3.60 \(\int \frac{\tan ^{-1}(a+b x)}{1+x^2} \, dx\)

Optimal. Leaf size=274 \[ -\frac{1}{4} \text{PolyLog}\left (2,-\frac{-a-b x+i}{a-i (1-b)}\right )+\frac{1}{4} \text{PolyLog}\left (2,-\frac{-a-b x+i}{a-i (b+1)}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+i}{a-i b+i}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+i}{a+i (b+1)}\right )+\frac{1}{4} \log \left (\frac{b (-x+i)}{a+i (b+1)}\right ) \log (-i a-i b x+1)-\frac{1}{4} \log \left (-\frac{b (x+i)}{a+i (1-b)}\right ) \log (-i a-i b x+1)-\frac{1}{4} \log \left (\frac{b (-x+i)}{a-i (1-b)}\right ) \log (i a+i b x+1)+\frac{1}{4} \log \left (-\frac{b (x+i)}{a-i (b+1)}\right ) \log (i a+i b x+1) \]

[Out]

(Log[(b*(I - x))/(a + I*(1 + b))]*Log[1 - I*a - I*b*x])/4 - (Log[-((b*(I + x))/(a + I*(1 - b)))]*Log[1 - I*a -
I*b*x])/4 - (Log[(b*(I - x))/(a - I*(1 - b))]*Log[1 + I*a + I*b*x])/4 + (Log[-((b*(I + x))/(a - I*(1 + b)))]*
Log[1 + I*a + I*b*x])/4 - PolyLog[2, -((I - a - b*x)/(a - I*(1 - b)))]/4 + PolyLog[2, -((I - a - b*x)/(a - I*(
1 + b)))]/4 - PolyLog[2, (I + a + b*x)/(I + a - I*b)]/4 + PolyLog[2, (I + a + b*x)/(a + I*(1 + b))]/4

________________________________________________________________________________________

Rubi [A]  time = 0.270686, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5051, 2409, 2394, 2393, 2391} \[ -\frac{1}{4} \text{PolyLog}\left (2,-\frac{-a-b x+i}{a-i (1-b)}\right )+\frac{1}{4} \text{PolyLog}\left (2,-\frac{-a-b x+i}{a-i (b+1)}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+i}{a-i b+i}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{a+b x+i}{a+i (b+1)}\right )+\frac{1}{4} \log \left (\frac{b (-x+i)}{a+i (b+1)}\right ) \log (-i a-i b x+1)-\frac{1}{4} \log \left (-\frac{b (x+i)}{a+i (1-b)}\right ) \log (-i a-i b x+1)-\frac{1}{4} \log \left (\frac{b (-x+i)}{a-i (1-b)}\right ) \log (i a+i b x+1)+\frac{1}{4} \log \left (-\frac{b (x+i)}{a-i (b+1)}\right ) \log (i a+i b x+1) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/(1 + x^2),x]

[Out]

(Log[(b*(I - x))/(a + I*(1 + b))]*Log[1 - I*a - I*b*x])/4 - (Log[-((b*(I + x))/(a + I*(1 - b)))]*Log[1 - I*a -
I*b*x])/4 - (Log[(b*(I - x))/(a - I*(1 - b))]*Log[1 + I*a + I*b*x])/4 + (Log[-((b*(I + x))/(a - I*(1 + b)))]*
Log[1 + I*a + I*b*x])/4 - PolyLog[2, -((I - a - b*x)/(a - I*(1 - b)))]/4 + PolyLog[2, -((I - a - b*x)/(a - I*(
1 + b)))]/4 - PolyLog[2, (I + a + b*x)/(I + a - I*b)]/4 + PolyLog[2, (I + a + b*x)/(a + I*(1 + b))]/4

