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

Optimal. Leaf size=46 \[ \frac {17}{25} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {1}{5} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {34}{175} \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

[Out]

-1/5*EllipticE(1/2*x*2^(1/2),I*2^(1/2))+17/25*EllipticF(1/2*x*2^(1/2),I*2^(1/2))-34/175*EllipticPi(1/2*x*2^(1/
2),-10/7,I*2^(1/2))

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Rubi [A]  time = 0.08, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1208, 1180, 524, 424, 419, 1212, 537} \[ \frac {17}{25} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {1}{5} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {34}{175} \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-EllipticE[ArcSin[x/Sqrt[2]], -2]/5 + (17*EllipticF[ArcSin[x/Sqrt[2]], -2])/25 - (34*EllipticPi[-10/7, ArcSin[
x/Sqrt[2]], -2])/175

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 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 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 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)
])

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 1208

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d -
b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/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] && IGtQ[p + 1/2, 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {2+x^2-x^4}}{7+5 x^2} \, dx &=-\left (\frac {1}{25} \int \frac {-12+5 x^2}{\sqrt {2+x^2-x^4}} \, dx\right )-\frac {34}{25} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx\\ &=-\left (\frac {2}{25} \int \frac {-12+5 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\right )-\frac {68}{25} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2} \left (7+5 x^2\right )} \, dx\\ &=-\frac {34}{175} \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {1}{5} \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx+\frac {34}{25} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\\ &=-\frac {1}{5} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {17}{25} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {34}{175} \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 51, normalized size = 1.11 \[ -\frac {1}{175} i \sqrt {2} \left (7 F\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+35 E\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )-17 \Pi \left (\frac {5}{7};i \sinh ^{-1}(x)|-\frac {1}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/175*I)*Sqrt[2]*(35*EllipticE[I*ArcSinh[x], -1/2] + 7*EllipticF[I*ArcSinh[x], -1/2] - 17*EllipticPi[5/7, I*
ArcSinh[x], -1/2])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-x^{4} + x^{2} + 2}}{5 \, x^{2} + 7}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-x^{4} + x^{2} + 2}}{5 \, x^{2} + 7}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 141, normalized size = 3.07 \[ -\frac {\sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{10 \sqrt {-x^{4}+x^{2}+2}}+\frac {17 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{50 \sqrt {-x^{4}+x^{2}+2}}-\frac {34 \sqrt {2}\, \sqrt {-\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {\sqrt {2}\, x}{2}, -\frac {10}{7}, i \sqrt {2}\right )}{175 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/50*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticF(1/2*2^(1/2)*x,I*2^(1/2))-1/10*2^(1/2
)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticE(1/2*2^(1/2)*x,I*2^(1/2))-34/175*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*2^(1/2)*x,-10/7,I*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-x^{4} + x^{2} + 2}}{5 \, x^{2} + 7}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {-x^4+x^2+2}}{5\,x^2+7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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