\(\int \frac {\arctan (a+b x)}{c+d x} \, dx\) [54]

Optimal result

Integrand size = 14, antiderivative size = 152 \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=-\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\arctan (a+b x) \log \left (\frac {2 b (c+d x)}{(b c+(i-a) d) (1-i (a+b x))}\right )}{d}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )}{2 d}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(b c+(i-a) d) (1-i (a+b x))}\right )}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.52 \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\frac {i \log (1-i (a+b x)) \log \left (-\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}-\frac {i (b c-a d)}{b}}\right )}{2 d}-\frac {i \log (1+i (a+b x)) \log \left (\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}+\frac {i (b c-a d)}{b}}\right )}{2 d}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i d (1-i (a+b x))}{b c-i d-a d}\right )}{2 d}-\frac {i \operatorname {PolyLog}\left (2,\frac {i d (1+i (a+b x))}{b c+i d-a d}\right )}{2 d} \] Input:

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

Output:

((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)*Po 
lyLog[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
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5570, 27, 5381, 2849, 2752, 2897}

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.

\(\displaystyle \int \frac {\arctan (a+b x)}{c+d x} \, dx\)

\(\Big \downarrow \) 5570

\(\displaystyle \frac {\int \frac {b \arctan (a+b x)}{b \left (c-\frac {a d}{b}\right )+d (a+b x)}d(a+b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\arctan (a+b x)}{d (a+b x)-a d+b c}d(a+b x)\)

\(\Big \downarrow \) 5381

\(\displaystyle -\frac {\int \frac {\log \left (\frac {2 (b c-a d+d (a+b x))}{(b c-a d+i d) (1-i (a+b x))}\right )}{(a+b x)^2+1}d(a+b x)}{d}+\frac {\int \frac {\log \left (\frac {2}{1-i (a+b x)}\right )}{(a+b x)^2+1}d(a+b x)}{d}+\frac {\arctan (a+b x) \log \left (\frac {2 (d (a+b x)-a d+b c)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}\)

\(\Big \downarrow \) 2849

\(\displaystyle -\frac {\int \frac {\log \left (\frac {2 (b c-a d+d (a+b x))}{(b c-a d+i d) (1-i (a+b x))}\right )}{(a+b x)^2+1}d(a+b x)}{d}+\frac {i \int \frac {\log \left (\frac {2}{1-i (a+b x)}\right )}{1-\frac {2}{1-i (a+b x)}}d\frac {1}{1-i (a+b x)}}{d}+\frac {\arctan (a+b x) \log \left (\frac {2 (d (a+b x)-a d+b c)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}\)

\(\Big \downarrow \) 2752

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

\(\Big \downarrow \) 2897

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2752

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-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 2897

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[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 5381

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Si 
mp[(-(a + b*ArcTan[c*x]))*(Log[2/(1 - I*c*x)]/e), x] + (Simp[(a + b*ArcTan[ 
c*x])*(Log[2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] + Simp[b*(c/e) 
   Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Simp[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]) /; FreeQ[{a 
, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]
 

rule 5570

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A 
rcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && I 
GtQ[p, 0]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {\frac {b \ln \left (a d -b c -d \left (b x +a \right )\right ) \arctan \left (b x +a \right )}{d}+b \left (-\frac {i \ln \left (a d -b c -d \left (b x +a \right )\right ) \left (\ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\operatorname {dilog}\left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}\right )}{b}\)	\(186\)
default	\(\frac {\frac {b \ln \left (a d -b c -d \left (b x +a \right )\right ) \arctan \left (b x +a \right )}{d}+b \left (-\frac {i \ln \left (a d -b c -d \left (b x +a \right )\right ) \left (\ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\operatorname {dilog}\left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}\right )}{b}\)	\(186\)
parts	\(\frac {\ln \left (d x +c \right ) \arctan \left (b x +a \right )}{d}-b \left (-\frac {i \ln \left (d x +c \right ) \left (\ln \left (\frac {i d -a d +b c -b \left (d x +c \right )}{-a d +b c +i d}\right )-\ln \left (\frac {i d +a d -b c +b \left (d x +c \right )}{a d -b c +i d}\right )\right )}{2 b d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d -a d +b c -b \left (d x +c \right )}{-a d +b c +i d}\right )-\operatorname {dilog}\left (\frac {i d +a d -b c +b \left (d x +c \right )}{a d -b c +i d}\right )\right )}{2 b d}\right )\)	\(194\)
risch	\(\frac {i \operatorname {dilog}\left (\frac {i a d -i b c +\left (-b x i-a i+1\right ) d -d}{i a d -i b c -d}\right )}{2 d}+\frac {i \ln \left (-b x i-a i+1\right ) \ln \left (\frac {i a d -i b c +\left (-b x i-a i+1\right ) d -d}{i a d -i b c -d}\right )}{2 d}-\frac {i \operatorname {dilog}\left (\frac {-i a d +i b c +\left (b x i+a i+1\right ) d -d}{-i a d +i b c -d}\right )}{2 d}-\frac {i \ln \left (b x i+a i+1\right ) \ln \left (\frac {-i a d +i b c +\left (b x i+a i+1\right ) d -d}{-i a d +i b c -d}\right )}{2 d}\)	\(230\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

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. 284 vs. \(2 (128) = 256\).

Time = 0.27 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.87 \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\frac {\arctan \left (b x + a\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b d^{2} x + b c d}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}, \frac {b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - \arctan \left (b x + a\right ) \log \left (\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) + i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a + 1\right )} d}{-i \, b c + {\left (i \, a + 1\right )} d}\right ) - i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a - 1\right )} d}{-i \, b c + {\left (i \, a - 1\right )} d}\right )}{2 \, d} \] Input:

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

Output:

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
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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