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

Optimal. Leaf size=100 \[ \frac{89 \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right ),-2\right )}{24276}+\frac{625 \sqrt{-x^4+x^2+2} x}{16184 \left (5 x^2+7\right )}+\frac{\left (580-287 x^2\right ) x}{10404 \sqrt{-x^4+x^2+2}}+\frac{5143 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{145656}-\frac{10825 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{113288} \]

[Out]

(x*(580 - 287*x^2))/(10404*Sqrt[2 + x^2 - x^4]) + (625*x*Sqrt[2 + x^2 - x^4])/(16184*(7 + 5*x^2)) + (5143*Elli
pticE[ArcSin[x/Sqrt[2]], -2])/145656 + (89*EllipticF[ArcSin[x/Sqrt[2]], -2])/24276 - (10825*EllipticPi[-10/7,
ArcSin[x/Sqrt[2]], -2])/113288

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Rubi [A]  time = 0.296641, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {1228, 1178, 1180, 524, 424, 419, 1223, 1716, 1212, 537} \[ \frac{625 \sqrt{-x^4+x^2+2} x}{16184 \left (5 x^2+7\right )}+\frac{\left (580-287 x^2\right ) x}{10404 \sqrt{-x^4+x^2+2}}+\frac{89 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{24276}+\frac{5143 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{145656}-\frac{10825 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{113288} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(580 - 287*x^2))/(10404*Sqrt[2 + x^2 - x^4]) + (625*x*Sqrt[2 + x^2 - x^4])/(16184*(7 + 5*x^2)) + (5143*Elli
pticE[ArcSin[x/Sqrt[2]], -2])/145656 + (89*EllipticF[ArcSin[x/Sqrt[2]], -2])/24276 - (10825*EllipticPi[-10/7,
ArcSin[x/Sqrt[2]], -2])/113288

Rule 1228

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{aa, bb, cc}, In
t[ExpandIntegrand[1/Sqrt[aa + bb*x^2 + cc*x^4], (d + e*x^2)^q*(aa + bb*x^2 + cc*x^4)^(p + 1/2), x] /. {aa -> a
, bb -> b, cc -> c}, 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[q, 0] && IntegerQ[p + 1/2]

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 1223

Int[((d_) + (e_.)*(x_)^2)^(q_)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Simp[(e^2*x*(d + e*x^2)
^(q + 1)*Sqrt[a + b*x^2 + c*x^4])/(2*d*(q + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(2*d*(q + 1)*(c*d^2 - b*d
*e + a*e^2)), Int[((d + e*x^2)^(q + 1)*Simp[a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1) - 2*e*(c*d*(q + 1) - b*e
*(q + 2))*x^2 + c*e^2*(2*q + 5)*x^4, x])/Sqrt[a + b*x^2 + c*x^4], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b
^2 - 4*a*c, 0] && ILtQ[q, -1]

Rule 1716

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[P4x,
x, 0], B = Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Dist[(e^2)^(-1), Int[(C*d - B*e - C*e*x^2)/Sqrt[a + b*x^
2 + c*x^4], x], x] + Dist[(C*d^2 - B*d*e + A*e^2)/e^2, Int[1/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /;
FreeQ[{a, b, c, d, e}, x] && PolyQ[P4x, x^2, 2] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && Ne
Q[c*d^2 - a*e^2, 0]

