\(\int -\frac {\arctan (a-x)}{a+x} \, dx\) [698]

Optimal result

Integrand size = 13, antiderivative size = 122 \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=\arctan (a-x) \log \left (\frac {2}{1-i (a-x)}\right )-\arctan (a-x) \log \left (-\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a-x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1+\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.86 \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=-\frac {1}{2} i \left (-\log (1+i (a-x)) \log \left (\frac {a+x}{-i+2 a}\right )+\log (1-i a+i x) \log \left (\frac {a+x}{i+2 a}\right )+\operatorname {PolyLog}\left (2,\frac {i+a-x}{i+2 a}\right )-\operatorname {PolyLog}\left (2,\frac {i-a+x}{i-2 a}\right )\right ) \] Input:

Integrate[-(ArcTan[a - x]/(a + x)),x]
 

Output:

(-1/2*I)*(-(Log[1 + I*(a - x)]*Log[(a + x)/(-I + 2*a)]) + Log[1 - I*a + I* 
x]*Log[(a + x)/(I + 2*a)] + PolyLog[2, (I + a - x)/(I + 2*a)] - PolyLog[2, 
 (I - a + x)/(I - 2*a)])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {25, 5570, 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.

\(\displaystyle \int -\frac {\arctan (a-x)}{a+x} \, dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\arctan (a-x)}{a+x}dx\)

\(\Big \downarrow \) 5570

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

\(\Big \downarrow \) 5381

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

\(\Big \downarrow \) 2897

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

Input:

Int[-(ArcTan[a - x]/(a + x)),x]
 

Output:

ArcTan[a - x]*Log[2/(1 - I*(a - x))] - ArcTan[a - x]*Log[(-2*(a + x))/((I 
- 2*a)*(1 - I*(a - x)))] - (I/2)*PolyLog[2, 1 - 2/(1 - I*(a - x))] + (I/2) 
*PolyLog[2, 1 + (2*(a + x))/((I - 2*a)*(1 - I*(a - x)))]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 = 102, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\ln \left (a +x \right ) \arctan \left (a -x \right )+\frac {i \ln \left (a +x \right ) \ln \left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \ln \left (a +x \right ) \ln \left (\frac {i-a +x}{i-2 a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-a +x}{i-2 a}\right )}{2}\)	\(102\)
default	\(-\ln \left (a +x \right ) \arctan \left (a -x \right )+\frac {i \ln \left (a +x \right ) \ln \left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \ln \left (a +x \right ) \ln \left (\frac {i-a +x}{i-2 a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-a +x}{i-2 a}\right )}{2}\)	\(102\)
parts	\(-\ln \left (a +x \right ) \arctan \left (a -x \right )+\frac {i \ln \left (a +x \right ) \ln \left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \ln \left (a +x \right ) \ln \left (\frac {i-a +x}{i-2 a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-a +x}{i-2 a}\right )}{2}\)	\(102\)
risch	\(-\frac {i \operatorname {dilog}\left (\frac {i a +i x}{2 i a -1}\right )}{2}-\frac {i \ln \left (-i a +i x +1\right ) \ln \left (\frac {i a +i x}{2 i a -1}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {-i a -i x}{-2 i a -1}\right )}{2}+\frac {i \ln \left (i a -i x +1\right ) \ln \left (\frac {-i a -i x}{-2 i a -1}\right )}{2}\)	\(112\)

Input:

int(-arctan(a-x)/(a+x),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

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

integrate(-arctan(a-x)/(a+x),x, algorithm="fricas")
 

Output:

integral(arctan(-a + x)/(a + x), x)
 

Sympy [F]

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

integrate(-atan(a-x)/(a+x),x)
 

Output:

-Integral(atan(a - x)/(a + x), x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97 \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=-\frac {1}{2} \, \arctan \left (\frac {a + x}{4 \, a^{2} + 1}, \frac {2 \, {\left (a^{2} + a x\right )}}{4 \, a^{2} + 1}\right ) \log \left (a^{2} - 2 \, a x + x^{2} + 1\right ) + \frac {1}{2} \, \arctan \left (-a + x\right ) \log \left (\frac {a^{2} + 2 \, a x + x^{2}}{4 \, a^{2} + 1}\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-\frac {-i \, a + i \, x + 1}{2 i \, a - 1}\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-\frac {-i \, a + i \, x - 1}{2 i \, a + 1}\right ) \] Input:

integrate(-arctan(a-x)/(a+x),x, algorithm="maxima")
 

Output:

-1/2*arctan2((a + x)/(4*a^2 + 1), 2*(a^2 + a*x)/(4*a^2 + 1))*log(a^2 - 2*a 
*x + x^2 + 1) + 1/2*arctan(-a + x)*log((a^2 + 2*a*x + x^2)/(4*a^2 + 1)) - 
1/2*I*dilog(-(-I*a + I*x + 1)/(2*I*a - 1)) + 1/2*I*dilog(-(-I*a + I*x - 1) 
/(2*I*a + 1))
 

Giac [F]

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

integrate(-arctan(a-x)/(a+x),x, algorithm="giac")
 

Output:

integrate(-arctan(a - x)/(a + x), x)
 

Mupad [F(-1)]

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

int(-atan(a - x)/(a + x),x)
 

Output:

-int(atan(a - x)/(a + x), x)
                                                                                    
                                                                                    
 

Reduce [F]

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

int(-atan(a-x)/(a+x),x)
 

Output:

 - int(atan(a - x)/(a + x),x)