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

Optimal. Leaf size=102 \[ -\frac{269 \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right ),-2\right )}{166600}-\frac{31 \sqrt{-x^4+x^2+2} x}{13328 \left (5 x^2+7\right )}+\frac{\sqrt{-x^4+x^2+2} x}{28 \left (5 x^2+7\right )^2}-\frac{31 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{66640}+\frac{16601 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{2332400} \]

[Out]

(x*Sqrt[2 + x^2 - x^4])/(28*(7 + 5*x^2)^2) - (31*x*Sqrt[2 + x^2 - x^4])/(13328*(7 + 5*x^2)) - (31*EllipticE[Ar
cSin[x/Sqrt[2]], -2])/66640 - (269*EllipticF[ArcSin[x/Sqrt[2]], -2])/166600 + (16601*EllipticPi[-10/7, ArcSin[
x/Sqrt[2]], -2])/2332400

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Rubi [A]  time = 0.41383, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {1228, 1223, 1696, 1716, 1180, 524, 424, 419, 1212, 537} \[ -\frac{31 \sqrt{-x^4+x^2+2} x}{13328 \left (5 x^2+7\right )}+\frac{\sqrt{-x^4+x^2+2} x}{28 \left (5 x^2+7\right )^2}-\frac{269 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{166600}-\frac{31 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{66640}+\frac{16601 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{2332400} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[2 + x^2 - x^4])/(28*(7 + 5*x^2)^2) - (31*x*Sqrt[2 + x^2 - x^4])/(13328*(7 + 5*x^2)) - (31*EllipticE[Ar
cSin[x/Sqrt[2]], -2])/66640 - (269*EllipticF[ArcSin[x/Sqrt[2]], -2])/166600 + (16601*EllipticPi[-10/7, ArcSin[
x/Sqrt[2]], -2])/2332400

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 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 1696

Int[((P4x_)*((d_) + (e_.)*(x_)^2)^(q_))/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]}, -Simp[((C*d^2 - B*d*e + A*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*d*(C*d - B*e) + A*(a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1)) - 2*((B*d
- A*e)*(b*e*(q + 2) - c*d*(q + 1)) - C*d*(b*d + a*e*(q + 1)))*x^2 + c*(C*d^2 - B*d*e + A*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] && PolyQ[P4x, x^2] && LeQ[Expon[P4x, x], 4] &
& NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 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 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{\sqrt{2+x^2-x^4}}{\left (7+5 x^2\right )^3} \, dx &=\int \left (-\frac{34}{25 \left (7+5 x^2\right )^3 \sqrt{2+x^2-x^4}}+\frac{19}{25 \left (7+5 x^2\right )^2 \sqrt{2+x^2-x^4}}-\frac{1}{25 \left (7+5 x^2\right ) \sqrt{2+x^2-x^4}}\right ) \, dx\\ &=-\left (\frac{1}{25} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx\right )+\frac{19}{25} \int \frac{1}{\left (7+5 x^2\right )^2 \sqrt{2+x^2-x^4}} \, dx-\frac{34}{25} \int \frac{1}{\left (7+5 x^2\right )^3 \sqrt{2+x^2-x^4}} \, dx\\ &=\frac{x \sqrt{2+x^2-x^4}}{28 \left (7+5 x^2\right )^2}-\frac{19 x \sqrt{2+x^2-x^4}}{476 \left (7+5 x^2\right )}-\frac{1}{700} \int \frac{186-190 x^2+25 x^4}{\left (7+5 x^2\right )^2 \sqrt{2+x^2-x^4}} \, dx+\frac{19 \int \frac{118-70 x^2-25 x^4}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{11900}-\frac{2}{25} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx\\ &=\frac{x \sqrt{2+x^2-x^4}}{28 \left (7+5 x^2\right )^2}-\frac{31 x \sqrt{2+x^2-x^4}}{13328 \left (7+5 x^2\right )}-\frac{1}{175} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{\int \frac{37698-32690 x^2-12525 x^4}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{333200}-\frac{19 \int \frac{175+125 x^2}{\sqrt{2+x^2-x^4}} \, dx}{297500}+\frac{3173 \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{11900}\\ &=\frac{x \sqrt{2+x^2-x^4}}{28 \left (7+5 x^2\right )^2}-\frac{31 x \sqrt{2+x^2-x^4}}{13328 \left (7+5 x^2\right )}-\frac{1}{175} \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{\int \frac{75775+62625 x^2}{\sqrt{2+x^2-x^4}} \, dx}{8330000}-\frac{19 \int \frac{175+125 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{148750}-\frac{11783 \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{66640}+\frac{3173 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx}{5950}\\ &=\frac{x \sqrt{2+x^2-x^4}}{28 \left (7+5 x^2\right )^2}-\frac{31 x \sqrt{2+x^2-x^4}}{13328 \left (7+5 x^2\right )}+\frac{2697 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{83300}+\frac{\int \frac{75775+62625 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{4165000}-\frac{19 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{2975}-\frac{19 \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx}{2380}-\frac{11783 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx}{33320}\\ &=\frac{x \sqrt{2+x^2-x^4}}{28 \left (7+5 x^2\right )^2}-\frac{31 x \sqrt{2+x^2-x^4}}{13328 \left (7+5 x^2\right )}-\frac{19 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{2380}-\frac{19 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{5950}+\frac{16601 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{2332400}+\frac{263 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{83300}+\frac{501 \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx}{66640}\\ &=\frac{x \sqrt{2+x^2-x^4}}{28 \left (7+5 x^2\right )^2}-\frac{31 x \sqrt{2+x^2-x^4}}{13328 \left (7+5 x^2\right )}-\frac{31 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{66640}-\frac{269 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{166600}+\frac{16601 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{2332400}\\ \end{align*}

