3.9.55 \(\int \frac {-1+2 x^2}{(1+x^2) \sqrt {-1-x^2+x^4}} \, dx\)

Optimal. Leaf size=65 \[ -\frac {1}{2} i \log \left (\frac {-2 i x^2+\sqrt {x^4-x^2-1} x-i}{-2 i x^2-\sqrt {x^4-x^2-1} x-i}\right ) \]

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Rubi [C]  time = 0.41, antiderivative size = 520, normalized size of antiderivative = 8.00, number of steps used = 10, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {1710, 1098, 1214, 1456, 540, 421, 419, 538, 537} \begin {gather*} \frac {3 \left (1+\sqrt {5}\right ) \sqrt {2 x^2+\sqrt {5}-1} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} F\left (\sin ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )|\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {2} \left (3+\sqrt {5}\right ) \sqrt {x^4-x^2-1}}-\frac {3 \sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x^2+2}{\left (1-\sqrt {5}\right ) x^2+2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (3+\sqrt {5}\right ) \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x^2+2}} \sqrt {x^4-x^2-1}}+\frac {\sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x^2+2}{\left (1-\sqrt {5}\right ) x^2+2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x^2+2}} \sqrt {x^4-x^2-1}}-\frac {3 \sqrt {2} \left (2+\sqrt {5}\right ) \sqrt {2 x^2+\sqrt {5}-1} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} \Pi \left (\frac {1}{2} \left (-1-\sqrt {5}\right );\sin ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )|\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\left (3+\sqrt {5}\right ) \sqrt {x^4-x^2-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*(1 + Sqrt[5])*Sqrt[-1 + Sqrt[5] + 2*x^2]*Sqrt[1 - (2*x^2)/(1 + Sqrt[5])]*EllipticF[ArcSin[Sqrt[2/(1 + Sqrt[
5])]*x], (-3 - Sqrt[5])/2])/(Sqrt[2]*(3 + Sqrt[5])*Sqrt[-1 - x^2 + x^4]) + (Sqrt[-2 - (1 - Sqrt[5])*x^2]*Sqrt[
(2 + (1 + Sqrt[5])*x^2)/(2 + (1 - Sqrt[5])*x^2)]*EllipticF[ArcSin[(Sqrt[2]*5^(1/4)*x)/Sqrt[-2 - (1 - Sqrt[5])*
x^2]], (5 - Sqrt[5])/10])/(5^(1/4)*Sqrt[(2 + (1 - Sqrt[5])*x^2)^(-1)]*Sqrt[-1 - x^2 + x^4]) - (3*Sqrt[-2 - (1
- Sqrt[5])*x^2]*Sqrt[(2 + (1 + Sqrt[5])*x^2)/(2 + (1 - Sqrt[5])*x^2)]*EllipticF[ArcSin[(Sqrt[2]*5^(1/4)*x)/Sqr
t[-2 - (1 - Sqrt[5])*x^2]], (5 - Sqrt[5])/10])/(5^(1/4)*(3 + Sqrt[5])*Sqrt[(2 + (1 - Sqrt[5])*x^2)^(-1)]*Sqrt[
-1 - x^2 + x^4]) - (3*Sqrt[2]*(2 + Sqrt[5])*Sqrt[-1 + Sqrt[5] + 2*x^2]*Sqrt[1 - (2*x^2)/(1 + Sqrt[5])]*Ellipti
cPi[(-1 - Sqrt[5])/2, ArcSin[Sqrt[2/(1 + Sqrt[5])]*x], (-3 - Sqrt[5])/2])/((3 + Sqrt[5])*Sqrt[-1 - x^2 + x^4])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 538

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d*x^2)/c]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[c, 0]

Rule 540

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[d/b, Int[1/
(Sqrt[c + d*x^2]*Sqrt[e + f*x^2]), x], x] + Dist[(b*c - a*d)/b, Int[1/((a + b*x^2)*Sqrt[c + d*x^2]*Sqrt[e + f*
x^2]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NegQ[d/c]

Rule 1098

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(Sqrt[(2*a +
(b - q)*x^2)/(2*a + (b + q)*x^2)]*Sqrt[(2*a + (b + q)*x^2)/q]*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q
)]], (b + q)/(2*q)])/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)*x^2)]), x]] /; FreeQ[{a, b, c}, x] && Gt
Q[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]

Rule 1214

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(2*c)/(2*c*d - e*(b - q)), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/(2*c*d - e*(b - q)), Int[
(b - q + 2*c*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a
*c, 0] &&  !LtQ[c, 0]

Rule 1456

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPar
t[p]), Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c*x^n)/e)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p,
q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p]

Rule 1710

Int[((A_.) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist
[B/e, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[(e*A - d*B)/e, Int[1/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]),
x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^
2 - a*e^2, 0] && NegQ[c/a]

