\(\int x \sqrt [4]{-1+x^2} \, dx\) [6]

Optimal result

Integrand size = 11, antiderivative size = 13 \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2}{5} \left (-1+x^2\right )^{5/4} \]

[Out]

2/5*(x^2-1)^(5/4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {267} \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2}{5} \left (x^2-1\right )^{5/4} \]

[In]

Int[x*(-1 + x^2)^(1/4),x]

[Out]

(2*(-1 + x^2)^(5/4))/5

Rule 267

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

Rubi steps \begin{align*} \text {integral}& = \frac {2}{5} \left (-1+x^2\right )^{5/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2}{5} \left (-1+x^2\right )^{5/4} \]

[In]

Integrate[x*(-1 + x^2)^(1/4),x]

[Out]

(2*(-1 + x^2)^(5/4))/5

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 \left (x^{2}-1\right )^{\frac {5}{4}}}{5}\)	\(10\)
default	\(\frac {2 \left (x^{2}-1\right )^{\frac {5}{4}}}{5}\)	\(10\)
risch	\(\frac {2 \left (x^{2}-1\right )^{\frac {5}{4}}}{5}\)	\(10\)
pseudoelliptic	\(\frac {2 \left (x^{2}-1\right )^{\frac {5}{4}}}{5}\)	\(10\)
gosper	\(\frac {2 \left (x -1\right ) \left (1+x \right ) \left (x^{2}-1\right )^{\frac {1}{4}}}{5}\)	\(16\)
trager	\(\left (\frac {2 x^{2}}{5}-\frac {2}{5}\right ) \left (x^{2}-1\right )^{\frac {1}{4}}\)	\(16\)
meijerg	\(\frac {\operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{2} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, 1\right ], \left [2\right ], x^{2}\right )}{2 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}}}\)	\(33\)

[In]

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

[Out]

2/5*(x^2-1)^(5/4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2}{5} \, {\left (x^{2} - 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

2/5*(x^2 - 1)^(5/4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.00 \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2 x^{2} \sqrt [4]{x^{2} - 1}}{5} - \frac {2 \sqrt [4]{x^{2} - 1}}{5} \]

[In]

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

[Out]

2*x**2*(x**2 - 1)**(1/4)/5 - 2*(x**2 - 1)**(1/4)/5

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2}{5} \, {\left (x^{2} - 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

2/5*(x^2 - 1)^(5/4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2}{5} \, {\left (x^{2} - 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

2/5*(x^2 - 1)^(5/4)

Mupad [B] (verification not implemented)

Time = 5.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \sqrt [4]{-1+x^2} \, dx=\frac {2\,{\left (x^2-1\right )}^{5/4}}{5} \]

[In]

int(x*(x^2 - 1)^(1/4),x)

[Out]

(2*(x^2 - 1)^(5/4))/5