\(\int \frac {\arctan (a+b x)}{1+x^2} \, dx\) [58]

Optimal result

Integrand size = 14, antiderivative size = 179 \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=-\frac {1}{2} i \arctan (a+b x) \log \left (\frac {2 b (i-x)}{(a-i (1-b)) (1-i (a+b x))}\right )+\frac {1}{2} i \arctan (a+b x) \log \left (-\frac {2 b (i+x)}{(a-i (1+b)) (1-i (a+b x))}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,1-\frac {2 b (i-x)}{(a-i (1-b)) (1-i (a+b x))}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,1+\frac {2 b (i+x)}{(a-i (1+b)) (1-i (a+b x))}\right ) \] Output:

-1/2*I*arctan(b*x+a)*ln(2*b*(I-x)/(a-I*(1-b))/(1-I*(b*x+a)))+1/2*I*arctan( 
b*x+a)*ln(-2*b*(I+x)/(a-I*(1+b))/(1-I*(b*x+a)))-1/4*polylog(2,1-2*b*(I-x)/ 
(a-I*(1-b))/(1-I*(b*x+a)))+1/4*polylog(2,1+2*b*(I+x)/(a-I*(1+b))/(1-I*(b*x 
+a)))
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.58 \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\frac {1}{4} \log \left (\frac {b (i-x)}{a+i (1+b)}\right ) \log (1-i a-i b x)-\frac {1}{4} \log \left (-\frac {b (i+x)}{a+i (1-b)}\right ) \log (1-i a-i b x)-\frac {1}{4} \log \left (\frac {b (i-x)}{a-i (1-b)}\right ) \log (1+i a+i b x)+\frac {1}{4} \log \left (-\frac {b (i+x)}{a-i (1+b)}\right ) \log (1+i a+i b x)-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-i a-i b x}{1-i a-b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-i a-i b x}{1-i a+b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+i a+i b x}{1+i a-b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+i a+i b x}{1+i a+b}\right ) \] Input:

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

Output:

(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)))]*Lo 
g[1 + I*a + I*b*x])/4 - PolyLog[2, (1 - I*a - I*b*x)/(1 - I*a - b)]/4 + Po 
lyLog[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
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.70, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5574, 2856, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\arctan (a+b x)}{x^2+1} \, dx\)

\(\Big \downarrow \) 5574

\(\displaystyle \frac {1}{2} i \int \frac {\log (-i a-i b x+1)}{x^2+1}dx-\frac {1}{2} i \int \frac {\log (i a+i b x+1)}{x^2+1}dx\)

\(\Big \downarrow \) 2856

\(\displaystyle \frac {1}{2} i \int \left (\frac {i \log (-i a-i b x+1)}{2 (i-x)}+\frac {i \log (-i a-i b x+1)}{2 (x+i)}\right )dx-\frac {1}{2} i \int \left (\frac {i \log (i a+i b x+1)}{2 (i-x)}+\frac {i \log (i a+i b x+1)}{2 (x+i)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {a+b x+i}{a-i b+i}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {a+b x+i}{a+i (b+1)}\right )-\frac {1}{2} i \log \left (\frac {b (-x+i)}{a+i (b+1)}\right ) \log (-i a-i b x+1)+\frac {1}{2} i \log \left (-\frac {b (x+i)}{a+i (1-b)}\right ) \log (-i a-i b x+1)\right )-\frac {1}{2} i \left (-\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {-a-b x+i}{a-i (1-b)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {-a-b x+i}{a-i (b+1)}\right )-\frac {1}{2} i \log \left (\frac {b (-x+i)}{a-i (1-b)}\right ) \log (i a+i b x+1)+\frac {1}{2} i \log \left (-\frac {b (x+i)}{a-i (b+1)}\right ) \log (i a+i b x+1)\right )\)

Input:

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

Output:

(-1/2*I)*((-1/2*I)*Log[(b*(I - x))/(a - I*(1 - b))]*Log[1 + I*a + I*b*x] + 
 (I/2)*Log[-((b*(I + x))/(a - I*(1 + b)))]*Log[1 + I*a + I*b*x] - (I/2)*Po 
lyLog[2, -((I - a - b*x)/(a - I*(1 - b)))] + (I/2)*PolyLog[2, -((I - a - b 
*x)/(a - I*(1 + b)))]) + (I/2)*((-1/2*I)*Log[(b*(I - x))/(a + I*(1 + b))]* 
Log[1 - I*a - I*b*x] + (I/2)*Log[-((b*(I + x))/(a + I*(1 - b)))]*Log[1 - I 
*a - I*b*x] + (I/2)*PolyLog[2, (I + a + b*x)/(I + a - I*b)] - (I/2)*PolyLo 
g[2, (I + a + b*x)/(a + I*(1 + b))])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2856

