Integrand size = 10, antiderivative size = 90 \[ \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx=-\frac {b}{2 \left (1-a^2\right ) x}-\frac {\coth ^{-1}(a+b x)}{2 x^2}+\frac {a b^2 \log (x)}{\left (1-a^2\right )^2}-\frac {b^2 \log (1-a-b x)}{4 (1-a)^2}+\frac {b^2 \log (1+a+b x)}{4 (1+a)^2} \]
-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
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx=\frac {1}{4} \left (-\frac {2 \coth ^{-1}(a+b x)}{x^2}+b \left (\frac {2}{\left (-1+a^2\right ) x}+\frac {4 a b \log (x)}{\left (-1+a^2\right )^2}-\frac {b \log (1-a-b x)}{(-1+a)^2}+\frac {b \log (1+a+b x)}{(1+a)^2}\right )\right ) \]
((-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
Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6660, 896, 477, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx\) |
\(\Big \downarrow \) 6660 |
\(\displaystyle \frac {1}{2} b \int \frac {1}{x^2 \left (1-(a+b x)^2\right )}dx-\frac {\coth ^{-1}(a+b x)}{2 x^2}\) |
\(\Big \downarrow \) 896 |
\(\displaystyle \frac {1}{2} b^2 \int \frac {1}{b^2 x^2 \left (1-(a+b x)^2\right )}d(a+b x)-\frac {\coth ^{-1}(a+b x)}{2 x^2}\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {1}{2} b^2 \int \left (\frac {2 a}{\left (1-a^2\right )^2 b x}+\frac {1}{2 (1-a)^2 (-a-b x+1)}+\frac {1}{2 (a+1)^2 (a+b x+1)}+\frac {1}{\left (1-a^2\right ) b^2 x^2}\right )d(a+b x)-\frac {\coth ^{-1}(a+b x)}{2 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} b^2 \left (-\frac {1}{\left (1-a^2\right ) b x}+\frac {2 a \log (-b x)}{\left (1-a^2\right )^2}-\frac {\log (-a-b x+1)}{2 (1-a)^2}+\frac {\log (a+b x+1)}{2 (a+1)^2}\right )-\frac {\coth ^{-1}(a+b x)}{2 x^2}\) |
-1/2*ArcCoth[a + b*x]/x^2 + (b^2*(-(1/((1 - a^2)*b*x)) + (2*a*Log[-(b*x)]) /(1 - a^2)^2 - Log[1 - a - b*x]/(2*(1 - a)^2) + Log[1 + a + b*x]/(2*(1 + a )^2)))/2
3.1.68.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(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]
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] - Simp[b*d*(p/(f*(m + 1))) Int[(e + f*x)^(m + 1)*((a + b*ArcCo th[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]
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87
method | result | size |
parts | \(-\frac {\operatorname {arccoth}\left (b x +a \right )}{2 x^{2}}-\frac {b \left (-\frac {1}{\left (-1+a \right ) \left (1+a \right ) x}-\frac {2 a b \ln \left (x \right )}{\left (-1+a \right )^{2} \left (1+a \right )^{2}}+\frac {b \ln \left (b x +a -1\right )}{2 \left (-1+a \right )^{2}}-\frac {b \ln \left (b x +a +1\right )}{2 \left (1+a \right )^{2}}\right )}{2}\) | \(78\) |
derivativedivides | \(b^{2} \left (-\frac {\operatorname {arccoth}\left (b x +a \right )}{2 b^{2} x^{2}}+\frac {\ln \left (b x +a +1\right )}{4 \left (1+a \right )^{2}}-\frac {\ln \left (b x +a -1\right )}{4 \left (-1+a \right )^{2}}+\frac {1}{2 \left (-1+a \right ) \left (1+a \right ) b x}+\frac {a \ln \left (-b x \right )}{\left (-1+a \right )^{2} \left (1+a \right )^{2}}\right )\) | \(83\) |
default | \(b^{2} \left (-\frac {\operatorname {arccoth}\left (b x +a \right )}{2 b^{2} x^{2}}+\frac {\ln \left (b x +a +1\right )}{4 \left (1+a \right )^{2}}-\frac {\ln \left (b x +a -1\right )}{4 \left (-1+a \right )^{2}}+\frac {1}{2 \left (-1+a \right ) \left (1+a \right ) b x}+\frac {a \ln \left (-b x \right )}{\left (-1+a \right )^{2} \left (1+a \right )^{2}}\right )\) | \(83\) |
