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

Optimal result

Integrand size = 14, antiderivative size = 117 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx=-\frac {b^2 c^2}{12 x^2}-\frac {b c (a+b \text {arctanh}(c x))}{6 x^3}-\frac {b c^3 (a+b \text {arctanh}(c x))}{2 x}+\frac {1}{4} c^4 (a+b \text {arctanh}(c x))^2-\frac {(a+b \text {arctanh}(c x))^2}{4 x^4}+\frac {2}{3} b^2 c^4 \log (x)-\frac {1}{3} b^2 c^4 \log \left (1-c^2 x^2\right ) \] Output:

-1/12*b^2*c^2/x^2-1/6*b*c*(a+b*arctanh(c*x))/x^3-1/2*b*c^3*(a+b*arctanh(c* 
x))/x+1/4*c^4*(a+b*arctanh(c*x))^2-1/4*(a+b*arctanh(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 = 0.05 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx=-\frac {3 a^2+2 a b c x+b^2 c^2 x^2+6 a b c^3 x^3+2 b \left (3 a+b c x+3 b c^3 x^3\right ) \text {arctanh}(c x)-3 b^2 \left (-1+c^4 x^4\right ) \text {arctanh}(c x)^2-8 b^2 c^4 x^4 \log (x)+3 a b c^4 x^4 \log (1-c x)+4 b^2 c^4 x^4 \log (1-c x)-3 a b c^4 x^4 \log (1+c x)+4 b^2 c^4 x^4 \log (1+c x)}{12 x^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 136, 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 = {6452, 6544, 6452, 243, 54, 2009, 6544, 6452, 243, 47, 14, 16, 6510}

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.

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 54

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 6510

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

Input:

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

Output:

-1/4*(a + b*ArcTanh[c*x])^2/x^4 + (b*c*(-1/3*(a + b*ArcTanh[c*x])/x^3 + c^ 
2*(-((a + b*ArcTanh[c*x])/x) + (c*(a + b*ArcTanh[c*x])^2)/(2*b) + (b*c*(Lo 
g[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 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[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 6510

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb 
ol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b 
, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
 

rule 6544

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( 
e_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x 
], x] - Simp[e/(d*f^2)   Int[(f*x)^(m + 2)*((a + b*ArcTanh[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 = 0.35 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.46

