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

Optimal. Leaf size=59 \[ -\frac {\sqrt {2 x^4+3 x^2+1} x}{2 \left (2 x^4+2 x^2+1\right )}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {2 x^4+3 x^2+1}}\right ) \]

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Rubi [C]  time = 2.43, antiderivative size = 253, normalized size of antiderivative = 4.29, number of steps used = 90, number of rules used = 10, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6742, 1226, 1189, 1100, 1136, 1214, 1456, 539, 1208, 6728} \begin {gather*} -\frac {i \sqrt {2 x^4+3 x^2+1} x}{-4 x^2-(2-2 i)}-\frac {i \sqrt {2 x^4+3 x^2+1} x}{4 x^2+(2+2 i)}-\frac {\left (x^2+1\right ) \sqrt {\frac {2 x^2+1}{x^2+1}} F\left (\left .\tan ^{-1}(x)\right |-1\right )}{2 \sqrt {2 x^4+3 x^2+1}}+\frac {i \left (x^2+1\right ) \Pi \left (\frac {1}{2}-\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {x^2+1}{2 x^2+1}} \sqrt {2 x^4+3 x^2+1}}-\frac {i \left (x^2+1\right ) \Pi \left (\frac {1}{2}+\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {x^2+1}{2 x^2+1}} \sqrt {2 x^4+3 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 539

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

Rule 1100

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

Rule 1136

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

Rule 1189

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

Rule 1208

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d -
b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/e^2, Int[(a + b*x^2 + c*x^4
)^(p - 1)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && IGtQ[p + 1/2, 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 1226

