3.292 \(\int \frac{\sqrt{2+3 x^2+x^4}}{(7+5 x^2)^3} \, dx\)

Optimal. Leaf size=237 \[ \frac{81 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{7840 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{11 \sqrt{x^4+3 x^2+2} x}{2352 \left (5 x^2+7\right )}+\frac{\sqrt{x^4+3 x^2+2} x}{28 \left (5 x^2+7\right )^2}-\frac{11 \left (x^2+2\right ) x}{11760 \sqrt{x^4+3 x^2+2}}+\frac{11 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5880 \sqrt{2} \sqrt{x^4+3 x^2+2}}-\frac{1201 \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{164640 \sqrt{2} \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

[Out]

(-11*x*(2 + x^2))/(11760*Sqrt[2 + 3*x^2 + x^4]) + (x*Sqrt[2 + 3*x^2 + x^4])/(28*(7 + 5*x^2)^2) + (11*x*Sqrt[2
+ 3*x^2 + x^4])/(2352*(7 + 5*x^2)) + (11*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(5880*
Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (81*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(7840*Sqrt
[2]*Sqrt[2 + 3*x^2 + x^4]) - (1201*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(164640*Sqrt[2]*Sqrt[(2 + x^2)/(
1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.595241, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {1228, 1223, 1696, 1716, 1189, 1099, 1135, 1214, 1456, 539} \[ \frac{11 \sqrt{x^4+3 x^2+2} x}{2352 \left (5 x^2+7\right )}+\frac{\sqrt{x^4+3 x^2+2} x}{28 \left (5 x^2+7\right )^2}-\frac{11 \left (x^2+2\right ) x}{11760 \sqrt{x^4+3 x^2+2}}+\frac{81 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{7840 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{11 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5880 \sqrt{2} \sqrt{x^4+3 x^2+2}}-\frac{1201 \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{164640 \sqrt{2} \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*x*(2 + x^2))/(11760*Sqrt[2 + 3*x^2 + x^4]) + (x*Sqrt[2 + 3*x^2 + x^4])/(28*(7 + 5*x^2)^2) + (11*x*Sqrt[2
+ 3*x^2 + x^4])/(2352*(7 + 5*x^2)) + (11*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(5880*
Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (81*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(7840*Sqrt
[2]*Sqrt[2 + 3*x^2 + x^4]) - (1201*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(164640*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 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 1696

