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

Optimal. Leaf size=428 \[ -\frac{\left (66+383 \sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right ),\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{10\ 2^{3/4} \left (2+3 \sqrt{2}\right ) \sqrt{2 x^4+2 x^2+1}}-\frac{1}{10} \left (2 x^2+9\right ) \sqrt{2 x^4+2 x^2+1} x-\frac{103 \sqrt{2 x^4+2 x^2+1} x}{10 \sqrt{2} \left (\sqrt{2} x^2+1\right )}+\frac{17}{8} \sqrt{\frac{17}{3}} \tanh ^{-1}\left (\frac{\sqrt{\frac{17}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )+\frac{103 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{10\ 2^{3/4} \sqrt{2 x^4+2 x^2+1}}-\frac{289 \left (3-\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12+11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{24\ 2^{3/4} \left (2+3 \sqrt{2}\right ) \sqrt{2 x^4+2 x^2+1}} \]

[Out]

-(x*(9 + 2*x^2)*Sqrt[1 + 2*x^2 + 2*x^4])/10 - (103*x*Sqrt[1 + 2*x^2 + 2*x^4])/(10*Sqrt[2]*(1 + Sqrt[2]*x^2)) +
(17*Sqrt[17/3]*ArcTanh[(Sqrt[17/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/8 + (103*(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]) - ((66 + 383*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])/(10*2^(3/4)*(2 + 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4]) - (289*(3 - Sqrt[2])*(1 + Sq
rt[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])/(24*2^(3/4)*(2 + 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4])

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Rubi [A]  time = 0.348961, antiderivative size = 602, normalized size of antiderivative = 1.41, number of steps used = 12, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {1208, 1176, 1197, 1103, 1195, 1216, 1706} \[ -\frac{1}{10} \left (2 x^2+9\right ) \sqrt{2 x^4+2 x^2+1} x-\frac{103 \sqrt{2 x^4+2 x^2+1} x}{10 \sqrt{2} \left (\sqrt{2} x^2+1\right )}+\frac{17}{8} \sqrt{\frac{17}{3}} \tanh ^{-1}\left (\frac{\sqrt{\frac{17}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )-\frac{\left (9+8 \sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{10\ 2^{3/4} \sqrt{2 x^4+2 x^2+1}}-\frac{17 \left (5+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{8 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}+\frac{289 \left (3-\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{56 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}+\frac{103 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{10\ 2^{3/4} \sqrt{2 x^4+2 x^2+1}}-\frac{289 \left (11-6 \sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12+11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{336 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(9 + 2*x^2)*Sqrt[1 + 2*x^2 + 2*x^4])/10 - (103*x*Sqrt[1 + 2*x^2 + 2*x^4])/(10*Sqrt[2]*(1 + Sqrt[2]*x^2)) +
(17*Sqrt[17/3]*ArcTanh[(Sqrt[17/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/8 + (103*(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]) + (289*(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])/(56*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4]) - (17*(5 + 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])/(8*2^(1/4)*Sqrt[1 + 2*
x^2 + 2*x^4]) - ((9 + 8*Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*A
rcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(10*2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) - (289*(11 - 6*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])/(336*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])

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 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

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 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 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{\left (1+2 x^2+2 x^4\right )^{3/2}}{3-2 x^2} \, dx &=-\left (\frac{1}{4} \int \left (10+4 x^2\right ) \sqrt{1+2 x^2+2 x^4} \, dx\right )+\frac{17}{2} \int \frac{\sqrt{1+2 x^2+2 x^4}}{3-2 x^2} \, dx\\ &=-\frac{1}{10} x \left (9+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}-\frac{1}{120} \int \frac{192+216 x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx-\frac{17}{8} \int \frac{10+4 x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx+\frac{289}{4} \int \frac{1}{\left (3-2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{1}{10} x \left (9+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}+\frac{9 \int \frac{1-\sqrt{2} x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx}{5 \sqrt{2}}+\frac{17 \int \frac{1-\sqrt{2} x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx}{2 \sqrt{2}}-\frac{1}{28} \left (289 \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}{28} \left (289 \left (3-\sqrt{2}\right )\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx-\frac{1}{4} \left (17 \left (5+\sqrt{2}\right )\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx-\frac{1}{10} \left (16+9 \sqrt{2}\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{1}{10} x \left (9+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}-\frac{103 x \sqrt{1+2 x^2+2 x^4}}{10 \sqrt{2} \left (1+\sqrt{2} x^2\right )}+\frac{17}{8} \sqrt{\frac{17}{3}} \tanh ^{-1}\left (\frac{\sqrt{\frac{17}{3}} x}{\sqrt{1+2 x^2+2 x^4}}\right )+\frac{103 \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{289 \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 )}{56 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}-\frac{17 \left (5+\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 )}{8 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}-\frac{\left (9+8 \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 )}{10\ 2^{3/4} \sqrt{1+2 x^2+2 x^4}}-\frac{289 \left (11-6 \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 )}{336 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.159486, size = 209, normalized size = 0.49 \[ \frac{-(1371-753 i) \sqrt{1-i} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{1-i} x\right ),i\right )-48 x^7-264 x^5-240 x^3+618 i \sqrt{1-i} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} E\left (\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )+1445 (1-i)^{3/2} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} \Pi \left (-\frac{1}{3}-\frac{i}{3};\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )-108 x}{120 \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-108*x - 240*x^3 - 264*x^5 - 48*x^7 + (618*I)*Sqrt[1 - I]*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*Ellipti
cE[I*ArcSinh[Sqrt[1 - I]*x], I] - (1371 - 753*I)*Sqrt[1 - I]*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*Ellip
ticF[I*ArcSinh[Sqrt[1 - I]*x], I] + 1445*(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])/(120*Sqrt[1 + 2*x^2 + 2*x^4])

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Maple [C]  time = 0.007, size = 377, normalized size = 0.9 \begin{align*} -{\frac{{x}^{3}}{5}\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}-{\frac{9\,x}{10}\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}-{\frac{177\,{\it EllipticF} \left ( x\sqrt{-1+i},1/2\,\sqrt{2}+i/2\sqrt{2} \right ) }{10\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{{\frac{103\,i}{20}}{\it EllipticF} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{103\,{\it EllipticE} \left ( x\sqrt{-1+i},1/2\,\sqrt{2}+i/2\sqrt{2} \right ) }{20\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}+{\frac{{\frac{103\,i}{20}}{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}+{\frac{289}{12\,\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*x^3*(2*x^4+2*x^2+1)^(1/2)-9/10*x*(2*x^4+2*x^2+1)^(1/2)-177/10/(-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)*EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-103/20*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)*EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2)
)-103/20/(-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(x*(-1+I)^(1/2),
1/2*2^(1/2)+1/2*I*2^(1/2))+103/20*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(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))+289/12/(-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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{\frac{3}{2}}}{2 \, x^{2} - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{\frac{3}{2}}}{2 \, x^{2} - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\sqrt{2 x^{4} + 2 x^{2} + 1}}{2 x^{2} - 3}\, dx - \int \frac{2 x^{2} \sqrt{2 x^{4} + 2 x^{2} + 1}}{2 x^{2} - 3}\, dx - \int \frac{2 x^{4} \sqrt{2 x^{4} + 2 x^{2} + 1}}{2 x^{2} - 3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{\frac{3}{2}}}{2 \, x^{2} - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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