3.67 \(\int \frac {\coth ^{-1}(a+b x)}{x^2} \, dx\)

Optimal. Leaf size=64 \[ \frac {b \log (x)}{1-a^2}-\frac {b \log (-a-b x+1)}{2 (1-a)}-\frac {b \log (a+b x+1)}{2 (a+1)}-\frac {\coth ^{-1}(a+b x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6110, 371, 706, 31, 633} \[ \frac {b \log (x)}{1-a^2}-\frac {b \log (-a-b x+1)}{2 (1-a)}-\frac {b \log (a+b x+1)}{2 (a+1)}-\frac {\coth ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]/x^2,x]

[Out]

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

Rule 31

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

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 6110

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcCoth[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcCoth[c + d*x])^(p - 1))/(1 - (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -
1]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(a+b x)}{x^2} \, dx &=-\frac {\coth ^{-1}(a+b x)}{x}+b \int \frac {1}{x \left (1-(a+b x)^2\right )} \, dx\\ &=-\frac {\coth ^{-1}(a+b x)}{x}+b \operatorname {Subst}\left (\int \frac {1}{(-a+x) \left (1-x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac {\coth ^{-1}(a+b x)}{x}+\frac {b \operatorname {Subst}\left (\int \frac {1}{-a+x} \, dx,x,a+b x\right )}{1-a^2}+\frac {b \operatorname {Subst}\left (\int \frac {a+x}{1-x^2} \, dx,x,a+b x\right )}{1-a^2}\\ &=-\frac {\coth ^{-1}(a+b x)}{x}+\frac {b \log (x)}{1-a^2}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,a+b x\right )}{2 (1-a)}+\frac {b \operatorname {Subst}\left (\int \frac {1}{-1-x} \, dx,x,a+b x\right )}{2 (1+a)}\\ &=-\frac {\coth ^{-1}(a+b x)}{x}+\frac {b \log (x)}{1-a^2}-\frac {b \log (1-a-b x)}{2 (1-a)}-\frac {b \log (1+a+b x)}{2 (1+a)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 55, normalized size = 0.86 \[ \frac {b ((a+1) \log (-a-b x+1)-(a-1) \log (a+b x+1)-2 \log (x))}{2 \left (a^2-1\right )}-\frac {\coth ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a + b*x]/x^2,x]

[Out]

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

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fricas [A]  time = 0.68, size = 68, normalized size = 1.06 \[ -\frac {{\left (a - 1\right )} b x \log \left (b x + a + 1\right ) - {\left (a + 1\right )} b x \log \left (b x + a - 1\right ) + 2 \, b x \log \relax (x) + {\left (a^{2} - 1\right )} \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{2 \, {\left (a^{2} - 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (b x + a\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(b*x + a)/x^2, x)

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maple [A]  time = 0.04, size = 63, normalized size = 0.98 \[ -\frac {\mathrm {arccoth}\left (b x +a \right )}{x}+\frac {b \ln \left (b x +a -1\right )}{2 a -2}-\frac {b \ln \left (b x +a +1\right )}{2+2 a}-\frac {b \ln \left (b x \right )}{\left (a -1\right ) \left (1+a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a)/x^2,x)

[Out]

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

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maxima [A]  time = 0.31, size = 54, normalized size = 0.84 \[ -\frac {1}{2} \, b {\left (\frac {\log \left (b x + a + 1\right )}{a + 1} - \frac {\log \left (b x + a - 1\right )}{a - 1} + \frac {2 \, \log \relax (x)}{a^{2} - 1}\right )} - \frac {\operatorname {arcoth}\left (b x + a\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.73, size = 62, normalized size = 0.97 \[ -\frac {\mathrm {acoth}\left (a+b\,x\right )}{x}-\frac {b\,x\,\ln \relax (x)-\frac {b\,x\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )}{2}+a\,b\,x\,\mathrm {acoth}\left (a+b\,x\right )}{x\,\left (a^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)/x^2,x)

[Out]

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

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sympy [A]  time = 1.48, size = 144, normalized size = 2.25 \[ \begin {cases} \frac {b \operatorname {acoth}{\left (b x - 1 \right )}}{2} - \frac {\operatorname {acoth}{\left (b x - 1 \right )}}{x} - \frac {1}{2 x} & \text {for}\: a = -1 \\- \frac {b \operatorname {acoth}{\left (b x + 1 \right )}}{2} - \frac {\operatorname {acoth}{\left (b x + 1 \right )}}{x} + \frac {1}{2 x} & \text {for}\: a = 1 \\- \frac {a^{2} \operatorname {acoth}{\left (a + b x \right )}}{a^{2} x - x} - \frac {a b x \operatorname {acoth}{\left (a + b x \right )}}{a^{2} x - x} - \frac {b x \log {\relax (x )}}{a^{2} x - x} + \frac {b x \log {\left (a + b x + 1 \right )}}{a^{2} x - x} - \frac {b x \operatorname {acoth}{\left (a + b x \right )}}{a^{2} x - x} + \frac {\operatorname {acoth}{\left (a + b x \right )}}{a^{2} x - x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((b*acoth(b*x - 1)/2 - acoth(b*x - 1)/x - 1/(2*x), Eq(a, -1)), (-b*acoth(b*x + 1)/2 - acoth(b*x + 1)/
x + 1/(2*x), Eq(a, 1)), (-a**2*acoth(a + b*x)/(a**2*x - x) - a*b*x*acoth(a + b*x)/(a**2*x - x) - b*x*log(x)/(a
**2*x - x) + b*x*log(a + b*x + 1)/(a**2*x - x) - b*x*acoth(a + b*x)/(a**2*x - x) + acoth(a + b*x)/(a**2*x - x)
, True))

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