\(\int (a^2-x^2)^{5/2} \, dx\) [39]

Optimal result

Integrand size = 13, antiderivative size = 84 \[ \int \left (a^2-x^2\right )^{5/2} \, dx=\frac {5}{16} a^4 x \sqrt {a^2-x^2}+\frac {5}{24} a^2 x \left (a^2-x^2\right )^{3/2}+\frac {1}{6} x \left (a^2-x^2\right )^{5/2}+\frac {5}{16} a^6 \arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right ) \]

[Out]

5/24*a^2*x*(a^2-x^2)^(3/2)+1/6*x*(a^2-x^2)^(5/2)+5/16*a^6*arctan(x/(a^2-x^2)^(1/2))+5/16*a^4*x*(a^2-x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {201, 223, 209} \[ \int \left (a^2-x^2\right )^{5/2} \, dx=\frac {5}{24} a^2 x \left (a^2-x^2\right )^{3/2}+\frac {1}{6} x \left (a^2-x^2\right )^{5/2}+\frac {5}{16} a^6 \arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right )+\frac {5}{16} a^4 x \sqrt {a^2-x^2} \]

[In]

Int[(a^2 - x^2)^(5/2),x]

[Out]

(5*a^4*x*Sqrt[a^2 - x^2])/16 + (5*a^2*x*(a^2 - x^2)^(3/2))/24 + (x*(a^2 - x^2)^(5/2))/6 + (5*a^6*ArcTan[x/Sqrt
[a^2 - x^2]])/16

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} x \left (a^2-x^2\right )^{5/2}+\frac {1}{6} \left (5 a^2\right ) \int \left (a^2-x^2\right )^{3/2} \, dx \\ & = \frac {5}{24} a^2 x \left (a^2-x^2\right )^{3/2}+\frac {1}{6} x \left (a^2-x^2\right )^{5/2}+\frac {1}{8} \left (5 a^4\right ) \int \sqrt {a^2-x^2} \, dx \\ & = \frac {5}{16} a^4 x \sqrt {a^2-x^2}+\frac {5}{24} a^2 x \left (a^2-x^2\right )^{3/2}+\frac {1}{6} x \left (a^2-x^2\right )^{5/2}+\frac {1}{16} \left (5 a^6\right ) \int \frac {1}{\sqrt {a^2-x^2}} \, dx \\ & = \frac {5}{16} a^4 x \sqrt {a^2-x^2}+\frac {5}{24} a^2 x \left (a^2-x^2\right )^{3/2}+\frac {1}{6} x \left (a^2-x^2\right )^{5/2}+\frac {1}{16} \left (5 a^6\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {a^2-x^2}}\right ) \\ & = \frac {5}{16} a^4 x \sqrt {a^2-x^2}+\frac {5}{24} a^2 x \left (a^2-x^2\right )^{3/2}+\frac {1}{6} x \left (a^2-x^2\right )^{5/2}+\frac {5}{16} a^6 \arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \left (a^2-x^2\right )^{5/2} \, dx=\frac {1}{48} \sqrt {a^2-x^2} \left (33 a^4 x-26 a^2 x^3+8 x^5\right )+\frac {5}{16} a^6 \arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right ) \]

[In]

Integrate[(a^2 - x^2)^(5/2),x]

[Out]

(Sqrt[a^2 - x^2]*(33*a^4*x - 26*a^2*x^3 + 8*x^5))/48 + (5*a^6*ArcTan[x/Sqrt[a^2 - x^2]])/16

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {x \left (33 a^{4}-26 a^{2} x^{2}+8 x^{4}\right ) \sqrt {a^{2}-x^{2}}}{48}+\frac {5 a^{6} \arctan \left (\frac {x}{\sqrt {a^{2}-x^{2}}}\right )}{16}\)	\(54\)
pseudoelliptic	\(-\frac {5 \arctan \left (\frac {\sqrt {a^{2}-x^{2}}}{x}\right ) a^{6}}{16}+\frac {11 \sqrt {a^{2}-x^{2}}\, \left (a^{4}-\frac {26}{33} a^{2} x^{2}+\frac {8}{33} x^{4}\right ) x}{16}\)	\(54\)
default	\(\frac {x \left (a^{2}-x^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 a^{2} \left (\frac {\left (a^{2}-x^{2}\right )^{\frac {3}{2}} x}{4}+\frac {3 a^{2} \left (\frac {x \sqrt {a^{2}-x^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {x}{\sqrt {a^{2}-x^{2}}}\right )}{2}\right )}{4}\right )}{6}\)	\(75\)

