\(\int \frac {\arctan (x) \log (1+x^2)}{x} \, dx\) [1278]

Optimal result

Integrand size = 12, antiderivative size = 189 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x} \, dx=-\frac {1}{2} i \log ^2(1+i x) \log (-i x)+\frac {1}{2} i \log ^2(1-i x) \log (i x)+i \log (1-i x) \operatorname {PolyLog}(2,1-i x)-i \log (1+i x) \operatorname {PolyLog}(2,1+i x)-\frac {1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \operatorname {PolyLog}(2,-i x)+\frac {1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \operatorname {PolyLog}(2,i x)-i \operatorname {PolyLog}(3,1-i x)+i \operatorname {PolyLog}(3,1+i x) \] Output:

-1/2*I*ln(1+I*x)^2*ln(-I*x)+1/2*I*ln(1-I*x)^2*ln(I*x)+I*ln(1-I*x)*polylog( 
2,1-I*x)-I*ln(1+I*x)*polylog(2,1+I*x)-1/2*I*(ln(1-I*x)+ln(1+I*x)-ln(x^2+1) 
)*polylog(2,-I*x)+1/2*I*(ln(1-I*x)+ln(1+I*x)-ln(x^2+1))*polylog(2,I*x)-I*p 
olylog(3,1-I*x)+I*polylog(3,1+I*x)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.61 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x} \, dx=\frac {1}{2} i \left (-\log ^2(1-i x) \log (x)+\log ^2(1+i x) \log (x)-2 \log (1+i x) \log (x) \log (-i+x)-\log (-i x) \log ^2(-i+x)+\log (x) \log ^2(-i+x)+2 \log (1-i x) \log (x) \log (i+x)+\log (i x) \log ^2(i+x)-\log (x) \log ^2(i+x)+2 \log (i+x) \operatorname {PolyLog}(2,1-i x)-2 \log (-i+x) \operatorname {PolyLog}(2,1+i x)-\log (1-i x) \operatorname {PolyLog}(2,-i x)+\log (1+i x) \operatorname {PolyLog}(2,-i x)-2 \log (-i+x) \operatorname {PolyLog}(2,-i x)+\log \left (1+x^2\right ) \operatorname {PolyLog}(2,-i x)-\log (1-i x) \operatorname {PolyLog}(2,i x)+\log (1+i x) \operatorname {PolyLog}(2,i x)+2 \log (i+x) \operatorname {PolyLog}(2,i x)-\log \left (1+x^2\right ) \operatorname {PolyLog}(2,i x)-2 \operatorname {PolyLog}(3,1-i x)+2 \operatorname {PolyLog}(3,1+i x)\right ) \] Input:

Integrate[(ArcTan[x]*Log[1 + x^2])/x,x]
 

Output:

(I/2)*(-(Log[1 - I*x]^2*Log[x]) + Log[1 + I*x]^2*Log[x] - 2*Log[1 + I*x]*L 
og[x]*Log[-I + x] - Log[(-I)*x]*Log[-I + x]^2 + Log[x]*Log[-I + x]^2 + 2*L 
og[1 - I*x]*Log[x]*Log[I + x] + Log[I*x]*Log[I + x]^2 - Log[x]*Log[I + x]^ 
2 + 2*Log[I + x]*PolyLog[2, 1 - I*x] - 2*Log[-I + x]*PolyLog[2, 1 + I*x] - 
 Log[1 - I*x]*PolyLog[2, (-I)*x] + Log[1 + I*x]*PolyLog[2, (-I)*x] - 2*Log 
[-I + x]*PolyLog[2, (-I)*x] + Log[1 + x^2]*PolyLog[2, (-I)*x] - Log[1 - I* 
x]*PolyLog[2, I*x] + Log[1 + I*x]*PolyLog[2, I*x] + 2*Log[I + x]*PolyLog[2 
, I*x] - Log[1 + x^2]*PolyLog[2, I*x] - 2*PolyLog[3, 1 - I*x] + 2*PolyLog[ 
3, 1 + I*x])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5546, 2843, 2881, 2821, 5355, 2838, 7143}

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.

