3.2.3 \(\int \frac {\cot ^{-1}(a+b x)}{x} \, dx\) [103]

Optimal. Leaf size=120 \[ -\cot ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )+\cot ^{-1}(a+b x) \log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )-\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )+\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right ) \]

[Out]

-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)))

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Rubi [A]
time = 0.09, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5156, 4967, 2449, 2352, 2497} \begin {gather*} -\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )+\frac {1}{2} i \text {Li}_2\left (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) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCot[a + b*x]/x,x]

[Out]

-(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)))]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-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]

Rule 2497

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[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4967

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcCot[c*x]))*(Log[2/(1
- I*c*x)]/e), x] + (-Dist[b*(c/e), Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] + Dist[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] + Simp[(a + b*ArcCot[c*x])*(Log[2*c*((d + e*x)/((c
*d + I*e)*(1 - I*c*x)))]/e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 5156

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

Rubi steps

\begin {align*} \int \frac {\cot ^{-1}(a+b x)}{x} \, dx &=\frac {\text {Subst}\left (\int \frac {\cot ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=-\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 )-\text {Subst}\left (\int \frac {\log \left (\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,a+b x\right )+\text {Subst}\left (\int \frac {\log \left (\frac {2 \left (-\frac {a}{b}+\frac {x}{b}\right )}{\left (\frac {i}{b}-\frac {a}{b}\right ) (1-i x)}\right )}{1+x^2} \, dx,x,a+b x\right )\\ &=-\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 \text {Li}_2\left (1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right )-i \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i (a+b x)}\right )\\ &=-\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 \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )+\frac {1}{2} i \text {Li}_2\left (1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right )\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(251\) vs. \(2(120)=240\).
time = 0.02, size = 251, normalized size = 2.09 \begin {gather*} -\frac {1}{2} i \log \left (\frac {-i+a+b x}{\left (-\frac {i}{b}+\frac {a}{b}\right ) b}\right ) \log \left (-\frac {a}{b}+\frac {a+b x}{b}\right )+\frac {1}{2} i \log \left (\frac {-i+a+b x}{a+b x}\right ) \log \left (-\frac {a}{b}+\frac {a+b x}{b}\right )+\frac {1}{2} i \log \left (\frac {i+a+b x}{\left (\frac {i}{b}+\frac {a}{b}\right ) b}\right ) \log \left (-\frac {a}{b}+\frac {a+b x}{b}\right )-\frac {1}{2} i \log \left (\frac {i+a+b x}{a+b x}\right ) \log \left (-\frac {a}{b}+\frac {a+b x}{b}\right )-\frac {1}{2} i \text {PolyLog}\left (2,-\frac {b \left (-\frac {a}{b}+\frac {a+b x}{b}\right )}{-i+a}\right )+\frac {1}{2} i \text {PolyLog}\left (2,-\frac {b \left (-\frac {a}{b}+\frac {a+b x}{b}\right )}{i+a}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[a + b*x]/x,x]

[Out]

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

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Maple [A]
time = 0.06, size = 106, normalized size = 0.88

method	result	size
risch	\(\frac {\pi \ln \left (-i b x \right )}{2}-\frac {i \ln \left (-\frac {i x b}{i a -1}\right ) \ln \left (-i b x -i a +1\right )}{2}-\frac {i \dilog \left (-\frac {i x b}{i a -1}\right )}{2}+\frac {i \ln \left (\frac {i x b}{-i a -1}\right ) \ln \left (i b x +i a +1\right )}{2}+\frac {i \dilog \left (\frac {i x b}{-i a -1}\right )}{2}\)	\(103\)
derivativedivides	\(\ln \left (-b x \right ) \mathrm {arccot}\left (b x +a \right )+\frac {i \ln \left (-b x \right ) \ln \left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \ln \left (-b x \right ) \ln \left (\frac {-b x -a +i}{i-a}\right )}{2}+\frac {i \dilog \left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \dilog \left (\frac {-b x -a +i}{i-a}\right )}{2}\)	\(106\)
default	\(\ln \left (-b x \right ) \mathrm {arccot}\left (b x +a \right )+\frac {i \ln \left (-b x \right ) \ln \left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \ln \left (-b x \right ) \ln \left (\frac {-b x -a +i}{i-a}\right )}{2}+\frac {i \dilog \left (\frac {b x +a +i}{i+a}\right )}{2}-\frac {i \dilog \left (\frac {-b x -a +i}{i-a}\right )}{2}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 133, normalized size = 1.11 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccot(b*x + a)/x, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccot(b*x + a)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {acot}\left (a+b\,x\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a + b*x)/x,x)

[Out]

int(acot(a + b*x)/x, x)

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