parallelrisch | \(\frac {x^{2} \operatorname {arccoth}\left (b x +a \right ) a^{2} b^{2}+2 \ln \left (x \right ) a \,b^{2} x^{2}-2 \ln \left (b x +a -1\right ) x^{2} a \,b^{2}-2 x^{2} \operatorname {arccoth}\left (b x +a \right ) a \,b^{2}+\operatorname {arccoth}\left (b x +a \right ) b^{2} x^{2}-2 a \,b^{2} x^{2}-\operatorname {arccoth}\left (b x +a \right ) a^{4}+a^{2} b x +2 \,\operatorname {arccoth}\left (b x +a \right ) a^{2}-b x -\operatorname {arccoth}\left (b x +a \right )}{2 x^{2} \left (a^{2}+2 a +1\right ) \left (a^{2}-2 a +1\right )}\) | \(147\) |
risch | \(-\frac {\ln \left (b x +a +1\right )}{4 x^{2}}-\frac {\ln \left (-b x -a +1\right ) a^{2} b^{2} x^{2}-\ln \left (-b x -a -1\right ) a^{2} b^{2} x^{2}+2 \ln \left (-b x -a +1\right ) a \,b^{2} x^{2}+2 \ln \left (-b x -a -1\right ) a \,b^{2} x^{2}-4 \ln \left (x \right ) a \,b^{2} x^{2}+\ln \left (-b x -a +1\right ) b^{2} x^{2}-\ln \left (-b x -a -1\right ) b^{2} x^{2}-a^{4} \ln \left (b x +a -1\right )-2 a^{2} b x +2 \ln \left (b x +a -1\right ) a^{2}+2 b x -\ln \left (b x +a -1\right )}{4 x^{2} \left (a^{4}-2 a^{2}+1\right )}\) | \(201\) |
-1/2*arccoth(b*x+a)/x^2-1/2*b*(-1/(-1+a)/(1+a)/x-2*a*b/(-1+a)^2/(1+a)^2*ln (x)+1/2*b/(-1+a)^2*ln(b*x+a-1)-1/2*b*ln(b*x+a+1)/(1+a)^2)
Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23 \[ \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx=\frac {{\left (a^{2} - 2 \, a + 1\right )} b^{2} x^{2} \log \left (b x + a + 1\right ) - {\left (a^{2} + 2 \, a + 1\right )} b^{2} x^{2} \log \left (b x + a - 1\right ) + 4 \, a b^{2} x^{2} \log \left (x\right ) + 2 \, {\left (a^{2} - 1\right )} b x - {\left (a^{4} - 2 \, a^{2} + 1\right )} \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{4 \, {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}} \]
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*lo g(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)
Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (73) = 146\).
Time = 0.78 (sec) , antiderivative size = 410, normalized size of antiderivative = 4.56 \[ \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx=\begin {cases} \frac {b^{2} \operatorname {acoth}{\left (b x - 1 \right )}}{8} - \frac {b}{8 x} - \frac {\operatorname {acoth}{\left (b x - 1 \right )}}{2 x^{2}} - \frac {1}{8 x^{2}} & \text {for}\: a = -1 \\\frac {b^{2} \operatorname {acoth}{\left (b x + 1 \right )}}{8} - \frac {b}{8 x} - \frac {\operatorname {acoth}{\left (b x + 1 \right )}}{2 x^{2}} + \frac {1}{8 x^{2}} & \text {for}\: a = 1 \\- \frac {a^{4} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {a^{2} b^{2} x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {a^{2} b x}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {2 a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {2 a b^{2} x^{2} \log {\left (x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac {2 a b^{2} x^{2} \log {\left (a + b x + 1 \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {2 a b^{2} x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} + \frac {b^{2} x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac {b x}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{2 a^{4} x^{2} - 4 a^{2} x^{2} + 2 x^{2}} & \text {otherwise} \end {cases} \]
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))
Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx=\frac {1}{4} \, {\left (\frac {4 \, a b \log \left (x\right )}{a^{4} - 2 \, a^{2} + 1} + \frac {b \log \left (b x + a + 1\right )}{a^{2} + 2 \, a + 1} - \frac {b \log \left (b x + a - 1\right )}{a^{2} - 2 \, a + 1} + \frac {2}{{\left (a^{2} - 1\right )} x}\right )} b - \frac {\operatorname {arcoth}\left (b x + a\right )}{2 \, x^{2}} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (76) = 152\).
