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

Optimal. Leaf size=45 \[ \frac {i \text {Li}_2\left (\frac {i}{a+b x}\right )}{2 b}-\frac {i \text {Li}_2\left (-\frac {i}{a+b x}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5044, 4849, 2391} \[ \frac {i \text {PolyLog}\left (2,\frac {i}{a+b x}\right )}{2 b}-\frac {i \text {PolyLog}\left (2,-\frac {i}{a+b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[a + b*x]/(a + b*x),x]

[Out]

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

Rule 2391

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 4849

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

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Mathematica [A]  time = 0.01, size = 38, normalized size = 0.84 \[ -\frac {i \left (\text {Li}_2\left (-\frac {i}{a+b x}\right )-\text {Li}_2\left (\frac {i}{a+b x}\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccot}\left (b x + a\right )}{b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.55, size = 100, normalized size = 2.22 \[ -\frac {\arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \arctan \left (\frac {1}{b x + a}\right ) + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )}{8 \, b^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/(b*x+a),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^2*tan(1/2*arct
an(1/(b*x + a)))^2)

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maple [B]  time = 0.06, size = 98, normalized size = 2.18 \[ \frac {\ln \left (b x +a \right ) \mathrm {arccot}\left (b x +a \right )}{b}-\frac {i \ln \left (b x +a \right ) \ln \left (1+i \left (b x +a \right )\right )}{2 b}+\frac {i \ln \left (b x +a \right ) \ln \left (1-i \left (b x +a \right )\right )}{2 b}-\frac {i \dilog \left (1+i \left (b x +a \right )\right )}{2 b}+\frac {i \dilog \left (1-i \left (b x +a \right )\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.48, size = 112, normalized size = 2.49 \[ \frac {\operatorname {arccot}\left (b x + a\right ) \log \left (b x + a\right )}{b} + \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (b x + a\right )}{b} + \frac {\arctan \left (b x + a, 0\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, \arctan \left (b x + a\right ) \log \left ({\left | b x + a \right |}\right ) + i \, {\rm Li}_2\left (i \, b x + i \, a + 1\right ) - i \, {\rm Li}_2\left (-i \, b x - i \, a + 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acot}\left (a+b\,x\right )}{a+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a + b*x)/(a + b*x),x)

[Out]

int(acot(a + b*x)/(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acot}{\left (a + b x \right )}}{a + b x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acot(a + b*x)/(a + b*x), x)

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