\(\int \frac {\arctan (a+b x)}{c+\frac {d}{x}} \, dx\) [55]

Optimal result

Integrand size = 16, antiderivative size = 244 \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x}} \, dx=-\frac {(1+i a+i b x) \log (1+i a+i b x)}{2 b c}-\frac {(1-i a-i b x) \log (-i (i+a+b x))}{2 b c}-\frac {i d \log (1-i a-i b x) \log \left (-\frac {b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac {i d \log (1+i a+i b x) \log \left (\frac {b (d+c x)}{(i-a) c+b d}\right )}{2 c^2}+\frac {i d \operatorname {PolyLog}\left (2,\frac {c (i-a-b x)}{i c-a c+b d}\right )}{2 c^2}-\frac {i d \operatorname {PolyLog}\left (2,\frac {c (i+a+b x)}{(i+a) c-b d}\right )}{2 c^2} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(771\) vs. \(2(244)=488\).

Time = 7.96 (sec) , antiderivative size = 771, normalized size of antiderivative = 3.16 \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x}} \, dx =\text {Too large to display} \] Input:

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

Output:

(-2*a^2*c^2*ArcTan[a + b*x] + 2*a*b*c*d*ArcTan[a + b*x] + I*a*b*c*d*Pi*Arc 
Tan[a + b*x] - I*b^2*d^2*Pi*ArcTan[a + b*x] - 2*a*b*c^2*x*ArcTan[a + b*x] 
+ 2*b^2*c*d*x*ArcTan[a + b*x] + (2*I)*a*b*c*d*ArcTan[a - (b*d)/c]*ArcTan[a 
 + b*x] - (2*I)*b^2*d^2*ArcTan[a - (b*d)/c]*ArcTan[a + b*x] - b*c*d*ArcTan 
[a + b*x]^2 + I*a*b*c*d*ArcTan[a + b*x]^2 - I*b^2*d^2*ArcTan[a + b*x]^2 + 
(b*c*d*Sqrt[1 + a^2 - (2*a*b*d)/c + (b^2*d^2)/c^2]*ArcTan[a + b*x]^2)/E^(I 
*ArcTan[a - (b*d)/c]) + a*b*c*d*Pi*Log[1 + E^((-2*I)*ArcTan[a + b*x])] - b 
^2*d^2*Pi*Log[1 + E^((-2*I)*ArcTan[a + b*x])] - 2*a*b*c*d*ArcTan[a + b*x]* 
Log[1 + E^((2*I)*ArcTan[a + b*x])] + 2*b^2*d^2*ArcTan[a + b*x]*Log[1 + E^( 
(2*I)*ArcTan[a + b*x])] - 2*a*b*c*d*ArcTan[a - (b*d)/c]*Log[1 - E^((2*I)*( 
-ArcTan[a - (b*d)/c] + ArcTan[a + b*x]))] + 2*b^2*d^2*ArcTan[a - (b*d)/c]* 
Log[1 - E^((2*I)*(-ArcTan[a - (b*d)/c] + ArcTan[a + b*x]))] + 2*a*b*c*d*Ar 
cTan[a + b*x]*Log[1 - E^((2*I)*(-ArcTan[a - (b*d)/c] + ArcTan[a + b*x]))] 
- 2*b^2*d^2*ArcTan[a + b*x]*Log[1 - E^((2*I)*(-ArcTan[a - (b*d)/c] + ArcTa 
n[a + b*x]))] - 2*a*c^2*Log[1/Sqrt[1 + (a + b*x)^2]] + 2*b*c*d*Log[1/Sqrt[ 
1 + (a + b*x)^2]] - a*b*c*d*Pi*Log[1/Sqrt[1 + (a + b*x)^2]] + b^2*d^2*Pi*L 
og[1/Sqrt[1 + (a + b*x)^2]] + 2*a*b*c*d*ArcTan[a - (b*d)/c]*Log[Sin[ArcTan 
[(-(a*c) + b*d)/c] + ArcTan[a + b*x]]] - 2*b^2*d^2*ArcTan[a - (b*d)/c]*Log 
[Sin[ArcTan[(-(a*c) + b*d)/c] + ArcTan[a + b*x]]] + I*b*d*(a*c - b*d)*Poly 
Log[2, -E^((2*I)*ArcTan[a + b*x])] + I*b*d*(-(a*c) + b*d)*PolyLog[2, E^...
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, 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.

