3.3.62 \(\int \frac {(c+a^2 c x^2) \arctan (a x)^2}{x} \, dx\) [262]

3.3.62.1 Optimal result

Integrand size = 20, antiderivative size = 169 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)^2}{x} \, dx=-a c x \arctan (a x)+\frac {1}{2} c \arctan (a x)^2+\frac {1}{2} a^2 c x^2 \arctan (a x)^2+2 c \arctan (a x)^2 \text {arctanh}\left (1-\frac {2}{1+i a x}\right )+\frac {1}{2} c \log \left (1+a^2 x^2\right )-i c \arctan (a x) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )+i c \arctan (a x) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i a x}\right )-\frac {1}{2} c \operatorname {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )+\frac {1}{2} c \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i a x}\right ) \]

output

-a*c*x*arctan(a*x)+1/2*c*arctan(a*x)^2+1/2*a^2*c*x^2*arctan(a*x)^2-2*c*arc 
tan(a*x)^2*arctanh(-1+2/(1+I*a*x))+1/2*c*ln(a^2*x^2+1)-I*c*arctan(a*x)*pol 
ylog(2,1-2/(1+I*a*x))+I*c*arctan(a*x)*polylog(2,-1+2/(1+I*a*x))-1/2*c*poly 
log(3,1-2/(1+I*a*x))+1/2*c*polylog(3,-1+2/(1+I*a*x))
 

3.3.62.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.08 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)^2}{x} \, dx=-a c x \arctan (a x)+\frac {1}{2} c \left (1+a^2 x^2\right ) \arctan (a x)^2+\frac {2}{3} i c \arctan (a x)^3+c \arctan (a x)^2 \log \left (1-e^{-2 i \arctan (a x)}\right )-c \arctan (a x)^2 \log \left (1+e^{2 i \arctan (a x)}\right )+\frac {1}{2} c \log \left (1+a^2 x^2\right )+i c \arctan (a x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (a x)}\right )+i c \arctan (a x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (a x)}\right )+\frac {1}{2} c \operatorname {PolyLog}\left (3,e^{-2 i \arctan (a x)}\right )-\frac {1}{2} c \operatorname {PolyLog}\left (3,-e^{2 i \arctan (a x)}\right ) \]

input

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

output

-(a*c*x*ArcTan[a*x]) + (c*(1 + a^2*x^2)*ArcTan[a*x]^2)/2 + ((2*I)/3)*c*Arc 
Tan[a*x]^3 + c*ArcTan[a*x]^2*Log[1 - E^((-2*I)*ArcTan[a*x])] - c*ArcTan[a* 
x]^2*Log[1 + E^((2*I)*ArcTan[a*x])] + (c*Log[1 + a^2*x^2])/2 + I*c*ArcTan[ 
a*x]*PolyLog[2, E^((-2*I)*ArcTan[a*x])] + I*c*ArcTan[a*x]*PolyLog[2, -E^(( 
2*I)*ArcTan[a*x])] + (c*PolyLog[3, E^((-2*I)*ArcTan[a*x])])/2 - (c*PolyLog 
[3, -E^((2*I)*ArcTan[a*x])])/2
 

3.3.62.3 Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5485, 5357, 5361, 5451, 5345, 240, 5419, 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.

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

\(\Big \downarrow \) 5485

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

\(\Big \downarrow \) 5357

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5451

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

\(\Big \downarrow \) 5345

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

\(\Big \downarrow \) 240

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

\(\Big \downarrow \) 5419

\(\displaystyle a^2 c \left (\frac {1}{2} x^2 \arctan (a x)^2-a \left (\frac {x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}}{a^2}-\frac {\arctan (a x)^2}{2 a^3}\right )\right )+c \left (2 \arctan (a x)^2 \text {arctanh}\left (1-\frac {2}{1+i a x}\right )-4 a \int \frac {\arctan (a x) \text {arctanh}\left (1-\frac {2}{i a x+1}\right )}{a^2 x^2+1}dx\right )\)

\(\Big \downarrow \) 5523

\(\displaystyle a^2 c \left (\frac {1}{2} x^2 \arctan (a x)^2-a \left (\frac {x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}}{a^2}-\frac {\arctan (a x)^2}{2 a^3}\right )\right )+c \left (2 \arctan (a x)^2 \text {arctanh}\left (1-\frac {2}{1+i a x}\right )-4 a \left (\frac {1}{2} \int \frac {\arctan (a x) \log \left (2-\frac {2}{i a x+1}\right )}{a^2 x^2+1}dx-\frac {1}{2} \int \frac {\arctan (a x) \log \left (\frac {2}{i a x+1}\right )}{a^2 x^2+1}dx\right )\right )\)

