Integrand size = 14, antiderivative size = 47 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\cot ^{-1}(a+b x)}{b (a+b x)}-\frac {\log (a+b x)}{b}+\frac {\log \left (1+(a+b x)^2\right )}{2 b} \] Output:
-arccot(b*x+a)/b/(b*x+a)-ln(b*x+a)/b+1/2*ln(1+(b*x+a)^2)/b
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {-\frac {\cot ^{-1}(a+b x)}{a+b x}-\log (a+b x)+\frac {1}{2} \log \left (1+(a+b x)^2\right )}{b} \] Input:
Integrate[ArcCot[a + b*x]/(a + b*x)^2,x]
Output:
(-(ArcCot[a + b*x]/(a + b*x)) - Log[a + b*x] + Log[1 + (a + b*x)^2]/2)/b
Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5567, 5362, 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.
\(\displaystyle \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx\) |
\(\Big \downarrow \) 5567 |
\(\displaystyle \frac {\int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2}d(a+b x)}{b}\) |
\(\Big \downarrow \) 5362 |
\(\displaystyle \frac {-\int \frac {1}{(a+b x) \left ((a+b x)^2+1\right )}d(a+b x)-\frac {\cot ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {-\frac {1}{2} \int \frac {1}{(a+b x)^2 \left ((a+b x)^2+1\right )}d(a+b x)^2-\frac {\cot ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {\frac {1}{2} \left (\int \frac {1}{(a+b x)^2+1}d(a+b x)^2-\int \frac {1}{(a+b x)^2}d(a+b x)^2\right )-\frac {\cot ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {1}{2} \left (\int \frac {1}{(a+b x)^2+1}d(a+b x)^2-\log \left ((a+b x)^2\right )\right )-\frac {\cot ^{-1}(a+b x)}{a+b x}}{b}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {1}{2} \left (\log \left ((a+b x)^2+1\right )-\log \left ((a+b x)^2\right )\right )-\frac {\cot ^{-1}(a+b x)}{a+b x}}{b}\) |
Input:
Int[ArcCot[a + b*x]/(a + b*x)^2,x]
Output:
(-(ArcCot[a + b*x]/(a + b*x)) + (-Log[(a + b*x)^2] + Log[1 + (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_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCot[c*x^n])^p/(m + 1)), x] + Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcCot[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_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcCot[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 = 0.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(\frac {-\frac {\operatorname {arccot}\left (b x +a \right )}{b x +a}-\ln \left (b x +a \right )+\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}}{b}\) | \(41\) |
default | \(\frac {-\frac {\operatorname {arccot}\left (b x +a \right )}{b x +a}-\ln \left (b x +a \right )+\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}}{b}\) | \(41\) |
parts | \(-\frac {\operatorname {arccot}\left (b x +a \right )}{b \left (b x +a \right )}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b}-\frac {\ln \left (b x +a \right )}{b}\) | \(54\) |
parallelrisch | \(-\frac {6 \ln \left (b x +a \right ) x a \,b^{2}-3 b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a x +6 \ln \left (b x +a \right ) a^{2} b -3 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2} b +6 \,\operatorname {arccot}\left (b x +a \right ) a b}{6 \left (b x +a \right ) b^{2} a}\) | \(101\) |
risch | \(-\frac {i \ln \left (1+i \left (b x +a \right )\right )}{2 b \left (b x +a \right )}-\frac {2 \ln \left (-b x -a \right ) b x -\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) b x +2 \ln \left (-b x -a \right ) a -a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )-i \ln \left (1-i \left (b x +a \right )\right )+\pi }{2 b \left (b x +a \right )}\) | \(122\) |
Input:
int(arccot(b*x+a)/(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(-arccot(b*x+a)/(b*x+a)-ln(b*x+a)+1/2*ln(1+(b*x+a)^2))
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, {\left (b x + a\right )} \log \left (b x + a\right ) - 2 \, \operatorname {arccot}\left (b x + a\right )}{2 \, {\left (b^{2} x + a b\right )}} \] Input:
integrate(arccot(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) - 2*arccot(b*x + a))/(b^2*x + a*b)
Result contains complex when optimal does not.
