Integrand size = 24, antiderivative size = 81 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=-\frac {2 \left (-1+x^2\right ) \sqrt [4]{-x^2+x^4}}{5 x^3}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^2+x^4}}\right )}{2^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^2+x^4}}\right )}{2^{3/4}} \] Output:
-2/5*(x^2-1)*(x^4-x^2)^(1/4)/x^3-1/2*arctan(2^(1/4)*x/(x^4-x^2)^(1/4))*2^( 1/4)+1/2*arctanh(2^(1/4)*x/(x^4-x^2)^(1/4))*2^(1/4)
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=-\frac {\sqrt [4]{x^2 \left (-1+x^2\right )} \left (4 \left (-1+x^2\right )^{5/4}+5 \sqrt [4]{2} x^{5/2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )-5 \sqrt [4]{2} x^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )\right )}{10 x^3 \sqrt [4]{-1+x^2}} \] Input:
Integrate[(-x^2 + x^4)^(1/4)/(x^4*(-1 + x^4)),x]
Output:
-1/10*((x^2*(-1 + x^2))^(1/4)*(4*(-1 + x^2)^(5/4) + 5*2^(1/4)*x^(5/2)*ArcT an[(2^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)] - 5*2^(1/4)*x^(5/2)*ArcTanh[(2^(1/4 )*Sqrt[x])/(-1 + x^2)^(1/4)]))/(x^3*(-1 + x^2)^(1/4))
Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2467, 25, 1388, 368, 25, 996, 961, 2009}
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.
\(\displaystyle \int \frac {\sqrt [4]{x^4-x^2}}{x^4 \left (x^4-1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \int -\frac {\sqrt [4]{x^2-1}}{x^{7/2} \left (1-x^4\right )}dx}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^2} \int \frac {\sqrt [4]{x^2-1}}{x^{7/2} \left (1-x^4\right )}dx}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^2} \int \frac {1}{x^{7/2} \left (-x^2-1\right ) \left (x^2-1\right )^{3/4}}dx}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 368 |
\(\displaystyle -\frac {2 \sqrt [4]{x^4-x^2} \int -\frac {1}{x^3 \left (x^2-1\right )^{3/4} \left (x^2+1\right )}d\sqrt {x}}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \sqrt [4]{x^4-x^2} \int \frac {1}{x^3 \left (x^2-1\right )^{3/4} \left (x^2+1\right )}d\sqrt {x}}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 996 |
\(\displaystyle \frac {2 \sqrt [4]{x^4-x^2} \int \frac {\left (1-x^2\right )^2}{x^3 \left (1-2 x^2\right )}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 961 |
\(\displaystyle \frac {2 \sqrt [4]{x^4-x^2} \int \left (\frac {1}{x^3}-\frac {x}{2 x^2-1}\right )d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt [4]{x^4-x^2} \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2\ 2^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2\ 2^{3/4}}-\frac {1}{5 x^{5/2}}\right )}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
Input:
Int[(-x^2 + x^4)^(1/4)/(x^4*(-1 + x^4)),x]
Output:
(2*(-x^2 + x^4)^(1/4)*(-1/5*1/x^(5/2) - ArcTan[(2^(1/4)*Sqrt[x])/(-1 + x^2 )^(1/4)]/(2*2^(3/4)) + ArcTanh[(2^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)]/(2*2^(3 /4))))/(Sqrt[x]*(-1 + x^2)^(1/4))
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^( n_)), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.) , x_Symbol] :> With[{k = Denominator[p]}, Simp[k*(a^(p + (m + 1)/n)/n) Su bst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p + q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && RationalQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 3.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19
method | result | size |
pseudoelliptic | \(\frac {\left (-8 x^{2}+8\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}+5 x^{3} 2^{\frac {1}{4}} \left (\ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{4}-x^{2}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (x^{4}-x^{2}\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right )\right )}{20 x^{3}}\) | \(96\) |
trager | \(-\frac {2 \left (x^{2}-1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{3}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x +4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \sqrt {x^{4}-x^{2}}\, x -4 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}}{x \left (x^{2}+1\right )}\right )}{4}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{3}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x +4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \sqrt {x^{4}-x^{2}}\, x -4 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}}{x \left (x^{2}+1\right )}\right )}{4}\) | \(268\) |
risch | \(-\frac {2 \left (x^{4}-2 x^{2}+1\right ) \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}}}{5 x^{3} \left (x^{2}-1\right )}+\frac {\left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{4}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{6}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}+7 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{4}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}}-5 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {3}{4}}+4 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}\, x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}-4 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}}{\left (-1+x \right )^{2} \left (1+x \right )^{2} \left (x^{2}+1\right )}\right )}{4}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{4}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{6}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} x^{2}+7 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{4}+2 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3}-5 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}+4 \left (x^{8}-3 x^{6}+3 x^{4}-x^{2}\right )^{\frac {3}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )-4 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}\, x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}+4 \sqrt {x^{8}-3 x^{6}+3 x^{4}-x^{2}}}{\left (-1+x \right )^{2} \left (1+x \right )^{2} \left (x^{2}+1\right )}\right )}{4}\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 )}\) | \(629\) |
Input:
int((x^4-x^2)^(1/4)/x^4/(x^4-1),x,method=_RETURNVERBOSE)
Output:
1/20*((-8*x^2+8)*(x^4-x^2)^(1/4)+5*x^3*2^(1/4)*(ln((2^(1/4)*x+(x^4-x^2)^(1 /4))/(-2^(1/4)*x+(x^4-x^2)^(1/4)))+2*arctan(1/2*(x^4-x^2)^(1/4)/x*2^(3/4)) ))/x^3
Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (66) = 132\).
