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

Optimal. Leaf size=422 \[ \frac {\sqrt {2 x^4+2 x^2+1} x}{10 \sqrt {2} \left (\sqrt {2} x^2+1\right )}-\frac {\left (x^2+2\right ) x}{10 \sqrt {2 x^4+2 x^2+1}}+\frac {3}{20} \sqrt {\frac {3}{5}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )+\frac {\left (2+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{4\ 2^{3/4} \left (3 \sqrt {2}-2\right ) \sqrt {2 x^4+2 x^2+1}}-\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{10\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}+\frac {3 \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{20\ 2^{3/4} \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}} \]

[Out]

3/100*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)-1/10*x*(x^2+2)/(2*x^4+2*x^2+1)^(1/2)+1/20*x*(2*x^4
+2*x^2+1)^(1/2)*2^(1/2)/(1+x^2*2^(1/2))-1/20*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*Ellip
ticE(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)+3/40*(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)+1/8*(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))*(2+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 = 0.24, antiderivative size = 501, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1313, 1178, 1197, 1103, 1195, 1216, 1706} \[ \frac {\sqrt {2 x^4+2 x^2+1} x}{10 \sqrt {2} \left (\sqrt {2} x^2+1\right )}-\frac {\left (x^2+2\right ) x}{10 \sqrt {2 x^4+2 x^2+1}}+\frac {3}{20} \sqrt {\frac {3}{5}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )+\frac {9 \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{140 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}+\frac {\left (1-\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{20\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}-\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{10\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}-\frac {3 \left (3+\sqrt {2}\right )^2 \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{280 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(2 + x^2))/(10*Sqrt[1 + 2*x^2 + 2*x^4]) + (x*Sqrt[1 + 2*x^2 + 2*x^4])/(10*Sqrt[2]*(1 + Sqrt[2]*x^2)) + (3*
Sqrt[3/5]*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/20 - ((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])/(10*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])/(20*2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + (9*(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])/(140*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])
- (3*(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])/(280*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 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), 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] && LtQ[p, -1] && IntegerQ[2*p]

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 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 1313

Int[(((f_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> -Dist
[f^4/(c*d^2 - b*d*e + a*e^2), Int[(f*x)^(m - 4)*(a*d + (b*d - a*e)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] + Dist[(
d^2*f^4)/(c*d^2 - b*d*e + a*e^2), Int[((f*x)^(m - 4)*(a + b*x^2 + c*x^4)^(p + 1))/(d + e*x^2), x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 2]

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 ) \left (1+2 x^2+2 x^4\right )^{3/2}} \, dx &=-\left (\frac {1}{10} \int \frac {3+4 x^2}{\left (1+2 x^2+2 x^4\right )^{3/2}} \, dx\right )+\frac {9}{10} \int \frac {1}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {x \left (2+x^2\right )}{10 \sqrt {1+2 x^2+2 x^4}}-\frac {1}{40} \int \frac {4-4 x^2}{\sqrt {1+2 x^2+2 x^4}} \, dx+\frac {1}{70} \left (9 \left (3+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{70} \left (9 \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 \left (2+x^2\right )}{10 \sqrt {1+2 x^2+2 x^4}}+\frac {3}{20} \sqrt {\frac {3}{5}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )+\frac {9 \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 )}{140 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}-\frac {3 \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 )}{280 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}-\frac {\int \frac {1-\sqrt {2} x^2}{\sqrt {1+2 x^2+2 x^4}} \, dx}{10 \sqrt {2}}-\frac {1}{20} \left (2-\sqrt {2}\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {x \left (2+x^2\right )}{10 \sqrt {1+2 x^2+2 x^4}}+\frac {x \sqrt {1+2 x^2+2 x^4}}{10 \sqrt {2} \left (1+\sqrt {2} x^2\right )}+\frac {3}{20} \sqrt {\frac {3}{5}} \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 )}{10\ 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 )}{20\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {9 \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 )}{140 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}-\frac {3 \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 )}{280 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 199, normalized size = 0.47 \[ -\frac {2 x^3+(1-2 i) \sqrt {1-i} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} F\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+i \sqrt {1-i} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} E\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )-3 (1-i)^{3/2} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \Pi \left (\frac {1}{3}+\frac {i}{3};\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+4 x}{20 \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/20*(4*x + 2*x^3 + I*Sqrt[1 - I]*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticE[I*ArcSinh[Sqrt[1 - I]
*x], I] + (1 - 2*I)*Sqrt[1 - I]*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticF[I*ArcSinh[Sqrt[1 - I]*x]
, I] - 3*(1 - I)^(3/2)*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 + 2*x^2 + 2*x^4]

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fricas [F]  time = 1.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, x^{4} + 2 \, x^{2} + 1} x^{4}}{8 \, x^{10} + 28 \, x^{8} + 40 \, x^{6} + 32 \, x^{4} + 14 \, x^{2} + 3}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*x^4 + 2*x^2 + 1)*x^4/(8*x^10 + 28*x^8 + 40*x^6 + 32*x^4 + 14*x^2 + 3), x)

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

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.01, size = 561, normalized size = 1.33 \[ \frac {3 x^{3}}{4 \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {27 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{40 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {27 i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{40 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {\left (1-i\right ) x^{2}+1}\, \sqrt {\left (1+i\right ) x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {9 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{40 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {27 i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{40 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {3 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (\sqrt {-1+i}\, x , \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{10 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {2 \left (-\frac {1}{4} x^{3}-\frac {1}{8} x \right )}{\sqrt {2 x^{4}+2 x^{2}+1}}+\frac {\left (\frac {5}{8}-\frac {5 i}{8}\right ) \sqrt {\left (1-i\right ) x^{2}+1}\, \sqrt {\left (1+i\right ) x^{2}+1}\, \left (-\EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )+\EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {9 \left (\frac {3}{20} x^{3}+\frac {1}{20} x \right )}{\sqrt {2 x^{4}+2 x^{2}+1}} \]

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)^(3/2),x)

[Out]

-2*(-1/4*x^3-1/8*x)/(2*x^4+2*x^2+1)^(1/2)-1/(-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))+(5/8-5/8*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/(2*x^4+2*x^2+1)^(1/2)*x^3-9*(3/20*x^3+1/20*x)/(2*x^4+2*x^2+1)^(1/2)+9/40
/(-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))+27/40*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))+27/40/(-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))-27/40*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))+3/10/(-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 \[ \int \frac {x^{4}}{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{\frac {3}{2}} {\left (2 \, x^{2} + 3\right )}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

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

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

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)^(3/2)),x)

[Out]

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

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

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)**(3/2),x)

[Out]

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

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