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

Optimal. Leaf size=41 \[ \frac {1}{4} a^4 \tanh ^{-1}(a x)-\frac {a^3}{4 x}-\frac {\coth ^{-1}(a x)}{4 x^4}-\frac {a}{12 x^3} \]

[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)

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5917, 325, 206} \[ -\frac {a^3}{4 x}+\frac {1}{4} a^4 \tanh ^{-1}(a x)-\frac {a}{12 x^3}-\frac {\coth ^{-1}(a x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 325

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 5917

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

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(a x)}{x^5} \, dx &=-\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 \tanh ^{-1}(a x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 57, normalized size = 1.39 \[ -\frac {1}{8} a^4 \log (1-a x)+\frac {1}{8} a^4 \log (a x+1)-\frac {a^3}{4 x}-\frac {\coth ^{-1}(a x)}{4 x^4}-\frac {a}{12 x^3} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.52, size = 43, normalized size = 1.05 \[ -\frac {6 \, a^{3} x^{3} + 2 \, a x - 3 \, {\left (a^{4} x^{4} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )}{24 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (a x\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x)/x^5, x)

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maple [A]  time = 0.04, size = 47, normalized size = 1.15 \[ -\frac {\mathrm {arccoth}\left (a x \right )}{4 x^{4}}-\frac {a}{12 x^{3}}-\frac {a^{3}}{4 x}-\frac {a^{4} \ln \left (a x -1\right )}{8}+\frac {a^{4} \ln \left (a x +1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*arccoth(a*x)/x^4-1/12*a/x^3-1/4*a^3/x-1/8*a^4*ln(a*x-1)+1/8*a^4*ln(a*x+1)

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maxima [A]  time = 0.31, size = 51, normalized size = 1.24 \[ \frac {1}{24} \, {\left (3 \, a^{3} \log \left (a x + 1\right ) - 3 \, a^{3} \log \left (a x - 1\right ) - \frac {2 \, {\left (3 \, a^{2} x^{2} + 1\right )}}{x^{3}}\right )} a - \frac {\operatorname {arcoth}\left (a x\right )}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.61, size = 60, normalized size = 1.46 \[ \frac {\ln \left (1-\frac {1}{a\,x}\right )}{8\,x^4}-\frac {\ln \left (\frac {1}{a\,x}+1\right )}{8\,x^4}-\frac {a^3\,x^2+\frac {a}{3}}{4\,x^3}-\frac {a^4\,\mathrm {atan}\left (a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.16, size = 32, normalized size = 0.78 \[ \frac {a^{4} \operatorname {acoth}{\left (a x \right )}}{4} - \frac {a^{3}}{4 x} - \frac {a}{12 x^{3}} - \frac {\operatorname {acoth}{\left (a x \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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