3.13.27 \(\int \frac {-2+x+x^2}{x^2 (-1+x^2)^{3/4}} \, dx\)

Optimal. Leaf size=90 \[ -\frac {2 \sqrt [4]{x^2-1}}{x}+\frac {\tan ^{-1}\left (\frac {\frac {\sqrt {x^2-1}}{\sqrt {2}}-\frac {1}{\sqrt {2}}}{\sqrt [4]{x^2-1}}\right )}{\sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^2-1}}{\sqrt {x^2-1}+1}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 138, normalized size of antiderivative = 1.53, number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1807, 266, 63, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {2 \sqrt [4]{x^2-1}}{x}-\frac {\log \left (\sqrt {x^2-1}-\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}}+\frac {\log \left (\sqrt {x^2-1}+\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + x + x^2)/(x^2*(-1 + x^2)^(3/4)),x]

[Out]

(-2*(-1 + x^2)^(1/4))/x - ArcTan[1 - Sqrt[2]*(-1 + x^2)^(1/4)]/Sqrt[2] + ArcTan[1 + Sqrt[2]*(-1 + x^2)^(1/4)]/
Sqrt[2] - Log[1 - Sqrt[2]*(-1 + x^2)^(1/4) + Sqrt[-1 + x^2]]/(2*Sqrt[2]) + Log[1 + Sqrt[2]*(-1 + x^2)^(1/4) +
Sqrt[-1 + x^2]]/(2*Sqrt[2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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
]]))

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rubi steps

\begin {align*} \int \frac {-2+x+x^2}{x^2 \left (-1+x^2\right )^{3/4}} \, dx &=-\frac {2 \sqrt [4]{-1+x^2}}{x}+\int \frac {1}{x \left (-1+x^2\right )^{3/4}} \, dx\\ &=-\frac {2 \sqrt [4]{-1+x^2}}{x}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{3/4} x} \, dx,x,x^2\right )\\ &=-\frac {2 \sqrt [4]{-1+x^2}}{x}+2 \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt [4]{-1+x^2}\right )\\ &=-\frac {2 \sqrt [4]{-1+x^2}}{x}+\operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^2}\right )+\operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^2}\right )\\ &=-\frac {2 \sqrt [4]{-1+x^2}}{x}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )}{2 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )}{2 \sqrt {2}}\\ &=-\frac {2 \sqrt [4]{-1+x^2}}{x}-\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{2 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+x^2}\right )}{\sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{-1+x^2}\right )}{\sqrt {2}}\\ &=-\frac {2 \sqrt [4]{-1+x^2}}{x}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{-1+x^2}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt [4]{-1+x^2}\right )}{\sqrt {2}}-\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 138, normalized size = 1.53 \begin {gather*} -\frac {2 \sqrt [4]{x^2-1}}{x}-\frac {\log \left (\sqrt {x^2-1}-\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}}+\frac {\log \left (\sqrt {x^2-1}+\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + x + x^2)/(x^2*(-1 + x^2)^(3/4)),x]

[Out]

(-2*(-1 + x^2)^(1/4))/x - ArcTan[1 - Sqrt[2]*(-1 + x^2)^(1/4)]/Sqrt[2] + ArcTan[1 + Sqrt[2]*(-1 + x^2)^(1/4)]/
Sqrt[2] - Log[1 - Sqrt[2]*(-1 + x^2)^(1/4) + Sqrt[-1 + x^2]]/(2*Sqrt[2]) + Log[1 + Sqrt[2]*(-1 + x^2)^(1/4) +
Sqrt[-1 + x^2]]/(2*Sqrt[2])

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IntegrateAlgebraic [A]  time = 19.95, size = 86, normalized size = 0.96 \begin {gather*} -\frac {2 \sqrt [4]{-1+x^2}}{x}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{-1+\sqrt {-1+x^2}}\right )}{\sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{1+\sqrt {-1+x^2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-2 + x + x^2)/(x^2*(-1 + x^2)^(3/4)),x]

[Out]

