3.1.78 \(\int \frac {(a+b \arctan (c x^2))^2}{x} \, dx\) [78]

3.1.78.1 Optimal result

Integrand size = 16, antiderivative size = 151 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x} \, dx=\left (a+b \arctan \left (c x^2\right )\right )^2 \text {arctanh}\left (1-\frac {2}{1+i c x^2}\right )-\frac {1}{2} i b \left (a+b \arctan \left (c x^2\right )\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right )+\frac {1}{2} i b \left (a+b \arctan \left (c x^2\right )\right ) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x^2}\right )-\frac {1}{4} b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x^2}\right )+\frac {1}{4} b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x^2}\right ) \]

output

-(a+b*arctan(c*x^2))^2*arctanh(-1+2/(1+I*c*x^2))-1/2*I*b*(a+b*arctan(c*x^2 
))*polylog(2,1-2/(1+I*c*x^2))+1/2*I*b*(a+b*arctan(c*x^2))*polylog(2,-1+2/( 
1+I*c*x^2))-1/4*b^2*polylog(3,1-2/(1+I*c*x^2))+1/4*b^2*polylog(3,-1+2/(1+I 
*c*x^2))
 

3.1.78.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x} \, dx=a^2 \log (x)+\frac {1}{2} i a b \left (\operatorname {PolyLog}\left (2,-i c x^2\right )-\operatorname {PolyLog}\left (2,i c x^2\right )\right )+\frac {1}{48} b^2 \left (-i \pi ^3+16 i \arctan \left (c x^2\right )^3+24 \arctan \left (c x^2\right )^2 \log \left (1-e^{-2 i \arctan \left (c x^2\right )}\right )-24 \arctan \left (c x^2\right )^2 \log \left (1+e^{2 i \arctan \left (c x^2\right )}\right )+24 i \arctan \left (c x^2\right ) \operatorname {PolyLog}\left (2,e^{-2 i \arctan \left (c x^2\right )}\right )+24 i \arctan \left (c x^2\right ) \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (c x^2\right )}\right )+12 \operatorname {PolyLog}\left (3,e^{-2 i \arctan \left (c x^2\right )}\right )-12 \operatorname {PolyLog}\left (3,-e^{2 i \arctan \left (c x^2\right )}\right )\right ) \]

input

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

output

a^2*Log[x] + (I/2)*a*b*(PolyLog[2, (-I)*c*x^2] - PolyLog[2, I*c*x^2]) + (b 
^2*((-I)*Pi^3 + (16*I)*ArcTan[c*x^2]^3 + 24*ArcTan[c*x^2]^2*Log[1 - E^((-2 
*I)*ArcTan[c*x^2])] - 24*ArcTan[c*x^2]^2*Log[1 + E^((2*I)*ArcTan[c*x^2])] 
+ (24*I)*ArcTan[c*x^2]*PolyLog[2, E^((-2*I)*ArcTan[c*x^2])] + (24*I)*ArcTa 
n[c*x^2]*PolyLog[2, -E^((2*I)*ArcTan[c*x^2])] + 12*PolyLog[3, E^((-2*I)*Ar 
cTan[c*x^2])] - 12*PolyLog[3, -E^((2*I)*ArcTan[c*x^2])]))/48
 

3.1.78.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5359, 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.

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

\(\Big \downarrow \) 5359

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

\(\Big \downarrow \) 5357

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

\(\Big \downarrow \) 5523

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

\(\Big \downarrow \) 5529

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

\(\Big \downarrow \) 7164

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

input

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

output

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

3.1.78.3.1 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 5359

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[1 
/n   Subst[Int[(a + b*ArcTan[c*x])^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, 
n}, x] && IGtQ[p, 0]
 

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]
 

3.1.78.4 Maple [F]

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

input

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

output

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

3.1.78.5 Fricas [F]

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

input

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

output

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

3.1.78.6 Sympy [F]

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

input

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

output

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

3.1.78.7 Maxima [F]

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

input

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

output

a^2*log(x) + 1/16*integrate((12*b^2*arctan(c*x^2)^2 + b^2*log(c^2*x^4 + 1) 
^2 + 32*a*b*arctan(c*x^2))/x, x)
 

3.1.78.8 Giac [F]

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

input

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

output

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

3.1.78.9 Mupad [F(-1)]

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

input

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

output

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