Rule 5051

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

Rule 2409

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a+b x)}{1+x^2} \, dx &=\frac{1}{2} i \int \frac{\log (1-i a-i b x)}{1+x^2} \, dx-\frac{1}{2} i \int \frac{\log (1+i a+i b x)}{1+x^2} \, dx\\ &=\frac{1}{2} i \int \left (\frac{i \log (1-i a-i b x)}{2 (i-x)}+\frac{i \log (1-i a-i b x)}{2 (i+x)}\right ) \, dx-\frac{1}{2} i \int \left (\frac{i \log (1+i a+i b x)}{2 (i-x)}+\frac{i \log (1+i a+i b x)}{2 (i+x)}\right ) \, dx\\ &=-\left (\frac{1}{4} \int \frac{\log (1-i a-i b x)}{i-x} \, dx\right )-\frac{1}{4} \int \frac{\log (1-i a-i b x)}{i+x} \, dx+\frac{1}{4} \int \frac{\log (1+i a+i b x)}{i-x} \, dx+\frac{1}{4} \int \frac{\log (1+i a+i b x)}{i+x} \, dx\\ &=\frac{1}{4} \log \left (\frac{b (i-x)}{a+i (1+b)}\right ) \log (1-i a-i b x)-\frac{1}{4} \log \left (-\frac{b (i+x)}{a+i (1-b)}\right ) \log (1-i a-i b x)-\frac{1}{4} \log \left (\frac{b (i-x)}{a-i (1-b)}\right ) \log (1+i a+i b x)+\frac{1}{4} \log \left (-\frac{b (i+x)}{a-i (1+b)}\right ) \log (1+i a+i b x)+\frac{1}{4} (i b) \int \frac{\log \left (\frac{i b (i-x)}{1+i a-b}\right )}{1+i a+i b x} \, dx+\frac{1}{4} (i b) \int \frac{\log \left (-\frac{i b (i-x)}{1-i a+b}\right )}{1-i a-i b x} \, dx-\frac{1}{4} (i b) \int \frac{\log \left (\frac{i b (i+x)}{-1-i a-b}\right )}{1+i a+i b x} \, dx-\frac{1}{4} (i b) \int \frac{\log \left (-\frac{i b (i+x)}{-1+i a+b}\right )}{1-i a-i b x} \, dx\\ &=\frac{1}{4} \log \left (\frac{b (i-x)}{a+i (1+b)}\right ) \log (1-i a-i b x)-\frac{1}{4} \log \left (-\frac{b (i+x)}{a+i (1-b)}\right ) \log (1-i a-i b x)-\frac{1}{4} \log \left (\frac{b (i-x)}{a-i (1-b)}\right ) \log (1+i a+i b x)+\frac{1}{4} \log \left (-\frac{b (i+x)}{a-i (1+b)}\right ) \log (1+i a+i b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{-1-i a-b}\right )}{x} \, dx,x,1+i a+i b x\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{1+i a-b}\right )}{x} \, dx,x,1+i a+i b x\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{1-i a+b}\right )}{x} \, dx,x,1-i a-i b x\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{-1+i a+b}\right )}{x} \, dx,x,1-i a-i b x\right )\\ &=\frac{1}{4} \log \left (\frac{b (i-x)}{a+i (1+b)}\right ) \log (1-i a-i b x)-\frac{1}{4} \log \left (-\frac{b (i+x)}{a+i (1-b)}\right ) \log (1-i a-i b x)-\frac{1}{4} \log \left (\frac{b (i-x)}{a-i (1-b)}\right ) \log (1+i a+i b x)+\frac{1}{4} \log \left (-\frac{b (i+x)}{a-i (1+b)}\right ) \log (1+i a+i b x)-\frac{1}{4} \text{Li}_2\left (-\frac{i-a-b x}{a-i (1-b)}\right )+\frac{1}{4} \text{Li}_2\left (-\frac{i-a-b x}{a-i (1+b)}\right )-\frac{1}{4} \text{Li}_2\left (\frac{i+a+b x}{i+a-i b}\right )+\frac{1}{4} \text{Li}_2\left (\frac{i+a+b x}{a+i (1+b)}\right )\\ \end{align*}

