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

Optimal. Leaf size=418 \[ \frac {\sqrt {2 x^4+2 x^2+1} x}{2 \sqrt {2} \left (\sqrt {2} x^2+1\right )}-\frac {3 \sqrt {\frac {3}{10}} \left (3-\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{4 \left (2-3 \sqrt {2}\right )}+\frac {\left (1-3 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{2\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}}-\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{2\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}+\frac {3 \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{8\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}} \]

[Out]

-3/40*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*30^(1/2)*(3-2^(1/2))/(2-3*2^(1/2))+1/4*x*(2*x^4+2*x^2+1)^(1
/2)*2^(1/2)/(1+x^2*2^(1/2))-1/4*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticE(sin(2*ar
ctan(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)+1/4*(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))*(1-3*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)+3/16*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticP
i(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)

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Rubi [A]  time = 0.18, antiderivative size = 418, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1325, 1103, 1195, 1706} \[ \frac {\sqrt {2 x^4+2 x^2+1} x}{2 \sqrt {2} \left (\sqrt {2} x^2+1\right )}-\frac {3 \sqrt {\frac {3}{10}} \left (3-\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{4 \left (2-3 \sqrt {2}\right )}+\frac {\left (1-3 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{2\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}}-\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{2\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}+\frac {3 \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{8\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 + 2*x^2 + 2*x^4])/(2*Sqrt[2]*(1 + Sqrt[2]*x^2)) - (3*Sqrt[3/10]*(3 - Sqrt[2])*ArcTan[(Sqrt[5/3]*x)/S
qrt[1 + 2*x^2 + 2*x^4]])/(4*(2 - 3*Sqrt[2])) - ((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*2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + ((1 - 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])/(
2*2^(3/4)*(2 - 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4]) + (3*(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])/(8*2^(3/4)*(2
- 3*Sqrt[2])*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 1325

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

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 {x^4}{\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}{2 \sqrt {2}}+\frac {9 \int \frac {1+\sqrt {2} x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx}{2 \left (2-3 \sqrt {2}\right )}-\frac {\left (12-2 \sqrt {2}\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx}{4 \left (2-3 \sqrt {2}\right )}\\ &=\frac {x \sqrt {1+2 x^2+2 x^4}}{2 \sqrt {2} \left (1+\sqrt {2} x^2\right )}-\frac {3 \sqrt {\frac {3}{10}} \left (3-\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )}{4 \left (2-3 \sqrt {2}\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\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {\left (1-3 \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 )}{2\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {1+2 x^2+2 x^4}}+\frac {3 \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}} \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 )}{8\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {1+2 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 127, normalized size = 0.30 \[ -\frac {\sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \left (-(1+4 i) F\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+(1+i) E\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+3 i \Pi \left (\frac {1}{3}+\frac {i}{3};\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )\right )}{4 \sqrt {1-i} \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*((1 + I)*EllipticE[I*ArcSinh[Sqrt[1 - I]*x], I] - (1 + 4*I)*
EllipticF[I*ArcSinh[Sqrt[1 - I]*x], I] + (3*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 = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} x^{4}}{4 \, x^{6} + 10 \, x^{4} + 8 \, x^{2} + 3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(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)*x^4/(4*x^6 + 10*x^4 + 8*x^2 + 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 222, normalized size = 0.53 \[ -\frac {3 \sqrt {\left (1-i\right ) x^{2}+1}\, \sqrt {\left (1+i\right ) x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {3 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (\sqrt {-1+i}\, x , \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{4 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {\left (-\frac {1}{4}+\frac {i}{4}\right ) \sqrt {\left (1-i\right ) x^{2}+1}\, \sqrt {\left (1+i\right ) x^{2}+1}\, \left (-\EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )+\EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/4+1/4*I)/(-1+I)^(1/2)*((1-I)*x^2+1)^(1/2)*((1+I)*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))-EllipticE((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2)))-3/4/(-1+I)^(1/2)*((1-I)*x^
2+1)^(1/2)*((1+I)*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))+3/4/(
-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 \[ \int \frac {x^{4}}{\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} {\left (2 \, x^{2} + 3\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(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 \[ \int \frac {x^4}{\left (2\,x^2+3\right )\,\sqrt {2\,x^4+2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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