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

Optimal. Leaf size=17 \[ \frac{1}{7} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

[Out]

EllipticPi[-10/7, ArcSin[x/Sqrt[2]], -2]/7

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Rubi [A]  time = 0.0330613, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1212, 537} \[ \frac{1}{7} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticPi[-10/7, ArcSin[x/Sqrt[2]], -2]/7

Rule 1212

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*Sqrt[-c], Int[1/((d + e*x^2)*Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a,
b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[c, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rubi steps

\begin{align*} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx &=2 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx\\ &=\frac{1}{7} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )\\ \end{align*}

Mathematica [C]  time = 0.0972269, size = 24, normalized size = 1.41 \[ -\frac{i \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )}{7 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/7)*EllipticPi[5/7, I*ArcSinh[x], -1/2])/Sqrt[2]

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Maple [B]  time = 0.01, size = 48, normalized size = 2.8 \begin{align*}{\frac{\sqrt{2}}{7}\sqrt{1-{\frac{{x}^{2}}{2}}}\sqrt{{x}^{2}+1}{\it EllipticPi} \left ({\frac{x\sqrt{2}}{2}},-{\frac{10}{7}},i\sqrt{2} \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{4} + 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(1/(5*x^2+7)/(-x^4+x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-x^4 + 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} + x^{2} + 2}}{5 \, x^{6} + 2 \, x^{4} - 17 \, x^{2} - 14}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-x^4 + x^2 + 2)/(5*x^6 + 2*x^4 - 17*x^2 - 14), x)

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{4} + 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(1/(5*x^2+7)/(-x^4+x^2+2)^(1/2),x, algorithm="giac")

[Out]

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