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

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fricas [B]  time = 1.46, size = 451, normalized size = 5.01 \begin {gather*} -\frac {4 \, \sqrt {2} x \arctan \left (\frac {x^{4} + 4 \, \sqrt {x^{2} - 1} x^{2} + 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} {\left (x^{2} - 4\right )} + 2 \, \sqrt {2} {\left (3 \, x^{2} - 4\right )} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + {\left (4 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {2} \sqrt {x^{2} - 1} {\left (x^{2} - 4\right )} + \sqrt {2} {\left (x^{4} - 10 \, x^{2} + 8\right )} + 16 \, {\left (x^{2} - 1\right )}^{\frac {3}{4}}\right )} \sqrt {\frac {x^{2} + 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} + 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{2} - 1}}{x^{2}}}}{x^{4} - 16 \, x^{2} + 16}\right ) - 4 \, \sqrt {2} x \arctan \left (\frac {x^{4} + 4 \, \sqrt {x^{2} - 1} x^{2} - 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} {\left (x^{2} - 4\right )} - 2 \, \sqrt {2} {\left (3 \, x^{2} - 4\right )} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + {\left (4 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}} x^{2} - 2 \, \sqrt {2} \sqrt {x^{2} - 1} {\left (x^{2} - 4\right )} - \sqrt {2} {\left (x^{4} - 10 \, x^{2} + 8\right )} + 16 \, {\left (x^{2} - 1\right )}^{\frac {3}{4}}\right )} \sqrt {\frac {x^{2} - 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} - 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{2} - 1}}{x^{2}}}}{x^{4} - 16 \, x^{2} + 16}\right ) - \sqrt {2} x \log \left (\frac {4 \, {\left (x^{2} + 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} + 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{2} - 1}\right )}}{x^{2}}\right ) + \sqrt {2} x \log \left (\frac {4 \, {\left (x^{2} - 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} - 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{2} - 1}\right )}}{x^{2}}\right ) + 16 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x-2)/x^2/(x^2-1)^(3/4),x, algorithm="fricas")

[Out]

-1/8*(4*sqrt(2)*x*arctan((x^4 + 4*sqrt(x^2 - 1)*x^2 + 2*sqrt(2)*(x^2 - 1)^(3/4)*(x^2 - 4) + 2*sqrt(2)*(3*x^2 -
4)*(x^2 - 1)^(1/4) + (4*(x^2 - 1)^(1/4)*x^2 + 2*sqrt(2)*sqrt(x^2 - 1)*(x^2 - 4) + sqrt(2)*(x^4 - 10*x^2 + 8)
+ 16*(x^2 - 1)^(3/4))*sqrt((x^2 + 2*sqrt(2)*(x^2 - 1)^(3/4) + 2*sqrt(2)*(x^2 - 1)^(1/4) + 4*sqrt(x^2 - 1))/x^2
))/(x^4 - 16*x^2 + 16)) - 4*sqrt(2)*x*arctan((x^4 + 4*sqrt(x^2 - 1)*x^2 - 2*sqrt(2)*(x^2 - 1)^(3/4)*(x^2 - 4)
- 2*sqrt(2)*(3*x^2 - 4)*(x^2 - 1)^(1/4) + (4*(x^2 - 1)^(1/4)*x^2 - 2*sqrt(2)*sqrt(x^2 - 1)*(x^2 - 4) - sqrt(2)
*(x^4 - 10*x^2 + 8) + 16*(x^2 - 1)^(3/4))*sqrt((x^2 - 2*sqrt(2)*(x^2 - 1)^(3/4) - 2*sqrt(2)*(x^2 - 1)^(1/4) +
4*sqrt(x^2 - 1))/x^2))/(x^4 - 16*x^2 + 16)) - sqrt(2)*x*log(4*(x^2 + 2*sqrt(2)*(x^2 - 1)^(3/4) + 2*sqrt(2)*(x^
2 - 1)^(1/4) + 4*sqrt(x^2 - 1))/x^2) + sqrt(2)*x*log(4*(x^2 - 2*sqrt(2)*(x^2 - 1)^(3/4) - 2*sqrt(2)*(x^2 - 1)^
(1/4) + 4*sqrt(x^2 - 1))/x^2) + 16*(x^2 - 1)^(1/4))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + x - 2}{{\left (x^{2} - 1\right )}^{\frac {3}{4}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x-2)/x^2/(x^2-1)^(3/4),x, algorithm="giac")