\(\displaystyle \int \frac {\arctan (x) \log \left (x^2+1\right )}{x} \, dx\)

\(\Big \downarrow \) 5546

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

\(\Big \downarrow \) 2843

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

\(\Big \downarrow \) 2881

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

\(\Big \downarrow \) 2821

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

\(\Big \downarrow \) 5355

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 7143

\(\displaystyle -\left (\left (\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x)\right ) \left (-\log \left (x^2+1\right )+\log (1-i x)+\log (1+i x)\right )\right )+\frac {1}{2} i \left (\log ^2(1-i x) \log (i x)-2 (\operatorname {PolyLog}(3,1-i x)-\operatorname {PolyLog}(2,1-i x) \log (1-i x))\right )-\frac {1}{2} i \left (\log ^2(1+i x) \log (-i x)-2 (\operatorname {PolyLog}(3,i x+1)-\operatorname {PolyLog}(2,i x+1) \log (1+i x))\right )\)

Input:

Int[(ArcTan[x]*Log[1 + x^2])/x,x]
 

Output:

-((Log[1 - I*x] + Log[1 + I*x] - Log[1 + x^2])*((I/2)*PolyLog[2, (-I)*x] - 
 (I/2)*PolyLog[2, I*x])) + (I/2)*(Log[1 - I*x]^2*Log[I*x] - 2*(-(Log[1 - I 
*x]*PolyLog[2, 1 - I*x]) + PolyLog[3, 1 - I*x])) - (I/2)*(Log[1 + I*x]^2*L 
og[(-I)*x] - 2*(-(Log[1 + I*x]*PolyLog[2, 1 + I*x]) + PolyLog[3, 1 + I*x]) 
)
 

Defintions of rubi rules used

rule 2821

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b 
_.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c 
*x^n])^p/m), x] + Simp[b*n*(p/m)   Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c 
*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 
0] && EqQ[d*e, 1]
 

rule 2838

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 2843

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_. 
)*(x_)), x_Symbol] :> Simp[Log[e*((f + g*x)/(e*f - d*g))]*((a + b*Log[c*(d 
+ e*x)^n])^p/g), x] - Simp[b*e*n*(p/g)   Int[Log[(e*(f + g*x))/(e*f - d*g)] 
*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, 
d, e, f, g, n, p}, x] && NeQ[e*f - d*g, 0] && IGtQ[p, 1]
 

rule 2881

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log 
[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Sym 
bol] :> Simp[1/e   Subst[Int[(k*(x/d))^r*(a + b*Log[c*x^n])^p*(f + g*Log[h* 
((e*i - d*j)/e + j*(x/e))^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, 
 f, g, h, i, j, k, l, n, p, r}, x] && EqQ[e*k - d*l, 0]
 

rule 5355

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] 
 + (Simp[I*(b/2)   Int[Log[1 - I*c*x]/x, x], x] - Simp[I*(b/2)   Int[Log[1 
+ I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]
 

rule 5546

Int[(ArcTan[(c_.)*(x_)]*Log[(f_.) + (g_.)*(x_)^2])/(x_), x_Symbol] :> Simp[ 
(Log[f + g*x^2] - Log[1 - I*c*x] - Log[1 + I*c*x])   Int[ArcTan[c*x]/x, x], 
 x] + (Simp[I/2   Int[Log[1 - I*c*x]^2/x, x], x] - Simp[I/2   Int[Log[1 + I 
*c*x]^2/x, x], x]) /; FreeQ[{c, f, g}, x] && EqQ[g, c^2*f]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.86 (sec) , antiderivative size = 2966, normalized size of antiderivative = 15.69

method	result	size
risch	\(\text {Expression too large to display}\)	\(2966\)

Input:

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

Output:

