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

Optimal. Leaf size=381 \[ \frac{2^{3/4} \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right ),\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{\left (3 \sqrt{2}-2\right ) \sqrt{2 x^4+2 x^2+1}}+\frac{\sqrt{2 x^4+2 x^2+1} x}{\sqrt{2} \left (\sqrt{2} x^2+1\right )}+\frac{1}{4} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )-\frac{\left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2^{3/4} \sqrt{2 x^4+2 x^2+1}}+\frac{5 \left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{12\ 2^{3/4} \left (2-3 \sqrt{2}\right ) \sqrt{2 x^4+2 x^2+1}} \]

[Out]

(x*Sqrt[1 + 2*x^2 + 2*x^4])/(Sqrt[2]*(1 + Sqrt[2]*x^2)) + (Sqrt[5/3]*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x
^4]])/4 - ((1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticE[2*ArcTan[2^(1/4)*x], (2 -
Sqrt[2])/4])/(2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + (2^(3/4)*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqr
t[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/((-2 + 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4]) + (5*
(3 + Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticPi[(12 - 11*Sqrt[2])/24,
2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(12*2^(3/4)*(2 - 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4])

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Rubi [A]  time = 0.202387, antiderivative size = 470, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {1208, 1197, 1103, 1195, 1216, 1706} \[ \frac{\sqrt{2 x^4+2 x^2+1} x}{\sqrt{2} \left (\sqrt{2} x^2+1\right )}+\frac{1}{4} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )+\frac{5 \left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{28 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}-\frac{\left (1-\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{4 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}-\frac{\left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2^{3/4} \sqrt{2 x^4+2 x^2+1}}-\frac{5 \left (3+\sqrt{2}\right )^2 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{168 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 + 2*x^2 + 2*x^4])/(Sqrt[2]*(1 + Sqrt[2]*x^2)) + (Sqrt[5/3]*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x
^4]])/4 - ((1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticE[2*ArcTan[2^(1/4)*x], (2 -
Sqrt[2])/4])/(2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) - ((1 - Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1
+ Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(4*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + (5*(
3 + Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2
- Sqrt[2])/4])/(28*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4]) - (5*(3 + Sqrt[2])^2*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 +
2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticPi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(168*2^(1/
4)*Sqrt[1 + 2*x^2 + 2*x^4])

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 1197

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

Rule 1103

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

Rule 1195

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

Rule 1216

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

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), 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] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{1+2 x^2+2 x^4}}{3+2 x^2} \, dx &=-\left (\frac{1}{4} \int \frac{2-4 x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx\right )+\frac{5}{2} \int \frac{1}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{\int \frac{1-\sqrt{2} x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx}{\sqrt{2}}-\frac{1}{2} \left (1-\sqrt{2}\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx+\frac{1}{14} \left (5 \left (3+\sqrt{2}\right )\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx-\frac{1}{14} \left (5 \left (2+3 \sqrt{2}\right )\right ) \int \frac{1+\sqrt{2} x^2}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx\\ &=\frac{x \sqrt{1+2 x^2+2 x^4}}{\sqrt{2} \left (1+\sqrt{2} x^2\right )}+\frac{1}{4} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{1+2 x^2+2 x^4}}\right )-\frac{\left (1+\sqrt{2} x^2\right ) \sqrt{\frac{1+2 x^2+2 x^4}{\left (1+\sqrt{2} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2^{3/4} \sqrt{1+2 x^2+2 x^4}}-\frac{\left (1-\sqrt{2}\right ) \left (1+\sqrt{2} x^2\right ) \sqrt{\frac{1+2 x^2+2 x^4}{\left (1+\sqrt{2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{4 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}+\frac{5 \left (3+\sqrt{2}\right ) \left (1+\sqrt{2} x^2\right ) \sqrt{\frac{1+2 x^2+2 x^4}{\left (1+\sqrt{2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{28 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}-\frac{5 \left (3+\sqrt{2}\right )^2 \left (1+\sqrt{2} x^2\right ) \sqrt{\frac{1+2 x^2+2 x^4}{\left (1+\sqrt{2} x^2\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{168 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.111975, size = 127, normalized size = 0.33 \[ -\frac{\sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} \left (-(3+6 i) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{1-i} x\right ),i\right )+(3+3 i) E\left (\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )+5 i \Pi \left (\frac{1}{3}+\frac{i}{3};\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )\right )}{6 \sqrt{1-i} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*((3 + 3*I)*EllipticE[I*ArcSinh[Sqrt[1 - I]*x], I] - (3 + 6*I)*El
lipticF[I*ArcSinh[Sqrt[1 - I]*x], I] + (5*I)*EllipticPi[1/3 + I/3, I*ArcSinh[Sqrt[1 - I]*x], I]))/(6*Sqrt[1 -
I]*Sqrt[1 + 2*x^2 + 2*x^4])

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Maple [C]  time = 0.003, size = 341, normalized size = 0.9 \begin{align*} -{\frac{{\it EllipticF} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}+{\frac{{\frac{i}{2}}{\it EllipticF} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}+{\frac{{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{2\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{{\frac{i}{2}}{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}+{\frac{5}{6\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\it EllipticPi} \left ( x\sqrt{-1+i},{\frac{1}{3}}+{\frac{i}{3}},{\frac{\sqrt{-1-i}}{\sqrt{-1+i}}} \right ){\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticF(x*(-1+I)^(1/2),1/2*2^
(1/2)+1/2*I*2^(1/2))+1/2*I/(-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*Ellipti
cF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))+1/2/(-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+
2*x^2+1)^(1/2)*EllipticE(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-1/2*I/(-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*
x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticE(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))+5/6/(-1+I)^(1/2)*(-
I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticPi(x*(-1+I)^(1/2),1/3+1/3*I,(-1-I)^(1/2)/
(-1+I)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x^{4} + 2 \, x^{2} + 1}}{2 \, x^{2} + 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2 \, x^{4} + 2 \, x^{2} + 1}}{2 \, x^{2} + 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 x^{4} + 2 x^{2} + 1}}{2 x^{2} + 3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(2*x**4 + 2*x**2 + 1)/(2*x**2 + 3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x^{4} + 2 \, x^{2} + 1}}{2 \, x^{2} + 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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