\(\int \frac {(c+a^2 c x^2) \arctan (a x)}{x^3} \, dx\) [155]

Optimal result

Integrand size = 18, antiderivative size = 70 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {a c}{2 x}-\frac {1}{2} a^2 c \arctan (a x)-\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \] Output:

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

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {c \arctan (a x)}{2 x^2}-\frac {a c \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-a^2 x^2\right )}{2 x}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5485, 5355, 2838, 5361, 264, 216}

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.

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

\(\Big \downarrow \) 5485

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

\(\Big \downarrow \) 5355

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 5355

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

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 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]))
 

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36

method	result	size
parts	\(-\frac {c \arctan \left (a x \right )}{2 x^{2}}+c \arctan \left (a x \right ) a^{2} \ln \left (x \right )-\frac {a c \left (a \arctan \left (a x \right )+\frac {1}{x}+2 a^{2} \left (-\frac {i \ln \left (x \right ) \left (-\ln \left (-i a x +1\right )+\ln \left (i a x +1\right )\right )}{2 a}-\frac {i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{2 a}\right )\right )}{2}\)	\(95\)
derivativedivides	\(a^{2} \left (c \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c \arctan \left (a x \right )}{2 a^{2} x^{2}}-\frac {c \left (\frac {1}{a x}+\arctan \left (a x \right )-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\)	\(96\)
default	\(a^{2} \left (c \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c \arctan \left (a x \right )}{2 a^{2} x^{2}}-\frac {c \left (\frac {1}{a x}+\arctan \left (a x \right )-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\)	\(96\)
meijerg	\(\frac {c \,a^{2} \left (-\frac {2 i a x \operatorname {polylog}\left (2, i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}+\frac {2 i a x \operatorname {polylog}\left (2, -i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{4}+\frac {c \,a^{2} \left (-\frac {2}{a x}-\frac {2 \left (a^{2} x^{2}+1\right ) \arctan \left (a x \right )}{a^{2} x^{2}}\right )}{4}\)	\(101\)
risch	\(-\frac {c a}{2 x}+\frac {i c \,a^{2} \ln \left (-i a x \right )}{4}-\frac {i c \,a^{2} \ln \left (-i a x +1\right )}{4}-\frac {i c \ln \left (-i a x +1\right )}{4 x^{2}}-\frac {i c \,a^{2} \operatorname {dilog}\left (-i a x +1\right )}{2}-\frac {i c \,a^{2} \ln \left (i a x \right )}{4}+\frac {i c \,a^{2} \ln \left (i a x +1\right )}{4}+\frac {i c \ln \left (i a x +1\right )}{4 x^{2}}+\frac {i c \,a^{2} \operatorname {dilog}\left (i a x +1\right )}{2}\)	\(125\)

Input:

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

Output:

-1/2*c*arctan(a*x)/x^2+c*arctan(a*x)*a^2*ln(x)-1/2*a*c*(a*arctan(a*x)+1/x+ 
2*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)-dil 
og(1-I*a*x))/a))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {\pi a^{2} c x^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{2} c x^{2} \arctan \left (a x\right ) \log \left (a x\right ) + 2 i \, a^{2} c x^{2} {\rm Li}_2\left (i \, a x + 1\right ) - 2 i \, a^{2} c x^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + 2 \, a c x + 2 \, {\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{4 \, x^{2}} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=\left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ -\frac {c\,\mathrm {atan}\left (a\,x\right )}{2\,x^2}-\frac {c\,\left (a^3\,\mathrm {atan}\left (a\,x\right )+\frac {a^2}{x}\right )}{2\,a}-\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }a\neq 0 \end {array}\right . \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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