\(13\)
risch	\(\frac {x^{8}-2 x^{4}+1}{3 x^{3} \sqrt {x^{4}-1}}\)	\(23\)
gosper	\(\frac {\sqrt {x^{4}-1}\, \left (x -1\right ) \left (1+x \right ) \left (x^{2}+1\right )}{3 x^{3}}\)	\(24\)
default	\(\frac {x \sqrt {x^{4}-1}}{3}-\frac {\sqrt {x^{4}-1}}{3 x^{3}}\)	\(24\)
pseudoelliptic	\(-\frac {\left (1+x \right ) \left (x -1\right ) \left (i-x \right ) \sqrt {x^{4}-1}\, \left (i+x \right )}{3 x^{3}}\)	\(29\)
meijerg	\(\frac {\sqrt {-\operatorname {signum}\left (x^{4}-1\right )}\, x^{5} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {9}{4}\right ], x^{4}\right )}{5 \sqrt {\operatorname {signum}\left (x^{4}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (x^{4}-1\right )}\, \operatorname {hypergeom}\left (\left [-\frac {3}{4}, \frac {1}{2}\right ], \left [\frac {1}{4}\right ], x^{4}\right )}{3 \sqrt {\operatorname {signum}\left (x^{4}-1\right )}\, x^{3}}\)	\(66\)

[In]

int((x^8-1)/x^4/(x^4-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=\frac {{\left (x^{4} - 1\right )}^{\frac {3}{2}}}{3 \, x^{3}} \]

[In]

integrate((x^8-1)/x^4/(x^4-1)^(1/2),x, algorithm="fricas")

[Out]

1/3*(x^4 - 1)^(3/2)/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.94 (sec) , antiderivative size = 56, normalized size of antiderivative = 3.50 \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=- \frac {i x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {i \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

[In]

integrate((x**8-1)/x**4/(x**4-1)**(1/2),x)

[Out]

-I*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), x**4)/(4*gamma(9/4)) + I*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), x
**4)/(4*x**3*gamma(1/4))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=\frac {{\left (x^{4} - 1\right )} \sqrt {x^{2} + 1} \sqrt {x + 1} \sqrt {x - 1}}{3 \, x^{3}} \]

[In]

integrate((x^8-1)/x^4/(x^4-1)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (12) = 24\).

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=\frac {1}{3} \, \sqrt {x^{4} - 1} x - \frac {\sqrt {-\frac {1}{x^{4}} + 1}}{3 \, x} \]

[In]

integrate((x^8-1)/x^4/(x^4-1)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-1+x^8}{x^4 \sqrt {-1+x^4}} \, dx=\frac {{\left (x^4-1\right )}^{3/2}}{3\,x^3} \]

[In]

int((x^8 - 1)/(x^4*(x^4 - 1)^(1/2)),x)

[Out]

(x^4 - 1)^(3/2)/(3*x^3)