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

Optimal. Leaf size=207 \[ \frac{56 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{375 \sqrt{x^4+3 x^2+2}}+\frac{24 x \left (x^2+2\right )}{125 \sqrt{x^4+3 x^2+2}}+\frac{1}{75} x \left (3 x^2+11\right ) \sqrt{x^4+3 x^2+2}-\frac{24 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{125 \sqrt{x^4+3 x^2+2}}-\frac{9 \sqrt{2} \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{875 \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

[Out]

(24*x*(2 + x^2))/(125*Sqrt[2 + 3*x^2 + x^4]) + (x*(11 + 3*x^2)*Sqrt[2 + 3*x^2 + x^4])/75 - (24*Sqrt[2]*(1 + x^
2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(125*Sqrt[2 + 3*x^2 + x^4]) + (56*Sqrt[2]*(1 + x^2)*Sq
rt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(375*Sqrt[2 + 3*x^2 + x^4]) - (9*Sqrt[2]*(2 + x^2)*Elliptic
Pi[2/7, ArcTan[x], 1/2])/(875*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.199018, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1208, 1176, 1189, 1099, 1135, 1214, 1456, 539} \[ \frac{24 x \left (x^2+2\right )}{125 \sqrt{x^4+3 x^2+2}}+\frac{1}{75} x \left (3 x^2+11\right ) \sqrt{x^4+3 x^2+2}+\frac{56 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{375 \sqrt{x^4+3 x^2+2}}-\frac{24 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{125 \sqrt{x^4+3 x^2+2}}-\frac{9 \sqrt{2} \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{875 \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*x*(2 + x^2))/(125*Sqrt[2 + 3*x^2 + x^4]) + (x*(11 + 3*x^2)*Sqrt[2 + 3*x^2 + x^4])/75 - (24*Sqrt[2]*(1 + x^
2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(125*Sqrt[2 + 3*x^2 + x^4]) + (56*Sqrt[2]*(1 + x^2)*Sq
rt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(375*Sqrt[2 + 3*x^2 + x^4]) - (9*Sqrt[2]*(2 + x^2)*Elliptic
Pi[2/7, ArcTan[x], 1/2])/(875*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^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 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{\left (2+3 x^2+x^4\right )^{3/2}}{7+5 x^2} \, dx &=-\left (\frac{1}{25} \int \left (-8-5 x^2\right ) \sqrt{2+3 x^2+x^4} \, dx\right )-\frac{6}{25} \int \frac{\sqrt{2+3 x^2+x^4}}{7+5 x^2} \, dx\\ &=\frac{1}{75} x \left (11+3 x^2\right ) \sqrt{2+3 x^2+x^4}-\frac{1}{375} \int \frac{-130-90 x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{6}{625} \int \frac{-8-5 x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{36}{625} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{1}{75} x \left (11+3 x^2\right ) \sqrt{2+3 x^2+x^4}+\frac{18}{625} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{6}{125} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{9}{125} \int \frac{2+2 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx-\frac{48}{625} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{6}{25} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{26}{75} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{24 x \left (2+x^2\right )}{125 \sqrt{2+3 x^2+x^4}}+\frac{1}{75} x \left (11+3 x^2\right ) \sqrt{2+3 x^2+x^4}-\frac{24 \sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{125 \sqrt{2+3 x^2+x^4}}+\frac{56 \sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{375 \sqrt{2+3 x^2+x^4}}-\frac{\left (9 \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}{125 \sqrt{2+3 x^2+x^4}}\\ &=\frac{24 x \left (2+x^2\right )}{125 \sqrt{2+3 x^2+x^4}}+\frac{1}{75} x \left (11+3 x^2\right ) \sqrt{2+3 x^2+x^4}-\frac{24 \sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{125 \sqrt{2+3 x^2+x^4}}+\frac{56 \sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{375 \sqrt{2+3 x^2+x^4}}-\frac{9 \sqrt{2} \left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{875 \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.175629, size = 148, normalized size = 0.71 \[ \frac{-1022 i \sqrt{x^2+1} \sqrt{x^2+2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+525 x^7+3500 x^5+6825 x^3-2520 i \sqrt{x^2+1} \sqrt{x^2+2} E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )-108 i \sqrt{x^2+1} \sqrt{x^2+2} \Pi \left (\frac{10}{7};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+3850 x}{13125 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3850*x + 6825*x^3 + 3500*x^5 + 525*x^7 - (2520*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]],
2] - (1022*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] - (108*I)*Sqrt[1 + x^2]*Sqrt[2 +
x^2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2])/(13125*Sqrt[2 + 3*x^2 + x^4])

________________________________________________________________________________________

Maple [C]  time = 0.014, size = 170, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{25}\sqrt{{x}^{4}+3\,{x}^{2}+2}}+{\frac{11\,x}{75}\sqrt{{x}^{4}+3\,{x}^{2}+2}}-{{\frac{73\,i}{1875}}\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{12\,i}{125}}\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{36\,i}{4375}}\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)^(3/2)/(5*x^2+7),x)

[Out]

1/25*x^3*(x^4+3*x^2+2)^(1/2)+11/75*x*(x^4+3*x^2+2)^(1/2)-73/1875*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))-12/125*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))-36/4375*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{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}}{5 \, x^{2} + 7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((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{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}}{5 \, x^{2} + 7}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((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{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}}{5 \, x^{2} + 7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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