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

Optimal. Leaf size=59 \[ \frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c d}-\frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4854, 2402, 2315} \[ \frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c d}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/(d + I*c*d*x),x]

[Out]

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 60, normalized size = 1.02 \[ \frac {2 i \log \left (\frac {2 d}{d+i c d x}\right ) \left (a+b \tan ^{-1}(c x)\right )-b \text {Li}_2\left (\frac {c x+i}{c x-i}\right )}{2 c d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])/(d + I*c*d*x),x]

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c d x - 2 i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(-(c*x + I)/(c*x - I)) - 2*I*a)/(2*c*d*x - 2*I*d), 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))/(d+I*c*d*x),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.04, size = 142, normalized size = 2.41 \[ -\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 c d}+\frac {a \arctan \left (c x \right )}{c d}-\frac {i b \ln \left (i c x +1\right ) \arctan \left (c x \right )}{c d}-\frac {b \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (i c x +1\right )}{2 c d}+\frac {b \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (\frac {i c x}{2}+\frac {1}{2}\right )}{2 c d}+\frac {b \dilog \left (\frac {i c x}{2}+\frac {1}{2}\right )}{2 c d}+\frac {b \ln \left (i c x +1\right )^{2}}{4 c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/c*a/d*ln(c^2*x^2+1)+1/c*a/d*arctan(c*x)-I/c*b/d*ln(1+I*c*x)*arctan(c*x)-1/2/c*b/d*ln(1/2-1/2*I*c*x)*ln(
1+I*c*x)+1/2/c*b/d*ln(1/2-1/2*I*c*x)*ln(1/2*I*c*x+1/2)+1/2/c*b/d*dilog(1/2*I*c*x+1/2)+1/4/c*b/d*ln(1+I*c*x)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (8 i \, c^{2} d \mathit {sage}_{0} x - 4 \, \arctan \left (c x\right )^{2}\right )} b}{8 \, c d} - \frac {i \, a \log \left (i \, c d x + d\right )}{c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*I*c^2*d*integrate(x*arctan(c*x)/(c^2*d*x^2 + d), x) + 4*c^2*d*integrate(x*log(c^2*x^2 + 1)/(c^2*d*x^2
+ d), x) - 4*arctan(c*x)^2 - log(c^2*x^2 + 1)^2)*b/(c*d) - I*a*log(I*c*d*x + d)/(c*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(d + c*d*x*1i),x)

[Out]

int((a + b*atan(c*x))/(d + c*d*x*1i), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b \log {\left (- i c x + 1 \right )} \log {\left (i c x + 1 \right )}}{2 c d} - \frac {i \left (\int \frac {i a}{c^{2} x^{2} + 1}\, dx + \int \frac {a c x}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {i b c x \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(-I*c*x + 1)*log(I*c*x + 1)/(2*c*d) - I*(Integral(I*a/(c**2*x**2 + 1), x) + Integral(a*c*x/(c**2*x**2 + 1
), x) + Integral(-I*b*c*x*log(I*c*x + 1)/(c**2*x**2 + 1), x))/d

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