\(\int \frac {(c+a^2 c x^2)^2 \arctan (a x)}{x} \, dx\) [161]

Optimal result

Integrand size = 20, antiderivative size = 99 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x} \, dx=-\frac {3}{4} a c^2 x-\frac {1}{12} a^3 c^2 x^3+\frac {3}{4} c^2 \arctan (a x)+a^2 c^2 x^2 \arctan (a x)+\frac {1}{4} a^4 c^2 x^4 \arctan (a x)+\frac {1}{2} i c^2 \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i c^2 \operatorname {PolyLog}(2,i a x) \] Output:

-3/4*a*c^2*x-1/12*a^3*c^2*x^3+3/4*c^2*arctan(a*x)+a^2*c^2*x^2*arctan(a*x)+ 
1/4*a^4*c^2*x^4*arctan(a*x)+1/2*I*c^2*polylog(2,-I*a*x)-1/2*I*c^2*polylog( 
2,I*a*x)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x} \, dx=-\frac {3}{4} a c^2 x-\frac {1}{12} a^3 c^2 x^3+\frac {3}{4} c^2 \arctan (a x)+a^2 c^2 x^2 \arctan (a x)+\frac {1}{4} a^4 c^2 x^4 \arctan (a x)+\frac {1}{2} i c^2 \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i c^2 \operatorname {PolyLog}(2,i a x) \] Input:

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

Output:

(-3*a*c^2*x)/4 - (a^3*c^2*x^3)/12 + (3*c^2*ArcTan[a*x])/4 + a^2*c^2*x^2*Ar 
cTan[a*x] + (a^4*c^2*x^4*ArcTan[a*x])/4 + (I/2)*c^2*PolyLog[2, (-I)*a*x] - 
 (I/2)*c^2*PolyLog[2, I*a*x]
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5483, 2009}

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.

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

\(\Big \downarrow \) 5483

\(\displaystyle \int \left (a^4 c^2 x^3 \arctan (a x)+2 a^2 c^2 x \arctan (a x)+\frac {c^2 \arctan (a x)}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} a^4 c^2 x^4 \arctan (a x)-\frac {1}{12} a^3 c^2 x^3+a^2 c^2 x^2 \arctan (a x)+\frac {3}{4} c^2 \arctan (a x)+\frac {1}{2} i c^2 \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i c^2 \operatorname {PolyLog}(2,i a x)-\frac {3}{4} a c^2 x\)

Input:

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

Output:

(-3*a*c^2*x)/4 - (a^3*c^2*x^3)/12 + (3*c^2*ArcTan[a*x])/4 + a^2*c^2*x^2*Ar 
cTan[a*x] + (a^4*c^2*x^4*ArcTan[a*x])/4 + (I/2)*c^2*PolyLog[2, (-I)*a*x] - 
 (I/2)*c^2*PolyLog[2, I*a*x]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5483

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {a^{4} c^{2} x^{4} \arctan \left (a x \right )}{4}+a^{2} c^{2} x^{2} \arctan \left (a x \right )+c^{2} \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c^{2} \left (\frac {a^{3} x^{3}}{3}+3 a x -3 \arctan \left (a x \right )-2 i \ln \left (a x \right ) \ln \left (i a x +1\right )+2 i \ln \left (a x \right ) \ln \left (-i a x +1\right )-2 i \operatorname {dilog}\left (i a x +1\right )+2 i \operatorname {dilog}\left (-i a x +1\right )\right )}{4}\)	\(119\)
default	\(\frac {a^{4} c^{2} x^{4} \arctan \left (a x \right )}{4}+a^{2} c^{2} x^{2} \arctan \left (a x \right )+c^{2} \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c^{2} \left (\frac {a^{3} x^{3}}{3}+3 a x -3 \arctan \left (a x \right )-2 i \ln \left (a x \right ) \ln \left (i a x +1\right )+2 i \ln \left (a x \right ) \ln \left (-i a x +1\right )-2 i \operatorname {dilog}\left (i a x +1\right )+2 i \operatorname {dilog}\left (-i a x +1\right )\right )}{4}\)	\(119\)
parts	\(\frac {a^{4} c^{2} x^{4} \arctan \left (a x \right )}{4}+a^{2} c^{2} x^{2} \arctan \left (a x \right )+c^{2} \arctan \left (a x \right ) \ln \left (x \right )-\frac {a \,c^{2} \left (\frac {a^{2} x^{3}}{3}+3 x -\frac {3 \arctan \left (a x \right )}{a}-\frac {2 i \ln \left (x \right ) \left (-\ln \left (-i a x +1\right )+\ln \left (i a x +1\right )\right )}{a}-\frac {2 i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{a}\right )}{4}\)	\(120\)
risch	\(\frac {i c^{2} \ln \left (-i a x +1\right ) x^{4} a^{4}}{8}+\frac {i c^{2} \ln \left (-i a x +1\right ) x^{2} a^{2}}{2}+\frac {3 c^{2} \arctan \left (a x \right )}{4}-\frac {a^{3} c^{2} x^{3}}{12}-\frac {3 a x \,c^{2}}{4}-\frac {i c^{2} \operatorname {dilog}\left (-i a x +1\right )}{2}-\frac {i c^{2} \ln \left (i a x +1\right ) x^{4} a^{4}}{8}-\frac {i c^{2} \ln \left (i a x +1\right ) x^{2} a^{2}}{2}+\frac {i c^{2} \operatorname {dilog}\left (i a x +1\right )}{2}\)	\(137\)
meijerg	\(\frac {c^{2} \left (\frac {a x \left (-5 a^{2} x^{2}+15\right )}{15}-\frac {a x \left (-5 a^{4} x^{4}+5\right ) \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}\right )}{4}+\frac {c^{2} \left (-2 a x +\frac {2 \left (3 a^{2} x^{2}+3\right ) \arctan \left (a x \right )}{3}\right )}{2}+\frac {c^{2} \left (-\frac {2 i a x \operatorname {polylog}\left (2, i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}+\frac {2 i a x \operatorname {polylog}\left (2, -i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{4}\)	\(143\)