Int[Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*Sqrt[a + b*x^2 + c*
x^4])/(2*d*(d + e*x^2)), x] + (Dist[c/(2*d*e^2), Int[(d - e*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[(c*d^2
- a*e^2)/(2*d*e^2), Int[1/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[
b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\left (-1+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx &=\int \left (-\frac {2 \left (1+x^2\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2}+\frac {\sqrt {1+3 x^2+2 x^4}}{1+2 x^2+2 x^4}\right ) \, dx\\ &=-\left (2 \int \frac {\left (1+x^2\right ) \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx\right )+\int \frac {\sqrt {1+3 x^2+2 x^4}}{1+2 x^2+2 x^4} \, dx\\ &=-\left (2 \int \left (\frac {\sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2}+\frac {x^2 \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2}\right ) \, dx\right )+\int \left (\frac {2 i \sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2}+\frac {2 i \sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2}\right ) \, dx\\ &=2 i \int \frac {\sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2} \, dx+2 i \int \frac {\sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2} \, dx-2 \int \frac {\sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx-2 \int \frac {x^2 \sqrt {1+3 x^2+2 x^4}}{\left (1+2 x^2+2 x^4\right )^2} \, dx\\ &=(-1-i) \int \frac {1}{\left ((-2+2 i)-4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx-\frac {1}{8} i \int \frac {(-8+4 i)-8 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx-\frac {1}{8} i \int \frac {(8+4 i)+8 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx+(1-i) \int \frac {1}{\left ((2+2 i)+4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx-2 \int \left (\frac {(2-2 i) \sqrt {1+3 x^2+2 x^4}}{\left ((-2+2 i)-4 x^2\right )^2}-\frac {i \sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2}+\frac {(2+2 i) \sqrt {1+3 x^2+2 x^4}}{\left ((2+2 i)+4 x^2\right )^2}-\frac {i \sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2}\right ) \, dx-2 \int \left (-\frac {4 \sqrt {1+3 x^2+2 x^4}}{\left ((-2+2 i)-4 x^2\right )^2}+\frac {2 i \sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2}-\frac {4 \sqrt {1+3 x^2+2 x^4}}{\left ((2+2 i)+4 x^2\right )^2}+\frac {2 i \sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2}\right ) \, dx\\ &=\left (-\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx+\left (-\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx+\left (-\frac {1}{2}+\frac {i}{2}\right ) \int \frac {2+4 x^2}{\left ((-2+2 i)-4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx-\left (-\frac {1}{2}-i\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx-\left (-\frac {1}{2}+i\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx+2 i \int \frac {\sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2} \, dx+2 i \int \frac {\sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2} \, dx-4 i \int \frac {\sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2} \, dx-4 i \int \frac {\sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2} \, dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {2+4 x^2}{\left ((2+2 i)+4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx-(4-4 i) \int \frac {\sqrt {1+3 x^2+2 x^4}}{\left ((-2+2 i)-4 x^2\right )^2} \, dx-(4+4 i) \int \frac {\sqrt {1+3 x^2+2 x^4}}{\left ((2+2 i)+4 x^2\right )^2} \, dx+8 \int \frac {\sqrt {1+3 x^2+2 x^4}}{\left ((-2+2 i)-4 x^2\right )^2} \, dx+8 \int \frac {\sqrt {1+3 x^2+2 x^4}}{\left ((2+2 i)+4 x^2\right )^2} \, dx\\ &=-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2}-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2}-(-2-2 i) \int \frac {1}{\left ((-2+2 i)-4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx+\left (-\frac {1}{8}-\frac {i}{8}\right ) \int \frac {(-2+2 i)+4 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx-\frac {1}{8} i \int \frac {(-8+4 i)-8 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx-\frac {1}{8} i \int \frac {(8+4 i)+8 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx+\frac {1}{4} i \int \frac {(-8+4 i)-8 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx+\frac {1}{4} i \int \frac {(8+4 i)+8 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx-2 i \int \frac {1}{\left ((-2+2 i)-4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx-2 i \int \frac {1}{\left ((2+2 i)+4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx-\frac {1}{8} \int \frac {(2+2 i)-4 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx+\frac {1}{8} \int \frac {(-2+2 i)+4 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx+\left (\frac {1}{8}-\frac {i}{8}\right ) \int \frac {(2+2 i)-4 x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx-(2-2 i) \int \frac {1}{\left ((2+2 i)+4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx+-\frac {\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\frac {1}{2}+\frac {x^2}{2}} \sqrt {2+4 x^2}\right ) \int \frac {\sqrt {2+4 x^2}}{\left ((-2+2 i)-4 x^2\right ) \sqrt {\frac {1}{2}+\frac {x^2}{2}}} \, dx}{\sqrt {1+3 x^2+2 x^4}}+\frac {\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {1}{2}+\frac {x^2}{2}} \sqrt {2+4 x^2}\right ) \int \frac {\sqrt {2+4 x^2}}{\sqrt {\frac {1}{2}+\frac {x^2}{2}} \left ((2+2 i)+4 x^2\right )} \, dx}{\sqrt {1+3 x^2+2 x^4}}\\ &=-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2}-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2}+\frac {\left (1+x^2\right ) \Pi \left (\frac {1}{2}-\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {1+x^2}{1+2 x^2}} \sqrt {1+3 x^2+2 x^4}}+\frac {\left (1+x^2\right ) \Pi \left (\frac {1}{2}+\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {1+x^2}{1+2 x^2}} \sqrt {1+3 x^2+2 x^4}}-(-1-i) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx-(-1+i) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx-(-1+i) \int \frac {2+4 x^2}{\left ((-2+2 i)-4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx+(-1-2 i) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx+(-1+2 i) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx+\left (-\frac {1}{2}-\frac {i}{2}\right ) \int \frac {x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx+\left (-\frac {1}{2}+\frac {i}{2}\right ) \int \frac {x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx-\left (-\frac {1}{2}-i\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx-\left (-\frac {1}{2}+i\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx+\left (-\frac {1}{4}+\frac {i}{4}\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx-\left (\frac {1}{4}+\frac {i}{4}\right ) \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx+2 \left (\frac {1}{2} \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx\right )+2 \left (\frac {1}{2} \int \frac {x^2}{\sqrt {1+3 x^2+2 x^4}} \, dx\right )-(1+i) \int \frac {2+4 x^2}{\left ((2+2 i)+4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx-2 \int \frac {1}{\sqrt {1+3 x^2+2 x^4}} \, dx-\int \frac {2+4 x^2}{\left ((-2+2 i)-4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx+\int \frac {2+4 x^2}{\left ((2+2 i)+4 x^2\right ) \sqrt {1+3 x^2+2 x^4}} \, dx\\ &=-\frac {x \left (1+2 x^2\right )}{2 \sqrt {1+3 x^2+2 x^4}}-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2}-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+2 x^2}{1+x^2}} E\left (\left .\tan ^{-1}(x)\right |-1\right )}{2 \sqrt {1+3 x^2+2 x^4}}+2 \left (\frac {x \left (1+2 x^2\right )}{4 \sqrt {1+3 x^2+2 x^4}}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+2 x^2}{1+x^2}} E\left (\left .\tan ^{-1}(x)\right |-1\right )}{4 \sqrt {1+3 x^2+2 x^4}}\right )-\frac {\left (1+x^2\right ) \sqrt {\frac {1+2 x^2}{1+x^2}} F\left (\left .\tan ^{-1}(x)\right |-1\right )}{2 \sqrt {1+3 x^2+2 x^4}}+\frac {\left (1+x^2\right ) \Pi \left (\frac {1}{2}-\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {1+x^2}{1+2 x^2}} \sqrt {1+3 x^2+2 x^4}}+\frac {\left (1+x^2\right ) \Pi \left (\frac {1}{2}+\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {1+x^2}{1+2 x^2}} \sqrt {1+3 x^2+2 x^4}}--\frac {\left ((1-i) \sqrt {\frac {1}{2}+\frac {x^2}{2}} \sqrt {2+4 x^2}\right ) \int \frac {\sqrt {2+4 x^2}}{\left ((-2+2 i)-4 x^2\right ) \sqrt {\frac {1}{2}+\frac {x^2}{2}}} \, dx}{\sqrt {1+3 x^2+2 x^4}}-\frac {\left (\sqrt {\frac {1}{2}+\frac {x^2}{2}} \sqrt {2+4 x^2}\right ) \int \frac {\sqrt {2+4 x^2}}{\left ((-2+2 i)-4 x^2\right ) \sqrt {\frac {1}{2}+\frac {x^2}{2}}} \, dx}{\sqrt {1+3 x^2+2 x^4}}+\frac {\left (\sqrt {\frac {1}{2}+\frac {x^2}{2}} \sqrt {2+4 x^2}\right ) \int \frac {\sqrt {2+4 x^2}}{\sqrt {\frac {1}{2}+\frac {x^2}{2}} \left ((2+2 i)+4 x^2\right )} \, dx}{\sqrt {1+3 x^2+2 x^4}}-\frac {\left ((1+i) \sqrt {\frac {1}{2}+\frac {x^2}{2}} \sqrt {2+4 x^2}\right ) \int \frac {\sqrt {2+4 x^2}}{\sqrt {\frac {1}{2}+\frac {x^2}{2}} \left ((2+2 i)+4 x^2\right )} \, dx}{\sqrt {1+3 x^2+2 x^4}}\\ &=-\frac {x \left (1+2 x^2\right )}{2 \sqrt {1+3 x^2+2 x^4}}-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(-2+2 i)-4 x^2}-\frac {i x \sqrt {1+3 x^2+2 x^4}}{(2+2 i)+4 x^2}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+2 x^2}{1+x^2}} E\left (\left .\tan ^{-1}(x)\right |-1\right )}{2 \sqrt {1+3 x^2+2 x^4}}+2 \left (\frac {x \left (1+2 x^2\right )}{4 \sqrt {1+3 x^2+2 x^4}}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+2 x^2}{1+x^2}} E\left (\left .\tan ^{-1}(x)\right |-1\right )}{4 \sqrt {1+3 x^2+2 x^4}}\right )-\frac {\left (1+x^2\right ) \sqrt {\frac {1+2 x^2}{1+x^2}} F\left (\left .\tan ^{-1}(x)\right |-1\right )}{2 \sqrt {1+3 x^2+2 x^4}}+\frac {i \left (1+x^2\right ) \Pi \left (\frac {1}{2}-\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {1+x^2}{1+2 x^2}} \sqrt {1+3 x^2+2 x^4}}-\frac {i \left (1+x^2\right ) \Pi \left (\frac {1}{2}+\frac {i}{2};\tan ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {\frac {1+x^2}{1+2 x^2}} \sqrt {1+3 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.87, size = 199, normalized size = 3.37 \begin {gather*} \frac {-i \sqrt {2} \sqrt {x^2+1} \sqrt {2 x^2+1} F\left (i \sinh ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )+i \sqrt {2} \sqrt {x^2+1} \sqrt {2 x^2+1} \Pi \left (\frac {1}{2}-\frac {i}{2};i \sinh ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )+i \sqrt {2} \sqrt {x^2+1} \sqrt {2 x^2+1} \Pi \left (\frac {1}{2}+\frac {i}{2};i \sinh ^{-1}\left (\sqrt {2} x\right )|\frac {1}{2}\right )-\frac {2 x \left (2 x^4+3 x^2+1\right )}{2 x^4+2 x^2+1}}{4 \sqrt {2 x^4+3 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.64, size = 59, normalized size = 1.00 \begin {gather*} -\frac {x \sqrt {1+3 x^2+2 x^4}}{2 \left (1+2 x^2+2 x^4\right )}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {1+3 x^2+2 x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(x*Sqrt[1 + 3*x^2 + 2*x^4])/(1 + 2*x^2 + 2*x^4) - ArcTanh[x/Sqrt[1 + 3*x^2 + 2*x^4]]/2

