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

Optimal. Leaf size=47 \[ -\frac{\log (a+b x)}{b}+\frac{\log \left ((a+b x)^2+1\right )}{2 b}-\frac{\cot ^{-1}(a+b x)}{b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0316369, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5044, 4853, 266, 36, 29, 31} \[ -\frac{\log (a+b x)}{b}+\frac{\log \left ((a+b x)^2+1\right )}{2 b}-\frac{\cot ^{-1}(a+b x)}{b (a+b x)} \]

Antiderivative was successfully verified.

[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 5044

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]

Rule 4853

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
t[c*x])^p)/(d*(m + 1)), x] + Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCot[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cot ^{-1}(a+b x)}{(a+b x)^2} \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,(a+b x)^2\right )}{2 b}+\frac{\operatorname{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]  time = 0.0174963, size = 40, normalized size = 0.85 \[ -\frac{\log (a+b x)-\frac{1}{2} \log \left ((a+b x)^2+1\right )+\frac{\cot ^{-1}(a+b x)}{a+b x}}{b} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.046, size = 46, normalized size = 1. \begin{align*} -{\frac{{\rm arccot} \left (bx+a\right )}{b \left ( bx+a \right ) }}-{\frac{\ln \left ( bx+a \right ) }{b}}+{\frac{\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00666, size = 72, normalized size = 1.53 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.22231, size = 150, normalized size = 3.19 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 1.96155, size = 150, normalized size = 3.19 \begin{align*} \begin{cases} - \frac{2 a \log{\left (\frac{a}{b} + x \right )}}{2 a b + 2 b^{2} x} + \frac{a \log{\left (\frac{a^{2}}{b^{2}} + \frac{2 a x}{b} + x^{2} + \frac{1}{b^{2}} \right )}}{2 a b + 2 b^{2} x} - \frac{2 b x \log{\left (\frac{a}{b} + x \right )}}{2 a b + 2 b^{2} x} + \frac{b x \log{\left (\frac{a^{2}}{b^{2}} + \frac{2 a x}{b} + x^{2} + \frac{1}{b^{2}} \right )}}{2 a b + 2 b^{2} x} - \frac{2 \operatorname{acot}{\left (a + b x \right )}}{2 a b + 2 b^{2} x} & \text{for}\: b \neq 0 \\\frac{x \operatorname{acot}{\left (a \right )}}{a^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10879, size = 49, normalized size = 1.04 \begin{align*} \frac{\log \left (\frac{1}{{\left (b x + a\right )}^{2}} + 1\right )}{2 \, b} - \frac{\arctan \left (\frac{1}{b x + a}\right )}{{\left (b x + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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