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

Optimal. Leaf size=55 \[ \frac {x \left (4897 x^2+4945\right )}{18 \sqrt {-x^4+x^2+2}}+\frac {1763}{6} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {7147}{18} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

[Out]

-7147/18*EllipticE(1/2*x*2^(1/2),I*2^(1/2))+1763/6*EllipticF(1/2*x*2^(1/2),I*2^(1/2))+1/18*x*(4897*x^2+4945)/(
-x^4+x^2+2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1205, 1180, 524, 424, 419} \[ \frac {x \left (4897 x^2+4945\right )}{18 \sqrt {-x^4+x^2+2}}+\frac {1763}{6} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {7147}{18} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(4945 + 4897*x^2))/(18*Sqrt[2 + x^2 - x^4]) - (7147*EllipticE[ArcSin[x/Sqrt[2]], -2])/18 + (1763*EllipticF[
ArcSin[x/Sqrt[2]], -2])/6

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 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 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]

Rubi steps

\begin {align*} \int \frac {\left (7+5 x^2\right )^3}{\left (2+x^2-x^4\right )^{3/2}} \, dx &=\frac {x \left (4945+4897 x^2\right )}{18 \sqrt {2+x^2-x^4}}-\frac {1}{18} \int \frac {1858+7147 x^2}{\sqrt {2+x^2-x^4}} \, dx\\ &=\frac {x \left (4945+4897 x^2\right )}{18 \sqrt {2+x^2-x^4}}-\frac {1}{9} \int \frac {1858+7147 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\\ &=\frac {x \left (4945+4897 x^2\right )}{18 \sqrt {2+x^2-x^4}}-\frac {7147}{18} \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx+\frac {1763}{3} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\\ &=\frac {x \left (4945+4897 x^2\right )}{18 \sqrt {2+x^2-x^4}}-\frac {7147}{18} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {1763}{6} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )\\ \end {align*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (125 \, x^{6} + 525 \, x^{4} + 735 \, x^{2} + 343\right )} \sqrt {-x^{4} + x^{2} + 2}}{x^{8} - 2 \, x^{6} - 3 \, x^{4} + 4 \, x^{2} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((125*x^6 + 525*x^4 + 735*x^2 + 343)*sqrt(-x^4 + x^2 + 2)/(x^8 - 2*x^6 - 3*x^4 + 4*x^2 + 4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{3}}{{\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 202, normalized size = 3.67 \[ -\frac {929 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{18 \sqrt {-x^{4}+x^{2}+2}}+\frac {\frac {625}{9} x^{3}+\frac {250}{9} x}{\sqrt {-x^{4}+x^{2}+2}}+\frac {7147 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )+\EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )\right )}{36 \sqrt {-x^{4}+x^{2}+2}}+\frac {\frac {175}{3} x^{3}+\frac {700}{3} x}{\sqrt {-x^{4}+x^{2}+2}}+\frac {\frac {490}{3} x^{3}-\frac {245}{3} x}{\sqrt {-x^{4}+x^{2}+2}}+\frac {-\frac {343}{18} x^{3}+\frac {1715}{18} x}{\sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

250*(5/18*x^3+1/9*x)/(-x^4+x^2+2)^(1/2)-929/18*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*Ellip
ticF(1/2*2^(1/2)*x,I*2^(1/2))+7147/36*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*(EllipticF(1/2
*2^(1/2)*x,I*2^(1/2))-EllipticE(1/2*2^(1/2)*x,I*2^(1/2)))+1050*(1/18*x^3+2/9*x)/(-x^4+x^2+2)^(1/2)+1470*(1/9*x
^3-1/18*x)/(-x^4+x^2+2)^(1/2)+686*(-1/36*x^3+5/36*x)/(-x^4+x^2+2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{3}}{{\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (5\,x^2+7\right )}^3}{{\left (-x^4+x^2+2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 x^{2} + 7\right )^{3}}{\left (- \left (x^{2} - 2\right ) \left (x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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