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

Optimal. Leaf size=142 \[ -\frac {132300}{143} \left (-x^4+x^2+2\right )^{5/2} x-\frac {\left (69817-1581440 x^2\right ) \left (-x^4+x^2+2\right )^{3/2} x}{1001}+\frac {3 \left (7837383 x^2+2193559\right ) \sqrt {-x^4+x^2+2} x}{5005}-\frac {125}{3} \left (-x^4+x^2+2\right )^{5/2} x^5-\frac {11750}{39} \left (-x^4+x^2+2\right )^{5/2} x^3-\frac {50794416 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{5005}+\frac {124141422 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{5005} \]

[Out]

-1/1001*x*(-1581440*x^2+69817)*(-x^4+x^2+2)^(3/2)-132300/143*x*(-x^4+x^2+2)^(5/2)-11750/39*x^3*(-x^4+x^2+2)^(5
/2)-125/3*x^5*(-x^4+x^2+2)^(5/2)[phone]/5005*EllipticE(1/2*x*2^(1/2),I*2^(1/2))-50794416/5005*EllipticF(1/2
*x*2^(1/2),I*2^(1/2))+3/5005*x*(7837383*x^2+2193559)*(-x^4+x^2+2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ -\frac {125}{3} \left (-x^4+x^2+2\right )^{5/2} x^5-\frac {11750}{39} \left (-x^4+x^2+2\right )^{5/2} x^3-\frac {132300}{143} \left (-x^4+x^2+2\right )^{5/2} x-\frac {\left (69817-1581440 x^2\right ) \left (-x^4+x^2+2\right )^{3/2} x}{1001}+\frac {3 \left (7837383 x^2+2193559\right ) \sqrt {-x^4+x^2+2} x}{5005}-\frac {50794416 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{5005}+\frac {124141422 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{5005} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*(2193559 + 7837383*x^2)*Sqrt[2 + x^2 - x^4])/5005 - (x*(69817 - 1581440*x^2)*(2 + x^2 - x^4)^(3/2))/1001
- (132300*x*(2 + x^2 - x^4)^(5/2))/143 - (11750*x^3*(2 + x^2 - x^4)^(5/2))/39 - (125*x^5*(2 + x^2 - x^4)^(5/2)
)/3 + (124141422*EllipticE[ArcSin[x/Sqrt[2]], -2])/5005 - (50794416*EllipticF[ArcSin[x/Sqrt[2]], -2])/5005

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]

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]

Rubi steps

\begin {align*} \int \left (7+5 x^2\right )^4 \left (2+x^2-x^4\right )^{3/2} \, dx &=-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}-\frac {1}{15} \int \left (2+x^2-x^4\right )^{3/2} \left (-36015-102900 x^2-116500 x^4-58750 x^6\right ) \, dx\\ &=-\frac {11750}{39} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}+\frac {1}{195} \int \left (2+x^2-x^4\right )^{3/2} \left (468195+1690200 x^2+1984500 x^4\right ) \, dx\\ &=-\frac {132300}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac {11750}{39} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}-\frac {\int \left (-9119145-30499200 x^2\right ) \left (2+x^2-x^4\right )^{3/2} \, dx}{2145}\\ &=-\frac {x \left (69817-1581440 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{1001}-\frac {132300}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac {11750}{39} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}+\frac {\int \left (389287620+1058046705 x^2\right ) \sqrt {2+x^2-x^4} \, dx}{45045}\\ &=\frac {3 x \left (2193559+7837383 x^2\right ) \sqrt {2+x^2-x^4}}{5005}-\frac {x \left (69817-1581440 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{1001}-\frac {132300}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac {11750}{39} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}-\frac {\int \frac {-9901845810-16759091970 x^2}{\sqrt {2+x^2-x^4}} \, dx}{675675}\\ &=\frac {3 x \left (2193559+7837383 x^2\right ) \sqrt {2+x^2-x^4}}{5005}-\frac {x \left (69817-1581440 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{1001}-\frac {132300}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac {11750}{39} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}-\frac {2 \int \frac {-9901845810-16759091970 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{675675}\\ &=\frac {3 x \left (2193559+7837383 x^2\right ) \sqrt {2+x^2-x^4}}{5005}-\frac {x \left (69817-1581440 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{1001}-\frac {132300}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac {11750}{39} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}-\frac {101588832 \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{5005}+\frac {124141422 \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx}{5005}\\ &=\frac {3 x \left (2193559+7837383 x^2\right ) \sqrt {2+x^2-x^4}}{5005}-\frac {x \left (69817-1581440 x^2\right ) \left (2+x^2-x^4\right )^{3/2}}{1001}-\frac {132300}{143} x \left (2+x^2-x^4\right )^{5/2}-\frac {11750}{39} x^3 \left (2+x^2-x^4\right )^{5/2}-\frac {125}{3} x^5 \left (2+x^2-x^4\right )^{5/2}+\frac {124141422 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{5005}-\frac {50794416 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{5005}\\ \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)^4*(2 + x^2 - x^4)^(3/2),x]

[Out]

$Aborted

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (625 \, x^{12} + 2875 \, x^{10} + 2600 \, x^{8} - 7490 \, x^{6} - 19159 \, x^{4} - 16121 \, x^{2} - 4802\right )} \sqrt {-x^{4} + x^{2} + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(625*x^12 + 2875*x^10 + 2600*x^8 - 7490*x^6 - 19159*x^4 - 16121*x^2 - 4802)*sqrt(-x^4 + x^2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 227, normalized size = 1.60 \[ -\frac {125 \sqrt {-x^{4}+x^{2}+2}\, x^{13}}{3}-\frac {8500 \sqrt {-x^{4}+x^{2}+2}\, x^{11}}{39}-\frac {84775 \sqrt {-x^{4}+x^{2}+2}\, x^{9}}{429}+\frac {432290 \sqrt {-x^{4}+x^{2}+2}\, x^{7}}{429}+\frac {833561 \sqrt {-x^{4}+x^{2}+2}\, x^{5}}{273}+\frac {43271392 \sqrt {-x^{4}+x^{2}+2}\, x^{3}}{15015}-\frac {12639493 \sqrt {-x^{4}+x^{2}+2}\, x}{5005}+\frac {36673503 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{5005 \sqrt {-x^{4}+x^{2}+2}}-\frac {62070711 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )+\EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )\right )}{5005 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

43271392/15015*(-x^4+x^2+2)^(1/2)*x^3-12639493/5005*(-x^4+x^2+2)^(1/2)*x+833561/273*(-x^4+x^2+2)^(1/2)*x^5+432
290/429*(-x^4+x^2+2)^(1/2)*x^7-84775/429*(-x^4+x^2+2)^(1/2)*x^9-8500/39*x^11*(-x^4+x^2+2)^(1/2)-125/3*x^13*(-x
^4+x^2+2)^(1/2)-62070711/5005*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)))[phone]/5005*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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- \left (x^{2} - 2\right ) \left (x^{2} + 1\right )\right )^{\frac {3}{2}} \left (5 x^{2} + 7\right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-(x**2 - 2)*(x**2 + 1))**(3/2)*(5*x**2 + 7)**4, x)

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