Time = 1.08 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.98 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=-\frac {10 \cdot 8^{\frac {3}{4}} x^{3} \arctan \left (\frac {8^{\frac {3}{4}} {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}} - 2 \cdot 8^{\frac {1}{4}} {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{8 \, {\left (x^{3} - x\right )}}\right ) - 5 \cdot 8^{\frac {3}{4}} x^{3} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 8^{\frac {3}{4}} \sqrt {x^{4} - x^{2}} x + 8^{\frac {1}{4}} {\left (3 \, x^{3} - x\right )} + 4 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x^{3} + x}\right ) + 5 \cdot 8^{\frac {3}{4}} x^{3} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} - 8^{\frac {3}{4}} \sqrt {x^{4} - x^{2}} x - 8^{\frac {1}{4}} {\left (3 \, x^{3} - x\right )} + 4 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x^{3} + x}\right ) + 64 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{160 \, x^{3}} \] Input:
integrate((x^4-x^2)^(1/4)/x^4/(x^4-1),x, algorithm="fricas")
Output:
-1/160*(10*8^(3/4)*x^3*arctan(1/8*(8^(3/4)*(x^4 - x^2)^(3/4) - 2*8^(1/4)*( x^4 - x^2)^(1/4)*(x^2 - 1))/(x^3 - x)) - 5*8^(3/4)*x^3*log((4*sqrt(2)*(x^4 - x^2)^(1/4)*x^2 + 8^(3/4)*sqrt(x^4 - x^2)*x + 8^(1/4)*(3*x^3 - x) + 4*(x ^4 - x^2)^(3/4))/(x^3 + x)) + 5*8^(3/4)*x^3*log((4*sqrt(2)*(x^4 - x^2)^(1/ 4)*x^2 - 8^(3/4)*sqrt(x^4 - x^2)*x - 8^(1/4)*(3*x^3 - x) + 4*(x^4 - x^2)^( 3/4))/(x^3 + x)) + 64*(x^4 - x^2)^(1/4)*(x^2 - 1))/x^3
\[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\int \frac {\sqrt [4]{x^{2} \left (x - 1\right ) \left (x + 1\right )}}{x^{4} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \] Input:
integrate((x**4-x**2)**(1/4)/x**4/(x**4-1),x)
Output:
Integral((x**2*(x - 1)*(x + 1))**(1/4)/(x**4*(x - 1)*(x + 1)*(x**2 + 1)), x)
\[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\int { \frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{4}}}{{\left (x^{4} - 1\right )} x^{4}} \,d x } \] Input:
integrate((x^4-x^2)^(1/4)/x^4/(x^4-1),x, algorithm="maxima")
Output:
integrate((x^4 - x^2)^(1/4)/((x^4 - 1)*x^4), x)
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=-\frac {2}{5} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} + \frac {1}{2} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} - {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) \] Input:
integrate((x^4-x^2)^(1/4)/x^4/(x^4-1),x, algorithm="giac")
Output:
-2/5*(-1/x^2 + 1)^(5/4) + 1/2*2^(1/4)*arctan(1/2*2^(3/4)*(-1/x^2 + 1)^(1/4 )) + 1/4*2^(1/4)*log(2^(1/4) + (-1/x^2 + 1)^(1/4)) - 1/4*2^(1/4)*log(2^(1/ 4) - (-1/x^2 + 1)^(1/4))
Timed out. \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=-\int \frac {{\left (x^4-x^2\right )}^{1/4}}{x^4-x^8} \,d x \] Input:
int((x^4 - x^2)^(1/4)/(x^4*(x^4 - 1)),x)
Output:
-int((x^4 - x^2)^(1/4)/(x^4 - x^8), x)
\[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\frac {8 \left (x^{2}-1\right )^{\frac {1}{4}} x^{2}+2 \left (x^{2}-1\right )^{\frac {1}{4}}-5 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (x^{2}-1\right )^{\frac {1}{4}}}{x^{6}-x^{2}}d x \right ) x^{2}}{5 \sqrt {x}\, x^{2}} \] Input:
int((x^4-x^2)^(1/4)/x^4/(x^4-1),x)
Output:
(8*(x**2 - 1)**(1/4)*x**2 + 2*(x**2 - 1)**(1/4) - 5*sqrt(x)*int((sqrt(x)*( x**2 - 1)**(1/4))/(x**6 - x**2),x)*x**2)/(5*sqrt(x)*x**2)