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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 47, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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 = 0.02, 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)

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 59, normalized size = 1.26 \[ \frac {{\left (b x + a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, {\left (b x + a\right )} \log \left (b x + a\right ) - 2 \, \operatorname {arccot}\left (b x + a\right )}{2 \, {\left (b^{2} x + a b\right )}} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.17, size = 238, normalized size = 5.06 \[ -\frac {\arctan \left (\frac {1}{b x + a}\right )^{2} - \frac {\arctan \left (\frac {1}{b x + a}\right )^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - \arctan \left (\frac {1}{b x + a}\right )^{2} + 4 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right )}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 1}}{2 \, b} \]

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*(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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 46, normalized size = 0.98 \[ -\frac {\mathrm {arccot}\left (b x +a \right )}{b \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{b}+\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2 b} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 53, normalized size = 1.13 \[ \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} - \frac {\log \left (b x + a\right )}{b} - \frac {\operatorname {arccot}\left (b x + a\right )}{{\left (b x + a\right )} b} \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.77, size = 57, normalized size = 1.21 \[ \frac {\ln \left (-a^2-2\,a\,b\,x-b^2\,x^2-1\right )}{2\,b}-\frac {\ln \left (a+b\,x\right )}{b}-\frac {\mathrm {acot}\left (a+b\,x\right )}{x\,b^2+a\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 6.14, size = 151, normalized size = 3.21 \[ \begin {cases} \frac {a \log {\left (\frac {a}{b} + x \right )}}{- a b - b^{2} x} - \frac {a \log {\left (\frac {a}{b} + x - \frac {i}{b} \right )}}{- a b - b^{2} x} - \frac {i a \operatorname {acot}{\left (a + b x \right )}}{- a b - b^{2} x} + \frac {b x \log {\left (\frac {a}{b} + x \right )}}{- a b - b^{2} x} - \frac {b x \log {\left (\frac {a}{b} + x - \frac {i}{b} \right )}}{- a b - b^{2} x} - \frac {i b x \operatorname {acot}{\left (a + b x \right )}}{- a b - b^{2} x} + \frac {\operatorname {acot}{\left (a + b x \right )}}{- a b - b^{2} x} & \text {for}\: b \neq 0 \\\frac {x \operatorname {acot}{\relax (a )}}{a^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________