3.314 \(\int \frac{1}{(7+5 x^2)^2 (2+3 x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=235 \[ -\frac{463 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{336 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{625 \sqrt{x^4+3 x^2+2} x}{504 \left (5 x^2+7\right )}-\frac{31 \left (x^2+2\right ) x}{56 \sqrt{x^4+3 x^2+2}}+\frac{\left (11 x^2+20\right ) x}{36 \sqrt{x^4+3 x^2+2}}+\frac{31 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{28 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{375 \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{784 \sqrt{2} \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

[Out]

(-31*x*(2 + x^2))/(56*Sqrt[2 + 3*x^2 + x^4]) + (x*(20 + 11*x^2))/(36*Sqrt[2 + 3*x^2 + x^4]) + (625*x*Sqrt[2 +
3*x^2 + x^4])/(504*(7 + 5*x^2)) + (31*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(28*Sqrt[
2]*Sqrt[2 + 3*x^2 + x^4]) - (463*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(336*Sqrt[2]*S
qrt[2 + 3*x^2 + x^4]) + (375*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(784*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]
*Sqrt[2 + 3*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.429144, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {1228, 1178, 1189, 1099, 1135, 1223, 1716, 1214, 1456, 539} \[ \frac{625 \sqrt{x^4+3 x^2+2} x}{504 \left (5 x^2+7\right )}-\frac{31 \left (x^2+2\right ) x}{56 \sqrt{x^4+3 x^2+2}}+\frac{\left (11 x^2+20\right ) x}{36 \sqrt{x^4+3 x^2+2}}-\frac{463 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{336 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{31 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{28 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{375 \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{784 \sqrt{2} \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((7 + 5*x^2)^2*(2 + 3*x^2 + x^4)^(3/2)),x]

[Out]

(-31*x*(2 + x^2))/(56*Sqrt[2 + 3*x^2 + x^4]) + (x*(20 + 11*x^2))/(36*Sqrt[2 + 3*x^2 + x^4]) + (625*x*Sqrt[2 +
3*x^2 + x^4])/(504*(7 + 5*x^2)) + (31*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(28*Sqrt[
2]*Sqrt[2 + 3*x^2 + x^4]) - (463*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(336*Sqrt[2]*S
qrt[2 + 3*x^2 + x^4]) + (375*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(784*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]
*Sqrt[2 + 3*x^2 + x^4])

Rule 1228

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{aa, bb, cc}, In
t[ExpandIntegrand[1/Sqrt[aa + bb*x^2 + cc*x^4], (d + e*x^2)^q*(aa + bb*x^2 + cc*x^4)^(p + 1/2), x] /. {aa -> a
, bb -> b, cc -> c}, 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]
&& ILtQ[q, 0] && IntegerQ[p + 1/2]

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 1189

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

Rule 1099

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b +
q)*x^2)*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)]
)/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1223

Int[((d_) + (e_.)*(x_)^2)^(q_)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Simp[(e^2*x*(d + e*x^2)
^(q + 1)*Sqrt[a + b*x^2 + c*x^4])/(2*d*(q + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(2*d*(q + 1)*(c*d^2 - b*d
*e + a*e^2)), Int[((d + e*x^2)^(q + 1)*Simp[a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1) - 2*e*(c*d*(q + 1) - b*e
*(q + 2))*x^2 + c*e^2*(2*q + 5)*x^4, x])/Sqrt[a + b*x^2 + c*x^4], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b
^2 - 4*a*c, 0] && ILtQ[q, -1]

Rule 1716

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

Rule 1214

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

Rule 1456

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPar
t[p]), Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c*x^n)/e)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p,
q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p]

