\(\int \frac {\coth ^{-1}(a x)}{x^2} \, dx\) [8]

Optimal result

Integrand size = 8, antiderivative size = 30 \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=-\frac {\coth ^{-1}(a x)}{x}+a \log (x)-\frac {1}{2} a \log \left (1-a^2 x^2\right ) \]

[Out]

-arccoth(a*x)/x+a*ln(x)-1/2*a*ln(-a^2*x^2+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6038, 272, 36, 29, 31} \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=-\frac {1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)-\frac {\coth ^{-1}(a x)}{x} \]

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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]]

Rule 6038

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCoth[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rubi steps \begin{align*} \text {integral}& = -\frac {\coth ^{-1}(a x)}{x}+a \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx \\ & = -\frac {\coth ^{-1}(a x)}{x}+\frac {1}{2} a \text {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {\coth ^{-1}(a x)}{x}+\frac {1}{2} a \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} a^3 \text {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right ) \\ & = -\frac {\coth ^{-1}(a x)}{x}+a \log (x)-\frac {1}{2} a \log \left (1-a^2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=-\frac {\coth ^{-1}(a x)}{x}+a \log (x)-\frac {1}{2} a \log \left (1-a^2 x^2\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17

method	result	size
parallelrisch	\(\frac {a \ln \left (x \right ) x -a \ln \left (a x -1\right ) x -a x \,\operatorname {arccoth}\left (a x \right )-\operatorname {arccoth}\left (a x \right )}{x}\)	\(35\)
parts	\(-\frac {\operatorname {arccoth}\left (a x \right )}{x}-a \left (\frac {\ln \left (a x +1\right )}{2}-\ln \left (x \right )+\frac {\ln \left (a x -1\right )}{2}\right )\)	\(35\)
derivativedivides	\(a \left (-\frac {\operatorname {arccoth}\left (a x \right )}{a x}-\frac {\ln \left (a x +1\right )}{2}-\frac {\ln \left (a x -1\right )}{2}+\ln \left (a x \right )\right )\)	\(36\)
default	\(a \left (-\frac {\operatorname {arccoth}\left (a x \right )}{a x}-\frac {\ln \left (a x +1\right )}{2}-\frac {\ln \left (a x -1\right )}{2}+\ln \left (a x \right )\right )\)	\(36\)
risch	\(-\frac {\ln \left (a x +1\right )}{2 x}+\frac {2 a \ln \left (x \right ) x -a \ln \left (a^{2} x^{2}-1\right ) x +\ln \left (a x -1\right )}{2 x}\)	\(45\)

[In]

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

[Out]

(a*ln(x)*x-a*ln(a*x-1)*x-a*x*arccoth(a*x)-arccoth(a*x))/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=-\frac {a x \log \left (a^{2} x^{2} - 1\right ) - 2 \, a x \log \left (x\right ) + \log \left (\frac {a x + 1}{a x - 1}\right )}{2 \, x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=a \log {\left (x \right )} - a \log {\left (a x + 1 \right )} + a \operatorname {acoth}{\left (a x \right )} - \frac {\operatorname {acoth}{\left (a x \right )}}{x} \]

[In]

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

[Out]

a*log(x) - a*log(a*x + 1) + a*acoth(a*x) - acoth(a*x)/x

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=-\frac {1}{2} \, a {\left (\log \left (a^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} - \frac {\operatorname {arcoth}\left (a x\right )}{x} \]

[In]

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

[Out]

-1/2*a*(log(a^2*x^2 - 1) - log(x^2)) - arccoth(a*x)/x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (28) = 56\).

Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 4.77 \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=a {\left (\frac {\log \left (-\frac {\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} + 1}{\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} - 1}\right )}{\frac {a x + 1}{a x - 1} + 1} - \log \left (\frac {{\left | a x + 1 \right |}}{{\left | a x - 1 \right |}}\right ) + \log \left ({\left | \frac {a x + 1}{a x - 1} + 1 \right |}\right )\right )} \]

[In]

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

[Out]

a*(log(-(((a*x + 1)*a/(a*x - 1) - a)/(a*((a*x + 1)/(a*x - 1) + 1)) + 1)/(((a*x + 1)*a/(a*x - 1) - a)/(a*((a*x
+ 1)/(a*x - 1) + 1)) - 1))/((a*x + 1)/(a*x - 1) + 1) - log(abs(a*x + 1)/abs(a*x - 1)) + log(abs((a*x + 1)/(a*x
- 1) + 1)))

Mupad [B] (verification not implemented)

Time = 4.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {\coth ^{-1}(a x)}{x^2} \, dx=a\,\ln \left (x\right )-\frac {a\,\ln \left (a^2\,x^2-1\right )}{2}-\frac {\mathrm {acoth}\left (a\,x\right )}{x} \]

[In]

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

[Out]

a*log(x) - (a*log(a^2*x^2 - 1))/2 - acoth(a*x)/x