\(\int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx\) [125]

Optimal result

Integrand size = 14, antiderivative size = 47 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\cot ^{-1}(a+b x)}{b (a+b x)}-\frac {\log (a+b x)}{b}+\frac {\log \left (1+(a+b x)^2\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5152, 4947, 272, 36, 29, 31} \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\log (a+b x)}{b}+\frac {\log \left ((a+b x)^2+1\right )}{2 b}-\frac {\cot ^{-1}(a+b x)}{b (a+b x)} \]

[In]

Int[ArcCot[a + b*x]/(a + b*x)^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4947

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] + Dist[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]

Rule 5152

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\cot ^{-1}(x)}{x^2} \, dx,x,a+b x\right )}{b} \\ & = -\frac {\cot ^{-1}(a+b x)}{b (a+b x)}-\frac {\text {Subst}\left (\int \frac {1}{x \left (1+x^2\right )} \, dx,x,a+b x\right )}{b} \\ & = -\frac {\cot ^{-1}(a+b x)}{b (a+b x)}-\frac {\text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,(a+b x)^2\right )}{2 b} \\ & = -\frac {\cot ^{-1}(a+b x)}{b (a+b x)}-\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,(a+b x)^2\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,(a+b x)^2\right )}{2 b} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {-\frac {\cot ^{-1}(a+b x)}{a+b x}-\log (a+b x)+\frac {1}{2} \log \left (1+(a+b x)^2\right )}{b} \]

[In]

Integrate[ArcCot[a + b*x]/(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.47 (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	\(-\frac {6 \ln \left (b x +a \right ) x a \,b^{2}-3 b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a x +6 \ln \left (b x +a \right ) a^{2} b -3 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2} b +6 \,\operatorname {arccot}\left (b x +a \right ) a b}{6 \left (b x +a \right ) a \,b^{2}}\)	\(101\)
risch	\(-\frac {i \ln \left (1+i \left (b x +a \right )\right )}{2 b \left (b x +a \right )}-\frac {2 \ln \left (-b x -a \right ) b x -\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) b x +2 \ln \left (-b x -a \right ) a -a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )-i \ln \left (1-i \left (b x +a \right )\right )+\pi }{2 \left (b x +a \right ) b}\)	\(122\)

[In]

int(arccot(b*x+a)/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, {\left (b x + a\right )} \log \left (b x + a\right ) - 2 \, \operatorname {arccot}\left (b x + a\right )}{2 \, {\left (b^{2} x + a b\right )}} \]

[In]

integrate(arccot(b*x+a)/(b*x+a)^2,x, algorithm="fricas")

[Out]

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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.96 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\begin {cases} - \frac {a \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} + \frac {a \log {\left (\frac {a}{b} + x - \frac {i}{b} \right )}}{a b + b^{2} x} + \frac {i a \operatorname {acot}{\left (a + b x \right )}}{a b + b^{2} x} - \frac {b x \log {\left (\frac {a}{b} + x \right )}}{a b + b^{2} x} + \frac {b x \log {\left (\frac {a}{b} + x - \frac {i}{b} \right )}}{a b + b^{2} x} + \frac {i b x \operatorname {acot}{\left (a + b x \right )}}{a b + b^{2} x} - \frac {\operatorname {acot}{\left (a + b x \right )}}{a b + b^{2} x} & \text {for}\: b \neq 0 \\\frac {x \operatorname {acot}{\left (a \right )}}{a^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(acot(b*x+a)/(b*x+a)**2,x)

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} - \frac {\log \left (b x + a\right )}{b} - \frac {\operatorname {arccot}\left (b x + a\right )}{{\left (b x + a\right )} b} \]

[In]

integrate(arccot(b*x+a)/(b*x+a)^2,x, algorithm="maxima")

[Out]

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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (45) = 90\).

Time = 0.33 (sec) , antiderivative size = 238, normalized size of antiderivative = 5.06 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=-\frac {\arctan \left (\frac {1}{b x + a}\right )^{2} - \frac {\arctan \left (\frac {1}{b x + a}\right )^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - \arctan \left (\frac {1}{b x + a}\right )^{2} + 4 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right )}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 1}}{2 \, b} \]

[In]

integrate(arccot(b*x+a)/(b*x+a)^2,x, algorithm="giac")

[Out]

-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

Mupad [B] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int \frac {\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx=\frac {\ln \left (-a^2-2\,a\,b\,x-b^2\,x^2-1\right )}{2\,b}-\frac {\ln \left (a+b\,x\right )}{b}-\frac {\mathrm {acot}\left (a+b\,x\right )}{x\,b^2+a\,b} \]

[In]

int(acot(a + b*x)/(a + b*x)^2,x)

[Out]

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)