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

Optimal. Leaf size=35 \[ \frac{1}{2} i b \text{PolyLog}(2,-i c x)-\frac{1}{2} i b \text{PolyLog}(2,i c x)+a \log (x) \]

[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 = 0.028752, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4848, 2391} \[ \frac{1}{2} i b \text{PolyLog}(2,-i c x)-\frac{1}{2} i b \text{PolyLog}(2,i c x)+a \log (x) \]

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 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}(c x)}{x} \, dx &=a \log (x)+\frac{1}{2} (i b) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{2} (i b) \int \frac{\log (1+i c x)}{x} \, dx\\ &=a \log (x)+\frac{1}{2} i b \text{Li}_2(-i c x)-\frac{1}{2} i b \text{Li}_2(i c x)\\ \end{align*}

Mathematica [A]  time = 0.0021458, size = 35, normalized size = 1. \[ \frac{1}{2} i b \text{PolyLog}(2,-i c x)-\frac{1}{2} i b \text{PolyLog}(2,i c x)+a \log (x) \]

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 = 0.015, size = 74, normalized size = 2.1 \begin{align*} a\ln \left ( cx \right ) +b\ln \left ( cx \right ) \arctan \left ( cx \right ) +{\frac{i}{2}}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -{\frac{i}{2}}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +{\frac{i}{2}}b{\it dilog} \left ( 1+icx \right ) -{\frac{i}{2}}b{\it dilog} \left ( 1-icx \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(arctan(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 (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 (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 (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)