\(\int \frac {(c+a^2 c x^2)^2 \arctan (a x)}{x^3} \, dx\) [163]

Optimal result

Integrand size = 20, antiderivative size = 90 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^3} \, dx=-\frac {a c^2}{2 x}-\frac {1}{2} a^3 c^2 x-\frac {c^2 \arctan (a x)}{2 x^2}+\frac {1}{2} a^4 c^2 x^2 \arctan (a x)+i a^2 c^2 \operatorname {PolyLog}(2,-i a x)-i a^2 c^2 \operatorname {PolyLog}(2,i a x) \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.33 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^3} \, dx=-\frac {1}{2} a^3 c^2 x+\frac {1}{2} a^2 c^2 \arctan (a x)-\frac {c^2 \arctan (a x)}{2 x^2}+\frac {1}{2} a^4 c^2 x^2 \arctan (a x)-\frac {a c^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-a^2 x^2\right )}{2 x}+i a^2 c^2 \operatorname {PolyLog}(2,-i a x)-i a^2 c^2 \operatorname {PolyLog}(2,i a x) \] Input:

Integrate[((c + a^2*c*x^2)^2*ArcTan[a*x])/x^3,x]
 

Output:

-1/2*(a^3*c^2*x) + (a^2*c^2*ArcTan[a*x])/2 - (c^2*ArcTan[a*x])/(2*x^2) + ( 
a^4*c^2*x^2*ArcTan[a*x])/2 - (a*c^2*Hypergeometric2F1[-1/2, 1, 1/2, -(a^2* 
x^2)])/(2*x) + I*a^2*c^2*PolyLog[2, (-I)*a*x] - I*a^2*c^2*PolyLog[2, I*a*x 
]
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 90, 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.

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

\(\Big \downarrow \) 5483

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

\(\Big \downarrow \) 2009

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

Input:

Int[((c + a^2*c*x^2)^2*ArcTan[a*x])/x^3,x]
 

Output:

-1/2*(a*c^2)/x - (a^3*c^2*x)/2 - (c^2*ArcTan[a*x])/(2*x^2) + (a^4*c^2*x^2* 
ArcTan[a*x])/2 + I*a^2*c^2*PolyLog[2, (-I)*a*x] - I*a^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 = 0.64 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.29

method	result	size
parts	\(\frac {a^{4} c^{2} x^{2} \arctan \left (a x \right )}{2}-\frac {c^{2} \arctan \left (a x \right )}{2 x^{2}}+2 c^{2} \arctan \left (a x \right ) a^{2} \ln \left (x \right )-\frac {a \,c^{2} \left (a^{2} x +\frac {1}{x}+4 a^{2} \left (-\frac {i \ln \left (x \right ) \left (-\ln \left (-i a x +1\right )+\ln \left (i a x +1\right )\right )}{2 a}-\frac {i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{2 a}\right )\right )}{2}\)	\(116\)
derivativedivides	\(a^{2} \left (\frac {a^{2} c^{2} x^{2} \arctan \left (a x \right )}{2}-\frac {c^{2} \arctan \left (a x \right )}{2 a^{2} x^{2}}+2 c^{2} \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c^{2} \left (a x +\frac {1}{a x}-2 i \ln \left (a x \right ) \ln \left (i a x +1\right )+2 i \ln \left (a x \right ) \ln \left (-i a x +1\right )-2 i \operatorname {dilog}\left (i a x +1\right )+2 i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\)	\(117\)
default	\(a^{2} \left (\frac {a^{2} c^{2} x^{2} \arctan \left (a x \right )}{2}-\frac {c^{2} \arctan \left (a x \right )}{2 a^{2} x^{2}}+2 c^{2} \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c^{2} \left (a x +\frac {1}{a x}-2 i \ln \left (a x \right ) \ln \left (i a x +1\right )+2 i \ln \left (a x \right ) \ln \left (-i a x +1\right )-2 i \operatorname {dilog}\left (i a x +1\right )+2 i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\)	\(117\)
meijerg	\(\frac {a^{2} c^{2} \left (-2 a x +\frac {2 \left (3 a^{2} x^{2}+3\right ) \arctan \left (a x \right )}{3}\right )}{4}+\frac {a^{2} c^{2} \left (-\frac {2 i a x \operatorname {polylog}\left (2, i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}+\frac {2 i a x \operatorname {polylog}\left (2, -i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{2}+\frac {a^{2} c^{2} \left (-\frac {2}{a x}-\frac {2 \left (a^{2} x^{2}+1\right ) \arctan \left (a x \right )}{a^{2} x^{2}}\right )}{4}\)	\(134\)
risch	\(\frac {i c^{2} a^{4} \ln \left (-i a x +1\right ) x^{2}}{4}-\frac {a^{3} c^{2} x}{2}-\frac {a \,c^{2}}{2 x}+\frac {i c^{2} a^{2} \ln \left (-i a x \right )}{4}-\frac {i c^{2} \ln \left (-i a x +1\right )}{4 x^{2}}-i c^{2} a^{2} \operatorname {dilog}\left (-i a x +1\right )-\frac {i c^{2} a^{4} \ln \left (i a x +1\right ) x^{2}}{4}-\frac {i c^{2} a^{2} \ln \left (i a x \right )}{4}+\frac {i c^{2} \ln \left (i a x +1\right )}{4 x^{2}}+i c^{2} a^{2} \operatorname {dilog}\left (i a x +1\right )\)	\(158\)

Input:

int((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

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

integrate((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x, algorithm="fricas")
 

Output:

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*arctan(a*x)/x^3, x)
 

Sympy [F]

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

integrate((a**2*c*x**2+c)**2*atan(a*x)/x**3,x)
 

Output:

c**2*(Integral(atan(a*x)/x**3, x) + Integral(2*a**2*atan(a*x)/x, x) + Inte 
gral(a**4*x*atan(a*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.33 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^3} \, dx=-\frac {a^{3} c^{2} x^{3} + \pi a^{2} c^{2} x^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{2} c^{2} x^{2} \arctan \left (a x\right ) \log \left (a x\right ) + 2 i \, a^{2} c^{2} x^{2} {\rm Li}_2\left (i \, a x + 1\right ) - 2 i \, a^{2} c^{2} x^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + a c^{2} x - {\left (a^{4} c^{2} x^{4} - c^{2}\right )} \arctan \left (a x\right )}{2 \, x^{2}} \] Input:

integrate((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x, algorithm="giac")
 

Output:

integrate((a^2*c*x^2 + c)^2*arctan(a*x)/x^3, x)
 

Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^3} \, dx=\left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ a^4\,c^2\,\mathrm {atan}\left (a\,x\right )\,\left (\frac {1}{2\,a^2}+\frac {x^2}{2}\right )-\frac {c^2\,\mathrm {atan}\left (a\,x\right )}{2\,x^2}-\frac {c^2\,\left (a^3\,\mathrm {atan}\left (a\,x\right )+\frac {a^2}{x}\right )}{2\,a}-\frac {a^3\,c^2\,x}{2}-a^2\,c^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+a^2\,c^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i} & \text {\ if\ \ }a\neq 0 \end {array}\right . \] Input:

int((atan(a*x)*(c + a^2*c*x^2)^2)/x^3,x)
 

Output:

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

Reduce [F]

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

int((a^2*c*x^2+c)^2*atan(a*x)/x^3,x)
 

Output:

(c**2*(atan(a*x)*a**4*x**4 - atan(a*x) + 4*int(atan(a*x)/x,x)*a**2*x**2 - 
a**3*x**3 - a*x))/(2*x**2)