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

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

[Out]

1/12*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)+1/2*x*(2*x^4+2*x^2+1)^(1/2)*2^(1/2)/(1+x^2*2^(1/2))
-1/2*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticE(sin(2*arctan(2^(1/4)*x)),1/2*(2-2^(
1/2))^(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^(1/4)/(2*x^4+2*x^2+1)^(1/2)+5/24*(cos
(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(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))*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(1/4)/(2-3*2^(1
/2))/(2*x^4+2*x^2+1)^(1/2)+2^(3/4)*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticF(sin(2
*arctan(2^(1/4)*x)),1/2*(2-2^(1/2))^(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*2^(
1/2))/(2*x^4+2*x^2+1)^(1/2)

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Rubi [A]  time = 0.20, 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 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 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 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 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*}

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Mathematica [C]  time = 0.11, size = 127, normalized size = 0.33 \[ -\frac {\sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \left (-(3+6 i) F\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+(3+3 i) E\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+5 i \Pi \left (\frac {1}{3}+\frac {i}{3};\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )\right )}{6 \sqrt {1-i} \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(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
)*EllipticF[I*ArcSinh[Sqrt[1 - I]*x], I] + (5*I)*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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fricas [F]  time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, x^{4} + 2 \, x^{2} + 1}}{2 \, x^{2} + 3}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {2 \, x^{4} + 2 \, x^{2} + 1}}{2 \, x^{2} + 3}\,{d x} \]

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)

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maple [C]  time = 0.01, size = 341, normalized size = 0.90 \[ \frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {5 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (\sqrt {-1+i}\, x , \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{6 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {2 \, x^{4} + 2 \, x^{2} + 1}}{2 \, x^{2} + 3}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {2\,x^4+2\,x^2+1}}{2\,x^2+3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {2 x^{4} + 2 x^{2} + 1}}{2 x^{2} + 3}\, dx \]

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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