Mathematica [C]  time = 0.35361, size = 244, normalized size = 2.39 \[ \frac{7021 i \sqrt{2} \left (5 x^2+7\right )^2 \sqrt{-x^4+x^2+2} \text{EllipticF}\left (i \sinh ^{-1}(x),-\frac{1}{2}\right )+54250 x^7-144900 x^5-17850 x^3-2170 i \sqrt{2} \left (5 x^2+7\right )^2 \sqrt{-x^4+x^2+2} E\left (i \sinh ^{-1}(x)|-\frac{1}{2}\right )-415025 i \sqrt{2} \sqrt{-x^4+x^2+2} x^4 \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )-1162070 i \sqrt{2} \sqrt{-x^4+x^2+2} x^2 \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )-813449 i \sqrt{2} \sqrt{-x^4+x^2+2} \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )+181300 x}{4664800 \left (5 x^2+7\right )^2 \sqrt{-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(181300*x - 17850*x^3 - 144900*x^5 + 54250*x^7 - (2170*I)*Sqrt[2]*(7 + 5*x^2)^2*Sqrt[2 + x^2 - x^4]*EllipticE[
I*ArcSinh[x], -1/2] + (7021*I)*Sqrt[2]*(7 + 5*x^2)^2*Sqrt[2 + x^2 - x^4]*EllipticF[I*ArcSinh[x], -1/2] - (8134
49*I)*Sqrt[2]*Sqrt[2 + x^2 - x^4]*EllipticPi[5/7, I*ArcSinh[x], -1/2] - (1162070*I)*Sqrt[2]*x^2*Sqrt[2 + x^2 -
x^4]*EllipticPi[5/7, I*ArcSinh[x], -1/2] - (415025*I)*Sqrt[2]*x^4*Sqrt[2 + x^2 - x^4]*EllipticPi[5/7, I*ArcSi
nh[x], -1/2])/(4664800*(7 + 5*x^2)^2*Sqrt[2 + x^2 - x^4])

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Maple [A]  time = 0.021, size = 189, normalized size = 1.9 \begin{align*}{\frac{x}{28\, \left ( 5\,{x}^{2}+7 \right ) ^{2}}\sqrt{-{x}^{4}+{x}^{2}+2}}-{\frac{31\,x}{66640\,{x}^{2}+93296}\sqrt{-{x}^{4}+{x}^{2}+2}}-{\frac{269\,\sqrt{2}}{333200}\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{31\,\sqrt{2}}{133280}\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{16601\,\sqrt{2}}{2332400}\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((-x^4+x^2+2)^(1/2)/(5*x^2+7)^3,x)

[Out]

1/28*x*(-x^4+x^2+2)^(1/2)/(5*x^2+7)^2-31/13328*x*(-x^4+x^2+2)^(1/2)/(5*x^2+7)-269/333200*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))-31/133280*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))+16601/2332400*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{\sqrt{-x^{4} + x^{2} + 2}}{{\left (5 \, x^{2} + 7\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-x^4 + x^2 + 2)/(5*x^2 + 7)^3, 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}}{125 \, x^{6} + 525 \, x^{4} + 735 \, x^{2} + 343}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-x^4 + x^2 + 2)/(125*x^6 + 525*x^4 + 735*x^2 + 343), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (x^{2} - 2\right ) \left (x^{2} + 1\right )}}{\left (5 x^{2} + 7\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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