Time = 0.29 (sec) , antiderivative size = 360, normalized size of antiderivative = 4.00 \[ \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx=-\frac {1}{2} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {a b \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a - 1 \right |}}\right )}{a^{4} - 2 \, a^{2} + 1} - \frac {a b \log \left ({\left | \frac {{\left (b x + a + 1\right )} a}{b x + a - 1} - a - \frac {b x + a + 1}{b x + a - 1} - 1 \right |}\right )}{a^{4} - 2 \, a^{2} + 1} + \frac {{\left (\frac {{\left (b x + a + 1\right )} a b}{b x + a - 1} - a b - \frac {{\left (b x + a + 1\right )} b}{b x + a - 1}\right )} \log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{{\left (a^{2} - 2 \, a + 1\right )} {\left (\frac {{\left (b x + a + 1\right )} a}{b x + a - 1} - a - \frac {b x + a + 1}{b x + a - 1} - 1\right )}^{2}} + \frac {a b + b}{{\left (\frac {{\left (b x + a + 1\right )} a}{b x + a - 1} - a - \frac {b x + a + 1}{b x + a - 1} - 1\right )} {\left (a + 1\right )}^{2} {\left (a - 1\right )}^{2}}\right )} \]
-1/2*((a + 1)*b - (a - 1)*b)*(a*b*log(abs(b*x + a + 1)/abs(b*x + a - 1))/( a^4 - 2*a^2 + 1) - a*b*log(abs((b*x + a + 1)*a/(b*x + a - 1) - a - (b*x + a + 1)/(b*x + a - 1) - 1))/(a^4 - 2*a^2 + 1) + ((b*x + a + 1)*a*b/(b*x + a - 1) - a*b - (b*x + a + 1)*b/(b*x + a - 1))*log(-(1/(a - ((b*x + a + 1)*( a - 1)/(b*x + a - 1) - a - 1)*b/((b*x + a + 1)*b/(b*x + a - 1) - b)) + 1)/ (1/(a - ((b*x + a + 1)*(a - 1)/(b*x + a - 1) - a - 1)*b/((b*x + a + 1)*b/( b*x + a - 1) - b)) - 1))/((a^2 - 2*a + 1)*((b*x + a + 1)*a/(b*x + a - 1) - a - (b*x + a + 1)/(b*x + a - 1) - 1)^2) + (a*b + b)/(((b*x + a + 1)*a/(b* x + a - 1) - a - (b*x + a + 1)/(b*x + a - 1) - 1)*(a + 1)^2*(a - 1)^2))
Time = 5.60 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.74 \[ \int \frac {\coth ^{-1}(a+b x)}{x^3} \, dx=\ln \left (x\right )\,\left (\frac {b^2}{4\,{\left (a-1\right )}^2}-\frac {b^2}{4\,{\left (a+1\right )}^2}\right )-\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )\,\left (\frac {b^2}{8\,{\left (a-1\right )}^2}-\frac {b^2}{8\,{\left (a+1\right )}^2}\right )-\frac {\mathrm {acoth}\left (a+b\,x\right )\,\left (\frac {a^2}{2}-\frac {1}{2}\right )-\frac {b\,x}{2}+\frac {b^2\,x^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}+\frac {x^3\,\left (3\,a^2\,b^3+b^3\right )}{2\,{\left (a^2-1\right )}^2}+\frac {a\,b^4\,x^4}{{\left (a^2-1\right )}^2}+a\,b\,x\,\mathrm {acoth}\left (a+b\,x\right )}{a^2\,x^2+2\,a\,b\,x^3+b^2\,x^4-x^2}-\frac {\mathrm {atan}\left (\frac {2\,x\,b^2+2\,a\,b}{2\,\sqrt {b^2\,\left (a^2-1\right )-a^2\,b^2}}\right )\,\left (a^2\,b^3+b^3\right )}{\sqrt {-b^2}\,\left (2\,a^4-4\,a^2+2\right )} \]
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))