\(\int \frac {(a+b \arctan (c x))^2}{x^5} \, dx\) [23]

Optimal result

Integrand size = 14, antiderivative size = 116 \[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=-\frac {b^2 c^2}{12 x^2}-\frac {b c (a+b \arctan (c x))}{6 x^3}+\frac {b c^3 (a+b \arctan (c x))}{2 x}+\frac {1}{4} c^4 (a+b \arctan (c x))^2-\frac {(a+b \arctan (c x))^2}{4 x^4}-\frac {2}{3} b^2 c^4 \log (x)+\frac {1}{3} b^2 c^4 \log \left (1+c^2 x^2\right ) \] Output:

-1/12*b^2*c^2/x^2-1/6*b*c*(a+b*arctan(c*x))/x^3+1/2*b*c^3*(a+b*arctan(c*x) 
)/x+1/4*c^4*(a+b*arctan(c*x))^2-1/4*(a+b*arctan(c*x))^2/x^4-2/3*b^2*c^4*ln 
(x)+1/3*b^2*c^4*ln(c^2*x^2+1)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=\frac {-3 a^2-2 a b c x-b^2 c^2 x^2+6 a b c^3 x^3+2 b \left (b c x \left (-1+3 c^2 x^2\right )+3 a \left (-1+c^4 x^4\right )\right ) \arctan (c x)+3 b^2 \left (-1+c^4 x^4\right ) \arctan (c x)^2-8 b^2 c^4 x^4 \log (x)+4 b^2 c^4 x^4 \log \left (1+c^2 x^2\right )}{12 x^4} \] Input:

Integrate[(a + b*ArcTan[c*x])^2/x^5,x]
 

Output:

(-3*a^2 - 2*a*b*c*x - b^2*c^2*x^2 + 6*a*b*c^3*x^3 + 2*b*(b*c*x*(-1 + 3*c^2 
*x^2) + 3*a*(-1 + c^4*x^4))*ArcTan[c*x] + 3*b^2*(-1 + c^4*x^4)*ArcTan[c*x] 
^2 - 8*b^2*c^4*x^4*Log[x] + 4*b^2*c^4*x^4*Log[1 + c^2*x^2])/(12*x^4)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.16, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {5361, 5453, 5361, 243, 54, 2009, 5453, 5361, 243, 47, 14, 16, 5419}

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.

\(\displaystyle \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx\)

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5453

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx\right )+\frac {1}{6} b c \int \frac {1}{x^4 \left (c^2 x^2+1\right )}dx^2-\frac {a+b \arctan (c x)}{3 x^3}\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 54

\(\displaystyle \frac {1}{2} b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx\right )+\frac {1}{6} b c \int \left (\frac {c^4}{c^2 x^2+1}-\frac {c^2}{x^2}+\frac {1}{x^4}\right )dx^2-\frac {a+b \arctan (c x)}{3 x^3}\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 5453

\(\displaystyle \frac {1}{2} b c \left (-\left (c^2 \left (\int \frac {a+b \arctan (c x)}{x^2}dx-c^2 \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{2} b c \left (-\left (c^2 \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+b c \int \frac {1}{x \left (c^2 x^2+1\right )}dx-\frac {a+b \arctan (c x)}{x}\right )\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} b c \left (-\left (c^2 \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \int \frac {1}{x^2 \left (c^2 x^2+1\right )}dx^2-\frac {a+b \arctan (c x)}{x}\right )\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 47

\(\displaystyle \frac {1}{2} b c \left (-\left (c^2 \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \left (\int \frac {1}{x^2}dx^2-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {1}{2} b c \left (-\left (c^2 \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \left (\log \left (x^2\right )-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} b c \left (-\left (c^2 \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 5419

\(\displaystyle \frac {1}{2} b c \left (-\left (c^2 \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )\right )-\frac {a+b \arctan (c x)}{3 x^3}+\frac {1}{6} b c \left (c^2 \left (-\log \left (x^2\right )\right )+c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )\right )-\frac {(a+b \arctan (c x))^2}{4 x^4}\)

Input:

Int[(a + b*ArcTan[c*x])^2/x^5,x]
 

Output:

