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

Optimal. Leaf size=152 \[ -\frac{i \text{PolyLog}\left (2,1-\frac{2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{2 d}+\frac{i \text{PolyLog}\left (2,1-\frac{2}{1-i (a+b x)}\right )}{2 d}+\frac{\tan ^{-1}(a+b x) \log \left (\frac{2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac{\log \left (\frac{2}{1-i (a+b x)}\right ) \tan ^{-1}(a+b x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.138446, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5047, 4856, 2402, 2315, 2447} \[ -\frac{i \text{PolyLog}\left (2,1-\frac{2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{2 d}+\frac{i \text{PolyLog}\left (2,1-\frac{2}{1-i (a+b x)}\right )}{2 d}+\frac{\tan ^{-1}(a+b x) \log \left (\frac{2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac{\log \left (\frac{2}{1-i (a+b x)}\right ) \tan ^{-1}(a+b x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/(c + d*x),x]

[Out]

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

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

Mathematica [A]  time = 0.0195464, size = 231, normalized size = 1.52 \[ \frac{i \text{PolyLog}\left (2,-\frac{i d (1-i (a+b x))}{-a d+b c-i d}\right )}{2 d}-\frac{i \text{PolyLog}\left (2,\frac{i d (1+i (a+b x))}{-a d+b c+i d}\right )}{2 d}+\frac{i \log (1-i (a+b x)) \log \left (-\frac{i \left (\frac{b c-a d}{b}+\frac{d (a+b x)}{b}\right )}{-\frac{d}{b}-\frac{i (b c-a d)}{b}}\right )}{2 d}-\frac{i \log (1+i (a+b x)) \log \left (\frac{i \left (\frac{b c-a d}{b}+\frac{d (a+b x)}{b}\right )}{-\frac{d}{b}+\frac{i (b c-a d)}{b}}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.054, size = 198, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( d \left ( bx+a \right ) -ad+bc \right ) \arctan \left ( bx+a \right ) }{d}}+{\frac{{\frac{i}{2}}\ln \left ( d \left ( bx+a \right ) -ad+bc \right ) }{d}\ln \left ({\frac{id-d \left ( bx+a \right ) }{bc+id-ad}} \right ) }-{\frac{{\frac{i}{2}}\ln \left ( d \left ( bx+a \right ) -ad+bc \right ) }{d}\ln \left ({\frac{id+d \left ( bx+a \right ) }{id+ad-bc}} \right ) }+{\frac{{\frac{i}{2}}}{d}{\it dilog} \left ({\frac{id-d \left ( bx+a \right ) }{bc+id-ad}} \right ) }-{\frac{{\frac{i}{2}}}{d}{\it dilog} \left ({\frac{id+d \left ( bx+a \right ) }{id+ad-bc}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.97429, size = 383, normalized size = 2.52 \begin{align*} \frac{\arctan \left (b x + a\right ) \log \left (d x + c\right )}{d} - \frac{\arctan \left (\frac{b^{2} x + a b}{b}\right ) \log \left (d x + c\right )}{d} - \frac{\arctan \left (\frac{b d^{2} x + b c d}{b^{2} c^{2} - 2 \, a b c d +{\left (a^{2} + 1\right )} d^{2}}, \frac{b^{2} c^{2} - a b c d +{\left (b^{2} c d - a b d^{2}\right )} x}{b^{2} c^{2} - 2 \, a b c d +{\left (a^{2} + 1\right )} d^{2}}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - \arctan \left (b x + a\right ) \log \left (\frac{b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}}{b^{2} c^{2} - 2 \, a b c d +{\left (a^{2} + 1\right )} d^{2}}\right ) + i \,{\rm Li}_2\left (\frac{i \, b d x +{\left (i \, a + 1\right )} d}{-i \, b c +{\left (i \, a + 1\right )} d}\right ) - i \,{\rm Li}_2\left (\frac{i \, b d x +{\left (i \, a - 1\right )} d}{-i \, b c +{\left (i \, a - 1\right )} d}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(b*x + a)*log(d*x + c)/d - arctan((b^2*x + a*b)/b)*log(d*x + c)/d - 1/2*(arctan2((b*d^2*x + b*c*d)/(b^2*
c^2 - 2*a*b*c*d + (a^2 + 1)*d^2), (b^2*c^2 - a*b*c*d + (b^2*c*d - a*b*d^2)*x)/(b^2*c^2 - 2*a*b*c*d + (a^2 + 1)
*d^2))*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - arctan(b*x + a)*log((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)/(b^2*c^2 -
2*a*b*c*d + (a^2 + 1)*d^2)) + I*dilog((I*b*d*x + (I*a + 1)*d)/(-I*b*c + (I*a + 1)*d)) - I*dilog((I*b*d*x + (I
*a - 1)*d)/(-I*b*c + (I*a - 1)*d)))/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a + b*x)/(c + d*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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