1/2*I*ln(x+I)^2*ln(1+I*(x+I))+I*ln(x+I)*polylog(2,-I*(x+I))-I*polylog(3,-I 
*(x+I))-1/2*I*ln(x-I)^2*ln(1-I*(x-I))-I*ln(x-I)*polylog(2,I*(x-I))+I*polyl 
og(3,I*(x-I))-1/8*I*Pi^2*csgn(I*(x-RootOf(_Z^2+1,index=1))*(x+RootOf(_Z^2+ 
1,index=1)))*(csgn(I*(x+RootOf(_Z^2+1,index=1)))*csgn(x+RootOf(_Z^2+1,inde 
x=1))*csgn(I*(x-RootOf(_Z^2+1,index=1))*(x+RootOf(_Z^2+1,index=1)))^2+csgn 
(I*(x-RootOf(_Z^2+1,index=1)))^2*csgn(I*(x+RootOf(_Z^2+1,index=1)))*csgn(x 
-RootOf(_Z^2+1,index=1))^2-csgn(I*(x-RootOf(_Z^2+1,index=1)))^2*csgn(I*(x- 
RootOf(_Z^2+1,index=1))*(x+RootOf(_Z^2+1,index=1)))*csgn(x-RootOf(_Z^2+1,i 
ndex=1))^2+csgn(I*(x-RootOf(_Z^2+1,index=1)))*csgn(I*(x+RootOf(_Z^2+1,inde 
x=1)))*csgn(x+RootOf(_Z^2+1,index=1))^2*csgn(I*(x-RootOf(_Z^2+1,index=1))* 
(x+RootOf(_Z^2+1,index=1)))+csgn(I*(x-RootOf(_Z^2+1,index=1)))*csgn(I*(x-R 
ootOf(_Z^2+1,index=1))*(x+RootOf(_Z^2+1,index=1)))^2*csgn(x-RootOf(_Z^2+1, 
index=1))^2+csgn(I*(x-RootOf(_Z^2+1,index=1)))*csgn(I*(x+RootOf(_Z^2+1,ind 
ex=1)))*csgn(I*(x-RootOf(_Z^2+1,index=1))*(x+RootOf(_Z^2+1,index=1)))*csgn 
(x-RootOf(_Z^2+1,index=1))-csgn(I*(x-RootOf(_Z^2+1,index=1)))*csgn(I*(x+Ro 
otOf(_Z^2+1,index=1)))*csgn(x-RootOf(_Z^2+1,index=1))^2+csgn(I*(x-RootOf(_ 
Z^2+1,index=1)))*csgn(I*(x-RootOf(_Z^2+1,index=1))*(x+RootOf(_Z^2+1,index= 
1)))*csgn(x-RootOf(_Z^2+1,index=1))^2+csgn(I*(x+RootOf(_Z^2+1,index=1)))*c 
sgn(I*(x-RootOf(_Z^2+1,index=1))*(x+RootOf(_Z^2+1,index=1)))*csgn(x-RootOf 
(_Z^2+1,index=1))^2-csgn(I*(x-RootOf(_Z^2+1,index=1)))*csgn(I*(x+RootOf...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate(atan(x)*ln(x**2+1)/x,x)
 

Output:

Integral(log(x**2 + 1)*atan(x)/x, x)
 

Maxima [F]

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

integrate(arctan(x)*log(x^2+1)/x,x, algorithm="maxima")
 

Output:

integrate(arctan(x)*log(x^2 + 1)/x, x)
 

Giac [F]

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

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

Output:

integrate(arctan(x)*log(x^2 + 1)/x, x)
 

Mupad [F(-1)]

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

int((log(x^2 + 1)*atan(x))/x,x)
 

Output:

int((log(x^2 + 1)*atan(x))/x, x)
 

Reduce [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x} \, dx=\int \frac {\mathit {atan} \left (x \right ) \mathrm {log}\left (x^{2}+1\right )}{x}d x \] Input:

int(atan(x)*log(x^2+1)/x,x)
 

Output:

int((atan(x)*log(x**2 + 1))/x,x)