Integrand size = 13, antiderivative size = 93 \[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{-1+x^2}}{2 x^2}-\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{-1+\sqrt {-1+x^2}}\right )}{4 \sqrt {2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{1+\sqrt {-1+x^2}}\right )}{4 \sqrt {2}} \] Output:
1/2*(x^2-1)^(1/4)/x^2-3/8*arctan(2^(1/2)*(x^2-1)^(1/4)/(-1+(x^2-1)^(1/2))) *2^(1/2)+3/8*arctanh(2^(1/2)*(x^2-1)^(1/4)/(1+(x^2-1)^(1/2)))*2^(1/2)
Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=\frac {1}{8} \left (\frac {4 \sqrt [4]{-1+x^2}}{x^2}+3 \sqrt {2} \arctan \left (\frac {-1+\sqrt {-1+x^2}}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )+3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{1+\sqrt {-1+x^2}}\right )\right ) \] Input:
Integrate[1/(x^3*(-1 + x^2)^(3/4)),x]
Output:
((4*(-1 + x^2)^(1/4))/x^2 + 3*Sqrt[2]*ArcTan[(-1 + Sqrt[-1 + x^2])/(Sqrt[2 ]*(-1 + x^2)^(1/4))] + 3*Sqrt[2]*ArcTanh[(Sqrt[2]*(-1 + x^2)^(1/4))/(1 + S qrt[-1 + x^2])])/8
Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.53, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {243, 52, 73, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int \frac {1}{x^3 \left (x^2-1\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (x^2-1\right )^{3/4}}dx^2\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} \int \frac {1}{x^2 \left (x^2-1\right )^{3/4}}dx^2+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (3 \int \frac {1}{x^8+1}d\sqrt [4]{x^2-1}+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 755 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{x^2-1}+\frac {1}{2} \int \frac {x^4+1}{x^8+1}d\sqrt [4]{x^2-1}\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4-\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}+\frac {1}{2} \int \frac {1}{x^4+\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}\right )+\frac {1}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{x^2-1}\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-x^4-1}d\left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x^4-1}d\left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{x^2-1}\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{x^2-1}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt [4]{x^2-1}}{x^4-\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{x^4+\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt [4]{x^2-1}}{x^4-\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{x^4+\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt [4]{x^2-1}}{x^4-\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt [4]{x^2-1}+1}{x^4+\sqrt {2} \sqrt [4]{x^2-1}+1}d\sqrt [4]{x^2-1}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x^4+\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^4-\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}}\right )\right )+\frac {\sqrt [4]{x^2-1}}{x^2}\right )\) |
Input:
Int[1/(x^3*(-1 + x^2)^(3/4)),x]
Output:
((-1 + x^2)^(1/4)/x^2 + 3*((-(ArcTan[1 - Sqrt[2]*(-1 + x^2)^(1/4)]/Sqrt[2] ) + ArcTan[1 + Sqrt[2]*(-1 + x^2)^(1/4)]/Sqrt[2])/2 + (-1/2*Log[1 + x^4 - Sqrt[2]*(-1 + x^2)^(1/4)]/Sqrt[2] + Log[1 + x^4 + Sqrt[2]*(-1 + x^2)^(1/4) ]/(2*Sqrt[2]))/2))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.99 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76
method | result | size |
meijerg | \(-\frac {{\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {3}{4}} \left (-\frac {21 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {11}{4}\right ], \left [2, 3\right ], x^{2}\right )}{32}-\frac {3 \left (\frac {1}{3}-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )}{4}+\frac {\Gamma \left (\frac {3}{4}\right )}{x^{2}}\right )}{2 \Gamma \left (\frac {3}{4}\right ) \operatorname {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}\) | \(71\) |
risch | \(\frac {\left (x^{2}-1\right )^{\frac {1}{4}}}{2 x^{2}}+\frac {3 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {3}{4}} \left (\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], x^{2}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{8 \Gamma \left (\frac {3}{4}\right ) \operatorname {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}\) | \(76\) |
