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

Optimal. Leaf size=422 \[ -\frac{\left (1+19 \sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right ),\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{63 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}-\frac{2 \sqrt{2} \sqrt{2 x^4+2 x^2+1} x}{3 \left (\sqrt{2} x^2+1\right )}+\frac{2 \sqrt{2 x^4+2 x^2+1}}{3 x}-\frac{\sqrt{2 x^4+2 x^2+1}}{9 x^3}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )}{9 \sqrt{15}}+\frac{2 \sqrt [4]{2} \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{3 \sqrt{2 x^4+2 x^2+1}}-\frac{\left (3+\sqrt{2}\right )^2 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{189 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

[Out]

-Sqrt[1 + 2*x^2 + 2*x^4]/(9*x^3) + (2*Sqrt[1 + 2*x^2 + 2*x^4])/(3*x) - (2*Sqrt[2]*x*Sqrt[1 + 2*x^2 + 2*x^4])/(
3*(1 + Sqrt[2]*x^2)) + (2*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/(9*Sqrt[15]) + (2*2^(1/4)*(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])/(3*Sqrt[
1 + 2*x^2 + 2*x^4]) - ((1 + 19*Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*Ellipt
icF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(63*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*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2
- Sqrt[2])/4])/(189*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])

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Rubi [A]  time = 0.517098, antiderivative size = 422, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1329, 1683, 1714, 1195, 1708, 1103, 1706} \[ -\frac{2 \sqrt{2} \sqrt{2 x^4+2 x^2+1} x}{3 \left (\sqrt{2} x^2+1\right )}+\frac{2 \sqrt{2 x^4+2 x^2+1}}{3 x}-\frac{\sqrt{2 x^4+2 x^2+1}}{9 x^3}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )}{9 \sqrt{15}}-\frac{\left (1+19 \sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{63 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}+\frac{2 \sqrt [4]{2} \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{3 \sqrt{2 x^4+2 x^2+1}}-\frac{\left (3+\sqrt{2}\right )^2 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{189 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + 2*x^2 + 2*x^4]/(9*x^3) + (2*Sqrt[1 + 2*x^2 + 2*x^4])/(3*x) - (2*Sqrt[2]*x*Sqrt[1 + 2*x^2 + 2*x^4])/(
3*(1 + Sqrt[2]*x^2)) + (2*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/(9*Sqrt[15]) + (2*2^(1/4)*(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])/(3*Sqrt[
1 + 2*x^2 + 2*x^4]) - ((1 + 19*Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*Ellipt
icF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(63*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*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2
- Sqrt[2])/4])/(189*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])

Rule 1329

Int[(x_)^(m_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Simp[(x^(m + 1)*S
qrt[a + b*x^2 + c*x^4])/(a*d*(m + 1)), x] - Dist[1/(a*d*(m + 1)), Int[(x^(m + 2)*Simp[a*e*(m + 1) + b*d*(m + 2
) + (b*e*(m + 2) + c*d*(m + 3))*x^2 + c*e*(m + 3)*x^4, x])/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; Fr
eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && ILtQ[m/2, 0]

Rule 1683

Int[((Px_)*(x_)^(m_))/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{A =
Coeff[Px, x, 0], B = Coeff[Px, x, 2], C = Coeff[Px, x, 4]}, Simp[(A*x^(m + 1)*Sqrt[a + b*x^2 + c*x^4])/(a*d*(
m + 1)), x] + Dist[1/(a*d*(m + 1)), Int[(x^(m + 2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]))*Simp[a*B*d*(m + 1) -
A*(a*e*(m + 1) + b*d*(m + 2)) + (a*C*d*(m + 1) - A*(b*e*(m + 2) + c*d*(m + 3)))*x^2 - A*c*e*(m + 3)*x^4, x],
x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Px, x^2, 2] && NeQ[b^2 - 4*a*c, 0] && ILtQ[m/2, 0]

Rule 1714

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]
, A = Coeff[P4x, x, 0], B = Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Dist[C/(e*q), Int[(1 - q*x^2)/Sqrt[a + b
*x^2 + c*x^4], x], x] + Dist[1/(c*e), Int[(A*c*e + a*C*d*q + (B*c*e - C*(c*d - a*e*q))*x^2)/((d + e*x^2)*Sqrt[
a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[P4x, x^2, 2] && 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] &&  !GtQ[b^2 - 4*a*c, 0]

