3.32 \(\int \frac{946+315 x^2}{(7+5 x^2) \sqrt{2+3 x^2+x^4}} \, dx\)

Optimal. Leaf size=106 \[ \frac{631 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{x^4+3 x^2+2}}-\frac{2525 \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{14 \sqrt{2} \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

[Out]

(631*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(2*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) - (2525*
(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(14*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

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Rubi [A]  time = 0.105343, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1704, 1099, 1456, 539} \[ \frac{631 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{x^4+3 x^2+2}}-\frac{2525 \left (x^2+2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{14 \sqrt{2} \sqrt{\frac{x^2+2}{x^2+1}} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[(946 + 315*x^2)/((7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4]),x]

[Out]

(631*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(2*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) - (2525*
(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(14*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

Rule 1704

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Sqrt[b^2 - 4*a*c]}, Dist[(2*a*B - A*(b + q))/(2*a*e - d*(b + q)), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] -
Dist[(B*d - A*e)/(2*a*e - d*(b + q)), Int[(2*a + (b + q)*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /
; RationalQ[q]] /; FreeQ[{a, b, c, d, e, A, B}, x] && GtQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&
NeQ[c*A^2 - b*A*B + a*B^2, 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 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{946+315 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx &=\frac{631}{2} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx-\frac{2525}{8} \int \frac{4+4 x^2}{\left (7+5 x^2\right ) \sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{631 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{\left (2525 \sqrt{\frac{1}{2}+\frac{x^2}{4}} \sqrt{4+4 x^2}\right ) \int \frac{\sqrt{4+4 x^2}}{\sqrt{\frac{1}{2}+\frac{x^2}{4}} \left (7+5 x^2\right )} \, dx}{8 \sqrt{2+3 x^2+x^4}}\\ &=\frac{631 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{2+3 x^2+x^4}}-\frac{2525 \left (2+x^2\right ) \Pi \left (\frac{2}{7};\tan ^{-1}(x)|\frac{1}{2}\right )}{14 \sqrt{2} \sqrt{\frac{2+x^2}{1+x^2}} \sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.182633, size = 74, normalized size = 0.7 \[ -\frac{i \sqrt{x^2+1} \sqrt{x^2+2} \left (441 \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+505 \Pi \left (\frac{10}{7};\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )\right )}{7 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(946 + 315*x^2)/((7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4]),x]

[Out]

((-I/7)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(441*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + 505*EllipticPi[10/7, I*ArcSinh[x
/Sqrt[2]], 2]))/Sqrt[2 + 3*x^2 + x^4]

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Maple [C]  time = 0.028, size = 93, normalized size = 0.9 \begin{align*}{-{\frac{63\,i}{2}}\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{505\,i}{7}}\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((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x)

[Out]

-63/2*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))-505/7*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{315 \, x^{2} + 946}{\sqrt{x^{4} + 3 \, x^{2} + 2}{\left (5 \, x^{2} + 7\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate((315*x^2 + 946)/(sqrt(x^4 + 3*x^2 + 2)*(5*x^2 + 7)), 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}{\left (315 \, x^{2} + 946\right )}}{5 \, x^{6} + 22 \, x^{4} + 31 \, x^{2} + 14}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)*(315*x^2 + 946)/(5*x^6 + 22*x^4 + 31*x^2 + 14), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{315 x^{2} + 946}{\sqrt{\left (x^{2} + 1\right ) \left (x^{2} + 2\right )} \left (5 x^{2} + 7\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x**2+946)/(5*x**2+7)/(x**4+3*x**2+2)**(1/2),x)

[Out]

Integral((315*x**2 + 946)/(sqrt((x**2 + 1)*(x**2 + 2))*(5*x**2 + 7)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{315 \, x^{2} + 946}{\sqrt{x^{4} + 3 \, x^{2} + 2}{\left (5 \, x^{2} + 7\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((315*x^2+946)/(5*x^2+7)/(x^4+3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate((315*x^2 + 946)/(sqrt(x^4 + 3*x^2 + 2)*(5*x^2 + 7)), x)