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_)^(r_))^(q_.), x_Symbol] :> Int[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] && I 
GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
 

rule 5574

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

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.26

method	result	size
risch	\(\frac {\ln \left (-b x i-a i+1\right ) \ln \left (\frac {-b x i-b}{a i-b -1}\right )}{4}+\frac {\operatorname {dilog}\left (\frac {-b x i-b}{a i-b -1}\right )}{4}-\frac {\ln \left (-b x i-a i+1\right ) \ln \left (\frac {-b x i+b}{a i+b -1}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {-b x i+b}{a i+b -1}\right )}{4}+\frac {\ln \left (b x i+a i+1\right ) \ln \left (\frac {b x i-b}{-a i-b -1}\right )}{4}+\frac {\operatorname {dilog}\left (\frac {b x i-b}{-a i-b -1}\right )}{4}-\frac {\ln \left (b x i+a i+1\right ) \ln \left (\frac {b x i+b}{-a i+b -1}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {b x i+b}{-a i+b -1}\right )}{4}\)	\(226\)
default	\(\arctan \left (x \right ) \arctan \left (b x +a \right )-b \left (-\frac {\ln \left (1-\frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) \arctan \left (x \right )}{2 \left (i b +a +i\right )}-\frac {\ln \left (1-\frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) \arctan \left (x \right )}{2 b \left (i b +a +i\right )}+\frac {i \ln \left (1-\frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) a \arctan \left (x \right )}{2 b \left (i b +a +i\right )}+\frac {i \arctan \left (x \right )^{2}}{2 i b +2 a +2 i}+\frac {i \operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right )}{4 i b +4 a +4 i}+\frac {i \arctan \left (x \right )^{2}}{2 b \left (i b +a +i\right )}+\frac {a \arctan \left (x \right )^{2}}{2 b \left (i b +a +i\right )}+\frac {i \operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right )}{4 b \left (i b +a +i\right )}+\frac {\operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) a}{4 b \left (i b +a +i\right )}-\frac {i \arctan \left (x \right ) \ln \left (1-\frac {\left (-i b +a -i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a +i\right )}\right )}{2 b}-\frac {\arctan \left (x \right )^{2}}{2 b}-\frac {\operatorname {polylog}\left (2, \frac {\left (-i b +a -i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a +i\right )}\right )}{4 b}\right )\)	\(501\)
parts	\(\arctan \left (x \right ) \arctan \left (b x +a \right )-b \left (-\frac {\ln \left (1-\frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) \arctan \left (x \right )}{2 \left (i b +a +i\right )}-\frac {\ln \left (1-\frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) \arctan \left (x \right )}{2 b \left (i b +a +i\right )}+\frac {i \ln \left (1-\frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) a \arctan \left (x \right )}{2 b \left (i b +a +i\right )}+\frac {i \arctan \left (x \right )^{2}}{2 i b +2 a +2 i}+\frac {i \operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right )}{4 i b +4 a +4 i}+\frac {i \arctan \left (x \right )^{2}}{2 b \left (i b +a +i\right )}+\frac {a \arctan \left (x \right )^{2}}{2 b \left (i b +a +i\right )}+\frac {i \operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right )}{4 b \left (i b +a +i\right )}+\frac {\operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a -i\right )}\right ) a}{4 b \left (i b +a +i\right )}-\frac {i \arctan \left (x \right ) \ln \left (1-\frac {\left (-i b +a -i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a +i\right )}\right )}{2 b}-\frac {\arctan \left (x \right )^{2}}{2 b}-\frac {\operatorname {polylog}\left (2, \frac {\left (-i b +a -i\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-i b -a +i\right )}\right )}{4 b}\right )\)	\(501\)
derivativedivides	\(\frac {b \arctan \left (x \right ) \arctan \left (b x +a \right )-b^{2} \left (-\frac {\arctan \left (b \left (\frac {b x +a}{b}-\frac {a}{b}\right )+a \right ) \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )}{b}-\frac {-\arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right ) \arctan \left (b \left (\frac {b x +a}{b}-\frac {a}{b}\right )+a \right )-b \left (\frac {\ln \left (1-\frac {\left (-i b +a +i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a -i\right )}\right ) \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )}{2 i b +2 a +2 i}+\frac {\ln \left (1-\frac {\left (-i b +a +i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a -i\right )}\right ) \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )}{2 b \left (i b +a +i\right )}-\frac {i \ln \left (1-\frac {\left (-i b +a +i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a -i\right )}\right ) a \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )}{2 b \left (i b +a +i\right )}+\frac {i \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )^{2}}{2 i b +2 a +2 i}+\frac {i \operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a -i\right )}\right )}{4 i b +4 a +4 i}+\frac {i \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )^{2}}{2 b \left (i b +a +i\right )}+\frac {a \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )^{2}}{2 b \left (i b +a +i\right )}+\frac {i \operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a -i\right )}\right )}{4 b \left (i b +a +i\right )}+\frac {\operatorname {polylog}\left (2, \frac {\left (-i b +a +i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a -i\right )}\right ) a}{4 b \left (i b +a +i\right )}+\frac {i \arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right ) \ln \left (1-\frac {\left (-i b +a -i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a +i\right )}\right )}{2 b}-\frac {\arctan \left (-\frac {b x +a}{b}+\frac {a}{b}\right )^{2}}{2 b}-\frac {\operatorname {polylog}\left (2, \frac {\left (-i b +a -i\right ) \left (1+i \left (\frac {b x +a}{b}-\frac {a}{b}\right )\right )^{2}}{\left (\left (\frac {b x +a}{b}-\frac {a}{b}\right )^{2}+1\right ) \left (-i b -a +i\right )}\right )}{4 b}\right )}{b}\right )}{b}\)	\(961\)

