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

Optimal. Leaf size=72 \[ \frac{1}{102} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right ),-2\right )+\frac{x \left (35-16 x^2\right )}{306 \sqrt{-x^4+x^2+2}}+\frac{8}{153} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{25}{238} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

[Out]

(x*(35 - 16*x^2))/(306*Sqrt[2 + x^2 - x^4]) + (8*EllipticE[ArcSin[x/Sqrt[2]], -2])/153 + EllipticF[ArcSin[x/Sq
rt[2]], -2]/102 - (25*EllipticPi[-10/7, ArcSin[x/Sqrt[2]], -2])/238

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Rubi [A]  time = 0.0895718, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1221, 1178, 1180, 524, 424, 419, 1212, 537} \[ \frac{x \left (35-16 x^2\right )}{306 \sqrt{-x^4+x^2+2}}+\frac{1}{102} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{8}{153} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{25}{238} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(35 - 16*x^2))/(306*Sqrt[2 + x^2 - x^4]) + (8*EllipticE[ArcSin[x/Sqrt[2]], -2])/153 + EllipticF[ArcSin[x/Sq
rt[2]], -2]/102 - (25*EllipticPi[-10/7, ArcSin[x/Sqrt[2]], -2])/238

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 1180

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[2*Sqrt[-c], Int[(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 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 419

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

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 ) \left (2+x^2-x^4\right )^{3/2}} \, dx &=-\left (\frac{1}{34} \int \frac{-12+5 x^2}{\left (2+x^2-x^4\right )^{3/2}} \, dx\right )-\frac{25}{34} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx\\ &=\frac{x \left (35-16 x^2\right )}{306 \sqrt{2+x^2-x^4}}+\frac{1}{612} \int \frac{38+32 x^2}{\sqrt{2+x^2-x^4}} \, dx-\frac{25}{17} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx\\ &=\frac{x \left (35-16 x^2\right )}{306 \sqrt{2+x^2-x^4}}-\frac{25}{238} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{1}{306} \int \frac{38+32 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx\\ &=\frac{x \left (35-16 x^2\right )}{306 \sqrt{2+x^2-x^4}}-\frac{25}{238} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{1}{51} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx+\frac{8}{153} \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx\\ &=\frac{x \left (35-16 x^2\right )}{306 \sqrt{2+x^2-x^4}}+\frac{8}{153} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{1}{102} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{25}{238} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )\\ \end{align*}

Mathematica [C]  time = 0.212313, size = 101, normalized size = 1.4 \[ \frac{-357 i \sqrt{2} \text{EllipticF}\left (i \sinh ^{-1}(x),-\frac{1}{2}\right )-\frac{224 x^3}{\sqrt{-x^4+x^2+2}}+\frac{490 x}{\sqrt{-x^4+x^2+2}}+224 i \sqrt{2} E\left (i \sinh ^{-1}(x)|-\frac{1}{2}\right )+225 i \sqrt{2} \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )}{4284} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((490*x)/Sqrt[2 + x^2 - x^4] - (224*x^3)/Sqrt[2 + x^2 - x^4] + (224*I)*Sqrt[2]*EllipticE[I*ArcSinh[x], -1/2] -
(357*I)*Sqrt[2]*EllipticF[I*ArcSinh[x], -1/2] + (225*I)*Sqrt[2]*EllipticPi[5/7, I*ArcSinh[x], -1/2])/4284

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Maple [B]  time = 0.014, size = 164, normalized size = 2.3 \begin{align*} 2\,{\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}} \left ( -{\frac{4\,{x}^{3}}{153}}+{\frac{35\,x}{612}} \right ) }+{\frac{\sqrt{2}}{204}\sqrt{-2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}}+{\frac{4\,\sqrt{2}}{153}\sqrt{-2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\it EllipticE} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}}-{\frac{25\,\sqrt{2}}{238}\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)^(3/2),x)

[Out]

2*(-4/153*x^3+35/612*x)/(-x^4+x^2+2)^(1/2)+1/204*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*Ell
ipticF(1/2*x*2^(1/2),I*2^(1/2))+4/153*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticE(1/2*
x*2^(1/2),I*2^(1/2))-25/238*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}{{\left (-x^{4} + 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+x^2+2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((-x^4 + x^2 + 2)^(3/2)*(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^{10} - 3 \, x^{8} - 29 \, x^{6} - x^{4} + 48 \, 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+x^2+2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-x^4 + x^2 + 2)/(5*x^10 - 3*x^8 - 29*x^6 - x^4 + 48*x^2 + 28), x)

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-x^{4} + 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+x^2+2)^(3/2),x, algorithm="giac")

[Out]

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