\(\int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx\) [132]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\frac {\sqrt {-16+x^2}}{32 x^2}+\frac {1}{128} \arctan \left (\frac {1}{4} \sqrt {-16+x^2}\right ) \]

[Out]

1/128*arctan(1/4*(x^2-16)^(1/2))+1/32*(x^2-16)^(1/2)/x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {272, 44, 65, 209} \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\frac {1}{128} \arctan \left (\frac {\sqrt {x^2-16}}{4}\right )+\frac {\sqrt {x^2-16}}{32 x^2} \]

[In]

Int[1/(x^3*Sqrt[-16 + x^2]),x]

[Out]

Sqrt[-16 + x^2]/(32*x^2) + ArcTan[Sqrt[-16 + x^2]/4]/128

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-16+x} x^2} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {-16+x^2}}{32 x^2}+\frac {1}{64} \text {Subst}\left (\int \frac {1}{\sqrt {-16+x} x} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {-16+x^2}}{32 x^2}+\frac {1}{32} \text {Subst}\left (\int \frac {1}{16+x^2} \, dx,x,\sqrt {-16+x^2}\right ) \\ & = \frac {\sqrt {-16+x^2}}{32 x^2}+\frac {1}{128} \arctan \left (\frac {1}{4} \sqrt {-16+x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\frac {\sqrt {-16+x^2}}{32 x^2}+\frac {1}{128} \arctan \left (\frac {1}{4} \sqrt {-16+x^2}\right ) \]

[In]

Integrate[1/(x^3*Sqrt[-16 + x^2]),x]

[Out]

Sqrt[-16 + x^2]/(32*x^2) + ArcTan[Sqrt[-16 + x^2]/4]/128

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\sqrt {x^{2}-16}}{32 x^{2}}-\frac {\arctan \left (\frac {4}{\sqrt {x^{2}-16}}\right )}{128}\)	\(26\)
risch	\(\frac {\sqrt {x^{2}-16}}{32 x^{2}}-\frac {\arctan \left (\frac {4}{\sqrt {x^{2}-16}}\right )}{128}\)	\(26\)
pseudoelliptic	\(\frac {\arctan \left (\frac {\sqrt {x^{2}-16}}{4}\right ) x^{2}+4 \sqrt {x^{2}-16}}{128 x^{2}}\)	\(30\)
trager	\(\frac {\sqrt {x^{2}-16}}{32 x^{2}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-\sqrt {x^{2}-16}}{x}\right )}{128}\)	\(46\)
meijerg	\(-\frac {\sqrt {-\operatorname {signum}\left (-1+\frac {x^{2}}{16}\right )}\, \left (\frac {16 \sqrt {\pi }}{x^{2}}-\frac {\left (1-6 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{2}-\frac {2 \sqrt {\pi }\, \left (-\frac {x^{2}}{4}+8\right )}{x^{2}}+\frac {16 \sqrt {\pi }\, \sqrt {1-\frac {x^{2}}{16}}}{x^{2}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {x^{2}}{16}}}{2}\right )\right )}{128 \sqrt {\pi }\, \sqrt {\operatorname {signum}\left (-1+\frac {x^{2}}{16}\right )}}\)	\(106\)

[In]

int(1/x^3/(x^2-16)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/32*(x^2-16)^(1/2)/x^2-1/128*arctan(4/(x^2-16)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\frac {x^{2} \arctan \left (-\frac {1}{4} \, x + \frac {1}{4} \, \sqrt {x^{2} - 16}\right ) + 2 \, \sqrt {x^{2} - 16}}{64 \, x^{2}} \]

[In]

integrate(1/x^3/(x^2-16)^(1/2),x, algorithm="fricas")

[Out]

1/64*(x^2*arctan(-1/4*x + 1/4*sqrt(x^2 - 16)) + 2*sqrt(x^2 - 16))/x^2

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.89 \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\begin {cases} \frac {i \operatorname {acosh}{\left (\frac {4}{x} \right )}}{128} - \frac {i}{32 x \sqrt {-1 + \frac {16}{x^{2}}}} + \frac {i}{2 x^{3} \sqrt {-1 + \frac {16}{x^{2}}}} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > \frac {1}{16} \\- \frac {\operatorname {asin}{\left (\frac {4}{x} \right )}}{128} + \frac {\sqrt {1 - \frac {16}{x^{2}}}}{32 x} & \text {otherwise} \end {cases} \]

[In]

integrate(1/x**3/(x**2-16)**(1/2),x)

[Out]

Piecewise((I*acosh(4/x)/128 - I/(32*x*sqrt(-1 + 16/x**2)) + I/(2*x**3*sqrt(-1 + 16/x**2)), 1/Abs(x**2) > 1/16)
, (-asin(4/x)/128 + sqrt(1 - 16/x**2)/(32*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\frac {\sqrt {x^{2} - 16}}{32 \, x^{2}} - \frac {1}{128} \, \arcsin \left (\frac {4}{{\left | x \right |}}\right ) \]

[In]

integrate(1/x^3/(x^2-16)^(1/2),x, algorithm="maxima")

[Out]

1/32*sqrt(x^2 - 16)/x^2 - 1/128*arcsin(4/abs(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\frac {\sqrt {x^{2} - 16}}{32 \, x^{2}} + \frac {1}{128} \, \arctan \left (\frac {1}{4} \, \sqrt {x^{2} - 16}\right ) \]

[In]

integrate(1/x^3/(x^2-16)^(1/2),x, algorithm="giac")

[Out]

1/32*sqrt(x^2 - 16)/x^2 + 1/128*arctan(1/4*sqrt(x^2 - 16))

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^3 \sqrt {-16+x^2}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {x^2-16}}{4}\right )}{128}+\frac {\sqrt {x^2-16}}{32\,x^2} \]

[In]

int(1/(x^3*(x^2 - 16)^(1/2)),x)

[Out]

atan((x^2 - 16)^(1/2)/4)/128 + (x^2 - 16)^(1/2)/(32*x^2)