-1/4*(a + b*ArcTan[c*x])^2/x^4 + (b*c*(-1/3*(a + b*ArcTan[c*x])/x^3 - c^2* 
(-((a + b*ArcTan[c*x])/x) - (c*(a + b*ArcTan[c*x])^2)/(2*b) + (b*c*(Log[x^ 
2] - Log[1 + c^2*x^2]))/2) + (b*c*(-x^(-2) - c^2*Log[x^2] + c^2*Log[1 + c^ 
2*x^2]))/6))/2
 

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

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

rule 47

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

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

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

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 5453

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.14

method	result	size
parts	\(-\frac {a^{2}}{4 x^{4}}+b^{2} c^{4} \left (-\frac {\arctan \left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )^{2}}{4}-\frac {\arctan \left (c x \right )}{6 c^{3} x^{3}}+\frac {\arctan \left (c x \right )}{2 c x}+\frac {\ln \left (c^{2} x^{2}+1\right )}{3}-\frac {1}{12 c^{2} x^{2}}-\frac {2 \ln \left (c x \right )}{3}\right )+2 a b \,c^{4} \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{4}-\frac {1}{12 c^{3} x^{3}}+\frac {1}{4 c x}\right )\)	\(132\)
derivativedivides	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\arctan \left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )^{2}}{4}-\frac {\arctan \left (c x \right )}{6 c^{3} x^{3}}+\frac {\arctan \left (c x \right )}{2 c x}+\frac {\ln \left (c^{2} x^{2}+1\right )}{3}-\frac {1}{12 c^{2} x^{2}}-\frac {2 \ln \left (c x \right )}{3}\right )+2 a b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{4}-\frac {1}{12 c^{3} x^{3}}+\frac {1}{4 c x}\right )\right )\)	\(133\)
default	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\arctan \left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )^{2}}{4}-\frac {\arctan \left (c x \right )}{6 c^{3} x^{3}}+\frac {\arctan \left (c x \right )}{2 c x}+\frac {\ln \left (c^{2} x^{2}+1\right )}{3}-\frac {1}{12 c^{2} x^{2}}-\frac {2 \ln \left (c x \right )}{3}\right )+2 a b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{4}-\frac {1}{12 c^{3} x^{3}}+\frac {1}{4 c x}\right )\right )\)	\(133\)
parallelrisch	\(-\frac {-3 b^{2} \arctan \left (c x \right )^{2} x^{4} c^{4}+8 b^{2} c^{4} \ln \left (x \right ) x^{4}-4 b^{2} c^{4} \ln \left (c^{2} x^{2}+1\right ) x^{4}-6 a b \arctan \left (c x \right ) x^{4} c^{4}-b^{2} c^{4} x^{4}-6 b^{2} \arctan \left (c x \right ) x^{3} c^{3}-6 a b \,c^{3} x^{3}+b^{2} c^{2} x^{2}+2 b^{2} \arctan \left (c x \right ) x c +2 a b c x +3 b^{2} \arctan \left (c x \right )^{2}+6 a b \arctan \left (c x \right )+3 a^{2}}{12 x^{4}}\)	\(159\)
risch	\(-\frac {b^{2} \left (c^{4} x^{4}-1\right ) \ln \left (i c x +1\right )^{2}}{16 x^{4}}+\frac {i b \left (-3 i b \,c^{4} x^{4} \ln \left (-i c x +1\right )-6 b \,c^{3} x^{3}+2 b c x +6 a +3 i b \ln \left (-i c x +1\right )\right ) \ln \left (i c x +1\right )}{24 x^{4}}-\frac {12 i \ln \left (\left (-4 i b c -a c \right ) x -4 b +i a \right ) a b \,c^{4} x^{4}-12 i \ln \left (\left (4 i b c -a c \right ) x -4 b -i a \right ) a b \,c^{4} x^{4}+3 c^{4} b^{2} x^{4} \ln \left (-i c x +1\right )^{2}-16 \ln \left (\left (-4 i b c -a c \right ) x -4 b +i a \right ) b^{2} c^{4} x^{4}-16 \ln \left (\left (4 i b c -a c \right ) x -4 b -i a \right ) b^{2} c^{4} x^{4}+32 b^{2} c^{4} \ln \left (-x \right ) x^{4}-12 i b^{2} c^{3} x^{3} \ln \left (-i c x +1\right )-24 a b \,c^{3} x^{3}+4 i b^{2} c x \ln \left (-i c x +1\right )+4 b^{2} c^{2} x^{2}+12 i a b \ln \left (-i c x +1\right )+8 a b c x -3 b^{2} \ln \left (-i c x +1\right )^{2}+12 a^{2}}{48 x^{4}}\)	\(358\)

