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

Optimal. Leaf size=106 \[ \frac {\left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {5 \left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{14 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

[Out]

-5/28*(x^2+2)*(1/(x^2+1))^(1/2)*(x^2+1)^(1/2)*EllipticPi(x/(x^2+1)^(1/2),2/7,1/2*2^(1/2))*2^(1/2)/((x^2+2)/(x^
2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+1/4*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(
1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1214, 1099, 1456, 539} \[ \frac {\left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {5 \left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{14 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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]) - (5*(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 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]

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx &=\frac {1}{2} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {5}{4} \int \frac {2+2 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {\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 (5 \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}{4 \sqrt {2+3 x^2+x^4}}\\ &=\frac {\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 {5 \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*}

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Mathematica [C]  time = 0.10, size = 55, normalized size = 0.52 \[ -\frac {i \sqrt {x^2+1} \sqrt {x^2+2} \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )}{7 \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 3 \, x^{2} + 2}}{5 \, x^{6} + 22 \, x^{4} + 31 \, x^{2} + 14}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 47, normalized size = 0.44 \[ -\frac {i \sqrt {2}\, \sqrt {\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{7 \sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*I*2^(1/2)*(1/2*x^2+1)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticPi(1/2*I*2^(1/2)*x,10/7,2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (5\,x^2+7\right )\,\sqrt {x^4+3\,x^2+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\left (x^{2} + 1\right ) \left (x^{2} + 2\right )} \left (5 x^{2} + 7\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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