\(\int \frac {\coth ^{-1}(a x)}{x^5} \, dx\) [11]

Optimal result

Integrand size = 8, antiderivative size = 41 \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=-\frac {a}{12 x^3}-\frac {a^3}{4 x}-\frac {\coth ^{-1}(a x)}{4 x^4}+\frac {1}{4} a^4 \text {arctanh}(a x) \]

[Out]

-1/12*a/x^3-1/4*a^3/x-1/4*arccoth(a*x)/x^4+1/4*a^4*arctanh(a*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, 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, 331, 212} \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=\frac {1}{4} a^4 \text {arctanh}(a x)-\frac {a^3}{4 x}-\frac {\coth ^{-1}(a x)}{4 x^4}-\frac {a}{12 x^3} \]

[In]

Int[ArcCoth[a*x]/x^5,x]

[Out]

-1/12*a/x^3 - a^3/(4*x) - ArcCoth[a*x]/(4*x^4) + (a^4*ArcTanh[a*x])/4

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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)}{4 x^4}+\frac {1}{4} a \int \frac {1}{x^4 \left (1-a^2 x^2\right )} \, dx \\ & = -\frac {a}{12 x^3}-\frac {\coth ^{-1}(a x)}{4 x^4}+\frac {1}{4} a^3 \int \frac {1}{x^2 \left (1-a^2 x^2\right )} \, dx \\ & = -\frac {a}{12 x^3}-\frac {a^3}{4 x}-\frac {\coth ^{-1}(a x)}{4 x^4}+\frac {1}{4} a^5 \int \frac {1}{1-a^2 x^2} \, dx \\ & = -\frac {a}{12 x^3}-\frac {a^3}{4 x}-\frac {\coth ^{-1}(a x)}{4 x^4}+\frac {1}{4} a^4 \text {arctanh}(a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=-\frac {a}{12 x^3}-\frac {a^3}{4 x}-\frac {\coth ^{-1}(a x)}{4 x^4}-\frac {1}{8} a^4 \log (1-a x)+\frac {1}{8} a^4 \log (1+a x) \]

[In]

Integrate[ArcCoth[a*x]/x^5,x]

[Out]

-1/12*a/x^3 - a^3/(4*x) - ArcCoth[a*x]/(4*x^4) - (a^4*Log[1 - a*x])/8 + (a^4*Log[1 + a*x])/8

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(-\frac {-3 a^{4} x^{4} \operatorname {arccoth}\left (a x \right )+3 a^{3} x^{3}+a x +3 \,\operatorname {arccoth}\left (a x \right )}{12 x^{4}}\)	\(36\)
parts	\(-\frac {\operatorname {arccoth}\left (a x \right )}{4 x^{4}}-\frac {a \left (-\frac {a^{3} \ln \left (a x +1\right )}{2}+\frac {1}{3 x^{3}}+\frac {a^{2}}{x}+\frac {a^{3} \ln \left (a x -1\right )}{2}\right )}{4}\)	\(49\)
derivativedivides	\(a^{4} \left (-\frac {\operatorname {arccoth}\left (a x \right )}{4 a^{4} x^{4}}-\frac {\ln \left (a x -1\right )}{8}+\frac {\ln \left (a x +1\right )}{8}-\frac {1}{12 a^{3} x^{3}}-\frac {1}{4 a x}\right )\)	\(50\)
default	\(a^{4} \left (-\frac {\operatorname {arccoth}\left (a x \right )}{4 a^{4} x^{4}}-\frac {\ln \left (a x -1\right )}{8}+\frac {\ln \left (a x +1\right )}{8}-\frac {1}{12 a^{3} x^{3}}-\frac {1}{4 a x}\right )\)	\(50\)
risch	\(-\frac {\ln \left (a x +1\right )}{8 x^{4}}+\frac {3 \ln \left (-a x -1\right ) a^{4} x^{4}-3 \ln \left (-a x +1\right ) a^{4} x^{4}-6 a^{3} x^{3}-2 a x +3 \ln \left (a x -1\right )}{24 x^{4}}\)	\(69\)

