Integrand size = 14, antiderivative size = 48 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\coth ^{-1}(a+b x)}{b (a+b x)}+\frac {\log (a+b x)}{b}-\frac {\log \left (1-(a+b x)^2\right )}{2 b} \] Output:
-arccoth(b*x+a)/b/(b*x+a)+ln(b*x+a)/b-1/2*ln(1-(b*x+a)^2)/b
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\frac {2 \coth ^{-1}(a+b x)}{a+b x}-2 \log (a+b x)+\log \left (1-(a+b x)^2\right )}{2 b} \] Input:
Integrate[ArcCoth[a + b*x]/(a + b*x)^2,x]
Output:
-1/2*((2*ArcCoth[a + b*x])/(a + b*x) - 2*Log[a + b*x] + Log[1 - (a + b*x)^ 2])/b
Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6658, 6453, 243, 47, 14, 16}
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.
\(\displaystyle \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx\) |
\(\Big \downarrow \) 6658 |
\(\displaystyle \frac {\int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2}d(a+b x)}{b}\) |
\(\Big \downarrow \) 6453 |
\(\displaystyle \frac {\int \frac {1}{(a+b x) \left (1-(a+b x)^2\right )}d(a+b x)-\frac {\coth ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{(-a-b x+1) (a+b x)^2}d(a+b x)^2-\frac {\coth ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {\frac {1}{2} \left (\int \frac {1}{-a-b x+1}d(a+b x)^2+\int \frac {1}{(a+b x)^2}d(a+b x)^2\right )-\frac {\coth ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {1}{2} \left (\int \frac {1}{-a-b x+1}d(a+b x)^2+\log \left ((a+b x)^2\right )\right )-\frac {\coth ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {1}{2} \left (\log \left ((a+b x)^2\right )-\log (-a-b x+1)\right )-\frac {\coth ^{-1}(a+b x)}{a+b x}}{b}\) |
Input:
Int[ArcCoth[a + b*x]/(a + b*x)^2,x]
Output:
(-(ArcCoth[a + b*x]/(a + b*x)) + (-Log[1 - a - b*x] + Log[(a + b*x)^2])/2) /b
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcCoth[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[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]
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {\operatorname {arccoth}\left (b x +a \right )}{b x +a}+\ln \left (b x +a \right )-\frac {\ln \left (b x +a +1\right )}{2}-\frac {\ln \left (b x +a -1\right )}{2}}{b}\) | \(45\) |
default | \(\frac {-\frac {\operatorname {arccoth}\left (b x +a \right )}{b x +a}+\ln \left (b x +a \right )-\frac {\ln \left (b x +a +1\right )}{2}-\frac {\ln \left (b x +a -1\right )}{2}}{b}\) | \(45\) |
parts | \(-\frac {\operatorname {arccoth}\left (b x +a \right )}{b \left (b x +a \right )}-\frac {\ln \left (b x +a -1\right )}{2 b}-\frac {\ln \left (b x +a +1\right )}{2 b}+\frac {\ln \left (b x +a \right )}{b}\) | \(54\) |
parallelrisch | \(\frac {3 \ln \left (b x +a \right ) x a \,b^{2}-3 \ln \left (b x +a -1\right ) x a \,b^{2}-3 x \,\operatorname {arccoth}\left (b x +a \right ) a \,b^{2}+3 \ln \left (b x +a \right ) a^{2} b -3 \ln \left (b x +a -1\right ) a^{2} b -3 \,\operatorname {arccoth}\left (b x +a \right ) a^{2} b -3 \,\operatorname {arccoth}\left (b x +a \right ) a b}{3 \left (b x +a \right ) a \,b^{2}}\) | \(104\) |
risch | \(-\frac {\ln \left (b x +a +1\right )}{2 b \left (b x +a \right )}+\frac {2 \ln \left (-b x -a \right ) b x -\ln \left (b^{2} x^{2}+2 b x a +a^{2}-1\right ) b x +2 \ln \left (-b x -a \right ) a -\ln \left (b^{2} x^{2}+2 b x a +a^{2}-1\right ) a +\ln \left (b x +a -1\right )}{2 b \left (b x +a \right )}\) | \(109\) |
Input:
int(arccoth(b*x+a)/(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(-arccoth(b*x+a)/(b*x+a)+ln(b*x+a)-1/2*ln(b*x+a+1)-1/2*ln(b*x+a-1))
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.40 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {{\left (b x + a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right ) - 2 \, {\left (b x + a\right )} \log \left (b x + a\right ) + \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{2 \, {\left (b^{2} x + a b\right )}} \] Input:
integrate(arccoth(b*x+a)/(b*x+a)^2,x, algorithm="fricas")
Output:
-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)
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (34) = 68\).
