3.68 \(\int \frac{\coth ^{-1}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=90 \[ \frac{a b^2 \log (x)}{\left (1-a^2\right )^2}-\frac{b}{2 \left (1-a^2\right ) x}-\frac{b^2 \log (-a-b x+1)}{4 (1-a)^2}+\frac{b^2 \log (a+b x+1)}{4 (a+1)^2}-\frac{\coth ^{-1}(a+b x)}{2 x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0997384, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6110, 371, 710, 801} \[ \frac{a b^2 \log (x)}{\left (1-a^2\right )^2}-\frac{b}{2 \left (1-a^2\right ) x}-\frac{b^2 \log (-a-b x+1)}{4 (1-a)^2}+\frac{b^2 \log (a+b x+1)}{4 (a+1)^2}-\frac{\coth ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]/x^3,x]

[Out]

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

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]

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 710

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(a+b x)}{x^3} \, dx &=-\frac{\coth ^{-1}(a+b x)}{2 x^2}+\frac{1}{2} b \int \frac{1}{x^2 \left (1-(a+b x)^2\right )} \, dx\\ &=-\frac{\coth ^{-1}(a+b x)}{2 x^2}+\frac{1}{2} b^2 \operatorname{Subst}\left (\int \frac{1}{(-a+x)^2 \left (1-x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac{b}{2 \left (1-a^2\right ) x}-\frac{\coth ^{-1}(a+b x)}{2 x^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{-a-x}{(-a+x) \left (1-x^2\right )} \, dx,x,a+b x\right )}{2 \left (1-a^2\right )}\\ &=-\frac{b}{2 \left (1-a^2\right ) x}-\frac{\coth ^{-1}(a+b x)}{2 x^2}-\frac{b^2 \operatorname{Subst}\left (\int \left (-\frac{2 a}{\left (-1+a^2\right ) (a-x)}+\frac{-1-a}{2 (-1+a) (-1+x)}+\frac{-1+a}{2 (1+a) (1+x)}\right ) \, dx,x,a+b x\right )}{2 \left (1-a^2\right )}\\ &=-\frac{b}{2 \left (1-a^2\right ) x}-\frac{\coth ^{-1}(a+b x)}{2 x^2}+\frac{a b^2 \log (x)}{\left (1-a^2\right )^2}-\frac{b^2 \log (1-a-b x)}{4 (1-a)^2}+\frac{b^2 \log (1+a+b x)}{4 (1+a)^2}\\ \end{align*}

