Integrand size = 13, antiderivative size = 49 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{4} x^3 \sqrt [4]{-1+x^4}+\frac {1}{8} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \] Output:
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))
Time = 0.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.92 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{8} \left (2 x^3 \sqrt [4]{-1+x^4}+\arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )\right ) \] Input:
Integrate[x^2*(-1 + x^4)^(1/4),x]
Output:
(2*x^3*(-1 + x^4)^(1/4) + ArcTan[x/(-1 + x^4)^(1/4)] - ArcTanh[x/(-1 + x^4 )^(1/4)])/8
Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {811, 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.
\(\displaystyle \int x^2 \sqrt [4]{x^4-1} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{x^4-1}-\frac {1}{4} \int \frac {x^2}{\left (x^4-1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{x^4-1}-\frac {1}{4} \int \frac {x^2}{\sqrt {x^4-1} \left (1-\frac {x^4}{x^4-1}\right )}d\frac {x}{\sqrt [4]{x^4-1}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {1}{\frac {x^2}{\sqrt {x^4-1}}+1}d\frac {x}{\sqrt [4]{x^4-1}}-\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{4} \sqrt [4]{x^4-1} x^3\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{4} \sqrt [4]{x^4-1} x^3\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )\right )+\frac {1}{4} \sqrt [4]{x^4-1} x^3\) |
Input:
Int[x^2*(-1 + x^4)^(1/4),x]
Output:
(x^3*(-1 + x^4)^(1/4))/4 + (ArcTan[x/(-1 + x^4)^(1/4)]/2 - ArcTanh[x/(-1 + x^4)^(1/4)]/2)/4
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_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[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] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, 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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.48 (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 | \(\frac {x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}}{4}-\frac {\ln \left (\frac {x +\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{16}-\frac {\arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{8}+\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right )}{16}\) | \(62\) |
trager | \(\frac {x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}}{4}-\frac {\ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x +2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}+2 x^{4}-1\right )}{16}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{16}\) | \(127\) |
Input:
int(x^2*(x^4-1)^(1/4),x,method=_RETURNVERBOSE)
Output:
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)
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.22 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{4} \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + \frac {1}{8} \, \arctan \left (\frac {x}{{\left (x^{4} - 1\right )}^{\frac {1}{4}}}\right ) - \frac {1}{16} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{16} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \] Input:
integrate(x^2*(x^4-1)^(1/4),x, algorithm="fricas")
Output:
1/4*(x^4 - 1)^(1/4)*x^3 + 1/8*arctan(x/(x^4 - 1)^(1/4)) - 1/16*log((x + (x ^4 - 1)^(1/4))/x) + 1/16*log(-(x - (x^4 - 1)^(1/4))/x)
Result contains complex when optimal does not.
Time = 0.59 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=- \frac {x^{3} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate(x**2*(x**4-1)**(1/4),x)
Output:
-x**3*exp(-3*I*pi/4)*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), x**4)/(4*gamma( 7/4))
Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.47 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{4 \, x {\left (\frac {x^{4} - 1}{x^{4}} - 1\right )}} - \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{16} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {1}{16} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \] Input:
integrate(x^2*(x^4-1)^(1/4),x, algorithm="maxima")
Output:
-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)
Time = 0.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.22 \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {1}{4} \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - \frac {1}{8} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{16} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {1}{16} \, \log \left (-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) \] Input:
integrate(x^2*(x^4-1)^(1/4),x, algorithm="giac")
Output:
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)
Timed out. \[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\int x^2\,{\left (x^4-1\right )}^{1/4} \,d x \] Input:
int(x^2*(x^4 - 1)^(1/4),x)
Output:
int(x^2*(x^4 - 1)^(1/4), x)
\[ \int x^2 \sqrt [4]{-1+x^4} \, dx=\frac {\left (x^{4}-1\right )^{\frac {1}{4}} x^{3}}{4}-\frac {\left (\int \frac {x^{2}}{\left (x^{4}-1\right )^{\frac {3}{4}}}d x \right )}{4} \] Input:
int(x^2*(x^4-1)^(1/4),x)
Output:
((x**4 - 1)**(1/4)*x**3 - int(((x**4 - 1)**(1/4)*x**2)/(x**4 - 1),x))/4