[In]

int((a^2-x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/48*x*(33*a^4-26*a^2*x^2+8*x^4)*(a^2-x^2)^(1/2)+5/16*a^6*arctan(x/(a^2-x^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \left (a^2-x^2\right )^{5/2} \, dx=-\frac {5}{8} \, a^{6} \arctan \left (-\frac {a - \sqrt {a^{2} - x^{2}}}{x}\right ) + \frac {1}{48} \, {\left (33 \, a^{4} x - 26 \, a^{2} x^{3} + 8 \, x^{5}\right )} \sqrt {a^{2} - x^{2}} \]

[In]

integrate((a^2-x^2)^(5/2),x, algorithm="fricas")

[Out]

-5/8*a^6*arctan(-(a - sqrt(a^2 - x^2))/x) + 1/48*(33*a^4*x - 26*a^2*x^3 + 8*x^5)*sqrt(a^2 - x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.40 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.14 \[ \int \left (a^2-x^2\right )^{5/2} \, dx=\begin {cases} - \frac {5 i a^{6} \operatorname {acosh}{\left (\frac {x}{a} \right )}}{16} - \frac {11 i a^{5} x}{16 \sqrt {-1 + \frac {x^{2}}{a^{2}}}} + \frac {59 i a^{3} x^{3}}{48 \sqrt {-1 + \frac {x^{2}}{a^{2}}}} - \frac {17 i a x^{5}}{24 \sqrt {-1 + \frac {x^{2}}{a^{2}}}} + \frac {i x^{7}}{6 a \sqrt {-1 + \frac {x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {x^{2}}{a^{2}}}\right | > 1 \\\frac {5 a^{6} \operatorname {asin}{\left (\frac {x}{a} \right )}}{16} + \frac {11 a^{5} x \sqrt {1 - \frac {x^{2}}{a^{2}}}}{16} - \frac {13 a^{3} x^{3} \sqrt {1 - \frac {x^{2}}{a^{2}}}}{24} + \frac {a x^{5} \sqrt {1 - \frac {x^{2}}{a^{2}}}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate((a**2-x**2)**(5/2),x)

[Out]

Piecewise((-5*I*a**6*acosh(x/a)/16 - 11*I*a**5*x/(16*sqrt(-1 + x**2/a**2)) + 59*I*a**3*x**3/(48*sqrt(-1 + x**2
/a**2)) - 17*I*a*x**5/(24*sqrt(-1 + x**2/a**2)) + I*x**7/(6*a*sqrt(-1 + x**2/a**2)), Abs(x**2/a**2) > 1), (5*a
**6*asin(x/a)/16 + 11*a**5*x*sqrt(1 - x**2/a**2)/16 - 13*a**3*x**3*sqrt(1 - x**2/a**2)/24 + a*x**5*sqrt(1 - x*
*2/a**2)/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \left (a^2-x^2\right )^{5/2} \, dx=\frac {5}{16} \, a^{6} \arcsin \left (\frac {x}{a}\right ) + \frac {5}{16} \, \sqrt {a^{2} - x^{2}} a^{4} x + \frac {5}{24} \, {\left (a^{2} - x^{2}\right )}^{\frac {3}{2}} a^{2} x + \frac {1}{6} \, {\left (a^{2} - x^{2}\right )}^{\frac {5}{2}} x \]

[In]

integrate((a^2-x^2)^(5/2),x, algorithm="maxima")

[Out]

5/16*a^6*arcsin(x/a) + 5/16*sqrt(a^2 - x^2)*a^4*x + 5/24*(a^2 - x^2)^(3/2)*a^2*x + 1/6*(a^2 - x^2)^(5/2)*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.60 \[ \int \left (a^2-x^2\right )^{5/2} \, dx=\frac {5}{16} \, a^{6} \arcsin \left (\frac {x}{a}\right ) \mathrm {sgn}\left (a\right ) + \frac {1}{48} \, {\left (33 \, a^{4} - 2 \, {\left (13 \, a^{2} - 4 \, x^{2}\right )} x^{2}\right )} \sqrt {a^{2} - x^{2}} x \]

[In]

integrate((a^2-x^2)^(5/2),x, algorithm="giac")

[Out]

5/16*a^6*arcsin(x/a)*sgn(a) + 1/48*(33*a^4 - 2*(13*a^2 - 4*x^2)*x^2)*sqrt(a^2 - x^2)*x

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \left (a^2-x^2\right )^{5/2} \, dx=\frac {x\,{\left (a^2-x^2\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ \frac {x^2}{a^2}\right )}{{\left (1-\frac {x^2}{a^2}\right )}^{5/2}} \]

[In]

int((a^2 - x^2)^(5/2),x)

[Out]

(x*(a^2 - x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, x^2/a^2))/(1 - x^2/a^2)^(5/2)