Time = 4.68 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.96 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\begin {cases} - \frac {a \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} + \frac {a \log {\left (\frac {a}{b} + x - \frac {i}{b} \right )}}{a b + b^{2} x} + \frac {i a \operatorname {acot}{\left (a + b x \right )}}{a b + b^{2} x} - \frac {b x \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} + \frac {b x \log {\left (\frac {a}{b} + x - \frac {i}{b} \right )}}{a b + b^{2} x} + \frac {i b x \operatorname {acot}{\left (a + b x \right )}}{a b + b^{2} x} - \frac {\operatorname {acot}{\left (a + b x \right )}}{a b + b^{2} x} & \text {for}\: b \neq 0 \\\frac {x \operatorname {acot}{\left (a \right )}}{a^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(acot(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 - I/b)/(a*b + b* *2*x) + I*a*acot(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 - I/b)/(a*b + b**2*x) + I*b*x*acot(a + b*x)/(a*b + b**2 *x) - acot(a + b*x)/(a*b + b**2*x), Ne(b, 0)), (x*acot(a)/a**2, True))
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} - \frac {\log \left (b x + a\right )}{b} - \frac {\operatorname {arccot}\left (b x + a\right )}{{\left (b x + a\right )} b} \] Input:
integrate(arccot(b*x+a)/(b*x+a)^2,x, algorithm="maxima")
Output:
1/2*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b - log(b*x + a)/b - arccot(b*x + a)/ ((b*x + a)*b)
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (45) = 90\).
Time = 0.16 (sec) , antiderivative size = 238, normalized size of antiderivative = 5.06 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\arctan \left (\frac {1}{b x + a}\right )^{2} - \frac {\arctan \left (\frac {1}{b x + a}\right )^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - \arctan \left (\frac {1}{b x + a}\right )^{2} + 4 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right )}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 1}}{2 \, b} \] Input:
integrate(arccot(b*x+a)/(b*x+a)^2,x, algorithm="giac")
Output:
-1/2*(arctan(1/(b*x + a))^2 - (arctan(1/(b*x + a))^2*tan(1/2*arctan(1/(b*x + a)))^2 - log(4*(tan(1/2*arctan(1/(b*x + a)))^4 - 2*tan(1/2*arctan(1/(b* x + a)))^2 + 1)/(tan(1/2*arctan(1/(b*x + a)))^4 + 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1))*tan(1/2*arctan(1/(b*x + a)))^2 - arctan(1/(b*x + a))^2 + 4* arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a))) + log(4*(tan(1/2*arctan(1 /(b*x + a)))^4 - 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)/(tan(1/2*arctan(1/( b*x + a)))^4 + 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1)))/(tan(1/2*arctan(1/( b*x + a)))^2 - 1))/b
Time = 0.91 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {\ln \left (-a^2-2\,a\,b\,x-b^2\,x^2-1\right )}{2\,b}-\frac {\ln \left (a+b\,x\right )}{b}-\frac {\mathrm {acot}\left (a+b\,x\right )}{x\,b^2+a\,b} \] Input:
int(acot(a + b*x)/(a + b*x)^2,x)
Output:
log(- a^2 - b^2*x^2 - 2*a*b*x - 1)/(2*b) - log(a + b*x)/b - acot(a + b*x)/ (a*b + b^2*x)
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.36 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {2 \mathit {acot} \left (b x +a \right ) b x +2 \mathit {atan} \left (b x +a \right ) a +2 \mathit {atan} \left (b x +a \right ) b x +\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}+\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a b x -2 \,\mathrm {log}\left (b x +a \right ) a^{2}-2 \,\mathrm {log}\left (b x +a \right ) a b x}{2 a b \left (b x +a \right )} \] Input:
int(acot(b*x+a)/(b*x+a)^2,x)
Output:
(2*acot(a + b*x)*b*x + 2*atan(a + b*x)*a + 2*atan(a + b*x)*b*x + log(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**2 + log(a**2 + 2*a*b*x + b**2*x**2 + 1)*a*b* x - 2*log(a + b*x)*a**2 - 2*log(a + b*x)*a*b*x)/(2*a*b*(a + b*x))