Int[((P4x_)*((d_) + (e_.)*(x_)^2)^(q_))/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]}, -Simp[((C*d^2 - B*d*e + A*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*d*(C*d - B*e) + A*(a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1)) - 2*((B*d
- A*e)*(b*e*(q + 2) - c*d*(q + 1)) - C*d*(b*d + a*e*(q + 1)))*x^2 + c*(C*d^2 - B*d*e + A*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] && PolyQ[P4x, x^2] && LeQ[Expon[P4x, x], 4] &
& NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 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 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 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{\sqrt{2+3 x^2+x^4}}{\left (7+5 x^2\right )^3} \, dx &=\int \left (-\frac{6}{25 \left (7+5 x^2\right )^3 \sqrt{2+3 x^2+x^4}}+\frac{1}{25 \left (7+5 x^2\right )^2 \sqrt{2+3 x^2+x^4}}+\frac{1}{25 \left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}}\right ) \, dx\\ &=\frac{1}{25} \int \frac{1}{\left (7+5 x^2\right )^2 \sqrt{2+3 x^2+x^4}} \, dx+\frac{1}{25} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx-\frac{6}{25} \int \frac{1}{\left (7+5 x^2\right )^3 \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \sqrt{2+3 x^2+x^4}}{28 \left (7+5 x^2\right )^2}-\frac{x \sqrt{2+3 x^2+x^4}}{84 \left (7+5 x^2\right )}+\frac{\int \frac{62+70 x^2+25 x^4}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx}{2100}-\frac{1}{700} \int \frac{74-10 x^2-25 x^4}{\left (7+5 x^2\right )^2 \sqrt{2+3 x^2+x^4}} \, dx+\frac{1}{50} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{1}{20} \int \frac{2+2 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \sqrt{2+3 x^2+x^4}}{28 \left (7+5 x^2\right )^2}+\frac{11 x \sqrt{2+3 x^2+x^4}}{2352 \left (7+5 x^2\right )}+\frac{\left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{50 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{\int \frac{2838+2310 x^2+975 x^4}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx}{58800}-\frac{\int \frac{-175-125 x^2}{\sqrt{2+3 x^2+x^4}} \, dx}{52500}+\frac{13 \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx}{2100}-\frac{\left (\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}{20 \sqrt{2+3 x^2+x^4}}\\ &=\frac{x \sqrt{2+3 x^2+x^4}}{28 \left (7+5 x^2\right )^2}+\frac{11 x \sqrt{2+3 x^2+x^4}}{2352 \left (7+5 x^2\right )}+\frac{\left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{50 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{\left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{70 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}+\frac{\int \frac{-4725-4875 x^2}{\sqrt{2+3 x^2+x^4}} \, dx}{1470000}+\frac{1}{420} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{13 \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx}{4200}+\frac{1}{300} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{13 \int \frac{2+2 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx}{1680}-\frac{101 \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx}{3920}\\ &=\frac{x \left (2+x^2\right )}{420 \sqrt{2+3 x^2+x^4}}+\frac{x \sqrt{2+3 x^2+x^4}}{28 \left (7+5 x^2\right )^2}+\frac{11 x \sqrt{2+3 x^2+x^4}}{2352 \left (7+5 x^2\right )}-\frac{\left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{210 \sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{37 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{1400 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{\left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{70 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}-\frac{9 \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx}{2800}-\frac{13 \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx}{3920}-\frac{101 \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx}{7840}+\frac{101 \int \frac{2+2 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx}{3136}-\frac{\left (13 \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}{1680 \sqrt{2+3 x^2+x^4}}\\ &=-\frac{11 x \left (2+x^2\right )}{11760 \sqrt{2+3 x^2+x^4}}+\frac{x \sqrt{2+3 x^2+x^4}}{28 \left (7+5 x^2\right )^2}+\frac{11 x \sqrt{2+3 x^2+x^4}}{2352 \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 )}{5880 \sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{81 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{7840 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{97 \left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{5880 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}+\frac{\left (101 \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}{3136 \sqrt{2+3 x^2+x^4}}\\ &=-\frac{11 x \left (2+x^2\right )}{11760 \sqrt{2+3 x^2+x^4}}+\frac{x \sqrt{2+3 x^2+x^4}}{28 \left (7+5 x^2\right )^2}+\frac{11 x \sqrt{2+3 x^2+x^4}}{2352 \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 )}{5880 \sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{81 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{7840 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{1201 \left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{164640 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.341789, size = 174, normalized size = 0.73 \[ \frac{-434 i \sqrt{x^2+1} \sqrt{x^2+2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+\frac{1925 x \left (x^4+3 x^2+2\right )}{5 x^2+7}+\frac{14700 x \left (x^4+3 x^2+2\right )}{\left (5 x^2+7\right )^2}+385 i \sqrt{x^2+1} \sqrt{x^2+2} E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )-1201 i \sqrt{x^2+1} \sqrt{x^2+2} \Pi \left (\frac{10}{7};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )}{411600 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((14700*x*(2 + 3*x^2 + x^4))/(7 + 5*x^2)^2 + (1925*x*(2 + 3*x^2 + x^4))/(7 + 5*x^2) + (385*I)*Sqrt[1 + x^2]*Sq
rt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] - (434*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqr
t[2]], 2] - (1201*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2])/(411600*Sqrt[2 + 3
*x^2 + x^4])

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Maple [C]  time = 0.018, size = 186, normalized size = 0.8 \begin{align*}{\frac{x}{28\, \left ( 5\,{x}^{2}+7 \right ) ^{2}}\sqrt{{x}^{4}+3\,{x}^{2}+2}}+{\frac{11\,x}{11760\,{x}^{2}+16464}\sqrt{{x}^{4}+3\,{x}^{2}+2}}-{{\frac{31\,i}{58800}}\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{11\,i}{23520}}\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{1201\,i}{411600}}\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((x^4+3*x^2+2)^(1/2)/(5*x^2+7)^3,x)

[Out]

1/28*x*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)^2+11/2352*x*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)-31/58800*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))+11/23520*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))-1201/411600*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{\sqrt{x^{4} + 3 \, x^{2} + 2}}{{\left (5 \, x^{2} + 7\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 3*x^2 + 2)/(5*x^2 + 7)^3, 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}}{125 \, x^{6} + 525 \, x^{4} + 735 \, x^{2} + 343}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)/(125*x^6 + 525*x^4 + 735*x^2 + 343), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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