Input:

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

Output:

1/4*ln(1-I*a-I*b*x)*ln((-I*b*x-b)/(I*a-b-1))+1/4*dilog((-I*b*x-b)/(I*a-b-1 
))-1/4*ln(1-I*a-I*b*x)*ln((-I*b*x+b)/(I*a+b-1))-1/4*dilog((-I*b*x+b)/(I*a+ 
b-1))+1/4*ln(1+I*a+I*b*x)*ln((I*b*x-b)/(-I*a-b-1))+1/4*dilog((I*b*x-b)/(-I 
*a-b-1))-1/4*ln(1+I*a+I*b*x)*ln((I*b*x+b)/(-I*a+b-1))-1/4*dilog((I*b*x+b)/ 
(-I*a+b-1))
 

Fricas [F]

\[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\int { \frac {\arctan \left (b x + a\right )}{x^{2} + 1} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\int \frac {\operatorname {atan}{\left (a + b x \right )}}{x^{2} + 1}\, dx \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (129) = 258\).

Time = 0.18 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.83 \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\frac {1}{8} \, b {\left (\frac {8 \, \arctan \left (x\right ) \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b} - \frac {4 \, \arctan \left (x\right ) \arctan \left (\frac {a b + {\left (b^{2} + b\right )} x}{a^{2} + b^{2} + 2 \, b + 1}, \frac {a b x + a^{2} + b + 1}{a^{2} + b^{2} + 2 \, b + 1}\right ) - 4 \, \arctan \left (x\right ) \arctan \left (\frac {a b + {\left (b^{2} - b\right )} x}{a^{2} + b^{2} - 2 \, b + 1}, \frac {a b x + a^{2} - b + 1}{a^{2} + b^{2} - 2 \, b + 1}\right ) + \log \left (x^{2} + 1\right ) \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{a^{2} + b^{2} + 2 \, b + 1}\right ) - \log \left (x^{2} + 1\right ) \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{a^{2} + b^{2} - 2 \, b + 1}\right ) + 2 \, {\rm Li}_2\left (-\frac {i \, b x - b}{i \, a + b + 1}\right ) - 2 \, {\rm Li}_2\left (-\frac {i \, b x - b}{i \, a + b - 1}\right ) + 2 \, {\rm Li}_2\left (\frac {i \, b x + b}{-i \, a + b + 1}\right ) - 2 \, {\rm Li}_2\left (\frac {i \, b x + b}{-i \, a + b - 1}\right )}{b}\right )} + \arctan \left (b x + a\right ) \arctan \left (x\right ) - \arctan \left (x\right ) \arctan \left (\frac {b^{2} x + a b}{b}\right ) \] Input:

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

Output:

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)
 

Giac [F]

\[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\int { \frac {\arctan \left (b x + a\right )}{x^{2} + 1} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )}{x^2+1} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\int \frac {\mathit {atan} \left (b x +a \right )}{x^{2}+1}d x \] Input:

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

Output:

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