Integrand size = 10, antiderivative size = 120 \[ \int \frac {\cot ^{-1}(a+b x)}{x} \, dx=-\cot ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )+\cot ^{-1}(a+b x) \log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right ) \]
-arccot(b*x+a)*ln(2/(1-I*(b*x+a)))+arccot(b*x+a)*ln(2*b*x/(I-a)/(1-I*(b*x+ a)))-1/2*I*polylog(2,1-2/(1-I*(b*x+a)))+1/2*I*polylog(2,1-2*b*x/(I-a)/(1-I *(b*x+a)))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(256\) vs. \(2(120)=240\).
Time = 0.15 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.13 \[ \int \frac {\cot ^{-1}(a+b x)}{x} \, dx=\left (\cot ^{-1}(a+b x)+\arctan (a+b x)\right ) \log (x)+\arctan (a+b x) \left (\log \left (\frac {1}{\sqrt {1+(a+b x)^2}}\right )-\log (-\sin (\arctan (a)-\arctan (a+b x)))\right )+\frac {1}{2} \left (\frac {1}{4} i (\pi -2 \arctan (a+b x))^2+i (\arctan (a)-\arctan (a+b x))^2-(\pi -2 \arctan (a+b x)) \log \left (1+e^{-2 i \arctan (a+b x)}\right )+2 (\arctan (a)-\arctan (a+b x)) \log \left (1-e^{2 i (-\arctan (a)+\arctan (a+b x))}\right )+(\pi -2 \arctan (a+b x)) \log \left (\frac {2}{\sqrt {1+(a+b x)^2}}\right )+2 (-\arctan (a)+\arctan (a+b x)) \log (-2 \sin (\arctan (a)-\arctan (a+b x)))+i \operatorname {PolyLog}\left (2,-e^{-2 i \arctan (a+b x)}\right )+i \operatorname {PolyLog}\left (2,e^{2 i (-\arctan (a)+\arctan (a+b x))}\right )\right ) \]
(ArcCot[a + b*x] + ArcTan[a + b*x])*Log[x] + ArcTan[a + b*x]*(Log[1/Sqrt[1 + (a + b*x)^2]] - Log[-Sin[ArcTan[a] - ArcTan[a + b*x]]]) + ((I/4)*(Pi - 2*ArcTan[a + b*x])^2 + I*(ArcTan[a] - ArcTan[a + b*x])^2 - (Pi - 2*ArcTan[ a + b*x])*Log[1 + E^((-2*I)*ArcTan[a + b*x])] + 2*(ArcTan[a] - ArcTan[a + b*x])*Log[1 - E^((2*I)*(-ArcTan[a] + ArcTan[a + b*x]))] + (Pi - 2*ArcTan[a + b*x])*Log[2/Sqrt[1 + (a + b*x)^2]] + 2*(-ArcTan[a] + ArcTan[a + b*x])*L og[-2*Sin[ArcTan[a] - ArcTan[a + b*x]]] + I*PolyLog[2, -E^((-2*I)*ArcTan[a + b*x])] + I*PolyLog[2, E^((2*I)*(-ArcTan[a] + ArcTan[a + b*x]))])/2
Time = 0.47 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5571, 25, 27, 5382, 2849, 2752, 2897}
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.
| \(\displaystyle \int \frac {\cot ^{-1}(a+b x)}{x} \, dx\) |
| \(\Big \downarrow \) 5571 |
| \(\displaystyle \frac {\int \frac {\cot ^{-1}(a+b x)}{x}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int -\frac {\cot ^{-1}(a+b x)}{x}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int -\frac {\cot ^{-1}(a+b x)}{b x}d(a+b x)\) |
| \(\Big \downarrow \) 5382 |
| \(\displaystyle -\int \frac {\log \left (\frac {2}{1-i (a+b x)}\right )}{(a+b x)^2+1}d(a+b x)+\int \frac {\log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )}{(a+b x)^2+1}d(a+b x)+\log \left (\frac {2}{1-i (a+b x)}\right ) \left (-\cot ^{-1}(a+b x)\right )+\log \left (\frac {2 b x}{(-a+i) (1-i (a+b x))}\right ) \cot ^{-1}(a+b x)\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle -i \int \frac {\log \left (\frac {2}{1-i (a+b x)}\right )}{1-\frac {2}{1-i (a+b x)}}d\frac {1}{1-i (a+b x)}+\int \frac {\log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )}{(a+b x)^2+1}d(a+b x)+\log \left (\frac {2}{1-i (a+b x)}\right ) \left (-\cot ^{-1}(a+b x)\right )+\log \left (\frac {2 b x}{(-a+i) (1-i (a+b x))}\right ) \cot ^{-1}(a+b x)\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle \int \frac {\log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )}{(a+b x)^2+1}d(a+b x)-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )+\log \left (\frac {2}{1-i (a+b x)}\right ) \left (-\cot ^{-1}(a+b x)\right )+\log \left (\frac {2 b x}{(-a+i) (1-i (a+b x))}\right ) \cot ^{-1}(a+b x)\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle -\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right )+\log \left (\frac {2}{1-i (a+b x)}\right ) \left (-\cot ^{-1}(a+b x)\right )+\log \left (\frac {2 b x}{(-a+i) (1-i (a+b x))}\right ) \cot ^{-1}(a+b x)\) |
