3.136 \(\int \frac{a+b \tan ^{-1}(\frac{c}{x})}{x} \, dx\)

Optimal. Leaf size=39 \[ -\frac{1}{2} i b \text{PolyLog}\left (2,-\frac{i c}{x}\right )+\frac{1}{2} i b \text{PolyLog}\left (2,\frac{i c}{x}\right )+a \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0448919, antiderivative size = 39, normalized size of antiderivative = 1., 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 = {5031, 4848, 2391} \[ -\frac{1}{2} i b \text{PolyLog}\left (2,-\frac{i c}{x}\right )+\frac{1}{2} i b \text{PolyLog}\left (2,\frac{i c}{x}\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c/x])/x,x]

[Out]

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

Rule 5031

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTan[c*x])^p
/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 4848

Int[((a_.) + ArcTan[(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 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]

Rubi steps

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

Mathematica [A]  time = 0.0056957, size = 39, normalized size = 1. \[ -\frac{1}{2} i b \text{PolyLog}\left (2,-\frac{i c}{x}\right )+\frac{1}{2} i b \text{PolyLog}\left (2,\frac{i c}{x}\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c/x])/x,x]

[Out]

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

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Maple [B]  time = 0.036, size = 94, normalized size = 2.4 \begin{align*} -a\ln \left ({\frac{c}{x}} \right ) -b\ln \left ({\frac{c}{x}} \right ) \arctan \left ({\frac{c}{x}} \right ) -{\frac{i}{2}}b\ln \left ({\frac{c}{x}} \right ) \ln \left ( 1+{\frac{ic}{x}} \right ) +{\frac{i}{2}}b\ln \left ({\frac{c}{x}} \right ) \ln \left ( 1-{\frac{ic}{x}} \right ) -{\frac{i}{2}}b{\it dilog} \left ( 1+{\frac{ic}{x}} \right ) +{\frac{i}{2}}b{\it dilog} \left ( 1-{\frac{ic}{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c/x))/x,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\arctan \left (c, x\right )}{x}\,{d x} + a \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(arctan2(c, x)/x, x) + a*log(x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arctan \left (\frac{c}{x}\right ) + a}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(c/x) + a)/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{atan}{\left (\frac{c}{x} \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c/x))/x,x)

[Out]

Integral((a + b*atan(c/x))/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arctan \left (\frac{c}{x}\right ) + a}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c/x) + a)/x, x)