Input:

int((a+b*arctan(c*x))^2/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/4*a^2/x^4+b^2*c^4*(-1/4/c^4/x^4*arctan(c*x)^2+1/4*arctan(c*x)^2-1/6/c^3 
/x^3*arctan(c*x)+1/2/c/x*arctan(c*x)+1/3*ln(c^2*x^2+1)-1/12/c^2/x^2-2/3*ln 
(c*x))+2*a*b*c^4*(-1/4/c^4/x^4*arctan(c*x)+1/4*arctan(c*x)-1/12/c^3/x^3+1/ 
4/c/x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=\frac {4 \, b^{2} c^{4} x^{4} \log \left (c^{2} x^{2} + 1\right ) - 8 \, b^{2} c^{4} x^{4} \log \left (x\right ) + 6 \, a b c^{3} x^{3} - b^{2} c^{2} x^{2} - 2 \, a b c x + 3 \, {\left (b^{2} c^{4} x^{4} - b^{2}\right )} \arctan \left (c x\right )^{2} - 3 \, a^{2} + 2 \, {\left (3 \, a b c^{4} x^{4} + 3 \, b^{2} c^{3} x^{3} - b^{2} c x - 3 \, a b\right )} \arctan \left (c x\right )}{12 \, x^{4}} \] Input:

integrate((a+b*arctan(c*x))^2/x^5,x, algorithm="fricas")
 

Output:

1/12*(4*b^2*c^4*x^4*log(c^2*x^2 + 1) - 8*b^2*c^4*x^4*log(x) + 6*a*b*c^3*x^ 
3 - b^2*c^2*x^2 - 2*a*b*c*x + 3*(b^2*c^4*x^4 - b^2)*arctan(c*x)^2 - 3*a^2 
+ 2*(3*a*b*c^4*x^4 + 3*b^2*c^3*x^3 - b^2*c*x - 3*a*b)*arctan(c*x))/x^4
 

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=\begin {cases} - \frac {a^{2}}{4 x^{4}} + \frac {a b c^{4} \operatorname {atan}{\left (c x \right )}}{2} + \frac {a b c^{3}}{2 x} - \frac {a b c}{6 x^{3}} - \frac {a b \operatorname {atan}{\left (c x \right )}}{2 x^{4}} - \frac {2 b^{2} c^{4} \log {\left (x \right )}}{3} + \frac {b^{2} c^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3} + \frac {b^{2} c^{4} \operatorname {atan}^{2}{\left (c x \right )}}{4} + \frac {b^{2} c^{3} \operatorname {atan}{\left (c x \right )}}{2 x} - \frac {b^{2} c^{2}}{12 x^{2}} - \frac {b^{2} c \operatorname {atan}{\left (c x \right )}}{6 x^{3}} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \] Input:

integrate((a+b*atan(c*x))**2/x**5,x)
 

Output:

Piecewise((-a**2/(4*x**4) + a*b*c**4*atan(c*x)/2 + a*b*c**3/(2*x) - a*b*c/ 
(6*x**3) - a*b*atan(c*x)/(2*x**4) - 2*b**2*c**4*log(x)/3 + b**2*c**4*log(x 
**2 + c**(-2))/3 + b**2*c**4*atan(c*x)**2/4 + b**2*c**3*atan(c*x)/(2*x) - 
b**2*c**2/(12*x**2) - b**2*c*atan(c*x)/(6*x**3) - b**2*atan(c*x)**2/(4*x** 
4), Ne(c, 0)), (-a**2/(4*x**4), True))
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=\frac {1}{6} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} a b + \frac {1}{12} \, {\left (2 \, {\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c \arctan \left (c x\right ) - \frac {{\left (3 \, c^{2} x^{2} \arctan \left (c x\right )^{2} - 4 \, c^{2} x^{2} \log \left (c^{2} x^{2} + 1\right ) + 8 \, c^{2} x^{2} \log \left (x\right ) + 1\right )} c^{2}}{x^{2}}\right )} b^{2} - \frac {b^{2} \arctan \left (c x\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \] Input:

integrate((a+b*arctan(c*x))^2/x^5,x, algorithm="maxima")
 

