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

Optimal. Leaf size=65 \[ -542 \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right ),-2\right )-25 \sqrt{-x^4+x^2+2} x^3-\frac{625}{3} \sqrt{-x^4+x^2+2} x+\frac{3905}{3} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

[Out]

(-625*x*Sqrt[2 + x^2 - x^4])/3 - 25*x^3*Sqrt[2 + x^2 - x^4] + (3905*EllipticE[ArcSin[x/Sqrt[2]], -2])/3 - 542*
EllipticF[ArcSin[x/Sqrt[2]], -2]

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Rubi [A]  time = 0.0763707, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1206, 1679, 1180, 524, 424, 419} \[ -25 \sqrt{-x^4+x^2+2} x^3-\frac{625}{3} \sqrt{-x^4+x^2+2} x-542 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{3905}{3} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-625*x*Sqrt[2 + x^2 - x^4])/3 - 25*x^3*Sqrt[2 + x^2 - x^4] + (3905*EllipticE[ArcSin[x/Sqrt[2]], -2])/3 - 542*
EllipticF[ArcSin[x/Sqrt[2]], -2]

Rule 1206

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

Rule 1679

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[(e*x^(2*q - 3)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(2*q + 4*p + 1)), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

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)])

Rubi steps

\begin{align*} \int \frac{\left (7+5 x^2\right )^3}{\sqrt{2+x^2-x^4}} \, dx &=-25 x^3 \sqrt{2+x^2-x^4}-\frac{1}{5} \int \frac{-1715-4425 x^2-3125 x^4}{\sqrt{2+x^2-x^4}} \, dx\\ &=-\frac{625}{3} x \sqrt{2+x^2-x^4}-25 x^3 \sqrt{2+x^2-x^4}+\frac{1}{15} \int \frac{11395+19525 x^2}{\sqrt{2+x^2-x^4}} \, dx\\ &=-\frac{625}{3} x \sqrt{2+x^2-x^4}-25 x^3 \sqrt{2+x^2-x^4}+\frac{2}{15} \int \frac{11395+19525 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx\\ &=-\frac{625}{3} x \sqrt{2+x^2-x^4}-25 x^3 \sqrt{2+x^2-x^4}-1084 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx+\frac{3905}{3} \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx\\ &=-\frac{625}{3} x \sqrt{2+x^2-x^4}-25 x^3 \sqrt{2+x^2-x^4}+\frac{3905}{3} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-542 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )\\ \end{align*}

Mathematica [C]  time = 0.112961, size = 97, normalized size = 1.49 \[ \frac{-10089 i \sqrt{-2 x^4+2 x^2+4} \text{EllipticF}\left (i \sinh ^{-1}(x),-\frac{1}{2}\right )+150 x^7+1100 x^5-1550 x^3+7810 i \sqrt{-2 x^4+2 x^2+4} E\left (i \sinh ^{-1}(x)|-\frac{1}{2}\right )-2500 x}{6 \sqrt{-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2500*x - 1550*x^3 + 1100*x^5 + 150*x^7 + (7810*I)*Sqrt[4 + 2*x^2 - 2*x^4]*EllipticE[I*ArcSinh[x], -1/2] - (1
0089*I)*Sqrt[4 + 2*x^2 - 2*x^4]*EllipticF[I*ArcSinh[x], -1/2])/(6*Sqrt[2 + x^2 - x^4])

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Maple [B]  time = 0.018, size = 142, normalized size = 2.2 \begin{align*} -25\,{x}^{3}\sqrt{-{x}^{4}+{x}^{2}+2}-{\frac{625\,x}{3}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{2279\,\sqrt{2}}{6}\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{3905\,\sqrt{2}}{6}\sqrt{-2\,{x}^{2}+4}\sqrt{{x}^{2}+1} \left ({\it EllipticF} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ) \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25*x^3*(-x^4+x^2+2)^(1/2)-625/3*x*(-x^4+x^2+2)^(1/2)+2279/6*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))-3905/6*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))-EllipticE(1/2*x*2^(1/2),I*2^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (125 \, x^{6} + 525 \, x^{4} + 735 \, x^{2} + 343\right )} \sqrt{-x^{4} + x^{2} + 2}}{x^{4} - x^{2} - 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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