\(\int \frac {\coth ^{-1}(a x)}{x^4} \, dx\) [10]

Optimal result

Integrand size = 8, antiderivative size = 47 \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=-\frac {a}{6 x^2}-\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1-a^2 x^2\right ) \]

[Out]

-1/6*a/x^2-1/3*arccoth(a*x)/x^3+1/3*a^3*ln(x)-1/6*a^3*ln(-a^2*x^2+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6038, 272, 46} \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=\frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1-a^2 x^2\right )-\frac {\coth ^{-1}(a x)}{3 x^3}-\frac {a}{6 x^2} \]

[In]

Int[ArcCoth[a*x]/x^4,x]

[Out]

-1/6*a/x^2 - ArcCoth[a*x]/(3*x^3) + (a^3*Log[x])/3 - (a^3*Log[1 - a^2*x^2])/6

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 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)}{3 x^3}+\frac {1}{3} a \int \frac {1}{x^3 \left (1-a^2 x^2\right )} \, dx \\ & = -\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{6} a \text {Subst}\left (\int \frac {1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{6} a \text {Subst}\left (\int \left (\frac {1}{x^2}+\frac {a^2}{x}-\frac {a^4}{-1+a^2 x}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a}{6 x^2}-\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1-a^2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=-\frac {a}{6 x^2}-\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1-a^2 x^2\right ) \]

[In]

Integrate[ArcCoth[a*x]/x^4,x]

[Out]

-1/6*a/x^2 - ArcCoth[a*x]/(3*x^3) + (a^3*Log[x])/3 - (a^3*Log[1 - a^2*x^2])/6

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(a^{3} \left (-\frac {\operatorname {arccoth}\left (a x \right )}{3 a^{3} x^{3}}-\frac {\ln \left (a x -1\right )}{6}-\frac {\ln \left (a x +1\right )}{6}-\frac {1}{6 a^{2} x^{2}}+\frac {\ln \left (a x \right )}{3}\right )\)	\(48\)
default	\(a^{3} \left (-\frac {\operatorname {arccoth}\left (a x \right )}{3 a^{3} x^{3}}-\frac {\ln \left (a x -1\right )}{6}-\frac {\ln \left (a x +1\right )}{6}-\frac {1}{6 a^{2} x^{2}}+\frac {\ln \left (a x \right )}{3}\right )\)	\(48\)
parts	\(-\frac {\operatorname {arccoth}\left (a x \right )}{3 x^{3}}-\frac {a \left (\frac {a^{2} \ln \left (a x +1\right )}{2}+\frac {1}{2 x^{2}}-a^{2} \ln \left (x \right )+\frac {a^{2} \ln \left (a x -1\right )}{2}\right )}{3}\)	\(49\)
risch	\(-\frac {\ln \left (a x +1\right )}{6 x^{3}}+\frac {2 a^{3} \ln \left (x \right ) x^{3}-\ln \left (a^{2} x^{2}-1\right ) a^{3} x^{3}-a x +\ln \left (a x -1\right )}{6 x^{3}}\)	\(57\)
parallelrisch	\(\frac {2 a^{3} \ln \left (x \right ) x^{3}-2 a^{3} \ln \left (a x -1\right ) x^{3}-2 a^{3} x^{3} \operatorname {arccoth}\left (a x \right )-a^{3} x^{3}-a x -2 \,\operatorname {arccoth}\left (a x \right )}{6 x^{3}}\)	\(61\)

[In]

int(arccoth(a*x)/x^4,x,method=_RETURNVERBOSE)

[Out]

a^3*(-1/3/a^3/x^3*arccoth(a*x)-1/6*ln(a*x-1)-1/6*ln(a*x+1)-1/6/a^2/x^2+1/3*ln(a*x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=-\frac {a^{3} x^{3} \log \left (a^{2} x^{2} - 1\right ) - 2 \, a^{3} x^{3} \log \left (x\right ) + a x + \log \left (\frac {a x + 1}{a x - 1}\right )}{6 \, x^{3}} \]

[In]

integrate(arccoth(a*x)/x^4,x, algorithm="fricas")

[Out]

-1/6*(a^3*x^3*log(a^2*x^2 - 1) - 2*a^3*x^3*log(x) + a*x + log((a*x + 1)/(a*x - 1)))/x^3

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=\frac {a^{3} \log {\left (x \right )}}{3} - \frac {a^{3} \log {\left (a x + 1 \right )}}{3} + \frac {a^{3} \operatorname {acoth}{\left (a x \right )}}{3} - \frac {a}{6 x^{2}} - \frac {\operatorname {acoth}{\left (a x \right )}}{3 x^{3}} \]

[In]

integrate(acoth(a*x)/x**4,x)

[Out]

a**3*log(x)/3 - a**3*log(a*x + 1)/3 + a**3*acoth(a*x)/3 - a/(6*x**2) - acoth(a*x)/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=-\frac {1}{6} \, {\left (a^{2} \log \left (a^{2} x^{2} - 1\right ) - a^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} a - \frac {\operatorname {arcoth}\left (a x\right )}{3 \, x^{3}} \]

[In]

integrate(arccoth(a*x)/x^4,x, algorithm="maxima")

[Out]

-1/6*(a^2*log(a^2*x^2 - 1) - a^2*log(x^2) + 1/x^2)*a - 1/3*arccoth(a*x)/x^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (39) = 78\).

Time = 0.27 (sec) , antiderivative size = 209, normalized size of antiderivative = 4.45 \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=-\frac {1}{3} \, {\left (a^{2} \log \left (\frac {{\left | a x + 1 \right |}}{{\left | a x - 1 \right |}}\right ) - a^{2} \log \left ({\left | \frac {a x + 1}{a x - 1} + 1 \right |}\right ) - \frac {2 \, {\left (a x + 1\right )} a^{2}}{{\left (a x - 1\right )} {\left (\frac {a x + 1}{a x - 1} + 1\right )}^{2}} - \frac {{\left (\frac {3 \, {\left (a x + 1\right )}^{2} a^{2}}{{\left (a x - 1\right )}^{2}} + a^{2}\right )} \log \left (-\frac {\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} + 1}{\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} - 1}\right )}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{3}}\right )} a \]

[In]

integrate(arccoth(a*x)/x^4,x, algorithm="giac")

[Out]

-1/3*(a^2*log(abs(a*x + 1)/abs(a*x - 1)) - a^2*log(abs((a*x + 1)/(a*x - 1) + 1)) - 2*(a*x + 1)*a^2/((a*x - 1)*
((a*x + 1)/(a*x - 1) + 1)^2) - (3*(a*x + 1)^2*a^2/(a*x - 1)^2 + a^2)*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)^3)*a

Mupad [B] (verification not implemented)

Time = 4.61 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {\coth ^{-1}(a x)}{x^4} \, dx=\frac {a^3\,\ln \left (x\right )}{3}-\frac {\frac {\mathrm {acoth}\left (a\,x\right )}{3}+\frac {a\,x}{6}}{x^3}-\frac {a^3\,\ln \left (a^2\,x^2-1\right )}{6} \]

[In]

int(acoth(a*x)/x^4,x)

[Out]

(a^3*log(x))/3 - (acoth(a*x)/3 + (a*x)/6)/x^3 - (a^3*log(a^2*x^2 - 1))/6