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fricas [A]  time = 0.66, size = 92, normalized size = 1.56 \begin {gather*} \frac {{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )} \log \left (\frac {2 \, x^{4} + 4 \, x^{2} - 2 \, \sqrt {2 \, x^{4} + 3 \, x^{2} + 1} x + 1}{2 \, x^{4} + 2 \, x^{2} + 1}\right ) - 2 \, \sqrt {2 \, x^{4} + 3 \, x^{2} + 1} x}{4 \, {\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*x^4 + 2*x^2 + 1)*log((2*x^4 + 4*x^2 - 2*sqrt(2*x^4 + 3*x^2 + 1)*x + 1)/(2*x^4 + 2*x^2 + 1)) - 2*sqrt(2
*x^4 + 3*x^2 + 1)*x)/(2*x^4 + 2*x^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.27, size = 74, normalized size = 1.25

method	result	size
elliptic	\(\frac {\left (-\frac {\sqrt {2 x^{4}+3 x^{2}+1}\, \sqrt {2}}{4 x \left (\frac {2 x^{4}+3 x^{2}+1}{2 x^{2}}-\frac {1}{2}\right )}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {2 x^{4}+3 x^{2}+1}}{x}\right )}{2}\right ) \sqrt {2}}{2}\)	\(74\)
trager	\(-\frac {x \sqrt {2 x^{4}+3 x^{2}+1}}{2 \left (2 x^{4}+2 x^{2}+1\right )}-\frac {\ln \left (-\frac {2 x^{4}+2 \sqrt {2 x^{4}+3 x^{2}+1}\, x +4 x^{2}+1}{2 x^{4}+2 x^{2}+1}\right )}{4}\)	\(81\)
risch	\(-\frac {x \sqrt {2 x^{4}+3 x^{2}+1}}{2 \left (2 x^{4}+2 x^{2}+1\right )}-\frac {i \sqrt {x^{2}+1}\, \sqrt {2 x^{2}+1}\, \EllipticF \left (i x , \sqrt {2}\right )}{2 \sqrt {2 x^{4}+3 x^{2}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (2 \textit {\_Z}^{4}+2 \textit {\_Z}^{2}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\left (4 \underline {\hspace {1.25 ex}}\alpha ^{2}+3\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{2}+5 x^{2}+4\right )}{10 \sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {2 x^{4}+3 x^{2}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {4 i \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}-\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {x^{2}+1}\, \sqrt {2 x^{2}+1}\, \EllipticPi \left (i x , 2 \underline {\hspace {1.25 ex}}\alpha ^{2}+2, i \sqrt {-2}\right )}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )\right )}{8}\)	\(201\)
default	\(-\frac {i \sqrt {x^{2}+1}\, \sqrt {2 x^{2}+1}\, \EllipticF \left (i x , \sqrt {2}\right )}{2 \sqrt {2 x^{4}+3 x^{2}+1}}-\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (2 \textit {\_Z}^{4}+2 \textit {\_Z}^{2}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (2 \underline {\hspace {1.25 ex}}\alpha ^{2}+1\right ) \left (-\frac {\arctanh \left (\frac {\left (4 \underline {\hspace {1.25 ex}}\alpha ^{2}+3\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{2}+5 x^{2}+4\right )}{10 \sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {2 x^{4}+3 x^{2}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {4 i \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}-\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {x^{2}+1}\, \sqrt {2 x^{2}+1}\, \EllipticPi \left (i x , 2 \underline {\hspace {1.25 ex}}\alpha ^{2}+2, i \sqrt {-2}\right )}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )\right )}{8}-\frac {x \sqrt {2 x^{4}+3 x^{2}+1}}{2 \left (2 x^{4}+2 x^{2}+1\right )}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (2 \textit {\_Z}^{4}+2 \textit {\_Z}^{2}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2}+1\right ) \left (-\frac {\arctanh \left (\frac {\left (4 \underline {\hspace {1.25 ex}}\alpha ^{2}+3\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{2}+5 x^{2}+4\right )}{10 \sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {2 x^{4}+3 x^{2}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {4 i \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}-\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {x^{2}+1}\, \sqrt {2 x^{2}+1}\, \EllipticPi \left (i x , 2 \underline {\hspace {1.25 ex}}\alpha ^{2}+2, i \sqrt {-2}\right )}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )\right )}{4}\)	\(340\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\left (2\,x^4-1\right )\,\sqrt {2\,x^4+3\,x^2+1}}{{\left (2\,x^4+2\,x^2+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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