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

Optimal. Leaf size=35 \[ a \log (x)+\frac {1}{2} i b \text {Li}_2(-i c x)-\frac {1}{2} i b \text {Li}_2(i c x) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.00, size = 35, normalized size = 1.00 \[ a \log (x)+\frac {1}{2} i b \text {Li}_2(-i c x)-\frac {1}{2} i b \text {Li}_2(i c x) \]

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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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x\right ) + a}{x}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.03, size = 74, normalized size = 2.11 \[ a \ln \left (c x \right )+b \ln \left (c x \right ) \arctan \left (c x \right )+\frac {i b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i b \dilog \left (i c x +1\right )}{2}-\frac {i b \dilog \left (-i c x +1\right )}{2} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {\arctan \left (c x\right )}{x}\,{d x} + a \log \relax (x) \]

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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mupad [B]  time = 0.29, size = 28, normalized size = 0.80 \[ a\,\ln \relax (x)-\frac {b\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/x,x)

[Out]

a*log(x) - (b*(dilog(1 - c*x*1i) - dilog(c*x*1i + 1))*1i)/2

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

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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