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]

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

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Rubi [A]  time = 0.10, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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]

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^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*}

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Mathematica [A]  time = 0.12, 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

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fricas [A]  time = 0.72, size = 111, normalized size = 1.23 \[ \frac {{\left (a^{2} - 2 \, a + 1\right )} b^{2} x^{2} \log \left (b x + a + 1\right ) - {\left (a^{2} + 2 \, a + 1\right )} b^{2} x^{2} \log \left (b x + a - 1\right ) + 4 \, a b^{2} x^{2} \log \relax (x) + 2 \, {\left (a^{2} - 1\right )} b x - {\left (a^{4} - 2 \, a^{2} + 1\right )} \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{4 \, {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}} \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (b x + a\right )}{x^{3}}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 82, normalized size = 0.91 \[ -\frac {\mathrm {arccoth}\left (b x +a \right )}{2 x^{2}}-\frac {b^{2} \ln \left (b x +a -1\right )}{4 \left (a -1\right )^{2}}+\frac {b^{2} \ln \left (b x +a +1\right )}{4 \left (1+a \right )^{2}}+\frac {b}{2 \left (a -1\right ) \left (1+a \right ) x}+\frac {b^{2} a \ln \left (b x \right )}{\left (a -1\right )^{2} \left (1+a \right )^{2}} \]

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

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maxima [A]  time = 0.30, size = 85, normalized size = 0.94 \[ \frac {1}{4} \, {\left (\frac {4 \, a b \log \relax (x)}{a^{4} - 2 \, a^{2} + 1} + \frac {b \log \left (b x + a + 1\right )}{a^{2} + 2 \, a + 1} - \frac {b \log \left (b x + a - 1\right )}{a^{2} - 2 \, a + 1} + \frac {2}{{\left (a^{2} - 1\right )} x}\right )} b - \frac {\operatorname {arcoth}\left (b x + a\right )}{2 \, x^{2}} \]

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

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mupad [B]  time = 1.95, size = 247, normalized size = 2.74 \[ \ln \relax (x)\,\left (\frac {b^2}{4\,{\left (a-1\right )}^2}-\frac {b^2}{4\,{\left (a+1\right )}^2}\right )-\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )\,\left (\frac {b^2}{8\,{\left (a-1\right )}^2}-\frac {b^2}{8\,{\left (a+1\right )}^2}\right )-\frac {\mathrm {acoth}\left (a+b\,x\right )\,\left (\frac {a^2}{2}-\frac {1}{2}\right )-\frac {b\,x}{2}+\frac {b^2\,x^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}+\frac {x^3\,\left (3\,a^2\,b^3+b^3\right )}{2\,{\left (a^2-1\right )}^2}+\frac {a\,b^4\,x^4}{{\left (a^2-1\right )}^2}+a\,b\,x\,\mathrm {acoth}\left (a+b\,x\right )}{a^2\,x^2+2\,a\,b\,x^3+b^2\,x^4-x^2}-\frac {\mathrm {atan}\left (\frac {2\,x\,b^2+2\,a\,b}{2\,\sqrt {b^2\,\left (a^2-1\right )-a^2\,b^2}}\right )\,\left (a^2\,b^3+b^3\right )}{\sqrt {-b^2}\,\left (2\,a^4-4\,a^2+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)/x^3,x)

[Out]

log(x)*(b^2/(4*(a - 1)^2) - b^2/(4*(a + 1)^2)) - log(a^2 + b^2*x^2 + 2*a*b*x - 1)*(b^2/(8*(a - 1)^2) - b^2/(8*
(a + 1)^2)) - (acoth(a + b*x)*(a^2/2 - 1/2) - (b*x)/2 + (b^2*x^2*acoth(a + b*x))/2 + (x^3*(b^3 + 3*a^2*b^3))/(
2*(a^2 - 1)^2) + (a*b^4*x^4)/(a^2 - 1)^2 + a*b*x*acoth(a + b*x))/(a^2*x^2 - x^2 + b^2*x^4 + 2*a*b*x^3) - (atan
((2*a*b + 2*b^2*x)/(2*(b^2*(a^2 - 1) - a^2*b^2)^(1/2)))*(b^3 + a^2*b^3))/((-b^2)^(1/2)*(2*a^4 - 4*a^2 + 2))

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sympy [A]  time = 2.43, size = 410, normalized size = 4.56 \[ \begin {cases} \frac {b^{2} \operatorname {acoth}{\left (b x - 1 \right )}}{8} - \frac {b}{8 x} - \frac {\operatorname {acoth}{\left (b x - 1 \right )}}{2 x^{2}} - \frac {1}{8 x^{2}} & \text {for}\: a = -1 \\\frac {b^{2} \operatorname {acoth}{\left (b x + 1 \right )}}{8} - \frac {b}{8 x} - \frac {\operatorname {acoth}{\left (b x + 1 \right )}}{2 x^{2}} + \frac {1}{8 x^{2}} & \text {for}\: a = 1 \\- \frac {a^{4} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {a^{2} b^{2} x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {a^{2} b x}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {2 a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {2 a b^{2} x^{2} \log {\relax (x )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac {2 a b^{2} x^{2} \log {\left (a + b x + 1 \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {2 a b^{2} x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {b^{2} x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac {b x}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} & \text {otherwise} \end {cases} \]

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))

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