3.5 \(\int x \coth ^{-1}(a x) \, dx\)

Optimal. Leaf size=31 \[ -\frac {\tanh ^{-1}(a x)}{2 a^2}+\frac {1}{2} x^2 \coth ^{-1}(a x)+\frac {x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5917, 321, 206} \[ -\frac {\tanh ^{-1}(a x)}{2 a^2}+\frac {1}{2} x^2 \coth ^{-1}(a x)+\frac {x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[a*x],x]

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 x \coth ^{-1}(a x) \, dx &=\frac {1}{2} x^2 \coth ^{-1}(a x)-\frac {1}{2} a \int \frac {x^2}{1-a^2 x^2} \, dx\\ &=\frac {x}{2 a}+\frac {1}{2} x^2 \coth ^{-1}(a x)-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{2 a}\\ &=\frac {x}{2 a}+\frac {1}{2} x^2 \coth ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 47, normalized size = 1.52 \[ \frac {\log (1-a x)}{4 a^2}-\frac {\log (a x+1)}{4 a^2}+\frac {1}{2} x^2 \coth ^{-1}(a x)+\frac {x}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[a*x],x]

[Out]

x/(2*a) + (x^2*ArcCoth[a*x])/2 + Log[1 - a*x]/(4*a^2) - Log[1 + a*x]/(4*a^2)

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fricas [A]  time = 0.58, size = 34, normalized size = 1.10 \[ \frac {2 \, a x + {\left (a^{2} x^{2} - 1\right )} \log \left (\frac {a x + 1}{a x - 1}\right )}{4 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccoth(a*x), x)

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maple [A]  time = 0.03, size = 39, normalized size = 1.26 \[ \frac {x^{2} \mathrm {arccoth}\left (a x \right )}{2}+\frac {x}{2 a}+\frac {\ln \left (a x -1\right )}{4 a^{2}}-\frac {\ln \left (a x +1\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(a*x),x)

[Out]

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

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maxima [A]  time = 0.30, size = 41, normalized size = 1.32 \[ \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (a x\right ) + \frac {1}{4} \, a {\left (\frac {2 \, x}{a^{2}} - \frac {\log \left (a x + 1\right )}{a^{3}} + \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.18, size = 26, normalized size = 0.84 \[ \frac {x^2\,\mathrm {acoth}\left (a\,x\right )}{2}-\frac {\frac {\mathrm {acoth}\left (a\,x\right )}{2}-\frac {a\,x}{2}}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acoth(a*x),x)

[Out]

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

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sympy [A]  time = 0.45, size = 32, normalized size = 1.03 \[ \begin {cases} \frac {x^{2} \operatorname {acoth}{\left (a x \right )}}{2} + \frac {x}{2 a} - \frac {\operatorname {acoth}{\left (a x \right )}}{2 a^{2}} & \text {for}\: a \neq 0 \\\frac {i \pi x^{2}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*acoth(a*x)/2 + x/(2*a) - acoth(a*x)/(2*a**2), Ne(a, 0)), (I*pi*x**2/4, True))

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