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{1}{\left (7+5 x^2\right )^2 \left (2+3 x^2+x^4\right )^{3/2}} \, dx &=\int \left (\frac{14+5 x^2}{36 \left (2+3 x^2+x^4\right )^{3/2}}-\frac{25}{6 \left (7+5 x^2\right )^2 \sqrt{2+3 x^2+x^4}}-\frac{25}{36 \left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}}\right ) \, dx\\ &=\frac{1}{36} \int \frac{14+5 x^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx-\frac{25}{36} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx-\frac{25}{6} \int \frac{1}{\left (7+5 x^2\right )^2 \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \left (20+11 x^2\right )}{36 \sqrt{2+3 x^2+x^4}}+\frac{625 x \sqrt{2+3 x^2+x^4}}{504 \left (7+5 x^2\right )}-\frac{1}{72} \int \frac{26+22 x^2}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{25}{504} \int \frac{62+70 x^2+25 x^4}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx-\frac{25}{72} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{125}{144} \int \frac{2+2 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \left (20+11 x^2\right )}{36 \sqrt{2+3 x^2+x^4}}+\frac{625 x \sqrt{2+3 x^2+x^4}}{504 \left (7+5 x^2\right )}-\frac{25 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{72 \sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{1}{504} \int \frac{-175-125 x^2}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{11}{36} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{13}{36} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{325}{504} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx+\frac{\left (125 \sqrt{1+\frac{x^2}{2}} \sqrt{2+2 x^2}\right ) \int \frac{\sqrt{2+2 x^2}}{\sqrt{1+\frac{x^2}{2}} \left (7+5 x^2\right )} \, dx}{144 \sqrt{2+3 x^2+x^4}}\\ &=-\frac{11 x \left (2+x^2\right )}{36 \sqrt{2+3 x^2+x^4}}+\frac{x \left (20+11 x^2\right )}{36 \sqrt{2+3 x^2+x^4}}+\frac{625 x \sqrt{2+3 x^2+x^4}}{504 \left (7+5 x^2\right )}+\frac{11 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{18 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{17 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{24 \sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{125 \left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{504 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}-\frac{125}{504} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{325 \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx}{1008}-\frac{25}{72} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{1625 \int \frac{2+2 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx}{2016}\\ &=-\frac{31 x \left (2+x^2\right )}{56 \sqrt{2+3 x^2+x^4}}+\frac{x \left (20+11 x^2\right )}{36 \sqrt{2+3 x^2+x^4}}+\frac{625 x \sqrt{2+3 x^2+x^4}}{504 \left (7+5 x^2\right )}+\frac{31 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{28 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{463 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{336 \sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{125 \left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{504 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}+\frac{\left (1625 \sqrt{1+\frac{x^2}{2}} \sqrt{2+2 x^2}\right ) \int \frac{\sqrt{2+2 x^2}}{\sqrt{1+\frac{x^2}{2}} \left (7+5 x^2\right )} \, dx}{2016 \sqrt{2+3 x^2+x^4}}\\ &=-\frac{31 x \left (2+x^2\right )}{56 \sqrt{2+3 x^2+x^4}}+\frac{x \left (20+11 x^2\right )}{36 \sqrt{2+3 x^2+x^4}}+\frac{625 x \sqrt{2+3 x^2+x^4}}{504 \left (7+5 x^2\right )}+\frac{31 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{28 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{463 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{336 \sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{375 \left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{784 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.274219, size = 208, normalized size = 0.89 \[ \frac{182 i \sqrt{x^2+1} \sqrt{x^2+2} \left (5 x^2+7\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+3255 x^5+10157 x^3+651 i \sqrt{x^2+1} \sqrt{x^2+2} \left (5 x^2+7\right ) E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+1125 i \sqrt{x^2+1} \sqrt{x^2+2} x^2 \Pi \left (\frac{10}{7};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+1575 i \sqrt{x^2+1} \sqrt{x^2+2} \Pi \left (\frac{10}{7};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+7490 x}{1176 \left (5 x^2+7\right ) \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((7 + 5*x^2)^2*(2 + 3*x^2 + x^4)^(3/2)),x]

[Out]

(7490*x + 10157*x^3 + 3255*x^5 + (651*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(7 + 5*x^2)*EllipticE[I*ArcSinh[x/Sqrt[2]
], 2] + (182*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(7 + 5*x^2)*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + (1575*I)*Sqrt[1 +
x^2]*Sqrt[2 + x^2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2] + (1125*I)*x^2*Sqrt[1 + x^2]*Sqrt[2 + x^2]*Ellip
ticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2])/(1176*(7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4])

________________________________________________________________________________________

Maple [C]  time = 0.019, size = 185, normalized size = 0.8 \begin{align*}{\frac{625\,x}{2520\,{x}^{2}+3528}\sqrt{{x}^{4}+3\,{x}^{2}+2}}-2\,{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}} \left ( -{\frac{11\,{x}^{3}}{72}}-{\frac{5\,x}{18}} \right ) }+{{\frac{13\,i}{168}}\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}+{{\frac{31\,i}{112}}\sqrt{2}{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}+{{\frac{75\,i}{392}}\sqrt{2}\sqrt{1+{\frac{{x}^{2}}{2}}}\sqrt{{x}^{2}+1}{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},{\frac{10}{7}},\sqrt{2} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x)

[Out]

625/504*x*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)-2*(-11/72*x^3-5/18*x)/(x^4+3*x^2+2)^(1/2)+13/168*I*2^(1/2)*(2*x^2+4)^(
1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*x*2^(1/2),2^(1/2))+31/112*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^
2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticE(1/2*I*x*2^(1/2),2^(1/2))+75/392*I*2^(1/2)*(1+1/2*x^2)^(1/2)*(x^2+1)^(
1/2)/(x^4+3*x^2+2)^(1/2)*EllipticPi(1/2*I*x*2^(1/2),10/7,2^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}{\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((x^4 + 3*x^2 + 2)^(3/2)*(5*x^2 + 7)^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 3 \, x^{2} + 2}}{25 \, x^{12} + 220 \, x^{10} + 794 \, x^{8} + 1504 \, x^{6} + 1577 \, x^{4} + 868 \, x^{2} + 196}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)/(25*x^12 + 220*x^10 + 794*x^8 + 1504*x^6 + 1577*x^4 + 868*x^2 + 196), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\left (x^{2} + 1\right ) \left (x^{2} + 2\right )\right )^{\frac{3}{2}} \left (5 x^{2} + 7\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x**2+7)**2/(x**4+3*x**2+2)**(3/2),x)

[Out]

Integral(1/(((x**2 + 1)*(x**2 + 2))**(3/2)*(5*x**2 + 7)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}{\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((x^4 + 3*x^2 + 2)^(3/2)*(5*x^2 + 7)^2), x)