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

Optimal. Leaf size=121 \[ -\frac{3199778 \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right ),-2\right )}{5005}-\frac{125}{13} \left (-x^4+x^2+2\right )^{5/2} x^3-\frac{7825}{143} \left (-x^4+x^2+2\right )^{5/2} x+\frac{\left (374045 x^2+33792\right ) \left (-x^4+x^2+2\right )^{3/2} x}{3003}+\frac{\left (5712051 x^2+2512273\right ) \sqrt{-x^4+x^2+2} x}{15015}+\frac{31072528 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{15015} \]

[Out]

(x*(2512273 + 5712051*x^2)*Sqrt[2 + x^2 - x^4])/15015 + (x*(33792 + 374045*x^2)*(2 + x^2 - x^4)^(3/2))/3003 -
(7825*x*(2 + x^2 - x^4)^(5/2))/143 - (125*x^3*(2 + x^2 - x^4)^(5/2))/13 + (31072528*EllipticE[ArcSin[x/Sqrt[2]
], -2])/15015 - (3199778*EllipticF[ArcSin[x/Sqrt[2]], -2])/5005

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Rubi [A]  time = 0.101206, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {1206, 1679, 1176, 1180, 524, 424, 419} \[ -\frac{125}{13} \left (-x^4+x^2+2\right )^{5/2} x^3-\frac{7825}{143} \left (-x^4+x^2+2\right )^{5/2} x+\frac{\left (374045 x^2+33792\right ) \left (-x^4+x^2+2\right )^{3/2} x}{3003}+\frac{\left (5712051 x^2+2512273\right ) \sqrt{-x^4+x^2+2} x}{15015}-\frac{3199778 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{5005}+\frac{31072528 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{15015} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2512273 + 5712051*x^2)*Sqrt[2 + x^2 - x^4])/15015 + (x*(33792 + 374045*x^2)*(2 + x^2 - x^4)^(3/2))/3003 -
(7825*x*(2 + x^2 - x^4)^(5/2))/143 - (125*x^3*(2 + x^2 - x^4)^(5/2))/13 + (31072528*EllipticE[ArcSin[x/Sqrt[2]
], -2])/15015 - (3199778*EllipticF[ArcSin[x/Sqrt[2]], -2])/5005

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 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 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 \left (7+5 x^2\right )^3 \left (2+x^2-x^4\right )^{3/2} \, dx &=-\frac{125}{13} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac{1}{13} \int \left (-4459-10305 x^2-7825 x^4\right ) \left (2+x^2-x^4\right )^{3/2} \, dx\\ &=-\frac{7825}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac{125}{13} x^3 \left (2+x^2-x^4\right )^{5/2}+\frac{1}{143} \int \left (64699+160305 x^2\right ) \left (2+x^2-x^4\right )^{3/2} \, dx\\ &=\frac{x \left (33792+374045 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{3003}-\frac{7825}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac{125}{13} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac{\int \left (-2649774-5712051 x^2\right ) \sqrt{2+x^2-x^4} \, dx}{3003}\\ &=\frac{x \left (2512273+5712051 x^2\right ) \sqrt{2+x^2-x^4}}{15015}+\frac{x \left (33792+374045 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{3003}-\frac{7825}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac{125}{13} x^3 \left (2+x^2-x^4\right )^{5/2}+\frac{\int \frac{64419582+93217584 x^2}{\sqrt{2+x^2-x^4}} \, dx}{45045}\\ &=\frac{x \left (2512273+5712051 x^2\right ) \sqrt{2+x^2-x^4}}{15015}+\frac{x \left (33792+374045 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{3003}-\frac{7825}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac{125}{13} x^3 \left (2+x^2-x^4\right )^{5/2}+\frac{2 \int \frac{64419582+93217584 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{45045}\\ &=\frac{x \left (2512273+5712051 x^2\right ) \sqrt{2+x^2-x^4}}{15015}+\frac{x \left (33792+374045 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{3003}-\frac{7825}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac{125}{13} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac{6399556 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{5005}+\frac{31072528 \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx}{15015}\\ &=\frac{x \left (2512273+5712051 x^2\right ) \sqrt{2+x^2-x^4}}{15015}+\frac{x \left (33792+374045 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{3003}-\frac{7825}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac{125}{13} x^3 \left (2+x^2-x^4\right )^{5/2}+\frac{31072528 E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{15015}-\frac{3199778 F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{5005}\\ \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)^3*(2 + x^2 - x^4)^(3/2),x]

[Out]

$Aborted

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Maple [A]  time = 0.007, size = 210, normalized size = 1.7 \begin{align*} -{\frac{436307\,x}{15015}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{65248\,{x}^{5}}{273}\sqrt{-{x}^{4}+{x}^{2}+2}}-{\frac{125\,{x}^{11}}{13}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{5757461\,{x}^{3}}{15015}\sqrt{-{x}^{4}+{x}^{2}+2}}-{\frac{5075\,{x}^{9}}{143}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{5890\,{x}^{7}}{429}\sqrt{-{x}^{4}+{x}^{2}+2}}-{\frac{15536264\,\sqrt{2}}{15015}\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}}}}+{\frac{10736597\,\sqrt{2}}{15015}\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}}}} \end{align*}

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]

-436307/15015*x*(-x^4+x^2+2)^(1/2)+65248/273*x^5*(-x^4+x^2+2)^(1/2)-125/13*x^11*(-x^4+x^2+2)^(1/2)+5757461/150
15*x^3*(-x^4+x^2+2)^(1/2)-5075/143*x^9*(-x^4+x^2+2)^(1/2)+5890/429*x^7*(-x^4+x^2+2)^(1/2)-15536264/15015*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)))[phone]/15015*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))

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (125 \, x^{10} + 400 \, x^{8} - 40 \, x^{6} - 1442 \, x^{4} - 1813 \, x^{2} - 686\right )} \sqrt{-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)^(3/2),x, algorithm="fricas")

[Out]

integral(-(125*x^10 + 400*x^8 - 40*x^6 - 1442*x^4 - 1813*x^2 - 686)*sqrt(-x^4 + x^2 + 2), x)

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

[Out]

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

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

[Out]

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