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

Optimal. Leaf size=245 \[ \frac{\left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right ),\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{14 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )}{2 \sqrt{15}}-\frac{\left (3+\sqrt{2}\right )^2 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{84 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

[Out]

ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]]/(2*Sqrt[15]) + ((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])/(14*2^(1/4)*Sqrt[1 + 2*x^2 + 2*
x^4]) - ((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*S
qrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(84*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])

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Rubi [A]  time = 0.110588, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1216, 1103, 1706} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )}{2 \sqrt{15}}+\frac{\left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{14 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}-\frac{\left (3+\sqrt{2}\right )^2 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{84 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]]/(2*Sqrt[15]) + ((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])/(14*2^(1/4)*Sqrt[1 + 2*x^2 + 2*
x^4]) - ((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*S
qrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(84*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])

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 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 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{1}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx &=\frac{1}{7} \left (3+\sqrt{2}\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx-\frac{1}{7} \left (2+3 \sqrt{2}\right ) \int \frac{1+\sqrt{2} x^2}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{1+2 x^2+2 x^4}}\right )}{2 \sqrt{15}}+\frac{\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 )}{14 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}-\frac{\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 )}{84 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0592903, size = 80, normalized size = 0.33 \[ -\frac{i \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} \Pi \left (\frac{1}{3}+\frac{i}{3};\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )}{3 \sqrt{1-i} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.003, size = 70, normalized size = 0.3 \begin{align*}{\frac{1}{3\,\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(1/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x)

[Out]

1/3/(-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{1}{\sqrt{2 \, x^{4} + 2 \, x^{2} + 1}{\left (2 \, x^{2} + 3\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(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}}{4 \, x^{6} + 10 \, x^{4} + 8 \, x^{2} + 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(2*x^4 + 2*x^2 + 1)/(4*x^6 + 10*x^4 + 8*x^2 + 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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