\(\int x^2 \sqrt [4]{-1+x^4} \, dx\) [621]

Optimal result

Integrand size = 13, antiderivative size = 49 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{4} x^3 \sqrt [4]{-1+x^4}+\frac {1}{8} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \]

[Out]

1/4*x^3*(x^4-1)^(1/4)+1/8*arctan(x/(x^4-1)^(1/4))-1/8*arctanh(x/(x^4-1)^(1/4))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {285, 338, 304, 209, 212} \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{8} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{4} \sqrt [4]{x^4-1} x^3 \]

[In]

Int[x^2*(-1 + x^4)^(1/4),x]

[Out]

(x^3*(-1 + x^4)^(1/4))/4 + ArcTan[x/(-1 + x^4)^(1/4)]/8 - ArcTanh[x/(-1 + x^4)^(1/4)]/8

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 285

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

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^3 \sqrt [4]{-1+x^4}-\frac {1}{4} \int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx \\ & = \frac {1}{4} x^3 \sqrt [4]{-1+x^4}-\frac {1}{4} \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right ) \\ & = \frac {1}{4} x^3 \sqrt [4]{-1+x^4}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right ) \\ & = \frac {1}{4} x^3 \sqrt [4]{-1+x^4}+\frac {1}{8} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.92 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{8} \left (2 x^3 \sqrt [4]{-1+x^4}+\arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )\right ) \]

[In]

Integrate[x^2*(-1 + x^4)^(1/4),x]

[Out]

(2*x^3*(-1 + x^4)^(1/4) + ArcTan[x/(-1 + x^4)^(1/4)] - ArcTanh[x/(-1 + x^4)^(1/4)])/8

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.96 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.67

method	result	size
meijerg	\(\frac {\operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{4}\right )}{3 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\)	\(33\)
risch	\(\frac {x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}}{4}-\frac {{\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {3}{4}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{4}\right )}{12 \operatorname {signum}\left (x^{4}-1\right )^{\frac {3}{4}}}\)	\(46\)
pseudoelliptic	\(\frac {x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}}{4}+\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right )}{16}-\frac {\arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{8}-\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}+x}{x}\right )}{16}\)	\(62\)
trager	\(\frac {x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{16}-\frac {\ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x +2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}+2 x^{4}-1\right )}{16}\)	\(127\)

[In]

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

[Out]

1/3*signum(x^4-1)^(1/4)/(-signum(x^4-1))^(1/4)*x^3*hypergeom([-1/4,3/4],[7/4],x^4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.27 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{4} \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{16} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{16} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

1/4*(x^4 - 1)^(1/4)*x^3 - 1/8*arctan((x^4 - 1)^(1/4)/x) - 1/16*log((x + (x^4 - 1)^(1/4))/x) + 1/16*log(-(x - (
x^4 - 1)^(1/4))/x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=- \frac {x^{3} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

-x**3*exp(-3*I*pi/4)*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), x**4)/(4*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.47 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{4 \, x {\left (\frac {x^{4} - 1}{x^{4}} - 1\right )}} - \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{16} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {1}{16} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \]

[In]

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

[Out]

-1/4*(x^4 - 1)^(1/4)/(x*((x^4 - 1)/x^4 - 1)) - 1/8*arctan((x^4 - 1)^(1/4)/x) - 1/16*log((x^4 - 1)^(1/4)/x + 1)
+ 1/16*log((x^4 - 1)^(1/4)/x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.22 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{4} \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{16} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {1}{16} \, \log \left (-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) \]

[In]

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

[Out]

1/4*(x^4 - 1)^(1/4)*x^3 - 1/8*arctan((x^4 - 1)^(1/4)/x) - 1/16*log((x^4 - 1)^(1/4)/x + 1) + 1/16*log(-(x^4 - 1
)^(1/4)/x + 1)

Mupad [F(-1)]

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

[In]

int(x^2*(x^4 - 1)^(1/4),x)

[Out]

int(x^2*(x^4 - 1)^(1/4), x)