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

Optimal. Leaf size=48 \[ \frac {\log (a+b x)}{b}-\frac {\log \left (1-(a+b x)^2\right )}{2 b}-\frac {\coth ^{-1}(a+b x)}{b (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6108, 5917, 266, 36, 31, 29} \[ \frac {\log (a+b x)}{b}-\frac {\log \left (1-(a+b x)^2\right )}{2 b}-\frac {\coth ^{-1}(a+b x)}{b (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]/(a + b*x)^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

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

Rule 266

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 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]

Rule 6108

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 43, normalized size = 0.90 \[ -\frac {-2 \log (a+b x)+\log \left (1-(a+b x)^2\right )+\frac {2 \coth ^{-1}(a+b x)}{a+b x}}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a + b*x]/(a + b*x)^2,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 67, normalized size = 1.40 \[ -\frac {{\left (b x + a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right ) - 2 \, {\left (b x + a\right )} \log \left (b x + a\right ) + \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{2 \, {\left (b^{2} x + a b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 54, normalized size = 1.12 \[ -\frac {\mathrm {arccoth}\left (b x +a \right )}{b \left (b x +a \right )}+\frac {\ln \left (b x +a \right )}{b}-\frac {\ln \left (b x +a -1\right )}{2 b}-\frac {\ln \left (b x +a +1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 53, normalized size = 1.10 \[ -\frac {\log \left (b x + a + 1\right )}{2 \, b} + \frac {\log \left (b x + a\right )}{b} - \frac {\log \left (b x + a - 1\right )}{2 \, b} - \frac {\operatorname {arcoth}\left (b x + a\right )}{{\left (b x + a\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.41, size = 93, normalized size = 1.94 \[ \frac {\ln \left (a+b\,x\right )}{b}-\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )}{2\,b}-\frac {\ln \left (\frac {a+b\,x+1}{a+b\,x}\right )}{2\,\left (x\,b^2+a\,b\right )}+\frac {\ln \left (\frac {a+b\,x-1}{a+b\,x}\right )}{2\,x\,b^2+2\,a\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)/(a + b*x)^2,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.40, size = 136, normalized size = 2.83 \[ \begin {cases} \frac {a \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} - \frac {a \log {\left (\frac {a}{b} + x + \frac {1}{b} \right )}}{a b + b^{2} x} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{a b + b^{2} x} + \frac {b x \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} - \frac {b x \log {\left (\frac {a}{b} + x + \frac {1}{b} \right )}}{a b + b^{2} x} + \frac {b x \operatorname {acoth}{\left (a + b x \right )}}{a b + b^{2} x} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{a b + b^{2} x} & \text {for}\: b \neq 0 \\\frac {x \operatorname {acoth}{\relax (a )}}{a^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*log(a/b + x)/(a*b + b**2*x) - a*log(a/b + x + 1/b)/(a*b + b**2*x) + a*acoth(a + b*x)/(a*b + b**2*
x) + b*x*log(a/b + x)/(a*b + b**2*x) - b*x*log(a/b + x + 1/b)/(a*b + b**2*x) + b*x*acoth(a + b*x)/(a*b + b**2*
x) - acoth(a + b*x)/(a*b + b**2*x), Ne(b, 0)), (x*acoth(a)/a**2, True))

________________________________________________________________________________________