Integrand size = 17, antiderivative size = 67 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{16} \left (-x+4 x^3\right ) \sqrt [4]{-x^2+x^4}+\frac {3}{32} \arctan \left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\frac {3}{32} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right ) \]
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))
Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.36 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {x^{3/2} \left (-1+x^2\right )^{3/4} \left (2 x^{3/2} \sqrt [4]{-1+x^2} \left (-1+4 x^2\right )+3 \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )-3 \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )\right )}{32 \left (x^2 \left (-1+x^2\right )\right )^{3/4}} \]
(x^(3/2)*(-1 + x^2)^(3/4)*(2*x^(3/2)*(-1 + x^2)^(1/4)*(-1 + 4*x^2) + 3*Arc Tan[Sqrt[x]/(-1 + x^2)^(1/4)] - 3*ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)]))/(32* (x^2*(-1 + x^2))^(3/4))
Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.73, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {1426, 1429, 1431, 266, 854, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^2 \sqrt [4]{x^4-x^2} \, dx\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {1}{4} x^3 \sqrt [4]{x^4-x^2}-\frac {1}{8} \int \frac {x^4}{\left (x^4-x^2\right )^{3/4}}dx\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {1}{8} \left (-\frac {3}{4} \int \frac {x^2}{\left (x^4-x^2\right )^{3/4}}dx-\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {1}{8} \left (-\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \int \frac {\sqrt {x}}{\left (x^2-1\right )^{3/4}}dx}{4 \left (x^4-x^2\right )^{3/4}}-\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {1}{8} \left (-\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \int \frac {x}{\left (x^2-1\right )^{3/4}}d\sqrt {x}}{2 \left (x^4-x^2\right )^{3/4}}-\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {1}{8} \left (-\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \int \frac {x}{1-x^2}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}}{2 \left (x^4-x^2\right )^{3/4}}-\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {1}{8} \left (-\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \left (\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}-\frac {1}{2} \int \frac {1}{x+1}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \left (x^4-x^2\right )^{3/4}}-\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{8} \left (-\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \left (\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}-\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )\right )}{2 \left (x^4-x^2\right )^{3/4}}-\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{8} \left (-\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \left (\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )-\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )\right )}{2 \left (x^4-x^2\right )^{3/4}}-\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )+\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\) |
(x^3*(-x^2 + x^4)^(1/4))/4 + (-1/2*(x*(-x^2 + x^4)^(1/4)) - (3*x^(3/2)*(-1 + x^2)^(3/4)*(-1/2*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)] + ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)]/2))/(2*(-x^2 + x^4)^(3/4)))/8
3.9.78.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 *(m + 4*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[m + 2*p - 1, 0] && NeQ[m + 4*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
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 | \(\frac {x \left (4 x^{2}-1\right ) \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}}}{16}+\frac {\left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{4}-2 x^{6}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {3}{4}}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}+2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}\, x^{2}+5 x^{4}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}}-2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}-4 x^{2}+1}{\left (x -1\right )^{2} \left (1+x \right )^{2}}\right )}{64}+\frac {3 \ln \left (-\frac {-2 x^{6}+2 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{4}-2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}\, x^{2}+5 x^{4}+2 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {3}{4}}-4 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}+2 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}-4 x^{2}+2 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}}+1}{\left (x -1\right )^{2} \left (1+x \right )^{2}}\right )}{64}\right ) \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}} \left (x^{2} \left (x^{2}-1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x^{2}-1\right )}\) | \(446\) |
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 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{16} \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (4 \, x^{3} - x\right )} - \frac {3}{64} \, \arctan \left (\frac {2 \, {\left ({\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) + \frac {3}{64} \, \log \left (-\frac {2 \, x^{3} - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} - x^{2}} x - x - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \]
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*s qrt(x^4 - x^2)*x - x - 2*(x^4 - x^2)^(3/4))/x)
\[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\int x^{2} \sqrt [4]{x^{2} \left (x - 1\right ) \left (x + 1\right )}\, dx \]
\[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\int { {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{16} \, {\left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right )} x^{4} - \frac {3}{32} \, \arctan \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{64} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{64} \, \log \left (-{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) \]
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)
Timed out. \[ \int x^2 \sqrt [4]{-x^2+x^4} \, dx=\int x^2\,{\left (x^4-x^2\right )}^{1/4} \,d x \]