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

Optimal. Leaf size=138 \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.0832176, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4856, 2402, 2315, 2447} \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/(d + e*x),x]

[Out]

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

Rule 4856

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

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

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

Mathematica [A]  time = 0.0599263, size = 138, normalized size = 1. \[ \frac{i b \text{PolyLog}\left (2,\frac{e (1-i c x)}{e+i c d}\right )-i b \text{PolyLog}\left (2,-\frac{e (c x-i)}{c d+i e}\right )+2 a \log (d+e x)+i b \log (1-i c x) \log \left (\frac{c (d+e x)}{c d-i e}\right )-i b \log (1+i c x) \log \left (\frac{c (d+e x)}{c d+i e}\right )}{2 e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 168, normalized size = 1.2 \begin{align*}{\frac{a\ln \left ( ecx+dc \right ) }{e}}+{\frac{b\ln \left ( ecx+dc \right ) \arctan \left ( cx \right ) }{e}}+{\frac{{\frac{i}{2}}b\ln \left ( ecx+dc \right ) }{e}\ln \left ({\frac{ie-ecx}{dc+ie}} \right ) }-{\frac{{\frac{i}{2}}b\ln \left ( ecx+dc \right ) }{e}\ln \left ({\frac{ie+ecx}{ie-dc}} \right ) }+{\frac{{\frac{i}{2}}b}{e}{\it dilog} \left ({\frac{ie-ecx}{dc+ie}} \right ) }-{\frac{{\frac{i}{2}}b}{e}{\it dilog} \left ({\frac{ie+ecx}{ie-dc}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, b \int \frac{\arctan \left (c x\right )}{2 \,{\left (e x + d\right )}}\,{d x} + \frac{a \log \left (e x + d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*integrate(1/2*arctan(c*x)/(e*x + d), x) + a*log(e*x + d)/e

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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}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(c*x) + a)/(e*x + d), 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 )}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c*x))/(d + e*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}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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