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 1708

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

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}{x^4 \left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx &=-\frac{\sqrt{1+2 x^2+2 x^4}}{9 x^3}+\frac{1}{9} \int \frac{-18-14 x^2-4 x^4}{x^2 \left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{\sqrt{1+2 x^2+2 x^4}}{9 x^3}+\frac{2 \sqrt{1+2 x^2+2 x^4}}{3 x}-\frac{1}{27} \int \frac{6+120 x^2+72 x^4}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{\sqrt{1+2 x^2+2 x^4}}{9 x^3}+\frac{2 \sqrt{1+2 x^2+2 x^4}}{3 x}-\frac{1}{108} \int \frac{24+216 \sqrt{2}+\left (480-72 \left (6-2 \sqrt{2}\right )\right ) x^2}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx+\frac{1}{3} \left (2 \sqrt{2}\right ) \int \frac{1-\sqrt{2} x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{\sqrt{1+2 x^2+2 x^4}}{9 x^3}+\frac{2 \sqrt{1+2 x^2+2 x^4}}{3 x}-\frac{2 \sqrt{2} x \sqrt{1+2 x^2+2 x^4}}{3 \left (1+\sqrt{2} x^2\right )}+\frac{2 \sqrt [4]{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}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{3 \sqrt{1+2 x^2+2 x^4}}-\frac{1}{63} \left (4 \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{1}{63} \left (2 \left (1+19 \sqrt{2}\right )\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{\sqrt{1+2 x^2+2 x^4}}{9 x^3}+\frac{2 \sqrt{1+2 x^2+2 x^4}}{3 x}-\frac{2 \sqrt{2} x \sqrt{1+2 x^2+2 x^4}}{3 \left (1+\sqrt{2} x^2\right )}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{1+2 x^2+2 x^4}}\right )}{9 \sqrt{15}}+\frac{2 \sqrt [4]{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}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{3 \sqrt{1+2 x^2+2 x^4}}-\frac{\left (1+19 \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 )}{63 \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 )}{189 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.19626, size = 219, normalized size = 0.52 \[ \frac{-(3+15 i) \sqrt{1-i} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} x^3 \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{1-i} x\right ),i\right )+36 x^6+30 x^4+12 x^2+18 i \sqrt{1-i} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} x^3 E\left (\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )+2 (1-i)^{3/2} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} x^3 \Pi \left (\frac{1}{3}+\frac{i}{3};\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )-3}{27 x^3 \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 + 12*x^2 + 30*x^4 + 36*x^6 + (18*I)*Sqrt[1 - I]*x^3*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticE[
I*ArcSinh[Sqrt[1 - I]*x], I] - (3 + 15*I)*Sqrt[1 - I]*x^3*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*Elliptic
F[I*ArcSinh[Sqrt[1 - I]*x], I] + 2*(1 - I)^(3/2)*x^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])/(27*x^3*Sqrt[1 + 2*x^2 + 2*x^4])

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Maple [C]  time = 0.016, size = 260, normalized size = 0.6 \begin{align*}{\frac{4}{27\,\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}}}}+{\frac{2}{3\,x}\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}+{\frac{ \left ({\frac{2}{3}}-{\frac{2\,i}{3}} \right ) \left ({\it EllipticF} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) -{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) \right ) }{\sqrt{-1+i}}\sqrt{1+ \left ( 1-i \right ){x}^{2}}\sqrt{1+ \left ( 1+i \right ){x}^{2}}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{1}{9\,{x}^{3}}\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}-{\frac{2\,{\it EllipticF} \left ( x\sqrt{-1+i},1/2\,\sqrt{2}+i/2\sqrt{2} \right ) }{9\,\sqrt{-1+i}}\sqrt{1+ \left ( 1-i \right ){x}^{2}}\sqrt{1+ \left ( 1+i \right ){x}^{2}}{\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/x^4/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x)

[Out]

4/27/(-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))+2/3*(2*x^4+2*x^2+1)^(1/2)/x+(2/3-2/3*I)/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*(1+
(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*(EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-EllipticE(x*(-1+I)
^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2)))-1/9*(2*x^4+2*x^2+1)^(1/2)/x^3-2/9/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*(1+(1+I)
*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(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 )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/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)/(4*x^10 + 10*x^8 + 8*x^6 + 3*x^4), x)

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

[Out]

Integral(1/(x**4*(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 )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(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^4), x)