3.4.39 \(\int \frac {1}{(3+2 x^2) \sqrt {1+2 x^2+2 x^4}} \, dx\) [339]

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

[Out]

1/30*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)+1/28*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arcta
n(2^(1/4)*x))*EllipticF(sin(2*arctan(2^(1/4)*x)),1/2*(2-2^(1/2))^(1/2))*(3+2^(1/2))*(1+x^2*2^(1/2))*((2*x^4+2*
x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(3/4)/(2*x^4+2*x^2+1)^(1/2)-1/168*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*a
rctan(2^(1/4)*x))*EllipticPi(sin(2*arctan(2^(1/4)*x)),1/2-11/24*2^(1/2),1/2*(2-2^(1/2))^(1/2))*(3+2^(1/2))^2*(
1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(3/4)/(2*x^4+2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 245, normalized size of antiderivative = 1.00, 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 = {1230, 1117, 1720} \begin {gather*} \frac {\text {ArcTan}\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 \text {ArcTan}\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 \text {ArcTan}\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}} \end {gather*}

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 1117

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)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1230

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 1720

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))]/(4*d*e*A*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1
/2 - b*(A/(4*a*B))], 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] Result contains complex when optimal does not.
time = 10.04, size = 80, normalized size = 0.33 \begin {gather*} -\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 {1+2 x^2+2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.12, size = 70, normalized size = 0.29

method	result	size
default	\(\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(70\)
elliptic	\(\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(70\)

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,method=_RETURNVERBOSE)

[Out]

1/3/(-1+I)^(1/2)*(1+x^2-I*x^2)^(1/2)*(1+x^2+I*x^2)^(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))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________