\(\int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx\) [127]

Optimal result

Integrand size = 19, antiderivative size = 45 \[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=-\frac {i \operatorname {PolyLog}\left (2,-\frac {i}{a+b x}\right )}{2 d}+\frac {i \operatorname {PolyLog}\left (2,\frac {i}{a+b x}\right )}{2 d} \]

[Out]

-1/2*I*polylog(2,-I/(b*x+a))/d+1/2*I*polylog(2,I/(b*x+a))/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5152, 12, 4941, 2438} \[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {i \operatorname {PolyLog}\left (2,\frac {i}{a+b x}\right )}{2 d}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i}{a+b x}\right )}{2 d} \]

[In]

Int[ArcCot[a + b*x]/((a*d)/b + d*x),x]

[Out]

((-1/2*I)*PolyLog[2, (-I)/(a + b*x)])/d + ((I/2)*PolyLog[2, I/(a + b*x)])/d

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4941

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Dist[I*(b/2), Int[Log[1 + I/(c
*x)]/x, x], x] + Dist[I*(b/2), Int[Log[1 - I/(c*x)]/x, x], x]) /; FreeQ[{a, b, c}, x]

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 {b \cot ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\cot ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d} \\ & = \frac {i \text {Subst}\left (\int \frac {\log \left (1-\frac {i}{x}\right )}{x} \, dx,x,a+b x\right )}{2 d}-\frac {i \text {Subst}\left (\int \frac {\log \left (1+\frac {i}{x}\right )}{x} \, dx,x,a+b x\right )}{2 d} \\ & = -\frac {i \operatorname {PolyLog}\left (2,-\frac {i}{a+b x}\right )}{2 d}+\frac {i \operatorname {PolyLog}\left (2,\frac {i}{a+b x}\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=-\frac {i \left (\operatorname {PolyLog}\left (2,-\frac {i}{a+b x}\right )-\operatorname {PolyLog}\left (2,\frac {i}{a+b x}\right )\right )}{2 d} \]

[In]

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

[Out]

((-1/2*I)*(PolyLog[2, (-I)/(a + b*x)] - PolyLog[2, I/(a + b*x)]))/d

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {\pi \ln \left (-i b x -i a \right )}{2 d}+\frac {i \operatorname {dilog}\left (-i b x -i a +1\right )}{2 d}-\frac {i \operatorname {dilog}\left (i b x +i a +1\right )}{2 d}\)	\(55\)
parts	\(\frac {\ln \left (b x +a \right ) \operatorname {arccot}\left (b x +a \right )}{d}+\frac {-\frac {i \ln \left (b x +a \right ) \ln \left (1+i \left (b x +a \right )\right )}{2}+\frac {i \ln \left (b x +a \right ) \ln \left (1-i \left (b x +a \right )\right )}{2}-\frac {i \operatorname {dilog}\left (1+i \left (b x +a \right )\right )}{2}+\frac {i \operatorname {dilog}\left (1-i \left (b x +a \right )\right )}{2}}{d}\)	\(91\)
derivativedivides	\(\frac {\frac {b \ln \left (b x +a \right ) \operatorname {arccot}\left (b x +a \right )}{d}+\frac {b \left (-\frac {i \ln \left (b x +a \right ) \ln \left (1+i \left (b x +a \right )\right )}{2}+\frac {i \ln \left (b x +a \right ) \ln \left (1-i \left (b x +a \right )\right )}{2}-\frac {i \operatorname {dilog}\left (1+i \left (b x +a \right )\right )}{2}+\frac {i \operatorname {dilog}\left (1-i \left (b x +a \right )\right )}{2}\right )}{d}}{b}\)	\(97\)
default	\(\frac {\frac {b \ln \left (b x +a \right ) \operatorname {arccot}\left (b x +a \right )}{d}+\frac {b \left (-\frac {i \ln \left (b x +a \right ) \ln \left (1+i \left (b x +a \right )\right )}{2}+\frac {i \ln \left (b x +a \right ) \ln \left (1-i \left (b x +a \right )\right )}{2}-\frac {i \operatorname {dilog}\left (1+i \left (b x +a \right )\right )}{2}+\frac {i \operatorname {dilog}\left (1-i \left (b x +a \right )\right )}{2}\right )}{d}}{b}\)	\(97\)

[In]

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

[Out]

1/2/d*Pi*ln(-I*a-I*b*x)+1/2*I/d*dilog(1-I*a-I*b*x)-1/2*I/d*dilog(1+I*a+I*b*x)

Fricas [F]

\[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arccot}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \]

[In]

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

[Out]

integral(b*arccot(b*x + a)/(b*d*x + a*d), x)

Sympy [F]

\[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {b \int \frac {\operatorname {acot}{\left (a + b x \right )}}{a + b x}\, dx}{d} \]

[In]

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

[Out]

b*Integral(acot(a + b*x)/(a + b*x), x)/d

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (31) = 62\).

Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.71 \[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {\operatorname {arccot}\left (b x + a\right ) \log \left (d x + \frac {a d}{b}\right )}{d} + \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (d x + \frac {a d}{b}\right )}{d} + \frac {\arctan \left (b x + a, 0\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, \arctan \left (b x + a\right ) \log \left ({\left | b x + a \right |}\right ) + i \, {\rm Li}_2\left (i \, b x + i \, a + 1\right ) - i \, {\rm Li}_2\left (-i \, b x - i \, a + 1\right )}{2 \, d} \]

[In]

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

[Out]

arccot(b*x + a)*log(d*x + a*d/b)/d + arctan((b^2*x + a*b)/b)*log(d*x + a*d/b)/d + 1/2*(arctan2(b*x + a, 0)*log
(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*arctan(b*x + a)*log(abs(b*x + a)) + I*dilog(I*b*x + I*a + 1) - I*dilog(-I*b*
x - I*a + 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (31) = 62\).

Time = 0.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.29 \[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=-\frac {\arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \arctan \left (\frac {1}{b x + a}\right ) + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )}{8 \, b d \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}} \]

[In]

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

[Out]

-1/8*(arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^4 + 2*arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^
2 - 2*tan(1/2*arctan(1/(b*x + a)))^3 + arctan(1/(b*x + a)) + 2*tan(1/2*arctan(1/(b*x + a))))/(b*d*tan(1/2*arct
an(1/(b*x + a)))^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int \frac {\mathrm {acot}\left (a+b\,x\right )}{d\,x+\frac {a\,d}{b}} \,d x \]

[In]

int(acot(a + b*x)/(d*x + (a*d)/b),x)

[Out]

int(acot(a + b*x)/(d*x + (a*d)/b), x)