\(\int \frac {1}{x \sqrt {16-x^4}} \, dx\) [964]

Optimal result

Integrand size = 15, antiderivative size = 20 \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=-\frac {1}{8} \text {arctanh}\left (\frac {\sqrt {16-x^4}}{4}\right ) \]

[Out]

-1/8*arctanh(1/4*(-x^4+16)^(1/2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {272, 65, 212} \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=-\frac {1}{8} \text {arctanh}\left (\frac {\sqrt {16-x^4}}{4}\right ) \]

[In]

Int[1/(x*Sqrt[16 - x^4]),x]

[Out]

-1/8*ArcTanh[Sqrt[16 - x^4]/4]

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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}{4} \text {Subst}\left (\int \frac {1}{\sqrt {16-x} x} \, dx,x,x^4\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{16-x^2} \, dx,x,\sqrt {16-x^4}\right )\right ) \\ & = -\frac {1}{8} \tanh ^{-1}\left (\frac {\sqrt {16-x^4}}{4}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=-\frac {1}{8} \text {arctanh}\left (\frac {\sqrt {16-x^4}}{4}\right ) \]

[In]

Integrate[1/(x*Sqrt[16 - x^4]),x]

[Out]

-1/8*ArcTanh[Sqrt[16 - x^4]/4]

Maple [A] (verified)

Time = 4.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
default	\(-\frac {\operatorname {arctanh}\left (\frac {4}{\sqrt {-x^{4}+16}}\right )}{8}\)	\(15\)
elliptic	\(-\frac {\operatorname {arctanh}\left (\frac {4}{\sqrt {-x^{4}+16}}\right )}{8}\)	\(15\)
pseudoelliptic	\(-\frac {\operatorname {arctanh}\left (\frac {4}{\sqrt {-x^{4}+16}}\right )}{8}\)	\(15\)
trager	\(\frac {\ln \left (\frac {\sqrt {-x^{4}+16}-4}{x^{2}}\right )}{8}\)	\(19\)
meijerg	\(\frac {\left (-6 \ln \left (2\right )+4 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {x^{4}}{16}}}{2}\right )}{16 \sqrt {\pi }}\)	\(43\)

[In]

int(1/x/(-x^4+16)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*arctanh(4/(-x^4+16)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (14) = 28\).

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=-\frac {1}{16} \, \log \left (\sqrt {-x^{4} + 16} + 4\right ) + \frac {1}{16} \, \log \left (\sqrt {-x^{4} + 16} - 4\right ) \]

[In]

integrate(1/x/(-x^4+16)^(1/2),x, algorithm="fricas")

[Out]

-1/16*log(sqrt(-x^4 + 16) + 4) + 1/16*log(sqrt(-x^4 + 16) - 4)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=\begin {cases} - \frac {\operatorname {acosh}{\left (\frac {4}{x^{2}} \right )}}{8} & \text {for}\: \frac {1}{\left |{x^{4}}\right |} > \frac {1}{16} \\\frac {i \operatorname {asin}{\left (\frac {4}{x^{2}} \right )}}{8} & \text {otherwise} \end {cases} \]

[In]

integrate(1/x/(-x**4+16)**(1/2),x)

[Out]

Piecewise((-acosh(4/x**2)/8, 1/Abs(x**4) > 1/16), (I*asin(4/x**2)/8, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (14) = 28\).

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=-\frac {1}{16} \, \log \left (\sqrt {-x^{4} + 16} + 4\right ) + \frac {1}{16} \, \log \left (\sqrt {-x^{4} + 16} - 4\right ) \]

[In]

integrate(1/x/(-x^4+16)^(1/2),x, algorithm="maxima")

[Out]

-1/16*log(sqrt(-x^4 + 16) + 4) + 1/16*log(sqrt(-x^4 + 16) - 4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=-\frac {1}{16} \, \log \left (\sqrt {-x^{4} + 16} + 4\right ) + \frac {1}{16} \, \log \left (-\sqrt {-x^{4} + 16} + 4\right ) \]

[In]

integrate(1/x/(-x^4+16)^(1/2),x, algorithm="giac")

[Out]

-1/16*log(sqrt(-x^4 + 16) + 4) + 1/16*log(-sqrt(-x^4 + 16) + 4)

Mupad [B] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x \sqrt {16-x^4}} \, dx=-\frac {\mathrm {atanh}\left (\frac {\sqrt {16-x^4}}{4}\right )}{8} \]

[In]

int(1/(x*(16 - x^4)^(1/2)),x)

[Out]

-atanh((16 - x^4)^(1/2)/4)/8