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

Optimal. Leaf size=173 \[ -\frac{9 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{2 x^2+2}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{4 \sqrt{x^4+3 x^2+2}}+\frac{x}{6 \sqrt{x^4+3 x^2+2}}+\frac{\sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{x^4+3 x^2+2}}+\frac{125 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{84 \sqrt{2} \sqrt{x^4+3 x^2+2}} \]

[Out]

x/(6*Sqrt[2 + 3*x^2 + x^4]) + (Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(3*Sqrt[
2 + 3*x^2 + x^4]) - (9*(1 + x^2)*Sqrt[(2 + x^2)/(2 + 2*x^2)]*EllipticF[ArcTan[x], 1/2])/(4*Sqrt[2 + 3*x^2 + x^
4]) + (125*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticPi[2/7, ArcTan[x], 1/2])/(84*Sqrt[2]*Sqrt[2 + 3*x^2 + x
^4])

________________________________________________________________________________________

Rubi [A]  time = 0.140555, antiderivative size = 207, normalized size of antiderivative = 1.2, number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1221, 1178, 1189, 1099, 1135, 1214, 1456, 539} \[ -\frac{x \left (x^2+2\right )}{3 \sqrt{x^4+3 x^2+2}}+\frac{x \left (2 x^2+5\right )}{6 \sqrt{x^4+3 x^2+2}}-\frac{9 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{4 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{\sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{x^4+3 x^2+2}}+\frac{125 \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{84 \sqrt{2} \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(2 + x^2))/(3*Sqrt[2 + 3*x^2 + x^4]) + (x*(5 + 2*x^2))/(6*Sqrt[2 + 3*x^2 + x^4]) + (Sqrt[2]*(1 + x^2)*Sqrt
[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(3*Sqrt[2 + 3*x^2 + x^4]) - (9*(1 + x^2)*Sqrt[(2 + x^2)/(1 +
x^2)]*EllipticF[ArcTan[x], 1/2])/(4*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (125*(2 + x^2)*EllipticPi[2/7, ArcTan[x],
1/2])/(84*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

Rule 1221

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

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 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 ) \left (2+3 x^2+x^4\right )^{3/2}} \, dx &=-\left (\frac{1}{6} \int \frac{-8-5 x^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx\right )-\frac{25}{6} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \left (5+2 x^2\right )}{6 \sqrt{2+3 x^2+x^4}}+\frac{1}{12} \int \frac{-2-4 x^2}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{25}{12} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{125}{24} \int \frac{2+2 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \left (5+2 x^2\right )}{6 \sqrt{2+3 x^2+x^4}}-\frac{25 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{12 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{1}{6} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{1}{3} \int \frac{x^2}{\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}{24 \sqrt{2+3 x^2+x^4}}\\ &=-\frac{x \left (2+x^2\right )}{3 \sqrt{2+3 x^2+x^4}}+\frac{x \left (5+2 x^2\right )}{6 \sqrt{2+3 x^2+x^4}}+\frac{\sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{2+3 x^2+x^4}}-\frac{9 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{4 \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 )}{84 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.17013, size = 138, normalized size = 0.8 \[ \frac{-7 i \sqrt{x^2+1} \sqrt{x^2+2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+14 x^3+14 i \sqrt{x^2+1} \sqrt{x^2+2} E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+25 i \sqrt{x^2+1} \sqrt{x^2+2} \Pi \left (\frac{10}{7};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+35 x}{42 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*x + 14*x^3 + (14*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] - (7*I)*Sqrt[1 + x^2]*S
qrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + (25*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticPi[10/7, I*ArcSin
h[x/Sqrt[2]], 2])/(42*Sqrt[2 + 3*x^2 + x^4])

________________________________________________________________________________________

Maple [C]  time = 0.015, size = 161, normalized size = 0.9 \begin{align*} -2\,{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}} \left ( -1/6\,{x}^{3}-{\frac{5\,x}{12}} \right ) }-{{\frac{i}{12}}\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{i}{6}}\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{25\,i}{42}}\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)/(x^4+3*x^2+2)^(3/2),x)

[Out]

-2*(-1/6*x^3-5/12*x)/(x^4+3*x^2+2)^(1/2)-1/12*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*Elli
pticF(1/2*I*x*2^(1/2),2^(1/2))+1/6*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))+25/42*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))

________________________________________________________________________________________

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 )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)/(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)), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 3 \, x^{2} + 2}}{5 \, x^{10} + 37 \, x^{8} + 107 \, x^{6} + 151 \, x^{4} + 104 \, x^{2} + 28}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)/(5*x^10 + 37*x^8 + 107*x^6 + 151*x^4 + 104*x^2 + 28), x)

________________________________________________________________________________________

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 )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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 )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)/(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)), x)