Mathematica [A]  time = 0.05279, size = 283, normalized size = 1.03 \[ -\frac{1}{4} \text{PolyLog}\left (2,\frac{-i a-i b x+1}{-i a-b+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{-i a-i b x+1}{-i a+b+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{i a+i b x+1}{i a-b+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{i a+i b x+1}{i a+b+1}\right )+\frac{1}{4} \log \left (\frac{b (-x+i)}{a+i (b+1)}\right ) \log (-i a-i b x+1)-\frac{1}{4} \log \left (-\frac{b (x+i)}{a+i (1-b)}\right ) \log (-i a-i b x+1)-\frac{1}{4} \log \left (\frac{b (-x+i)}{a-i (1-b)}\right ) \log (i a+i b x+1)+\frac{1}{4} \log \left (-\frac{b (x+i)}{a-i (b+1)}\right ) \log (i a+i b x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[a + b*x]/(1 + x^2),x]

[Out]

(Log[(b*(I - x))/(a + I*(1 + b))]*Log[1 - I*a - I*b*x])/4 - (Log[-((b*(I + x))/(a + I*(1 - b)))]*Log[1 - I*a -
I*b*x])/4 - (Log[(b*(I - x))/(a - I*(1 - b))]*Log[1 + I*a + I*b*x])/4 + (Log[-((b*(I + x))/(a - I*(1 + b)))]*
Log[1 + I*a + I*b*x])/4 - PolyLog[2, (1 - I*a - I*b*x)/(1 - I*a - b)]/4 + PolyLog[2, (1 - I*a - I*b*x)/(1 - I*
a + b)]/4 - PolyLog[2, (1 + I*a + I*b*x)/(1 + I*a - b)]/4 + PolyLog[2, (1 + I*a + I*b*x)/(1 + I*a + b)]/4

________________________________________________________________________________________

Maple [B]  time = 0.444, size = 833, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(b*x+a)/(x^2+1),x)

[Out]

arctan(x)*arctan(b*x+a)+1/2*I*arctan((b*x+a)/b-a/b)*ln(1-(-I*b+a-I)*(1+I*((b*x+a)/b-a/b))^2/(((b*x+a)/b-a/b)^2
+1)/(-I*b+I-a))+1/2*arctan((b*x+a)/b-a/b)^2+1/4*polylog(2,(-I*b+a-I)*(1+I*((b*x+a)/b-a/b))^2/(((b*x+a)/b-a/b)^
2+1)/(-I*b+I-a))+1/2*b/(I*b+I+a)*ln(1-(I+a-I*b)*(1+I*((b*x+a)/b-a/b))^2/(((b*x+a)/b-a/b)^2+1)/(-I*b-I-a))*arct
an((b*x+a)/b-a/b)+1/2/(I*b+I+a)*ln(1-(I+a-I*b)*(1+I*((b*x+a)/b-a/b))^2/(((b*x+a)/b-a/b)^2+1)/(-I*b-I-a))*arcta
n((b*x+a)/b-a/b)-1/2*I/(I*b+I+a)*ln(1-(I+a-I*b)*(1+I*((b*x+a)/b-a/b))^2/(((b*x+a)/b-a/b)^2+1)/(-I*b-I-a))*arct
an((b*x+a)/b-a/b)*a-1/2*I*b/(I*b+I+a)*arctan((b*x+a)/b-a/b)^2-1/4*I*b/(I*b+I+a)*polylog(2,(I+a-I*b)*(1+I*((b*x
+a)/b-a/b))^2/(((b*x+a)/b-a/b)^2+1)/(-I*b-I-a))-1/2*I/(I*b+I+a)*arctan((b*x+a)/b-a/b)^2-1/2/(I*b+I+a)*arctan((
b*x+a)/b-a/b)^2*a-1/4*I/(I*b+I+a)*polylog(2,(I+a-I*b)*(1+I*((b*x+a)/b-a/b))^2/(((b*x+a)/b-a/b)^2+1)/(-I*b-I-a)
)-1/4/(I*b+I+a)*polylog(2,(I+a-I*b)*(1+I*((b*x+a)/b-a/b))^2/(((b*x+a)/b-a/b)^2+1)/(-I*b-I-a))*a

