\(\int \cot ^{-1}(a+b x) \, dx\) [102]

Optimal result

Integrand size = 6, antiderivative size = 33 \[ \int \cot ^{-1}(a+b x) \, dx=\frac {(a+b x) \cot ^{-1}(a+b x)}{b}+\frac {\log \left (1+(a+b x)^2\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5148, 4931, 266} \[ \int \cot ^{-1}(a+b x) \, dx=\frac {\log \left ((a+b x)^2+1\right )}{2 b}+\frac {(a+b x) \cot ^{-1}(a+b x)}{b} \]

[In]

Int[ArcCot[a + b*x],x]

[Out]

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

Rule 266

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

Rule 4931

Int[((a_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x^n])^p, x] + Dist[b*c
*n*p, Int[x^n*((a + b*ArcCot[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 5148

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCot[x])^p, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \cot ^{-1}(x) \, dx,x,a+b x\right )}{b} \\ & = \frac {(a+b x) \cot ^{-1}(a+b x)}{b}+\frac {\text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{b} \\ & = \frac {(a+b x) \cot ^{-1}(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 = 44, normalized size of antiderivative = 1.33 \[ \int \cot ^{-1}(a+b x) \, dx=x \cot ^{-1}(a+b x)+\frac {-2 a \arctan (a+b x)+\log \left (1+a^2+2 a b x+b^2 x^2\right )}{2 b} \]

[In]

Integrate[ArcCot[a + b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\left (b x +a \right ) \operatorname {arccot}\left (b x +a \right )+\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}}{b}\)	\(30\)
default	\(\frac {\left (b x +a \right ) \operatorname {arccot}\left (b x +a \right )+\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}}{b}\)	\(30\)
parallelrisch	\(\frac {2 x \,\operatorname {arccot}\left (b x +a \right ) b^{2}+2 \,\operatorname {arccot}\left (b x +a \right ) a b +\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) b}{2 b^{2}}\)	\(49\)
parts	\(x \,\operatorname {arccot}\left (b x +a \right )+b \left (\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}-\frac {a \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b^{2}}\right )\)	\(59\)
risch	\(\frac {i x \ln \left (1+i \left (b x +a \right )\right )}{2}-\frac {i x \ln \left (1-i \left (b x +a \right )\right )}{2}+\frac {\pi x}{2}-\frac {a \arctan \left (b x +a \right )}{b}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \cot ^{-1}(a+b x) \, dx=\frac {2 \, b x \operatorname {arccot}\left (b x + a\right ) - 2 \, a \arctan \left (b x + a\right ) + \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \cot ^{-1}(a+b x) \, dx=\begin {cases} \frac {a \operatorname {acot}{\left (a + b x \right )}}{b} + x \operatorname {acot}{\left (a + b x \right )} + \frac {\log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b} & \text {for}\: b \neq 0 \\x \operatorname {acot}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*acot(a + b*x)/b + x*acot(a + b*x) + log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*b), Ne(b, 0)), (x*acot
(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \cot ^{-1}(a+b x) \, dx=\frac {2 \, {\left (b x + a\right )} \operatorname {arccot}\left (b x + a\right ) + \log \left ({\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (31) = 62\).

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.36 \[ \int \cot ^{-1}(a+b x) \, dx=-\frac {\arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + \log \left (\frac {16 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) - \arctan \left (\frac {1}{b x + a}\right )}{2 \, b \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )} \]

[In]

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

[Out]

-1/2*(arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^2 + log(16*tan(1/2*arctan(1/(b*x + a)))^2/(tan(1/2*arct
an(1/(b*x + a)))^4 + 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1))*tan(1/2*arctan(1/(b*x + a))) - arctan(1/(b*x + a))
)/(b*tan(1/2*arctan(1/(b*x + a))))

Mupad [B] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \cot ^{-1}(a+b x) \, dx=\frac {\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}+a\,\mathrm {acot}\left (a+b\,x\right )}{b}+x\,\mathrm {acot}\left (a+b\,x\right ) \]

[In]

int(acot(a + b*x),x)

[Out]

(log(a^2 + b^2*x^2 + 2*a*b*x + 1)/2 + a*acot(a + b*x))/b + x*acot(a + b*x)