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

\(\Big \downarrow \) 5574

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

\(\Big \downarrow \) 2856

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 0.35 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (b x +a \right ) \left (b x +a \right )}{c}-\frac {\arctan \left (b x +a \right ) d b \ln \left (a c -b d -c \left (b x +a \right )\right )}{c^{2}}+\frac {-\frac {\ln \left (a^{2} c^{2}-2 a b c d +d^{2} b^{2}-2 a c \left (a c -b d -c \left (b x +a \right )\right )+2 b d \left (a c -b d -c \left (b x +a \right )\right )+c^{2}+\left (a c -b d -c \left (b x +a \right )\right )^{2}\right )}{2}-b d \left (-\frac {i \ln \left (a c -b d -c \left (b x +a \right )\right ) \left (\ln \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )-\ln \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )\right )}{2 c}-\frac {i \left (\operatorname {dilog}\left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )-\operatorname {dilog}\left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )\right )}{2 c}\right )}{c}}{b}\)	\(295\)
default	\(\frac {\frac {\arctan \left (b x +a \right ) \left (b x +a \right )}{c}-\frac {\arctan \left (b x +a \right ) d b \ln \left (a c -b d -c \left (b x +a \right )\right )}{c^{2}}+\frac {-\frac {\ln \left (a^{2} c^{2}-2 a b c d +d^{2} b^{2}-2 a c \left (a c -b d -c \left (b x +a \right )\right )+2 b d \left (a c -b d -c \left (b x +a \right )\right )+c^{2}+\left (a c -b d -c \left (b x +a \right )\right )^{2}\right )}{2}-b d \left (-\frac {i \ln \left (a c -b d -c \left (b x +a \right )\right ) \left (\ln \left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )-\ln \left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )\right )}{2 c}-\frac {i \left (\operatorname {dilog}\left (\frac {i c +c \left (b x +a \right )}{a c -b d +i c}\right )-\operatorname {dilog}\left (\frac {i c -c \left (b x +a \right )}{-a c +b d +i c}\right )\right )}{2 c}\right )}{c}}{b}\)	\(295\)
parts	\(\frac {\arctan \left (b x +a \right ) x}{c}-\frac {\arctan \left (b x +a \right ) d \ln \left (c x +d \right )}{c^{2}}-\frac {b \left (\frac {\ln \left (a^{2} c^{2}-2 a b c d +2 a b c \left (c x +d \right )+d^{2} b^{2}-2 b^{2} d \left (c x +d \right )+b^{2} \left (c x +d \right )^{2}+c^{2}\right )}{2 b^{2}}-\frac {a \arctan \left (\frac {2 a b c -2 b^{2} d +2 b^{2} \left (c x +d \right )}{2 b c}\right )}{b^{2}}-d \left (-\frac {i \ln \left (c x +d \right ) \left (\ln \left (\frac {i c -a c +b d -b \left (c x +d \right )}{-a c +b d +i c}\right )-\ln \left (\frac {i c +a c -b d +b \left (c x +d \right )}{a c -b d +i c}\right )\right )}{2 b c}-\frac {i \left (\operatorname {dilog}\left (\frac {i c -a c +b d -b \left (c x +d \right )}{-a c +b d +i c}\right )-\operatorname {dilog}\left (\frac {i c +a c -b d +b \left (c x +d \right )}{a c -b d +i c}\right )\right )}{2 b c}\right )\right )}{c}\)	\(313\)
risch	\(\frac {i \ln \left (-b x i-a i+1\right ) x}{2 c}-\frac {i d \operatorname {dilog}\left (\frac {i a c -i b d +\left (-b x i-a i+1\right ) c -c}{i a c -i b d -c}\right )}{2 c^{2}}-\frac {i \ln \left (b x i+a i+1\right ) x}{2 c}-\frac {i \ln \left (b x i+a i+1\right ) a}{2 b c}-\frac {\ln \left (-b x i-a i+1\right )}{2 b c}+\frac {1}{b c}+\frac {i d \ln \left (b x i+a i+1\right ) \ln \left (\frac {-i a c +i b d +\left (b x i+a i+1\right ) c -c}{-i a c +i b d -c}\right )}{2 c^{2}}+\frac {i \ln \left (-b x i-a i+1\right ) a}{2 b c}+\frac {i d \operatorname {dilog}\left (\frac {-i a c +i b d +\left (b x i+a i+1\right ) c -c}{-i a c +i b d -c}\right )}{2 c^{2}}-\frac {i d \ln \left (-b x i-a i+1\right ) \ln \left (\frac {i a c -i b d +\left (-b x i-a i+1\right ) c -c}{i a c -i b d -c}\right )}{2 c^{2}}-\frac {\ln \left (b x i+a i+1\right )}{2 b c}\)	\(363\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.16 \[ \int \frac {\arctan (a+b x)}{c+\frac {d}{x}} \, dx=-\frac {b d \arctan \left (b x + a\right ) \log \left (-\frac {b^{2} c^{2} x^{2} + 2 \, b^{2} c d x + b^{2} d^{2}}{2 \, a b c d - b^{2} d^{2} - {\left (a^{2} + 1\right )} c^{2}}\right ) + i \, b d {\rm Li}_2\left (-\frac {i \, b c x + {\left (i \, a - 1\right )} c}{{\left (-i \, a + 1\right )} c + i \, b d}\right ) - i \, b d {\rm Li}_2\left (-\frac {i \, b c x + {\left (i \, a + 1\right )} c}{{\left (-i \, a - 1\right )} c + i \, b d}\right ) - 2 \, {\left (b c x + a c\right )} \arctan \left (b x + a\right ) - {\left (b d \arctan \left (-\frac {b c^{2} x + b c d}{2 \, a b c d - b^{2} d^{2} - {\left (a^{2} + 1\right )} c^{2}}, \frac {a b c d - b^{2} d^{2} + {\left (a b c^{2} - b^{2} c d\right )} x}{2 \, a b c d - b^{2} d^{2} - {\left (a^{2} + 1\right )} c^{2}}\right ) - c\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b c^{2}} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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