[In]

int(arccoth(a*x)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/12*(-3*a^4*x^4*arccoth(a*x)+3*a^3*x^3+a*x+3*arccoth(a*x))/x^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.05 \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=-\frac {6 \, a^{3} x^{3} + 2 \, a x - 3 \, {\left (a^{4} x^{4} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )}{24 \, x^{4}} \]

[In]

integrate(arccoth(a*x)/x^5,x, algorithm="fricas")

[Out]

-1/24*(6*a^3*x^3 + 2*a*x - 3*(a^4*x^4 - 1)*log((a*x + 1)/(a*x - 1)))/x^4

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=\frac {a^{4} \operatorname {acoth}{\left (a x \right )}}{4} - \frac {a^{3}}{4 x} - \frac {a}{12 x^{3}} - \frac {\operatorname {acoth}{\left (a x \right )}}{4 x^{4}} \]

[In]

integrate(acoth(a*x)/x**5,x)

[Out]

a**4*acoth(a*x)/4 - a**3/(4*x) - a/(12*x**3) - acoth(a*x)/(4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.24 \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=\frac {1}{24} \, {\left (3 \, a^{3} \log \left (a x + 1\right ) - 3 \, a^{3} \log \left (a x - 1\right ) - \frac {2 \, {\left (3 \, a^{2} x^{2} + 1\right )}}{x^{3}}\right )} a - \frac {\operatorname {arcoth}\left (a x\right )}{4 \, x^{4}} \]

[In]

integrate(arccoth(a*x)/x^5,x, algorithm="maxima")

[Out]

1/24*(3*a^3*log(a*x + 1) - 3*a^3*log(a*x - 1) - 2*(3*a^2*x^2 + 1)/x^3)*a - 1/4*arccoth(a*x)/x^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (33) = 66\).

Time = 0.28 (sec) , antiderivative size = 205, normalized size of antiderivative = 5.00 \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=\frac {1}{3} \, a {\left (\frac {\frac {3 \, {\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} + \frac {3 \, {\left (a x + 1\right )} a^{3}}{a x - 1} + 2 \, a^{3}}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{3}} + \frac {3 \, {\left (\frac {{\left (a x + 1\right )}^{3} a^{3}}{{\left (a x - 1\right )}^{3}} + \frac {{\left (a x + 1\right )} a^{3}}{a x - 1}\right )} \log \left (-\frac {\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} + 1}{\frac {\frac {{\left (a x + 1\right )} a}{a x - 1} - a}{a {\left (\frac {a x + 1}{a x - 1} + 1\right )}} - 1}\right )}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{4}}\right )} \]

[In]

integrate(arccoth(a*x)/x^5,x, algorithm="giac")

[Out]

1/3*a*((3*(a*x + 1)^2*a^3/(a*x - 1)^2 + 3*(a*x + 1)*a^3/(a*x - 1) + 2*a^3)/((a*x + 1)/(a*x - 1) + 1)^3 + 3*((a
*x + 1)^3*a^3/(a*x - 1)^3 + (a*x + 1)*a^3/(a*x - 1))*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)^4)

Mupad [B] (verification not implemented)

Time = 4.63 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.46 \[ \int \frac {\coth ^{-1}(a x)}{x^5} \, dx=\frac {\ln \left (1-\frac {1}{a\,x}\right )}{8\,x^4}-\frac {\ln \left (\frac {1}{a\,x}+1\right )}{8\,x^4}-\frac {a^3\,x^2+\frac {a}{3}}{4\,x^3}-\frac {a^4\,\mathrm {atan}\left (a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

[In]

int(acoth(a*x)/x^5,x)

[Out]

log(1 - 1/(a*x))/(8*x^4) - (a^4*atan(a*x*1i)*1i)/4 - log(1/(a*x) + 1)/(8*x^4) - (a/3 + a^3*x^2)/(4*x^3)