3.1.28 \(\int \frac {a+b \arctan (c+d x)}{e+f x} \, dx\) [28]

3.1.28.1 Optimal result

Integrand size = 18, antiderivative size = 162 \[ \int \frac {a+b \arctan (c+d x)}{e+f x} \, dx=-\frac {(a+b \arctan (c+d x)) \log \left (\frac {2}{1-i (c+d x)}\right )}{f}+\frac {(a+b \arctan (c+d x)) \log \left (\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right )}{2 f}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f} \]

output

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

3.1.28.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.99 \[ \int \frac {a+b \arctan (c+d x)}{e+f x} \, dx=\frac {2 a \log (d (e+f x))+i b \log \left (\frac {d (e+f x)}{d e-(i+c) f}\right ) \log (1-i (c+d x))-i b \log \left (\frac {d (e+f x)}{d e+i f-c f}\right ) \log (1+i (c+d x))-i b \operatorname {PolyLog}\left (2,\frac {f (-i+c+d x)}{-d e+(-i+c) f}\right )+i b \operatorname {PolyLog}\left (2,\frac {f (i+c+d x)}{-d e+(i+c) f}\right )}{2 f} \]

input

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

output

(2*a*Log[d*(e + f*x)] + I*b*Log[(d*(e + f*x))/(d*e - (I + c)*f)]*Log[1 - I 
*(c + d*x)] - I*b*Log[(d*(e + f*x))/(d*e + I*f - c*f)]*Log[1 + I*(c + d*x) 
] - I*b*PolyLog[2, (f*(-I + c + d*x))/(-(d*e) + (-I + c)*f)] + I*b*PolyLog 
[2, (f*(I + c + d*x))/(-(d*e) + (I + c)*f)])/(2*f)
 

3.1.28.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, 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.

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

\(\Big \downarrow \) 5570

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5381

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

\(\Big \downarrow \) 2897

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

input

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

output

-(((a + b*ArcTan[c + d*x])*Log[2/(1 - I*(c + d*x))])/f) + ((a + b*ArcTan[c 
 + d*x])*Log[(2*(d*e - c*f + f*(c + d*x)))/((d*e + I*f - c*f)*(1 - I*(c + 
d*x)))])/f + ((I/2)*b*PolyLog[2, 1 - 2/(1 - I*(c + d*x))])/f - ((I/2)*b*Po 
lyLog[2, 1 - (2*(d*e - c*f + f*(c + d*x)))/((d*e + I*f - c*f)*(1 - I*(c + 
d*x)))])/f
 

3.1.28.3.1 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]
 

3.1.28.4 Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.22

method	result	size
parts	\(\frac {a \ln \left (f x +e \right )}{f}+\frac {b \left (\frac {d \ln \left (f \left (d x +c \right )-c f +d e \right ) \arctan \left (d x +c \right )}{f}-d \left (-\frac {i \ln \left (f \left (d x +c \right )-c f +d e \right ) \left (\ln \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )-\ln \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )\right )}{2 f}-\frac {i \left (\operatorname {dilog}\left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )-\operatorname {dilog}\left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )\right )}{2 f}\right )\right )}{d}\)	\(198\)
derivativedivides	\(\frac {\frac {a d \ln \left (c f -d e -f \left (d x +c \right )\right )}{f}-b d \left (-\frac {\ln \left (c f -d e -f \left (d x +c \right )\right ) \arctan \left (d x +c \right )}{f}+\frac {i \ln \left (c f -d e -f \left (d x +c \right )\right ) \left (\ln \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )-\ln \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )\right )}{2 f}+\frac {i \left (\operatorname {dilog}\left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )-\operatorname {dilog}\left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )\right )}{2 f}\right )}{d}\)	\(211\)
default	\(\frac {\frac {a d \ln \left (c f -d e -f \left (d x +c \right )\right )}{f}-b d \left (-\frac {\ln \left (c f -d e -f \left (d x +c \right )\right ) \arctan \left (d x +c \right )}{f}+\frac {i \ln \left (c f -d e -f \left (d x +c \right )\right ) \left (\ln \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )-\ln \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )\right )}{2 f}+\frac {i \left (\operatorname {dilog}\left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )-\operatorname {dilog}\left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )\right )}{2 f}\right )}{d}\)	\(211\)
risch	\(\frac {a \ln \left (i c f -i d e +\left (-i d x -i c +1\right ) f -f \right )}{f}+\frac {i b \operatorname {dilog}\left (\frac {i c f -i d e +\left (-i d x -i c +1\right ) f -f}{i c f -i d e -f}\right )}{2 f}+\frac {i b \ln \left (-i d x -i c +1\right ) \ln \left (\frac {i c f -i d e +\left (-i d x -i c +1\right ) f -f}{i c f -i d e -f}\right )}{2 f}-\frac {i b \operatorname {dilog}\left (\frac {-i c f +i d e +\left (i d x +i c +1\right ) f -f}{-i c f +i d e -f}\right )}{2 f}-\frac {i b \ln \left (i d x +i c +1\right ) \ln \left (\frac {-i c f +i d e +\left (i d x +i c +1\right ) f -f}{-i c f +i d e -f}\right )}{2 f}\)	\(267\)

input

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

output

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

3.1.28.5 Fricas [F]

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

input

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

output

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

3.1.28.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.28.7 Maxima [F]

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

input

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

output

2*b*integrate(1/2*arctan(d*x + c)/(f*x + e), x) + a*log(f*x + e)/f
 

3.1.28.8 Giac [F]

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

input

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

output

sage0*x
 

3.1.28.9 Mupad [F(-1)]

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

input

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

output

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