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

Optimal. Leaf size=74 \[ -\frac {25}{7} x \left (-x^4+x^2+2\right )^{3/2}+\frac {1}{21} x \left (354 x^2+275\right ) \sqrt {-x^4+x^2+2}-\frac {79}{7} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {2045}{21} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

[Out]

-25/7*x*(-x^4+x^2+2)^(3/2)+2045/21*EllipticE(1/2*x*2^(1/2),I*2^(1/2))-79/7*EllipticF(1/2*x*2^(1/2),I*2^(1/2))+
1/21*x*(354*x^2+275)*(-x^4+x^2+2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1206, 1176, 1180, 524, 424, 419} \[ -\frac {25}{7} x \left (-x^4+x^2+2\right )^{3/2}+\frac {1}{21} x \left (354 x^2+275\right ) \sqrt {-x^4+x^2+2}-\frac {79}{7} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {2045}{21} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(275 + 354*x^2)*Sqrt[2 + x^2 - x^4])/21 - (25*x*(2 + x^2 - x^4)^(3/2))/7 + (2045*EllipticE[ArcSin[x/Sqrt[2]
], -2])/21 - (79*EllipticF[ArcSin[x/Sqrt[2]], -2])/7

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 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), 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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

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 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]

Rubi steps

\begin {align*} \int \left (7+5 x^2\right )^2 \sqrt {2+x^2-x^4} \, dx &=-\frac {25}{7} x \left (2+x^2-x^4\right )^{3/2}-\frac {1}{7} \int \left (-393-590 x^2\right ) \sqrt {2+x^2-x^4} \, dx\\ &=\frac {1}{21} x \left (275+354 x^2\right ) \sqrt {2+x^2-x^4}-\frac {25}{7} x \left (2+x^2-x^4\right )^{3/2}+\frac {1}{105} \int \frac {9040+10225 x^2}{\sqrt {2+x^2-x^4}} \, dx\\ &=\frac {1}{21} x \left (275+354 x^2\right ) \sqrt {2+x^2-x^4}-\frac {25}{7} x \left (2+x^2-x^4\right )^{3/2}+\frac {2}{105} \int \frac {9040+10225 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\\ &=\frac {1}{21} x \left (275+354 x^2\right ) \sqrt {2+x^2-x^4}-\frac {25}{7} x \left (2+x^2-x^4\right )^{3/2}-\frac {158}{7} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx+\frac {2045}{21} \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx\\ &=\frac {1}{21} x \left (275+354 x^2\right ) \sqrt {2+x^2-x^4}-\frac {25}{7} x \left (2+x^2-x^4\right )^{3/2}+\frac {2045}{21} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {79}{7} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 102, normalized size = 1.38 \[ \frac {-75 x^9-204 x^7+304 x^5+683 x^3-2949 i \sqrt {-2 x^4+2 x^2+4} F\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+2045 i \sqrt {-2 x^4+2 x^2+4} E\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+250 x}{21 \sqrt {-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(250*x + 683*x^3 + 304*x^5 - 204*x^7 - 75*x^9 + (2045*I)*Sqrt[4 + 2*x^2 - 2*x^4]*EllipticE[I*ArcSinh[x], -1/2]
- (2949*I)*Sqrt[4 + 2*x^2 - 2*x^4]*EllipticF[I*ArcSinh[x], -1/2])/(21*Sqrt[2 + x^2 - x^4])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (25 \, x^{4} + 70 \, x^{2} + 49\right )} \sqrt {-x^{4} + x^{2} + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((25*x^4 + 70*x^2 + 49)*sqrt(-x^4 + x^2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 159, normalized size = 2.15 \[ \frac {25 \sqrt {-x^{4}+x^{2}+2}\, x^{5}}{7}+\frac {93 \sqrt {-x^{4}+x^{2}+2}\, x^{3}}{7}+\frac {125 \sqrt {-x^{4}+x^{2}+2}\, x}{21}+\frac {904 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{21 \sqrt {-x^{4}+x^{2}+2}}-\frac {2045 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )+\EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )\right )}{42 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/7*(-x^4+x^2+2)^(1/2)*x^5+93/7*(-x^4+x^2+2)^(1/2)*x^3+125/21*(-x^4+x^2+2)^(1/2)*x+904/21*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))-2045/42*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)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (5\,x^2+7\right )}^2\,\sqrt {-x^4+x^2+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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