[Out]

integrate((x^2 + x - 2)/((x^2 - 1)^(3/4)*x^2), x)

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maple [C]  time = 2.70, size = 76, normalized size = 0.84

method	result	size
risch	\(-\frac {2 \left (x^{2}-1\right )^{\frac {1}{4}}}{x}+\frac {\left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {3}{4}} \left (\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], x^{2}\right )}{4}+\left (-3 \ln \relax (2)+\frac {\pi }{2}+2 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{2 \Gamma \left (\frac {3}{4}\right ) \mathrm {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}\)	\(76\)
meijerg	\(\frac {\left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], x^{2}\right )}{\mathrm {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}+\frac {\left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {3}{4}} \left (\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], x^{2}\right )}{4}+\left (-3 \ln \relax (2)+\frac {\pi }{2}+2 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{2 \Gamma \left (\frac {3}{4}\right ) \mathrm {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}+\frac {2 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {1}{2}\right ], x^{2}\right )}{\mathrm {signum}\left (x^{2}-1\right )^{\frac {3}{4}} x}\)	\(125\)
trager	\(-\frac {2 \left (x^{2}-1\right )^{\frac {1}{4}}}{x}-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {-2 \sqrt {x^{2}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \left (x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+\RootOf \left (\textit {\_Z}^{4}+1\right ) x^{2}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )}{x^{2}}\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {x^{2}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )+2 \left (x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}}{x^{2}}\right )}{2}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+x-2)/x^2/(x^2-1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-2*(x^2-1)^(1/4)/x+1/2/GAMMA(3/4)/signum(x^2-1)^(3/4)*(-signum(x^2-1))^(3/4)*(3/4*GAMMA(3/4)*x^2*hypergeom([1,
1,7/4],[2,2],x^2)+(-3*ln(2)+1/2*Pi+2*ln(x)+I*Pi)*GAMMA(3/4))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + x - 2}{{\left (x^{2} - 1\right )}^{\frac {3}{4}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x-2)/x^2/(x^2-1)^(3/4),x, algorithm="maxima")

[Out]

integrate((x^2 + x - 2)/((x^2 - 1)^(3/4)*x^2), x)

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mupad [B]  time = 1.33, size = 89, normalized size = 0.99 \begin {gather*} \frac {4\,{\left (\frac {1}{x^2}\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {5}{4};\ \frac {9}{4};\ \frac {1}{x^2}\right )}{5\,x}+\frac {x\,{\left (1-x^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {3}{2};\ x^2\right )}{{\left (x^2-1\right )}^{3/4}}+\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 {1}{2}+\frac {1}{2}{}\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 {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x^2 - 2)/(x^2*(x^2 - 1)^(3/4)),x)

[Out]

2^(1/2)*atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 - 1i/2))*(1/2 + 1i/2) + 2^(1/2)*atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 +
1i/2))*(1/2 - 1i/2) + (4*(1/x^2)^(3/4)*hypergeom([3/4, 5/4], 9/4, 1/x^2))/(5*x) + (x*(1 - x^2)^(3/4)*hypergeom
([1/2, 3/4], 3/2, x^2))/(x^2 - 1)^(3/4)

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sympy [C]  time = 2.90, size = 75, normalized size = 0.83 \begin {gather*} x e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {x^{2}} \right )} - \frac {2 e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {x^{2}} \right )}}{x} - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+x-2)/x**2/(x**2-1)**(3/4),x)

[Out]

x*exp(-3*I*pi/4)*hyper((1/2, 3/4), (3/2,), x**2) - 2*exp(I*pi/4)*hyper((-1/2, 3/4), (1/2,), x**2)/x - gamma(3/
4)*hyper((3/4, 3/4), (7/4,), exp_polar(2*I*pi)/x**2)/(2*x**(3/2)*gamma(7/4))

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