Time = 0.79 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.83 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=\begin {cases} \frac {a \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} - \frac {a \log {\left (\frac {a}{b} + x + \frac {1}{b} \right )}}{a b + b^{2} x} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{a b + b^{2} x} + \frac {b x \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} - \frac {b x \log {\left (\frac {a}{b} + x + \frac {1}{b} \right )}}{a b + b^{2} x} + \frac {b x \operatorname {acoth}{\left (a + b x \right )}}{a b + b^{2} x} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{a b + b^{2} x} & \text {for}\: b \neq 0 \\\frac {x \operatorname {acoth}{\left (a \right )}}{a^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(acoth(b*x+a)/(b*x+a)**2,x)
Output:
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))
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\log \left (b x + a + 1\right )}{2 \, b} + \frac {\log \left (b x + a\right )}{b} - \frac {\log \left (b x + a - 1\right )}{2 \, b} - \frac {\operatorname {arcoth}\left (b x + a\right )}{{\left (b x + a\right )} b} \] Input:
integrate(arccoth(b*x+a)/(b*x+a)^2,x, algorithm="maxima")
Output:
-1/2*log(b*x + a + 1)/b + log(b*x + a)/b - 1/2*log(b*x + a - 1)/b - arccot h(b*x + a)/((b*x + a)*b)
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (46) = 92\).
Time = 0.13 (sec) , antiderivative size = 198, normalized size of antiderivative = 4.12 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {1}{2} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {\log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a - 1 \right |}}\right )}{b^{2}} - \frac {\log \left ({\left | \frac {b x + a + 1}{b x + a - 1} + 1 \right |}\right )}{b^{2}} - \frac {\log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} + 1\right )}}\right )} \] Input:
integrate(arccoth(b*x+a)/(b*x+a)^2,x, algorithm="giac")
Output:
-1/2*((a + 1)*b - (a - 1)*b)*(log(abs(b*x + a + 1)/abs(b*x + a - 1))/b^2 - log(abs((b*x + a + 1)/(b*x + a - 1) + 1))/b^2 - 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))/(b^2*((b*x + a + 1)/(b*x + a - 1) + 1)))
Time = 4.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.94 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {\ln \left (a+b\,x\right )}{b}-\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )}{2\,b}-\frac {\ln \left (\frac {a+b\,x+1}{a+b\,x}\right )}{2\,\left (x\,b^2+a\,b\right )}+\frac {\ln \left (\frac {a+b\,x-1}{a+b\,x}\right )}{2\,x\,b^2+2\,a\,b} \] Input:
int(acoth(a + b*x)/(a + b*x)^2,x)
Output:
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)
Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.75 \[ \int \frac {\coth ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {2 \mathit {acoth} \left (b x +a \right ) b x +\mathrm {log}\left (b x +a -1\right ) a^{2}+\mathrm {log}\left (b x +a -1\right ) a b x -\mathrm {log}\left (b x +a -1\right ) a -\mathrm {log}\left (b x +a -1\right ) b x +\mathrm {log}\left (b x +a +1\right ) a^{2}+\mathrm {log}\left (b x +a +1\right ) a b x +\mathrm {log}\left (b x +a +1\right ) a +\mathrm {log}\left (b x +a +1\right ) b x -2 \,\mathrm {log}\left (b x +a \right ) a^{2}-2 \,\mathrm {log}\left (b x +a \right ) a b x}{2 a b \left (b x +a \right )} \] Input:
int(acoth(b*x+a)/(b*x+a)^2,x)
Output:
(2*acoth(a + b*x)*b*x + log(a + b*x - 1)*a**2 + log(a + b*x - 1)*a*b*x - l og(a + b*x - 1)*a - log(a + b*x - 1)*b*x + log(a + b*x + 1)*a**2 + log(a + b*x + 1)*a*b*x + log(a + b*x + 1)*a + log(a + b*x + 1)*b*x - 2*log(a + b* x)*a**2 - 2*log(a + b*x)*a*b*x)/(2*a*b*(a + b*x))