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

Optimal. Leaf size=81 \[ \frac {1}{63} x \left (35 x^2+48\right ) \left (-x^4+x^2+2\right )^{3/2}+\frac {1}{315} x \left (669 x^2+1087\right ) \sqrt {-x^4+x^2+2}+\frac {418}{105} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {4432}{315} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

[Out]

1/63*x*(35*x^2+48)*(-x^4+x^2+2)^(3/2)+4432/315*EllipticE(1/2*x*2^(1/2),I*2^(1/2))+418/105*EllipticF(1/2*x*2^(1
/2),I*2^(1/2))+1/315*x*(669*x^2+1087)*(-x^4+x^2+2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1176, 1180, 524, 424, 419} \[ \frac {1}{63} x \left (35 x^2+48\right ) \left (-x^4+x^2+2\right )^{3/2}+\frac {1}{315} x \left (669 x^2+1087\right ) \sqrt {-x^4+x^2+2}+\frac {418}{105} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {4432}{315} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1087 + 669*x^2)*Sqrt[2 + x^2 - x^4])/315 + (x*(48 + 35*x^2)*(2 + x^2 - x^4)^(3/2))/63 + (4432*EllipticE[Ar
cSin[x/Sqrt[2]], -2])/315 + (418*EllipticF[ArcSin[x/Sqrt[2]], -2])/105

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]

Rubi steps

\begin {align*} \int \left (7+5 x^2\right ) \left (2+x^2-x^4\right )^{3/2} \, dx &=\frac {1}{63} x \left (48+35 x^2\right ) \left (2+x^2-x^4\right )^{3/2}-\frac {1}{21} \int \left (-262-223 x^2\right ) \sqrt {2+x^2-x^4} \, dx\\ &=\frac {1}{315} x \left (1087+669 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{63} x \left (48+35 x^2\right ) \left (2+x^2-x^4\right )^{3/2}+\frac {1}{315} \int \frac {5686+4432 x^2}{\sqrt {2+x^2-x^4}} \, dx\\ &=\frac {1}{315} x \left (1087+669 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{63} x \left (48+35 x^2\right ) \left (2+x^2-x^4\right )^{3/2}+\frac {2}{315} \int \frac {5686+4432 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\\ &=\frac {1}{315} x \left (1087+669 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{63} x \left (48+35 x^2\right ) \left (2+x^2-x^4\right )^{3/2}+\frac {836}{105} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx+\frac {4432}{315} \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx\\ &=\frac {1}{315} x \left (1087+669 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{63} x \left (48+35 x^2\right ) \left (2+x^2-x^4\right )^{3/2}+\frac {4432}{315} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {418}{105} 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)*(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 \[ {\rm integral}\left (-{\left (5 \, x^{6} + 2 \, x^{4} - 17 \, x^{2} - 14\right )} \sqrt {-x^{4} + x^{2} + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(5*x^6 + 2*x^4 - 17*x^2 - 14)*sqrt(-x^4 + x^2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 176, normalized size = 2.17 \[ -\frac {5 \sqrt {-x^{4}+x^{2}+2}\, x^{7}}{9}-\frac {13 \sqrt {-x^{4}+x^{2}+2}\, x^{5}}{63}+\frac {1259 \sqrt {-x^{4}+x^{2}+2}\, x^{3}}{315}+\frac {1567 \sqrt {-x^{4}+x^{2}+2}\, x}{315}+\frac {2843 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{315 \sqrt {-x^{4}+x^{2}+2}}-\frac {2216 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )+\EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )\right )}{315 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/9*(-x^4+x^2+2)^(1/2)*x^7-13/63*(-x^4+x^2+2)^(1/2)*x^5+1259/315*(-x^4+x^2+2)^(1/2)*x^3+1567/315*(-x^4+x^2+2)
^(1/2)*x+2843/315*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))
-2216/315*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))-Ellipt
icE(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 \[ \int {\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}} {\left (5 \, x^{2} + 7\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (5\,x^2+7\right )\,{\left (-x^4+x^2+2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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