-(ArcCot[a + b*x]*Log[2/(1 - I*(a + b*x))]) + ArcCot[a + b*x]*Log[(2*b*x)/ ((I - a)*(1 - I*(a + b*x)))] - (I/2)*PolyLog[2, 1 - 2/(1 - I*(a + b*x))] + (I/2)*PolyLog[2, 1 - (2*b*x)/((I - a)*(1 - I*(a + b*x)))]
3.2.3.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-e/g Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ {c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Si mp[(-(a + b*ArcCot[c*x]))*(Log[2/(1 - I*c*x)]/e), x] + (Simp[(a + b*ArcCot[ c*x])*(Log[2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] - Simp[b*(c/e) Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] + Simp[b*(c/e) Int[Log[2* c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(1 + c^2*x^2), x], x]) /; FreeQ[{a , b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]
Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcCot[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && I GtQ[p, 0]
Time = 0.39 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86
| method | result | size |
| risch | \(\frac {\pi \ln \left (-i b x \right )}{2}-\frac {i \ln \left (-i b x -i a +1\right ) \ln \left (-\frac {i x b}{i a -1}\right )}{2}-\frac {i \operatorname {dilog}\left (-\frac {i x b}{i a -1}\right )}{2}+\frac {i \ln \left (i b x +i a +1\right ) \ln \left (\frac {i x b}{-i a -1}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i x b}{-i a -1}\right )}{2}\) | \(103\) |
| parts | \(\ln \left (x \right ) \operatorname {arccot}\left (b x +a \right )+b \left (-\frac {i \ln \left (x \right ) \left (\ln \left (\frac {-b x -a +i}{i-a}\right )-\ln \left (\frac {b x +a +i}{i+a}\right )\right )}{2 b}-\frac {i \left (\operatorname {dilog}\left (\frac {-b x -a +i}{i-a}\right )-\operatorname {dilog}\left (\frac {b x +a +i}{i+a}\right )\right )}{2 b}\right )\) | \(104\) |
| derivativedivides | \(\ln \left (-b x \right ) \operatorname {arccot}\left (b x +a \right )+\frac {i \ln \left (-b x \right ) \ln \left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \ln \left (-b x \right ) \ln \left (\frac {-b x -a +i}{i-a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {-b x -a +i}{i-a}\right )}{2}\) | \(106\) |
| default | \(\ln \left (-b x \right ) \operatorname {arccot}\left (b x +a \right )+\frac {i \ln \left (-b x \right ) \ln \left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \ln \left (-b x \right ) \ln \left (\frac {-b x -a +i}{i-a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {-b x -a +i}{i-a}\right )}{2}\) | \(106\) |
1/2*Pi*ln(-I*b*x)-1/2*I*ln(1-I*a-I*b*x)*ln(-I*x*b/(I*a-1))-1/2*I*dilog(-I* x*b/(I*a-1))+1/2*I*ln(1+I*a+I*b*x)*ln(I*x*b/(-I*a-1))+1/2*I*dilog(I*x*b/(- I*a-1))
\[ \int \frac {\cot ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arccot}\left (b x + a\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\cot ^{-1}(a+b x)}{x} \, dx=\text {Timed out} \]
Time = 0.34 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11 \[ \int \frac {\cot ^{-1}(a+b x)}{x} \, dx=\frac {1}{2} \, \arctan \left (\frac {b x}{a^{2} + 1}, -\frac {a b x}{a^{2} + 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - \frac {1}{2} \, \arctan \left (b x + a\right ) \log \left (\frac {b^{2} x^{2}}{a^{2} + 1}\right ) + \operatorname {arccot}\left (b x + a\right ) \log \left (x\right ) + \arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (x\right ) + \frac {1}{2} i \, {\rm Li}_2\left (\frac {i \, b x + i \, a + 1}{i \, a + 1}\right ) - \frac {1}{2} i \, {\rm Li}_2\left (\frac {i \, b x + i \, a - 1}{i \, a - 1}\right ) \]
1/2*arctan2(b*x/(a^2 + 1), -a*b*x/(a^2 + 1))*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - 1/2*arctan(b*x + a)*log(b^2*x^2/(a^2 + 1)) + arccot(b*x + a)*log(x) + arctan((b^2*x + a*b)/b)*log(x) + 1/2*I*dilog((I*b*x + I*a + 1)/(I*a + 1) ) - 1/2*I*dilog((I*b*x + I*a - 1)/(I*a - 1))
\[ \int \frac {\cot ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arccot}\left (b x + a\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\cot ^{-1}(a+b x)}{x} \, dx=\int \frac {\mathrm {acot}\left (a+b\,x\right )}{x} \,d x \]