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

Optimal. Leaf size=39 \[ a \log (x)-\frac {1}{2} i b \text {Li}_2\left (-\frac {i c}{x}\right )+\frac {1}{2} i b \text {Li}_2\left (\frac {i c}{x}\right ) \]

[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 = 0.04, antiderivative size = 39, 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 = {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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 1.00 \[ a \log (x)-\frac {1}{2} i b \text {Li}_2\left (-\frac {i c}{x}\right )+\frac {1}{2} i b \text {Li}_2\left (\frac {i c}{x}\right ) \]

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 = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (\frac {c}{x}\right ) + a}{x}, x\right ) \]

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 [B]  time = 3.96, size = 74, normalized size = 1.90 \[ \frac {{\left (\frac {b c^{6} i \log \left (\frac {c i}{x} + 1\right )}{x^{2}} - \frac {b c^{6} i \log \left (-\frac {c i}{x} + 1\right )}{x^{2}} - 2 \, b c^{4} \arctan \left (\frac {c}{x}\right ) - 2 \, a c^{4} - \frac {2 \, b c^{5}}{x}\right )} x^{2}}{4 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*c^6*i*log(c*i/x + 1)/x^2 - b*c^6*i*log(-c*i/x + 1)/x^2 - 2*b*c^4*arctan(c/x) - 2*a*c^4 - 2*b*c^5/x)*x^2
/c^5

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maple [B]  time = 0.05, size = 94, normalized size = 2.41 \[ -a \ln \left (\frac {c}{x}\right )-b \ln \left (\frac {c}{x}\right ) \arctan \left (\frac {c}{x}\right )-\frac {i b \ln \left (\frac {c}{x}\right ) \ln \left (1+\frac {i c}{x}\right )}{2}+\frac {i b \ln \left (\frac {c}{x}\right ) \ln \left (1-\frac {i c}{x}\right )}{2}-\frac {i b \dilog \left (1+\frac {i c}{x}\right )}{2}+\frac {i b \dilog \left (1-\frac {i c}{x}\right )}{2} \]

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(arctan2(c, x)/x, x) + a*log(x)

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mupad [B]  time = 0.34, size = 32, normalized size = 0.82 \[ a\,\ln \relax (x)+\frac {b\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-\frac {c\,1{}\mathrm {i}}{x}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+\frac {c\,1{}\mathrm {i}}{x}\right )\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c/x))/x,x)

[Out]

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

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

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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