\(\Big \downarrow \) 5529

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

\(\Big \downarrow \) 7164

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

input

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

output

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

3.3.62.3.1 Defintions of rubi rules used

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 5345

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a 
+ b*ArcTan[c*x^n])^p, x] - Simp[b*c*n*p   Int[x^n*((a + b*ArcTan[c*x^n])^(p 
 - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] && 
 (EqQ[n, 1] || EqQ[p, 1])
 

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 5361

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
 Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 
1))   Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], 
x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & 
& IntegerQ[m])) && NeQ[m, -1]
 

rule 5419

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, 
c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]
 

rule 5451

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

rule 5485

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(q_.), x_Symbol] :> Simp[d   Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + 
 b*ArcTan[c*x])^p, x], x] + Simp[c^2*(d/f^2)   Int[(f*x)^(m + 2)*(d + e*x^2 
)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] 
&& EqQ[e, c^2*d] && GtQ[q, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] 
&& IntegerQ[q]))
 

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]
 

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

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

Time = 18.39 (sec) , antiderivative size = 1055, normalized size of antiderivative = 6.24

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1055\)
default	\(\text {Expression too large to display}\)	\(1055\)
parts	\(\text {Expression too large to display}\)	\(1545\)

input

int((a^2*c*x^2+c)*arctan(a*x)^2/x,x,method=_RETURNVERBOSE)
 

output

1/2*a^2*c*x^2*arctan(a*x)^2+c*arctan(a*x)^2*ln(a*x)-c*(arctan(a*x)^2*ln((1 
+I*a*x)^2/(a^2*x^2+1)-1)-arctan(a*x)^2*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)+1 
/2*I*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2*ar 
ctan(a*x)^2-2*polylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-arctan(a*x)^2*ln(1-( 
1+I*a*x)/(a^2*x^2+1)^(1/2))-1/2*I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/(( 
1+I*a*x)^2/(a^2*x^2+1)+1))^3*arctan(a*x)^2-2*polylog(3,(1+I*a*x)/(a^2*x^2+ 
1)^(1/2))-1/2*I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^ 
2+1)+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))*arc 
tan(a*x)^2+1/2*polylog(3,-(1+I*a*x)^2/(a^2*x^2+1))+1/2*I*Pi*csgn(I*((1+I*a 
*x)^2/(a^2*x^2+1)-1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2 
*x^2+1)+1))^2*arctan(a*x)^2+ln((1+I*a*x)^2/(a^2*x^2+1)+1)-1/2*arctan(a*x)^ 
2+arctan(a*x)*(a*x-I)+1/2*I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a* 
x)^2/(a^2*x^2+1)+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^ 
2+1)+1))^2*arctan(a*x)^2-1/2*I*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a 
*x)^2/(a^2*x^2+1)+1))^3*arctan(a*x)^2+1/2*I*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^ 
2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2 
*arctan(a*x)^2+2*I*arctan(a*x)*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-1/2 
*I*Pi*arctan(a*x)^2+2*I*arctan(a*x)*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2)) 
-1/2*I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1))*csgn(I/((1+I*a*x)^2/(a^2*x^2 
+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))...
 

3.3.62.5 Fricas [F]

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

input

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

output

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

3.3.62.6 Sympy [F]

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

input

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

output

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

3.3.62.7 Maxima [F]

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

input

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

output

1/8*a^2*c*x^2*arctan(a*x)^2 - 1/32*a^2*c*x^2*log(a^2*x^2 + 1)^2 + 12*a^4*c 
*integrate(1/16*x^4*arctan(a*x)^2/(a^2*x^3 + x), x) + a^4*c*integrate(1/16 
*x^4*log(a^2*x^2 + 1)^2/(a^2*x^3 + x), x) + 2*a^4*c*integrate(1/16*x^4*log 
(a^2*x^2 + 1)/(a^2*x^3 + x), x) - 4*a^3*c*integrate(1/16*x^3*arctan(a*x)/( 
a^2*x^3 + x), x) + 24*a^2*c*integrate(1/16*x^2*arctan(a*x)^2/(a^2*x^3 + x) 
, x) + 1/48*c*log(a^2*x^2 + 1)^3 + 12*c*integrate(1/16*arctan(a*x)^2/(a^2* 
x^3 + x), x) + c*integrate(1/16*log(a^2*x^2 + 1)^2/(a^2*x^3 + x), x)
 

3.3.62.8 Giac [F]

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

input

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

output

sage0*x
 

3.3.62.9 Mupad [F(-1)]

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

input

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

output

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