Input:

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

Output:

1/4*a^4*c^2*x^4*arctan(a*x)+a^2*c^2*x^2*arctan(a*x)+c^2*arctan(a*x)*ln(a*x 
)-1/4*c^2*(1/3*a^3*x^3+3*a*x-3*arctan(a*x)-2*I*ln(a*x)*ln(1+I*a*x)+2*I*ln( 
a*x)*ln(1-I*a*x)-2*I*dilog(1+I*a*x)+2*I*dilog(1-I*a*x))
 

Fricas [F]

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

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

Output:

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*arctan(a*x)/x, x)
 

Sympy [F]

\[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x} \, dx=c^{2} \left (\int \frac {\operatorname {atan}{\left (a x \right )}}{x}\, dx + \int 2 a^{2} x \operatorname {atan}{\left (a x \right )}\, dx + \int a^{4} x^{3} \operatorname {atan}{\left (a x \right )}\, dx\right ) \] Input:

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

Output:

c**2*(Integral(atan(a*x)/x, x) + Integral(2*a**2*x*atan(a*x), x) + Integra 
l(a**4*x**3*atan(a*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x} \, dx=-\frac {1}{12} \, a^{3} c^{2} x^{3} - \frac {3}{4} \, a c^{2} x - \frac {1}{4} \, \pi c^{2} \log \left (a^{2} x^{2} + 1\right ) + c^{2} \arctan \left (a x\right ) \log \left (a x\right ) - \frac {1}{2} i \, c^{2} {\rm Li}_2\left (i \, a x + 1\right ) + \frac {1}{2} i \, c^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + \frac {1}{4} \, {\left (a^{4} c^{2} x^{4} + 4 \, a^{2} c^{2} x^{2} + 3 \, c^{2}\right )} \arctan \left (a x\right ) \] Input:

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

Output:

-1/12*a^3*c^2*x^3 - 3/4*a*c^2*x - 1/4*pi*c^2*log(a^2*x^2 + 1) + c^2*arctan 
(a*x)*log(a*x) - 1/2*I*c^2*dilog(I*a*x + 1) + 1/2*I*c^2*dilog(-I*a*x + 1) 
+ 1/4*(a^4*c^2*x^4 + 4*a^2*c^2*x^2 + 3*c^2)*arctan(a*x)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x} \, dx=\left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ 2\,a^2\,c^2\,\mathrm {atan}\left (a\,x\right )\,\left (\frac {1}{2\,a^2}+\frac {x^2}{2}\right )-a\,c^2\,x-\frac {c^2\,\left (3\,\mathrm {atan}\left (a\,x\right )-3\,a\,x+a^3\,x^3\right )}{12}+\frac {a^4\,c^2\,x^4\,\mathrm {atan}\left (a\,x\right )}{4}-\frac {c^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {c^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }a\neq 0 \end {array}\right . \] Input:

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

Output:

piecewise(a == 0, 0, a ~= 0, - (c^2*dilog(- a*x*1i + 1)*1i)/2 + (c^2*dilog 
(a*x*1i + 1)*1i)/2 - (c^2*(3*atan(a*x) - 3*a*x + a^3*x^3))/12 - a*c^2*x + 
2*a^2*c^2*atan(a*x)*(1/(2*a^2) + x^2/2) + (a^4*c^2*x^4*atan(a*x))/4)
 

Reduce [F]

\[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x} \, dx=\frac {c^{2} \left (3 \mathit {atan} \left (a x \right ) a^{4} x^{4}+12 \mathit {atan} \left (a x \right ) a^{2} x^{2}+9 \mathit {atan} \left (a x \right )+12 \left (\int \frac {\mathit {atan} \left (a x \right )}{x}d x \right )-a^{3} x^{3}-9 a x \right )}{12} \] Input:

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

Output:

(c**2*(3*atan(a*x)*a**4*x**4 + 12*atan(a*x)*a**2*x**2 + 9*atan(a*x) + 12*i 
nt(atan(a*x)/x,x) - a**3*x**3 - 9*a*x))/12