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

Optimal. Leaf size=93 \[ \frac{627857}{6} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right ),-2\right )+625 \sqrt{-x^4+x^2+2} x^3+\frac{27500}{3} \sqrt{-x^4+x^2+2} x+\frac{\left (1419793 x^2+1419985\right ) x}{18 \sqrt{-x^4+x^2+2}}-\frac{3482293}{18} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

[Out]

(x*(1419985 + 1419793*x^2))/(18*Sqrt[2 + x^2 - x^4]) + (27500*x*Sqrt[2 + x^2 - x^4])/3 + 625*x^3*Sqrt[2 + x^2
- x^4] - (3482293*EllipticE[ArcSin[x/Sqrt[2]], -2])/18 + (627857*EllipticF[ArcSin[x/Sqrt[2]], -2])/6

________________________________________________________________________________________

Rubi [A]  time = 0.100598, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1205, 1679, 1180, 524, 424, 419} \[ 625 \sqrt{-x^4+x^2+2} x^3+\frac{27500}{3} \sqrt{-x^4+x^2+2} x+\frac{\left (1419793 x^2+1419985\right ) x}{18 \sqrt{-x^4+x^2+2}}+\frac{627857}{6} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{3482293}{18} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1419985 + 1419793*x^2))/(18*Sqrt[2 + x^2 - x^4]) + (27500*x*Sqrt[2 + x^2 - x^4])/3 + 625*x^3*Sqrt[2 + x^2
- x^4] - (3482293*EllipticE[ArcSin[x/Sqrt[2]], -2])/18 + (627857*EllipticF[ArcSin[x/Sqrt[2]], -2])/6

Rule 1205

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coeff[Polynom
ialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x
^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2))/(
2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToS
um[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c
*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], 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[q, 1] && LtQ[p, -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 )^5}{\left (2+x^2-x^4\right )^{3/2}} \, dx &=\frac{x \left (1419985+1419793 x^2\right )}{18 \sqrt{2+x^2-x^4}}-\frac{1}{18} \int \frac{1268722+3084793 x^2+450000 x^4+56250 x^6}{\sqrt{2+x^2-x^4}} \, dx\\ &=\frac{x \left (1419985+1419793 x^2\right )}{18 \sqrt{2+x^2-x^4}}+625 x^3 \sqrt{2+x^2-x^4}+\frac{1}{90} \int \frac{-6343610-15761465 x^2-2475000 x^4}{\sqrt{2+x^2-x^4}} \, dx\\ &=\frac{x \left (1419985+1419793 x^2\right )}{18 \sqrt{2+x^2-x^4}}+\frac{27500}{3} x \sqrt{2+x^2-x^4}+625 x^3 \sqrt{2+x^2-x^4}-\frac{1}{270} \int \frac{23980830+52234395 x^2}{\sqrt{2+x^2-x^4}} \, dx\\ &=\frac{x \left (1419985+1419793 x^2\right )}{18 \sqrt{2+x^2-x^4}}+\frac{27500}{3} x \sqrt{2+x^2-x^4}+625 x^3 \sqrt{2+x^2-x^4}-\frac{1}{135} \int \frac{23980830+52234395 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx\\ &=\frac{x \left (1419985+1419793 x^2\right )}{18 \sqrt{2+x^2-x^4}}+\frac{27500}{3} x \sqrt{2+x^2-x^4}+625 x^3 \sqrt{2+x^2-x^4}-\frac{3482293}{18} \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx+\frac{627857}{3} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx\\ &=\frac{x \left (1419985+1419793 x^2\right )}{18 \sqrt{2+x^2-x^4}}+\frac{27500}{3} x \sqrt{2+x^2-x^4}+625 x^3 \sqrt{2+x^2-x^4}-\frac{3482293}{18} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{627857}{6} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )\\ \end{align*}

Mathematica [F]  time = 0, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

Maple [B]  time = 0.033, size = 280, normalized size = 3. \begin{align*} 6250\,{\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}} \left ({\frac{17\,{x}^{3}}{18}}+{\frac{7\,x}{9}} \right ) }+625\,{x}^{3}\sqrt{-{x}^{4}+{x}^{2}+2}+{\frac{27500\,x}{3}\sqrt{-{x}^{4}+{x}^{2}+2}}-{\frac{799361\,\sqrt{2}}{18}\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{3482293\,\sqrt{2}}{36}\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}}}}+43750\,{\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}} \left ({\frac{7\,{x}^{3}}{18}}+5/9\,x \right ) }+122500\,{\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}} \left ({\frac{5\,{x}^{3}}{18}}+x/9 \right ) }+171500\,{\frac{1/18\,{x}^{3}+2/9\,x}{\sqrt{-{x}^{4}+{x}^{2}+2}}}+120050\,{\frac{1/9\,{x}^{3}-x/18}{\sqrt{-{x}^{4}+{x}^{2}+2}}}+33614\,{\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}} \left ({\frac{5\,x}{36}}-1/36\,{x}^{3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6250*(17/18*x^3+7/9*x)/(-x^4+x^2+2)^(1/2)+625*x^3*(-x^4+x^2+2)^(1/2)+27500/3*x*(-x^4+x^2+2)^(1/2)-799361/18*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))+3482293/36*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)))+43750*(7/18*x^3+5/9*x)/(-x^4+x^2+2)^(1/2)+122500*(5/18*x^3+1/9*x)/(-x^4+x^2+2)^(1/2)+171500*(1/18*
x^3+2/9*x)/(-x^4+x^2+2)^(1/2)+120050*(1/9*x^3-1/18*x)/(-x^4+x^2+2)^(1/2)+33614*(5/36*x-1/36*x^3)/(-x^4+x^2+2)^
(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (3125 \, x^{10} + 21875 \, x^{8} + 61250 \, x^{6} + 85750 \, x^{4} + 60025 \, x^{2} + 16807\right )} \sqrt{-x^{4} + x^{2} + 2}}{x^{8} - 2 \, x^{6} - 3 \, x^{4} + 4 \, x^{2} + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3125*x^10 + 21875*x^8 + 61250*x^6 + 85750*x^4 + 60025*x^2 + 16807)*sqrt(-x^4 + x^2 + 2)/(x^8 - 2*x^6
- 3*x^4 + 4*x^2 + 4), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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