Rubi steps

\begin {align*} \int \frac {-1+2 x^2}{\left (1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx &=2 \int \frac {1}{\sqrt {-1-x^2+x^4}} \, dx-3 \int \frac {1}{\left (1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx\\ &=\frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}+\frac {3 \int \frac {-1-\sqrt {5}+2 x^2}{\left (1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx}{3+\sqrt {5}}-\frac {6 \int \frac {1}{\sqrt {-1-x^2+x^4}} \, dx}{3+\sqrt {5}}\\ &=\frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (3+\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}+\frac {\left (3 \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}\right ) \int \frac {\sqrt {-1-\sqrt {5}+2 x^2}}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \left (1+x^2\right )} \, dx}{\left (3+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}\\ &=\frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (3+\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}+\frac {\left (6 \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}} \, dx}{\left (3+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}+\frac {\left (3 \left (-3-\sqrt {5}\right ) \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \left (1+x^2\right ) \sqrt {-1-\sqrt {5}+2 x^2}} \, dx}{\left (3+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}\\ &=\frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (3+\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}+\frac {\left (6 \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}} \, dx}{\left (3+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}+\frac {\left (3 \left (-3-\sqrt {5}\right ) \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \left (1+x^2\right ) \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}} \, dx}{\left (3+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}\\ &=\frac {3 \left (1+\sqrt {5}\right ) \sqrt {-1+\sqrt {5}+2 x^2} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} F\left (\sin ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )|\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {2} \left (3+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}+\frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (3+\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \left (1+\sqrt {5}\right ) \sqrt {-1+\sqrt {5}+2 x^2} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} \Pi \left (\frac {1}{2} \left (-1-\sqrt {5}\right );\sin ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )|\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{2 \sqrt {2} \sqrt {-1-x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 131, normalized size = 2.02 \begin {gather*} -\frac {i \sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {-x^4+x^2+1} \left (2 F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )|\frac {1}{2} \left (-3+\sqrt {5}\right )\right )-3 \Pi \left (\frac {1}{2} \left (-1+\sqrt {5}\right );i \sinh ^{-1}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )|\frac {1}{2} \left (-3+\sqrt {5}\right )\right )\right )}{\sqrt {x^4-x^2-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*Sqrt[2/(1 + Sqrt[5])]*Sqrt[1 + x^2 - x^4]*(2*EllipticF[I*ArcSinh[Sqrt[2/(-1 + Sqrt[5])]*x], (-3 + Sqrt[5
])/2] - 3*EllipticPi[(-1 + Sqrt[5])/2, I*ArcSinh[Sqrt[2/(-1 + Sqrt[5])]*x], (-3 + Sqrt[5])/2]))/Sqrt[-1 - x^2
+ x^4]

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IntegrateAlgebraic [A]  time = 3.83, size = 65, normalized size = 1.00 \begin {gather*} -\frac {1}{2} i \log \left (\frac {-i-2 i x^2+x \sqrt {-1-x^2+x^4}}{-i-2 i x^2-x \sqrt {-1-x^2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 41, normalized size = 0.63 \begin {gather*} -\frac {1}{2} \, \arctan \left (\frac {2 \, \sqrt {x^{4} - x^{2} - 1} {\left (2 \, x^{3} + x\right )}}{x^{6} - 5 \, x^{4} - 5 \, x^{2} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctan(2*sqrt(x^4 - x^2 - 1)*(2*x^3 + x)/(x^6 - 5*x^4 - 5*x^2 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.48, size = 92, normalized size = 1.42

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{6}-5 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}-4 \sqrt {x^{4}-x^{2}-1}\, x^{3}-5 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-2 x \sqrt {x^{4}-x^{2}-1}-\RootOf \left (\textit {\_Z}^{2}+1\right )}{\left (x^{2}+1\right )^{3}}\right )}{2}\)	\(92\)
default	\(\frac {4 \sqrt {1-\left (-\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) x^{2}}\, \sqrt {1-\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-2-2 \sqrt {5}}}{2}, \frac {i \sqrt {5}}{2}-\frac {i}{2}\right )}{\sqrt {-2-2 \sqrt {5}}\, \sqrt {x^{4}-x^{2}-1}}-\frac {3 \sqrt {1+\frac {x^{2}}{2}+\frac {\sqrt {5}\, x^{2}}{2}}\, \sqrt {1-\frac {\sqrt {5}\, x^{2}}{2}+\frac {x^{2}}{2}}\, \EllipticPi \left (\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, x , -\frac {1}{-\frac {1}{2}-\frac {\sqrt {5}}{2}}, \frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}}\right )}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, \sqrt {x^{4}-x^{2}-1}}\)	\(178\)
elliptic	\(\frac {4 \sqrt {1+\frac {x^{2}}{2}+\frac {\sqrt {5}\, x^{2}}{2}}\, \sqrt {1-\frac {\sqrt {5}\, x^{2}}{2}+\frac {x^{2}}{2}}\, \EllipticF \left (\frac {x \sqrt {-2-2 \sqrt {5}}}{2}, \frac {i \sqrt {5}}{2}-\frac {i}{2}\right )}{\sqrt {-2-2 \sqrt {5}}\, \sqrt {x^{4}-x^{2}-1}}-\frac {3 \sqrt {1+\frac {x^{2}}{2}+\frac {\sqrt {5}\, x^{2}}{2}}\, \sqrt {1-\frac {\sqrt {5}\, x^{2}}{2}+\frac {x^{2}}{2}}\, \EllipticPi \left (\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, x , -\frac {1}{-\frac {1}{2}-\frac {\sqrt {5}}{2}}, \frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}}\right )}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, \sqrt {x^{4}-x^{2}-1}}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*RootOf(_Z^2+1)*ln(-(RootOf(_Z^2+1)*x^6-5*RootOf(_Z^2+1)*x^4-4*(x^4-x^2-1)^(1/2)*x^3-5*RootOf(_Z^2+1)*x^2-2
*x*(x^4-x^2-1)^(1/2)-RootOf(_Z^2+1))/(x^2+1)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {2\,x^2-1}{\left (x^2+1\right )\,\sqrt {x^4-x^2-1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((2*x**2 - 1)/((x**2 + 1)*sqrt(x**4 - x**2 - 1)), x)

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