3.1.19 \(\int \frac {(a+b \arctan (c x))^2}{x} \, dx\) [19]

3.1.19.1 Optimal result

Integrand size = 14, antiderivative size = 132 \[ \int \frac {(a+b \arctan (c x))^2}{x} \, dx=2 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]

output

-2*(a+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))-I*b*(a+b*arctan(c*x))*polyl 
og(2,1-2/(1+I*c*x))+I*b*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))-1/2*b^ 
2*polylog(3,1-2/(1+I*c*x))+1/2*b^2*polylog(3,-1+2/(1+I*c*x))
 

3.1.19.2 Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.36 \[ \int \frac {(a+b \arctan (c x))^2}{x} \, dx=a^2 \log (c x)+i a b (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+b^2 \left (-\frac {i \pi ^3}{24}+\frac {2}{3} i \arctan (c x)^3+\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-\arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right ) \]

input

Integrate[(a + b*ArcTan[c*x])^2/x,x]
 

output

a^2*Log[c*x] + I*a*b*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + b^2*((-1 
/24*I)*Pi^3 + ((2*I)/3)*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E^((-2*I)*Ar 
cTan[c*x])] - ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + I*ArcTan[c*x] 
*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, -E^((2*I)*A 
rcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2 - PolyLog[3, -E^((2*I) 
*ArcTan[c*x])]/2)
 

3.1.19.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5357, 5523, 5529, 7164}

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.

\(\displaystyle \int \frac {(a+b \arctan (c x))^2}{x} \, dx\)

\(\Big \downarrow \) 5357

\(\displaystyle 2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-4 b c \int \frac {(a+b \arctan (c x)) \text {arctanh}\left (1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\)

\(\Big \downarrow \) 5523

\(\displaystyle 2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-4 b c \left (\frac {1}{2} \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx-\frac {1}{2} \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )\)

\(\Big \downarrow \) 5529

\(\displaystyle 2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-4 b c \left (\frac {1}{2} \left (\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{2 c}\right )\right )\)

\(\Big \downarrow \) 7164

\(\displaystyle 2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-4 b c \left (\frac {1}{2} \left (\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}+\frac {b \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{4 c}\right )+\frac {1}{2} \left (-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{2 c}-\frac {b \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )}{4 c}\right )\right )\)

input

Int[(a + b*ArcTan[c*x])^2/x,x]
 

output

2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] - 4*b*c*((((I/2)*(a + b 
*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/c + (b*PolyLog[3, 1 - 2/(1 + 
I*c*x)])/(4*c))/2 + (((-1/2*I)*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + 
I*c*x)])/c - (b*PolyLog[3, -1 + 2/(1 + I*c*x)])/(4*c))/2)
 

3.1.19.3.1 Defintions of rubi rules used

rule 5357

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

rule 5523

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x 
_)^2), x_Symbol] :> Simp[1/2   Int[Log[1 + u]*((a + b*ArcTan[c*x])^p/(d + e 
*x^2)), x], x] - Simp[1/2   Int[Log[1 - u]*((a + b*ArcTan[c*x])^p/(d + e*x^ 
2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && 
EqQ[u^2 - (1 - 2*(I/(I - c*x)))^2, 0]
 

rule 5529

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 
), x_Symbol] :> Simp[(-I)*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) 
, x] + Simp[b*p*(I/2)   Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 - u]/ 
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c 
^2*d] && EqQ[(1 - u)^2 - (1 - 2*(I/(I - c*x)))^2, 0]
 

rule 7164

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, 
x]}, Simp[w*PolyLog[n + 1, v], x] /;  !FalseQ[w]] /; FreeQ[n, x]
 

3.1.19.4 Maple [C] (warning: unable to verify)

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

Time = 0.74 (sec) , antiderivative size = 1002, normalized size of antiderivative = 7.59

\[\text {Expression too large to display}\]

input

int((a+b*arctan(c*x))^2/x,x)
 

output

a^2*ln(c*x)+b^2*(ln(c*x)*arctan(c*x)^2+I*arctan(c*x)*polylog(2,-(1+I*c*x)^ 
2/(c^2*x^2+1))-1/2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-arctan(c*x)^2*ln((1 
+I*c*x)^2/(c^2*x^2+1)-1)+arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2 
*I*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*polylog(3,(1+I*c*x 
)/(c^2*x^2+1)^(1/2))+arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*a 
rctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*polylog(3,-(1+I*c*x)/ 
(c^2*x^2+1)^(1/2))+1/2*I*Pi*(csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c* 
x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^ 
2+1)))-csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2+csg 
n(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn( 
I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))-csgn(I*((1+I*c* 
x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^ 
2*x^2+1)))^2-csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2* 
x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2+csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1 
)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3-csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+ 
I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^ 
2*x^2+1)))^2+csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1))) 
^3+1)*arctan(c*x)^2)+2*a*b*(ln(c*x)*arctan(c*x)+1/2*I*ln(c*x)*ln(1+I*c*x)- 
1/2*I*ln(c*x)*ln(1-I*c*x)+1/2*I*dilog(1+I*c*x)-1/2*I*dilog(1-I*c*x))
 

3.1.19.5 Fricas [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]

input

integrate((a+b*arctan(c*x))^2/x,x, algorithm="fricas")
 

output

integral((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x) + a^2)/x, x)
 

3.1.19.6 Sympy [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{x}\, dx \]

input

integrate((a+b*atan(c*x))**2/x,x)
 

output

Integral((a + b*atan(c*x))**2/x, x)
 

3.1.19.7 Maxima [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]

input

integrate((a+b*arctan(c*x))^2/x,x, algorithm="maxima")
 

output

a^2*log(x) + 1/16*integrate((12*b^2*arctan(c*x)^2 + b^2*log(c^2*x^2 + 1)^2 
 + 32*a*b*arctan(c*x))/x, x)
 

3.1.19.8 Giac [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]

input

integrate((a+b*arctan(c*x))^2/x,x, algorithm="giac")
 

output

sage0*x
 

3.1.19.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x} \,d x \]

input

int((a + b*atan(c*x))^2/x,x)
 

output

int((a + b*atan(c*x))^2/x, x)