Output:

1/6*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*a*b 
+ 1/12*(2*(3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c*arctan(c*x) - (3*c^2 
*x^2*arctan(c*x)^2 - 4*c^2*x^2*log(c^2*x^2 + 1) + 8*c^2*x^2*log(x) + 1)*c^ 
2/x^2)*b^2 - 1/4*b^2*arctan(c*x)^2/x^4 - 1/4*a^2/x^4
 

Giac [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctan(c*x))^2/x^5,x, algorithm="giac")
 

Output:

integrate((b*arctan(c*x) + a)^2/x^5, x)
 

Mupad [B] (verification not implemented)

Time = 2.81 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=\frac {b^2\,c^4\,{\mathrm {atan}\left (c\,x\right )}^2}{4}-\frac {2\,b^2\,c^4\,\ln \left (x\right )}{3}-\frac {\frac {b^2\,{\mathrm {atan}\left (c\,x\right )}^2}{4}+\frac {a^2}{4}+x\,\left (\frac {c\,\mathrm {atan}\left (c\,x\right )\,b^2}{6}+\frac {a\,c\,b}{6}\right )-x^3\,\left (\frac {b^2\,c^3\,\mathrm {atan}\left (c\,x\right )}{2}+\frac {a\,b\,c^3}{2}\right )+\frac {b^2\,c^2\,x^2}{12}+\frac {a\,b\,\mathrm {atan}\left (c\,x\right )}{2}}{x^4}+\frac {b^2\,c^4\,\ln \left (c\,x+1{}\mathrm {i}\right )}{3}+\frac {b^2\,c^4\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )}{3}+\frac {a\,b\,c^4\,\ln \left (c\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}-\frac {a\,b\,c^4\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \] Input:

int((a + b*atan(c*x))^2/x^5,x)
 

Output:

(b^2*c^4*atan(c*x)^2)/4 - (2*b^2*c^4*log(x))/3 - ((b^2*atan(c*x)^2)/4 + a^ 
2/4 + x*((b^2*c*atan(c*x))/6 + (a*b*c)/6) - x^3*((b^2*c^3*atan(c*x))/2 + ( 
a*b*c^3)/2) + (b^2*c^2*x^2)/12 + (a*b*atan(c*x))/2)/x^4 + (b^2*c^4*log(c*x 
 + 1i))/3 + (b^2*c^4*log(c*x*1i + 1))/3 + (a*b*c^4*log(c*x + 1i)*1i)/4 - ( 
a*b*c^4*log(c*x*1i + 1)*1i)/4
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.28 \[ \int \frac {(a+b \arctan (c x))^2}{x^5} \, dx=\frac {3 \mathit {atan} \left (c x \right )^{2} b^{2} c^{4} x^{4}-3 \mathit {atan} \left (c x \right )^{2} b^{2}+6 \mathit {atan} \left (c x \right ) a b \,c^{4} x^{4}-6 \mathit {atan} \left (c x \right ) a b +6 \mathit {atan} \left (c x \right ) b^{2} c^{3} x^{3}-2 \mathit {atan} \left (c x \right ) b^{2} c x +4 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b^{2} c^{4} x^{4}-8 \,\mathrm {log}\left (x \right ) b^{2} c^{4} x^{4}-3 a^{2}+6 a b \,c^{3} x^{3}-2 a b c x -b^{2} c^{2} x^{2}}{12 x^{4}} \] Input:

int((a+b*atan(c*x))^2/x^5,x)
 

Output:

(3*atan(c*x)**2*b**2*c**4*x**4 - 3*atan(c*x)**2*b**2 + 6*atan(c*x)*a*b*c** 
4*x**4 - 6*atan(c*x)*a*b + 6*atan(c*x)*b**2*c**3*x**3 - 2*atan(c*x)*b**2*c 
*x + 4*log(c**2*x**2 + 1)*b**2*c**4*x**4 - 8*log(x)*b**2*c**4*x**4 - 3*a** 
2 + 6*a*b*c**3*x**3 - 2*a*b*c*x - b**2*c**2*x**2)/(12*x**4)