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 )^2 \left (2+x^2-x^4\right )^{3/2}} \, dx &=\int \left (\frac{194-95 x^2}{1156 \left (2+x^2-x^4\right )^{3/2}}-\frac{25}{34 \left (7+5 x^2\right )^2 \sqrt{2+x^2-x^4}}-\frac{475}{1156 \left (7+5 x^2\right ) \sqrt{2+x^2-x^4}}\right ) \, dx\\ &=\frac{\int \frac{194-95 x^2}{\left (2+x^2-x^4\right )^{3/2}} \, dx}{1156}-\frac{475 \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{1156}-\frac{25}{34} \int \frac{1}{\left (7+5 x^2\right )^2 \sqrt{2+x^2-x^4}} \, dx\\ &=\frac{x \left (580-287 x^2\right )}{10404 \sqrt{2+x^2-x^4}}+\frac{625 x \sqrt{2+x^2-x^4}}{16184 \left (7+5 x^2\right )}-\frac{\int \frac{-586-574 x^2}{\sqrt{2+x^2-x^4}} \, dx}{20808}-\frac{25 \int \frac{118-70 x^2-25 x^4}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{16184}-\frac{475}{578} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx\\ &=\frac{x \left (580-287 x^2\right )}{10404 \sqrt{2+x^2-x^4}}+\frac{625 x \sqrt{2+x^2-x^4}}{16184 \left (7+5 x^2\right )}-\frac{475 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{8092}+\frac{\int \frac{175+125 x^2}{\sqrt{2+x^2-x^4}} \, dx}{16184}-\frac{\int \frac{-586-574 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{10404}-\frac{4175 \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{16184}\\ &=\frac{x \left (580-287 x^2\right )}{10404 \sqrt{2+x^2-x^4}}+\frac{625 x \sqrt{2+x^2-x^4}}{16184 \left (7+5 x^2\right )}-\frac{475 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{8092}+\frac{\int \frac{175+125 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{8092}+\frac{1}{867} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx+\frac{287 \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx}{10404}-\frac{4175 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx}{8092}\\ &=\frac{x \left (580-287 x^2\right )}{10404 \sqrt{2+x^2-x^4}}+\frac{625 x \sqrt{2+x^2-x^4}}{16184 \left (7+5 x^2\right )}+\frac{287 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{10404}+\frac{F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{1734}-\frac{10825 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{113288}+\frac{25 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{4046}+\frac{125 \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx}{16184}\\ &=\frac{x \left (580-287 x^2\right )}{10404 \sqrt{2+x^2-x^4}}+\frac{625 x \sqrt{2+x^2-x^4}}{16184 \left (7+5 x^2\right )}+\frac{5143 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{145656}+\frac{89 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{24276}-\frac{10825 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{113288}\\ \end{align*}

Mathematica [C]  time = 0.325327, size = 196, normalized size = 1.96 \[ \frac{-111741 i \sqrt{2} \left (5 x^2+7\right ) \sqrt{-x^4+x^2+2} \text{EllipticF}\left (i \sinh ^{-1}(x),-\frac{1}{2}\right )-360010 x^5+253386 x^3+72002 i \sqrt{2} \left (5 x^2+7\right ) \sqrt{-x^4+x^2+2} E\left (i \sinh ^{-1}(x)|-\frac{1}{2}\right )+487125 i \sqrt{2} \sqrt{-x^4+x^2+2} x^2 \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )+681975 i \sqrt{2} \sqrt{-x^4+x^2+2} \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )+953260 x}{2039184 \left (5 x^2+7\right ) \sqrt{-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(953260*x + 253386*x^3 - 360010*x^5 + (72002*I)*Sqrt[2]*(7 + 5*x^2)*Sqrt[2 + x^2 - x^4]*EllipticE[I*ArcSinh[x]
, -1/2] - (111741*I)*Sqrt[2]*(7 + 5*x^2)*Sqrt[2 + x^2 - x^4]*EllipticF[I*ArcSinh[x], -1/2] + (681975*I)*Sqrt[2
]*Sqrt[2 + x^2 - x^4]*EllipticPi[5/7, I*ArcSinh[x], -1/2] + (487125*I)*Sqrt[2]*x^2*Sqrt[2 + x^2 - x^4]*Ellipti
cPi[5/7, I*ArcSinh[x], -1/2])/(2039184*(7 + 5*x^2)*Sqrt[2 + x^2 - x^4])

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Maple [B]  time = 0.02, size = 188, normalized size = 1.9 \begin{align*}{\frac{625\,x}{80920\,{x}^{2}+113288}\sqrt{-{x}^{4}+{x}^{2}+2}}+2\,{\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}} \left ( -{\frac{287\,{x}^{3}}{20808}}+{\frac{145\,x}{5202}} \right ) }+{\frac{89\,\sqrt{2}}{48552}\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{5143\,\sqrt{2}}{291312}\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{10825\,\sqrt{2}}{113288}\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)^2/(-x^4+x^2+2)^(3/2),x)

[Out]

625/16184*x*(-x^4+x^2+2)^(1/2)/(5*x^2+7)+2*(-287/20808*x^3+145/5202*x)/(-x^4+x^2+2)^(1/2)+89/48552*2^(1/2)*(-2
*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticF(1/2*x*2^(1/2),I*2^(1/2))+5143/291312*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))-10825/113288*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 )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((-x^4 + x^2 + 2)^(3/2)*(5*x^2 + 7)^2), 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}}{25 \, x^{12} + 20 \, x^{10} - 166 \, x^{8} - 208 \, x^{6} + 233 \, x^{4} + 476 \, x^{2} + 196}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-x^4 + x^2 + 2)/(25*x^12 + 20*x^10 - 166*x^8 - 208*x^6 + 233*x^4 + 476*x^2 + 196), 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 )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((-(x**2 - 2)*(x**2 + 1))**(3/2)*(5*x**2 + 7)**2), 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 )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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