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

Optimal. Leaf size=120 \[ \frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1-i (a+b x)}\right )-\frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2 b x}{(-a+i) (1-i (a+b x))}\right )+\log \left (\frac{2}{1-i (a+b x)}\right ) \left (-\tan ^{-1}(a+b x)\right )+\log \left (\frac{2 b x}{(-a+i) (1-i (a+b x))}\right ) \tan ^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.106247, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5047, 4856, 2402, 2315, 2447} \[ \frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1-i (a+b x)}\right )-\frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2 b x}{(-a+i) (1-i (a+b x))}\right )+\log \left (\frac{2}{1-i (a+b x)}\right ) \left (-\tan ^{-1}(a+b x)\right )+\log \left (\frac{2 b x}{(-a+i) (1-i (a+b x))}\right ) \tan ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/x,x]

[Out]

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

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

Mathematica [A]  time = 0.0081672, size = 171, normalized size = 1.42 \[ \frac{1}{2} i \text{PolyLog}\left (2,\frac{i (1-i (a+b x))}{a+i}\right )-\frac{1}{2} i \text{PolyLog}\left (2,-\frac{i (1+i (a+b x))}{a-i}\right )-\frac{1}{2} i \log (1+i (a+b x)) \log \left (\frac{i \left (\frac{a+b x}{b}-\frac{a}{b}\right )}{-\frac{1}{b}-\frac{i a}{b}}\right )+\frac{1}{2} i \log (1-i (a+b x)) \log \left (-\frac{i \left (\frac{a+b x}{b}-\frac{a}{b}\right )}{-\frac{1}{b}+\frac{i a}{b}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a + b*x]/x,x]

[Out]

(-I/2)*Log[1 + I*(a + b*x)]*Log[(I*(-(a/b) + (a + b*x)/b))/(-b^(-1) - (I*a)/b)] + (I/2)*Log[1 - I*(a + b*x)]*L
og[((-I)*(-(a/b) + (a + b*x)/b))/(-b^(-1) + (I*a)/b)] + (I/2)*PolyLog[2, (I*(1 - I*(a + b*x)))/(I + a)] - (I/2
)*PolyLog[2, ((-I)*(1 + I*(a + b*x)))/(-I + a)]

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Maple [A]  time = 0.047, size = 103, normalized size = 0.9 \begin{align*} \ln \left ( bx \right ) \arctan \left ( bx+a \right ) +{\frac{i}{2}}\ln \left ( bx \right ) \ln \left ({\frac{i-a-bx}{i-a}} \right ) -{\frac{i}{2}}\ln \left ( bx \right ) \ln \left ({\frac{i+a+bx}{i+a}} \right ) +{\frac{i}{2}}{\it dilog} \left ({\frac{i-a-bx}{i-a}} \right ) -{\frac{i}{2}}{\it dilog} \left ({\frac{i+a+bx}{i+a}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.71072, size = 181, normalized size = 1.51 \begin{align*} -\frac{1}{2} \, \arctan \left (\frac{b x}{a^{2} + 1}, -\frac{a b x}{a^{2} + 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) + \frac{1}{2} \, \arctan \left (b x + a\right ) \log \left (\frac{b^{2} x^{2}}{a^{2} + 1}\right ) + \arctan \left (b x + a\right ) \log \left (x\right ) - \arctan \left (\frac{b^{2} x + a b}{b}\right ) \log \left (x\right ) - \frac{1}{2} i \,{\rm Li}_2\left (\frac{i \, b x + i \, a + 1}{i \, a + 1}\right ) + \frac{1}{2} i \,{\rm Li}_2\left (\frac{i \, b x + i \, a - 1}{i \, a - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(b*x + a)/x, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(b*x + a)/x, x)