Mathematica [A]  time = 0.106819, size = 76, normalized size = 0.84 \[ \frac{1}{4} \left (b \left (\frac{4 a b \log (x)}{\left (a^2-1\right )^2}+\frac{2}{\left (a^2-1\right ) x}-\frac{b \log (-a-b x+1)}{(a-1)^2}+\frac{b \log (a+b x+1)}{(a+1)^2}\right )-\frac{2 \coth ^{-1}(a+b x)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a + b*x]/x^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.043, size = 82, normalized size = 0.9 \begin{align*} -{\frac{{\rm arccoth} \left (bx+a\right )}{2\,{x}^{2}}}-{\frac{{b}^{2}\ln \left ( bx+a-1 \right ) }{4\, \left ( a-1 \right ) ^{2}}}+{\frac{b}{ \left ( 2\,a-2 \right ) \left ( 1+a \right ) x}}+{\frac{a{b}^{2}\ln \left ( bx \right ) }{ \left ( a-1 \right ) ^{2} \left ( 1+a \right ) ^{2}}}+{\frac{{b}^{2}\ln \left ( bx+a+1 \right ) }{4\, \left ( 1+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a)/x^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.959889, size = 115, normalized size = 1.28 \begin{align*} \frac{1}{4} \,{\left (\frac{4 \, a b \log \left (x\right )}{a^{4} - 2 \, a^{2} + 1} + \frac{b \log \left (b x + a + 1\right )}{a^{2} + 2 \, a + 1} - \frac{b \log \left (b x + a - 1\right )}{a^{2} - 2 \, a + 1} + \frac{2}{{\left (a^{2} - 1\right )} x}\right )} b - \frac{\operatorname{arcoth}\left (b x + a\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)/x^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.75893, size = 279, normalized size = 3.1 \begin{align*} \frac{{\left (a^{2} - 2 \, a + 1\right )} b^{2} x^{2} \log \left (b x + a + 1\right ) -{\left (a^{2} + 2 \, a + 1\right )} b^{2} x^{2} \log \left (b x + a - 1\right ) + 4 \, a b^{2} x^{2} \log \left (x\right ) + 2 \,{\left (a^{2} - 1\right )} b x -{\left (a^{4} - 2 \, a^{2} + 1\right )} \log \left (\frac{b x + a + 1}{b x + a - 1}\right )}{4 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 3.45566, size = 410, normalized size = 4.56 \begin{align*} \begin{cases} \frac{b^{2} \operatorname{acoth}{\left (b x - 1 \right )}}{8} - \frac{b}{8 x} - \frac{\operatorname{acoth}{\left (b x - 1 \right )}}{2 x^{2}} - \frac{1}{8 x^{2}} & \text{for}\: a = -1 \\\frac{b^{2} \operatorname{acoth}{\left (b x + 1 \right )}}{8} - \frac{b}{8 x} - \frac{\operatorname{acoth}{\left (b x + 1 \right )}}{2 x^{2}} + \frac{1}{8 x^{2}} & \text{for}\: a = 1 \\- \frac{a^{4} \operatorname{acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac{a^{2} b^{2} x^{2} \operatorname{acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac{a^{2} b x}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac{2 a^{2} \operatorname{acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac{2 a b^{2} x^{2} \log{\left (x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac{2 a b^{2} x^{2} \log{\left (a + b x + 1 \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac{2 a b^{2} x^{2} \operatorname{acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac{b^{2} x^{2} \operatorname{acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac{b x}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac{\operatorname{acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(b*x+a)/x**3,x)

[Out]

Piecewise((b**2*acoth(b*x - 1)/8 - b/(8*x) - acoth(b*x - 1)/(2*x**2) - 1/(8*x**2), Eq(a, -1)), (b**2*acoth(b*x
+ 1)/8 - b/(8*x) - acoth(b*x + 1)/(2*x**2) + 1/(8*x**2), Eq(a, 1)), (-a**4*acoth(a + b*x)/(2*a**4*x**2 - 4*a*
*2*x**2 + 2*x**2) + a**2*b**2*x**2*acoth(a + b*x)/(2*a**4*x**2 - 4*a**2*x**2 + 2*x**2) + a**2*b*x/(2*a**4*x**2
- 4*a**2*x**2 + 2*x**2) + 2*a**2*acoth(a + b*x)/(2*a**4*x**2 - 4*a**2*x**2 + 2*x**2) + 2*a*b**2*x**2*log(x)/(
2*a**4*x**2 - 4*a**2*x**2 + 2*x**2) - 2*a*b**2*x**2*log(a + b*x + 1)/(2*a**4*x**2 - 4*a**2*x**2 + 2*x**2) + 2*
a*b**2*x**2*acoth(a + b*x)/(2*a**4*x**2 - 4*a**2*x**2 + 2*x**2) + b**2*x**2*acoth(a + b*x)/(2*a**4*x**2 - 4*a*
*2*x**2 + 2*x**2) - b*x/(2*a**4*x**2 - 4*a**2*x**2 + 2*x**2) - acoth(a + b*x)/(2*a**4*x**2 - 4*a**2*x**2 + 2*x
**2), True))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (b x + a\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)/x^3,x, algorithm="giac")

[Out]

integrate(arccoth(b*x + a)/x^3, x)