pseudoelliptic | \(\frac {3 \ln \left (\frac {-\left (x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {x^{2}-1}-1}{\left (x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {x^{2}-1}-1}\right ) \sqrt {2}\, x^{2}+6 \arctan \left (\left (x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}+1\right ) \sqrt {2}\, x^{2}+6 \arctan \left (\left (x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}-1\right ) \sqrt {2}\, x^{2}+8 \left (x^{2}-1\right )^{\frac {1}{4}}}{16 x^{2}}\) | \(117\) |
trager | \(\frac {\left (x^{2}-1\right )^{\frac {1}{4}}}{2 x^{2}}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {-2 \sqrt {x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \left (x^{2}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{2}}\right )}{8}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+2 \left (x^{2}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}}{x^{2}}\right )}{8}\) | \(170\) |
Input:
int(1/x^3/(x^2-1)^(3/4),x,method=_RETURNVERBOSE)
Output:
-1/2/GAMMA(3/4)/signum(x^2-1)^(3/4)*(-signum(x^2-1))^(3/4)*(-21/32*GAMMA(3 /4)*x^2*hypergeom([1,1,11/4],[2,3],x^2)-3/4*(1/3-3*ln(2)+1/2*Pi+2*ln(x)+I* Pi)*GAMMA(3/4)+GAMMA(3/4)/x^2)
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=\frac {6 \, \sqrt {2} x^{2} \arctan \left (\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + 6 \, \sqrt {2} x^{2} \arctan \left (\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) + 3 \, \sqrt {2} x^{2} \log \left (\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) - 3 \, \sqrt {2} x^{2} \log \left (-\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) + 8 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{16 \, x^{2}} \] Input:
integrate(1/x^3/(x^2-1)^(3/4),x, algorithm="fricas")
Output:
1/16*(6*sqrt(2)*x^2*arctan(sqrt(2)*(x^2 - 1)^(1/4) + 1) + 6*sqrt(2)*x^2*ar ctan(sqrt(2)*(x^2 - 1)^(1/4) - 1) + 3*sqrt(2)*x^2*log(sqrt(2)*(x^2 - 1)^(1 /4) + sqrt(x^2 - 1) + 1) - 3*sqrt(2)*x^2*log(-sqrt(2)*(x^2 - 1)^(1/4) + sq rt(x^2 - 1) + 1) + 8*(x^2 - 1)^(1/4))/x^2
Result contains complex when optimal does not.
Time = 0.67 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.37 \[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 x^{\frac {7}{2}} \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate(1/x**3/(x**2-1)**(3/4),x)
Output:
-gamma(7/4)*hyper((3/4, 7/4), (11/4,), exp_polar(2*I*pi)/x**2)/(2*x**(7/2) *gamma(11/4))
Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=\frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) + \frac {{\left (x^{2} - 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} \] Input:
integrate(1/x^3/(x^2-1)^(3/4),x, algorithm="maxima")
Output:
3/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^2 - 1)^(1/4))) + 3/8*sqrt(2 )*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(x^2 - 1)^(1/4))) + 3/16*sqrt(2)*log(sq rt(2)*(x^2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) - 3/16*sqrt(2)*log(-sqrt(2)*(x^ 2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) + 1/2*(x^2 - 1)^(1/4)/x^2
Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=\frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) + \frac {{\left (x^{2} - 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} \] Input:
integrate(1/x^3/(x^2-1)^(3/4),x, algorithm="giac")
Output:
3/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^2 - 1)^(1/4))) + 3/8*sqrt(2 )*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(x^2 - 1)^(1/4))) + 3/16*sqrt(2)*log(sq rt(2)*(x^2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) - 3/16*sqrt(2)*log(-sqrt(2)*(x^ 2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) + 1/2*(x^2 - 1)^(1/4)/x^2
Time = 8.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=\frac {{\left (x^2-1\right )}^{1/4}}{2\,x^2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^2-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{8}+\frac {3}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^2-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{8}-\frac {3}{8}{}\mathrm {i}\right ) \] Input:
int(1/(x^3*(x^2 - 1)^(3/4)),x)
Output:
2^(1/2)*atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 - 1i/2))*(3/8 + 3i/8) + 2^(1/2)* atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 + 1i/2))*(3/8 - 3i/8) + (x^2 - 1)^(1/4)/ (2*x^2)
\[ \int \frac {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx=\int \frac {1}{\left (x^{2}-1\right )^{\frac {3}{4}} x^{3}}d x \] Input:
int(1/x^3/(x^2-1)^(3/4),x)
Output:
int(1/((x**2 - 1)**(3/4)*x**3),x)