________________________________________________________________________________________

Maxima [A]  time = 1.8351, size = 443, normalized size = 1.62 \begin{align*} \frac{1}{8} \, b{\left (\frac{8 \, \arctan \left (x\right ) \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b} - \frac{4 \, \arctan \left (x\right ) \arctan \left (\frac{a b +{\left (b^{2} + b\right )} x}{a^{2} + b^{2} + 2 \, b + 1}, \frac{a b x + a^{2} + b + 1}{a^{2} + b^{2} + 2 \, b + 1}\right ) - 4 \, \arctan \left (x\right ) \arctan \left (\frac{a b +{\left (b^{2} - b\right )} x}{a^{2} + b^{2} - 2 \, b + 1}, \frac{a b x + a^{2} - b + 1}{a^{2} + b^{2} - 2 \, b + 1}\right ) + \log \left (x^{2} + 1\right ) \log \left (\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{a^{2} + b^{2} + 2 \, b + 1}\right ) - \log \left (x^{2} + 1\right ) \log \left (\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{a^{2} + b^{2} - 2 \, b + 1}\right ) + 2 \,{\rm Li}_2\left (-\frac{i \, b x - b}{i \, a + b + 1}\right ) - 2 \,{\rm Li}_2\left (-\frac{i \, b x - b}{i \, a + b - 1}\right ) + 2 \,{\rm Li}_2\left (\frac{i \, b x + b}{-i \, a + b + 1}\right ) - 2 \,{\rm Li}_2\left (\frac{i \, b x + b}{-i \, a + b - 1}\right )}{b}\right )} + \arctan \left (b x + a\right ) \arctan \left (x\right ) - \arctan \left (x\right ) \arctan \left (\frac{b^{2} x + a b}{b}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(b*x+a)/(x^2+1),x, algorithm="maxima")

[Out]

1/8*b*(8*arctan(x)*arctan((b^2*x + a*b)/b)/b - (4*arctan(x)*arctan2((a*b + (b^2 + b)*x)/(a^2 + b^2 + 2*b + 1),
(a*b*x + a^2 + b + 1)/(a^2 + b^2 + 2*b + 1)) - 4*arctan(x)*arctan2((a*b + (b^2 - b)*x)/(a^2 + b^2 - 2*b + 1),
(a*b*x + a^2 - b + 1)/(a^2 + b^2 - 2*b + 1)) + log(x^2 + 1)*log((b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^2 + b^2 + 2*
b + 1)) - log(x^2 + 1)*log((b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^2 + b^2 - 2*b + 1)) + 2*dilog(-(I*b*x - b)/(I*a +
b + 1)) - 2*dilog(-(I*b*x - b)/(I*a + b - 1)) + 2*dilog((I*b*x + b)/(-I*a + b + 1)) - 2*dilog((I*b*x + b)/(-I*
a + b - 1)))/b) + arctan(b*x + a)*arctan(x) - arctan(x)*arctan((b^2*x + a*b)/b)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (b x + a\right )}{x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(b*x+a)/(x^2+1),x, algorithm="fricas")

[Out]

integral(arctan(b*x + a)/(x^2 + 1), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atan}{\left (a + b x \right )}}{x^{2} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(b*x+a)/(x**2+1),x)

[Out]

Integral(atan(a + b*x)/(x**2 + 1), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (b x + a\right )}{x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(b*x+a)/(x^2+1),x, algorithm="giac")

[Out]

integrate(arctan(b*x + a)/(x^2 + 1), x)