\(\int \frac {\arctan (a+b x)}{1+x^2} \, dx\) [60]

Optimal result

Integrand size = 14, antiderivative size = 274 \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\frac {1}{4} \log \left (\frac {b (i-x)}{a+i (1+b)}\right ) \log (1-i a-i b x)-\frac {1}{4} \log \left (-\frac {b (i+x)}{a+i (1-b)}\right ) \log (1-i a-i b x)-\frac {1}{4} \log \left (\frac {b (i-x)}{a-i (1-b)}\right ) \log (1+i a+i b x)+\frac {1}{4} \log \left (-\frac {b (i+x)}{a-i (1+b)}\right ) \log (1+i a+i b x)-\frac {1}{4} \operatorname {PolyLog}\left (2,-\frac {i-a-b x}{a-i (1-b)}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,-\frac {i-a-b x}{a-i (1+b)}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {i+a+b x}{i+a-i b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {i+a+b x}{a+i (1+b)}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5159, 2456, 2441, 2440, 2438} \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=-\frac {1}{4} \operatorname {PolyLog}\left (2,-\frac {-a-b x+i}{a-i (1-b)}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,-\frac {-a-b x+i}{a-i (b+1)}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {a+b x+i}{a-i b+i}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {a+b x+i}{a+i (b+1)}\right )+\frac {1}{4} \log \left (\frac {b (-x+i)}{a+i (b+1)}\right ) \log (-i a-i b x+1)-\frac {1}{4} \log \left (-\frac {b (x+i)}{a+i (1-b)}\right ) \log (-i a-i b x+1)-\frac {1}{4} \log \left (\frac {b (-x+i)}{a-i (1-b)}\right ) \log (i a+i b x+1)+\frac {1}{4} \log \left (-\frac {b (x+i)}{a-i (b+1)}\right ) \log (i a+i b x+1) \]

[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 2438

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]

Rule 2440

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 2441

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 2456

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 5159

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]

Rubi steps \begin{align*} \text {integral}& = \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} \text {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} \text {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} \text {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} \text {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} \operatorname {PolyLog}\left (2,-\frac {i-a-b x}{a-i (1-b)}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,-\frac {i-a-b x}{a-i (1+b)}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {i+a+b x}{i+a-i b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {i+a+b x}{a+i (1+b)}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.03 \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\frac {1}{4} \log \left (\frac {b (i-x)}{a+i (1+b)}\right ) \log (1-i a-i b x)-\frac {1}{4} \log \left (-\frac {b (i+x)}{a+i (1-b)}\right ) \log (1-i a-i b x)-\frac {1}{4} \log \left (\frac {b (i-x)}{a-i (1-b)}\right ) \log (1+i a+i b x)+\frac {1}{4} \log \left (-\frac {b (i+x)}{a-i (1+b)}\right ) \log (1+i a+i b x)-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-i a-i b x}{1-i a-b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1-i a-i b x}{1-i a+b}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+i a+i b x}{1+i a-b}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {1+i a+i b x}{1+i a+b}\right ) \]

[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 [A] (verified)

Time = 0.78 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.82

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

[In]

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

[Out]

-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]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(atan(a + b*x)/(x**2 + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.20 \[ \int \frac {\arctan (a+b x)}{1+x^2} \, dx=\frac {1}{8} \, b {\left (\frac {8 \, \arctan \left (x\right ) \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b} - \frac {4 \, \arctan \left (x\right ) \arctan \left (\frac {a b + {\left (b^{2} + b\right )} x}{a^{2} + b^{2} + 2 \, b + 1}, \frac {a b x + a^{2} + b + 1}{a^{2} + b^{2} + 2 \, b + 1}\right ) - 4 \, \arctan \left (x\right ) \arctan \left (\frac {a b + {\left (b^{2} - b\right )} x}{a^{2} + b^{2} - 2 \, b + 1}, \frac {a b x + a^{2} - b + 1}{a^{2} + b^{2} - 2 \, b + 1}\right ) + \log \left (x^{2} + 1\right ) \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{a^{2} + b^{2} + 2 \, b + 1}\right ) - \log \left (x^{2} + 1\right ) \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{a^{2} + b^{2} - 2 \, b + 1}\right ) + 2 \, {\rm Li}_2\left (-\frac {i \, b x - b}{i \, a + b + 1}\right ) - 2 \, {\rm Li}_2\left (-\frac {i \, b x - b}{i \, a + b - 1}\right ) + 2 \, {\rm Li}_2\left (\frac {i \, b x + b}{-i \, a + b + 1}\right ) - 2 \, {\rm Li}_2\left (\frac {i \, b x + b}{-i \, a + b - 1}\right )}{b}\right )} + \arctan \left (b x + a\right ) \arctan \left (x\right ) - \arctan \left (x\right ) \arctan \left (\frac {b^{2} x + a b}{b}\right ) \]

[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)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(atan(a + b*x)/(x^2 + 1),x)

[Out]

int(atan(a + b*x)/(x^2 + 1), x)