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

Optimal. Leaf size=162 \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{2 f}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i (c+d x)}\right )}{2 f}+\frac{\left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac{\log \left (\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.150132, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {5047, 4856, 2402, 2315, 2447} \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{2 f}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i (c+d x)}\right )}{2 f}+\frac{\left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac{\log \left (\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5047

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]

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

Mathematica [A]  time = 0.104057, size = 160, normalized size = 0.99 \[ \frac{-i b \text{PolyLog}\left (2,\frac{f (c+d x-i)}{-d e+(c-i) f}\right )+i b \text{PolyLog}\left (2,\frac{f (c+d x+i)}{-d e+(c+i) f}\right )+2 a \log (d (e+f x))+i b \log (1-i (c+d x)) \log \left (\frac{d (e+f x)}{d e-(c+i) f}\right )-i b \log (1+i (c+d x)) \log \left (\frac{d (e+f x)}{-c f+d e+i f}\right )}{2 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.072, size = 224, normalized size = 1.4 \begin{align*}{\frac{a\ln \left ( f \left ( dx+c \right ) -cf+de \right ) }{f}}+{\frac{b\ln \left ( f \left ( dx+c \right ) -cf+de \right ) \arctan \left ( dx+c \right ) }{f}}+{\frac{{\frac{i}{2}}b\ln \left ( f \left ( dx+c \right ) -cf+de \right ) }{f}\ln \left ({\frac{if-f \left ( dx+c \right ) }{de+if-cf}} \right ) }-{\frac{{\frac{i}{2}}b\ln \left ( f \left ( dx+c \right ) -cf+de \right ) }{f}\ln \left ({\frac{if+f \left ( dx+c \right ) }{if+cf-de}} \right ) }+{\frac{{\frac{i}{2}}b}{f}{\it dilog} \left ({\frac{if-f \left ( dx+c \right ) }{de+if-cf}} \right ) }-{\frac{{\frac{i}{2}}b}{f}{\it dilog} \left ({\frac{if+f \left ( dx+c \right ) }{if+cf-de}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arctan \left (d x + c\right ) + a}{f x + e}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{atan}{\left (c + d x \right )}}{e + f x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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