\(\int \frac {\arccos (a x)}{x^2} \, dx\) [7]

Optimal result

Integrand size = 8, antiderivative size = 27 \[ \int \frac {\arccos (a x)}{x^2} \, dx=-\frac {\arccos (a x)}{x}+a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

-arccos(a*x)/x+a*arctanh((-a^2*x^2+1)^(1/2))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4724, 272, 65, 214} \[ \int \frac {\arccos (a x)}{x^2} \, dx=a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {\arccos (a x)}{x} \]

[In]

Int[ArcCos[a*x]/x^2,x]

[Out]

-(ArcCos[a*x]/x) + a*ArcTanh[Sqrt[1 - a^2*x^2]]

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 214

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

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]]

Rule 4724

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\arccos (a x)}{x}-a \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx \\ & = -\frac {\arccos (a x)}{x}-\frac {1}{2} a \text {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {\arccos (a x)}{x}+\frac {\text {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )}{a} \\ & = -\frac {\arccos (a x)}{x}+a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\arccos (a x)}{x^2} \, dx=-\frac {\arccos (a x)}{x}-a \log (x)+a \log \left (1+\sqrt {1-a^2 x^2}\right ) \]

[In]

Integrate[ArcCos[a*x]/x^2,x]

[Out]

-(ArcCos[a*x]/x) - a*Log[x] + a*Log[1 + Sqrt[1 - a^2*x^2]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

method	result	size
parts	\(-\frac {\arccos \left (a x \right )}{x}+a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(26\)
derivativedivides	\(a \left (-\frac {\arccos \left (a x \right )}{a x}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )\)	\(29\)
default	\(a \left (-\frac {\arccos \left (a x \right )}{a x}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )\)	\(29\)

[In]

int(arccos(a*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-arccos(a*x)/x+a*arctanh(1/(-a^2*x^2+1)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.04 \[ \int \frac {\arccos (a x)}{x^2} \, dx=\frac {a x \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - a x \log \left (\sqrt {-a^{2} x^{2} + 1} - 1\right ) + 2 \, {\left (x - 1\right )} \arccos \left (a x\right ) - 2 \, x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} a x}{a^{2} x^{2} - 1}\right )}{2 \, x} \]

[In]

integrate(arccos(a*x)/x^2,x, algorithm="fricas")

[Out]

1/2*(a*x*log(sqrt(-a^2*x^2 + 1) + 1) - a*x*log(sqrt(-a^2*x^2 + 1) - 1) + 2*(x - 1)*arccos(a*x) - 2*x*arctan(sq
rt(-a^2*x^2 + 1)*a*x/(a^2*x^2 - 1)))/x

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.98 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\arccos (a x)}{x^2} \, dx=- a \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {\operatorname {acos}{\left (a x \right )}}{x} \]

[In]

integrate(acos(a*x)/x**2,x)

[Out]

-a*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) - acos(a*x)/x

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\arccos (a x)}{x^2} \, dx=a \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {\arccos \left (a x\right )}{x} \]

[In]

integrate(arccos(a*x)/x^2,x, algorithm="maxima")

[Out]

a*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - arccos(a*x)/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {\arccos (a x)}{x^2} \, dx=\frac {1}{2} \, a {\left (\log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right )\right )} - \frac {\arccos \left (a x\right )}{x} \]

[In]

integrate(arccos(a*x)/x^2,x, algorithm="giac")

[Out]

1/2*a*(log(sqrt(-a^2*x^2 + 1) + 1) - log(-sqrt(-a^2*x^2 + 1) + 1)) - arccos(a*x)/x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\arccos (a x)}{x^2} \, dx=a\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-a^2\,x^2}}\right )-\frac {\mathrm {acos}\left (a\,x\right )}{x} \]

[In]

int(acos(a*x)/x^2,x)

[Out]

a*atanh(1/(1 - a^2*x^2)^(1/2)) - acos(a*x)/x