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

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

[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)

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Rubi [A]  time = 0.047172, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0140577, size = 60, normalized size = 1.02 \[ \frac{2 i \log \left (\frac{2 d}{d+i c d x}\right ) \left (a+b \tan ^{-1}(c x)\right )-b \text{PolyLog}\left (2,\frac{c x+i}{c x-i}\right )}{2 c d} \]

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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Maple [B]  time = 0.036, size = 142, normalized size = 2.4 \begin{align*}{\frac{-{\frac{i}{2}}a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{dc}}+{\frac{a\arctan \left ( cx \right ) }{dc}}-{\frac{ib\ln \left ( 1+icx \right ) \arctan \left ( cx \right ) }{dc}}-{\frac{b\ln \left ( 1+icx \right ) }{2\,dc}\ln \left ({\frac{1}{2}}-{\frac{i}{2}}cx \right ) }+{\frac{b}{2\,dc}\ln \left ({\frac{1}{2}}-{\frac{i}{2}}cx \right ) \ln \left ({\frac{i}{2}}cx+{\frac{1}{2}} \right ) }+{\frac{b}{2\,dc}{\it dilog} \left ({\frac{i}{2}}cx+{\frac{1}{2}} \right ) }+{\frac{b \left ( \ln \left ( 1+icx \right ) \right ) ^{2}}{4\,dc}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} -\frac{-4 \,{\left (-2 i \, c^{2} \int \frac{x \arctan \left (c x\right )}{c^{2} x^{2} + 1}\,{d x} + \arctan \left (c x\right )^{2}\right )} b}{8 \, c d} - \frac{i \, a \log \left (i \, c d x + d\right )}{c d} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

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]

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