\(\int x^2 \sqrt [4]{-x^2+x^4} \, dx\) [878]

Optimal result

Integrand size = 17, antiderivative size = 67 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{16} \left (-x+4 x^3\right ) \sqrt [4]{-x^2+x^4}+\frac {3}{32} \arctan \left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\frac {3}{32} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right ) \]

[Out]

1/16*(4*x^3-x)*(x^4-x^2)^(1/4)+3/32*arctan(x/(x^4-x^2)^(1/4))-3/32*arctanh(x/(x^4-x^2)^(1/4))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {2046, 2049, 2057, 335, 338, 304, 209, 212} \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{32 \left (x^4-x^2\right )^{3/4}}-\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{32 \left (x^4-x^2\right )^{3/4}}-\frac {1}{16} \sqrt [4]{x^4-x^2} x+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3 \]

[In]

Int[x^2*(-x^2 + x^4)^(1/4),x]

[Out]

-1/16*(x*(-x^2 + x^4)^(1/4)) + (x^3*(-x^2 + x^4)^(1/4))/4 + (3*x^(3/2)*(-1 + x^2)^(3/4)*ArcTan[Sqrt[x]/(-1 + x
^2)^(1/4)])/(32*(-x^2 + x^4)^(3/4)) - (3*x^(3/2)*(-1 + x^2)^(3/4)*ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)])/(32*(-x^2
+ x^4)^(3/4))

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 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 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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]

Rule 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2049

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^3 \sqrt [4]{-x^2+x^4}-\frac {1}{8} \int \frac {x^4}{\left (-x^2+x^4\right )^{3/4}} \, dx \\ & = -\frac {1}{16} x \sqrt [4]{-x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^2+x^4}-\frac {3}{32} \int \frac {x^2}{\left (-x^2+x^4\right )^{3/4}} \, dx \\ & = -\frac {1}{16} x \sqrt [4]{-x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^2+x^4}-\frac {\left (3 x^{3/2} \left (-1+x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (-1+x^2\right )^{3/4}} \, dx}{32 \left (-x^2+x^4\right )^{3/4}} \\ & = -\frac {1}{16} x \sqrt [4]{-x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^2+x^4}-\frac {\left (3 x^{3/2} \left (-1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{16 \left (-x^2+x^4\right )^{3/4}} \\ & = -\frac {1}{16} x \sqrt [4]{-x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^2+x^4}-\frac {\left (3 x^{3/2} \left (-1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{16 \left (-x^2+x^4\right )^{3/4}} \\ & = -\frac {1}{16} x \sqrt [4]{-x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^2+x^4}-\frac {\left (3 x^{3/2} \left (-1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \left (-x^2+x^4\right )^{3/4}}+\frac {\left (3 x^{3/2} \left (-1+x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \left (-x^2+x^4\right )^{3/4}} \\ & = -\frac {1}{16} x \sqrt [4]{-x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^2+x^4}+\frac {3 x^{3/2} \left (-1+x^2\right )^{3/4} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \left (-x^2+x^4\right )^{3/4}}-\frac {3 x^{3/2} \left (-1+x^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{32 \left (-x^2+x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.36 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {x^{3/2} \left (-1+x^2\right )^{3/4} \left (2 x^{3/2} \sqrt [4]{-1+x^2} \left (-1+4 x^2\right )+3 \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )-3 \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )\right )}{32 \left (x^2 \left (-1+x^2\right )\right )^{3/4}} \]

[In]

Integrate[x^2*(-x^2 + x^4)^(1/4),x]

[Out]

(x^(3/2)*(-1 + x^2)^(3/4)*(2*x^(3/2)*(-1 + x^2)^(1/4)*(-1 + 4*x^2) + 3*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)] - 3*Ar
cTanh[Sqrt[x]/(-1 + x^2)^(1/4)]))/(32*(x^2*(-1 + x^2))^(3/4))

Maple [C] (warning: unable to verify)

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

Time = 2.99 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.49

method	result	size
meijerg	\(\frac {2 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {7}{2}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], x^{2}\right )}{7 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}}}\)	\(33\)
pseudoelliptic	\(-\frac {x^{4} \left (4 \left (-4 x^{3}+x \right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}+6 \arctan \left (\frac {\left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{x}\right )-3 \ln \left (\frac {\left (x^{4}-x^{2}\right )^{\frac {1}{4}}-x}{x}\right )+3 \ln \left (\frac {\left (x^{4}-x^{2}\right )^{\frac {1}{4}}+x}{x}\right )\right )}{64 {\left (-\left (x^{4}-x^{2}\right )^{\frac {1}{4}}+x \right )}^{2} \left (x^{2}+\sqrt {x^{4}-x^{2}}\right )^{2} {\left (\left (x^{4}-x^{2}\right )^{\frac {1}{4}}+x \right )}^{2}}\)	\(136\)
trager	\(\frac {x \left (4 x^{2}-1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{16}+\frac {3 \ln \left (\frac {2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-x^{2}}\, x +2 x^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}-2 x^{3}+x}{x}\right )}{64}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-x^{2}}\, x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x}{x}\right )}{64}\)	\(162\)
risch	\(\frac {x \left (4 x^{2}-1\right ) \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}}}{16}+\frac {\left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{4}-2 x^{6}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {3}{4}}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}+2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}\, x^{2}+5 x^{4}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}}-2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}-4 x^{2}+1}{\left (x -1\right )^{2} \left (1+x \right )^{2}}\right )}{64}+\frac {3 \ln \left (-\frac {-2 x^{6}+2 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{4}-2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}\, x^{2}+5 x^{4}+2 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {3}{4}}-4 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}+2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}-4 x^{2}+2 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}}+1}{\left (x -1\right )^{2} \left (1+x \right )^{2}}\right )}{64}\right ) \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}} \left (x^{2} \left (x^{2}-1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x^{2}-1\right )}\)	\(446\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (55) = 110\).

Time = 0.83 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.76 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{16} \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (4 \, x^{3} - x\right )} - \frac {3}{64} \, \arctan \left (\frac {2 \, {\left ({\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) + \frac {3}{64} \, \log \left (-\frac {2 \, x^{3} - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} - x^{2}} x - x - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \]

[In]

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

[Out]

1/16*(x^4 - x^2)^(1/4)*(4*x^3 - x) - 3/64*arctan(2*((x^4 - x^2)^(1/4)*x^2 + (x^4 - x^2)^(3/4))/x) + 3/64*log(-
(2*x^3 - 2*(x^4 - x^2)^(1/4)*x^2 + 2*sqrt(x^4 - x^2)*x - x - 2*(x^4 - x^2)^(3/4))/x)

Sympy [F]

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

[In]

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

[Out]

Integral(x**2*(x**2*(x - 1)*(x + 1))**(1/4), x)

Maxima [F]

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

[In]

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

[Out]

integrate((x^4 - x^2)^(1/4)*x^2, x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{16} \, {\left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right )} x^{4} - \frac {3}{32} \, \arctan \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{64} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{64} \, \log \left (-{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) \]

[In]

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

[Out]

1/16*((-1/x^2 + 1)^(5/4) + 3*(-1/x^2 + 1)^(1/4))*x^4 - 3/32*arctan((-1/x^2 + 1)^(1/4)) - 3/64*log((-1/x^2 + 1)
^(1/4) + 1) + 3/64*log(-(-1/x^2 + 1)^(1/4) + 1)

Mupad [F(-1)]

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

[In]

int(x^2*(x^4 - x^2)^(1/4),x)

[Out]

int(x^2*(x^4 - x^2)^(1/4), x)