method	result	size
parallelrisch	\(\frac {3 b^{2} \operatorname {arctanh}\left (c x \right )^{2} x^{4} c^{4}+8 b^{2} c^{4} \ln \left (x \right ) x^{4}-8 \ln \left (c x -1\right ) x^{4} b^{2} c^{4}+6 a b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}-8 \,\operatorname {arctanh}\left (c x \right ) x^{4} b^{2} c^{4}-b^{2} c^{4} x^{4}-6 b^{2} \operatorname {arctanh}\left (c x \right ) x^{3} c^{3}-6 a b \,c^{3} x^{3}-b^{2} c^{2} x^{2}-2 b^{2} \operatorname {arctanh}\left (c x \right ) x c -2 a b c x -3 b^{2} \operatorname {arctanh}\left (c x \right )^{2}-6 \,\operatorname {arctanh}\left (c x \right ) a b -3 a^{2}}{12 x^{4}}\)	\(171\)
parts	\(-\frac {a^{2}}{4 x^{4}}+b^{2} c^{4} \left (-\frac {\operatorname {arctanh}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {\operatorname {arctanh}\left (c x \right )}{6 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{2 c x}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{4}+\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (c x -1\right )^{2}}{16}-\frac {\ln \left (c x +1\right )^{2}}{16}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (c x +1\right )}{3}-\frac {1}{12 c^{2} x^{2}}+\frac {2 \ln \left (c x \right )}{3}-\frac {\ln \left (c x -1\right )}{3}\right )+2 a b \,c^{4} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{8}-\frac {\ln \left (c x -1\right )}{8}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{4 c x}\right )\)	\(222\)
derivativedivides	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\operatorname {arctanh}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {\operatorname {arctanh}\left (c x \right )}{6 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{2 c x}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{4}+\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (c x -1\right )^{2}}{16}-\frac {\ln \left (c x +1\right )^{2}}{16}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (c x +1\right )}{3}-\frac {1}{12 c^{2} x^{2}}+\frac {2 \ln \left (c x \right )}{3}-\frac {\ln \left (c x -1\right )}{3}\right )+2 a b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{8}-\frac {\ln \left (c x -1\right )}{8}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{4 c x}\right )\right )\)	\(223\)
default	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\operatorname {arctanh}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {\operatorname {arctanh}\left (c x \right )}{6 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{2 c x}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{4}+\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (c x -1\right )^{2}}{16}-\frac {\ln \left (c x +1\right )^{2}}{16}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (c x +1\right )}{3}-\frac {1}{12 c^{2} x^{2}}+\frac {2 \ln \left (c x \right )}{3}-\frac {\ln \left (c x -1\right )}{3}\right )+2 a b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{8}-\frac {\ln \left (c x -1\right )}{8}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{4 c x}\right )\right )\)	\(223\)
risch	\(\frac {b^{2} \left (c^{4} x^{4}-1\right ) \ln \left (c x +1\right )^{2}}{16 x^{4}}-\frac {b \left (3 b \,x^{4} \ln \left (-c x +1\right ) c^{4}+6 b \,c^{3} x^{3}+2 b c x -3 b \ln \left (-c x +1\right )+6 a \right ) \ln \left (c x +1\right )}{24 x^{4}}-\frac {-3 b^{2} c^{4} x^{4} \ln \left (-c x +1\right )^{2}+12 a b \,c^{4} x^{4} \ln \left (-c x +1\right )+16 b^{2} c^{4} \ln \left (-c x +1\right ) x^{4}-12 b \,c^{4} \ln \left (-c x -1\right ) x^{4} a +16 b^{2} c^{4} \ln \left (-c x -1\right ) x^{4}-32 b^{2} c^{4} \ln \left (x \right ) x^{4}-12 b^{2} c^{3} x^{3} \ln \left (-c x +1\right )+24 a b \,c^{3} x^{3}+4 b^{2} c^{2} x^{2}-4 b^{2} c x \ln \left (-c x +1\right )+8 a b c x +3 b^{2} \ln \left (-c x +1\right )^{2}-12 b \ln \left (-c x +1\right ) a +12 a^{2}}{48 x^{4}}\)	\(281\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.48 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx=\frac {32 \, b^{2} c^{4} x^{4} \log \left (x\right ) + 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} c^{4} x^{4} \log \left (c x + 1\right ) - 4 \, {\left (3 \, a b + 4 \, b^{2}\right )} c^{4} x^{4} \log \left (c x - 1\right ) - 24 \, a b c^{3} x^{3} - 4 \, b^{2} c^{2} x^{2} - 8 \, a b c x + 3 \, {\left (b^{2} c^{4} x^{4} - b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 12 \, a^{2} - 4 \, {\left (3 \, b^{2} c^{3} x^{3} + b^{2} c x + 3 \, a b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{48 \, x^{4}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx=\begin {cases} - \frac {a^{2}}{4 x^{4}} + \frac {a b c^{4} \operatorname {atanh}{\left (c x \right )}}{2} - \frac {a b c^{3}}{2 x} - \frac {a b c}{6 x^{3}} - \frac {a b \operatorname {atanh}{\left (c x \right )}}{2 x^{4}} + \frac {2 b^{2} c^{4} \log {\left (x \right )}}{3} - \frac {2 b^{2} c^{4} \log {\left (x - \frac {1}{c} \right )}}{3} + \frac {b^{2} c^{4} \operatorname {atanh}^{2}{\left (c x \right )}}{4} - \frac {2 b^{2} c^{4} \operatorname {atanh}{\left (c x \right )}}{3} - \frac {b^{2} c^{3} \operatorname {atanh}{\left (c x \right )}}{2 x} - \frac {b^{2} c^{2}}{12 x^{2}} - \frac {b^{2} c \operatorname {atanh}{\left (c x \right )}}{6 x^{3}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a**2/(4*x**4) + a*b*c**4*atanh(c*x)/2 - a*b*c**3/(2*x) - a*b*c 
/(6*x**3) - a*b*atanh(c*x)/(2*x**4) + 2*b**2*c**4*log(x)/3 - 2*b**2*c**4*l 
og(x - 1/c)/3 + b**2*c**4*atanh(c*x)**2/4 - 2*b**2*c**4*atanh(c*x)/3 - b** 
2*c**3*atanh(c*x)/(2*x) - b**2*c**2/(12*x**2) - b**2*c*atanh(c*x)/(6*x**3) 
 - b**2*atanh(c*x)**2/(4*x**4), Ne(c, 0)), (-a**2/(4*x**4), True))
                                                                                    
                                                                                    
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (103) = 206\).

Time = 0.04 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.91 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx=\frac {1}{12} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} a b + \frac {1}{48} \, {\left ({\left (32 \, c^{2} \log \left (x\right ) - \frac {3 \, c^{2} x^{2} \log \left (c x + 1\right )^{2} + 3 \, c^{2} x^{2} \log \left (c x - 1\right )^{2} + 16 \, c^{2} x^{2} \log \left (c x - 1\right ) - 2 \, {\left (3 \, c^{2} x^{2} \log \left (c x - 1\right ) - 8 \, c^{2} x^{2}\right )} \log \left (c x + 1\right ) + 4}{x^{2}}\right )} c^{2} + 4 \, {\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c \operatorname {artanh}\left (c x\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \] Input:

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

Output:

1/12*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c 
- 6*arctanh(c*x)/x^4)*a*b + 1/48*((32*c^2*log(x) - (3*c^2*x^2*log(c*x + 1) 
^2 + 3*c^2*x^2*log(c*x - 1)^2 + 16*c^2*x^2*log(c*x - 1) - 2*(3*c^2*x^2*log 
(c*x - 1) - 8*c^2*x^2)*log(c*x + 1) + 4)/x^2)*c^2 + 4*(3*c^3*log(c*x + 1) 
- 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c*arctanh(c*x))*b^2 - 1/4*b^ 
2*arctanh(c*x)^2/x^4 - 1/4*a^2/x^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (103) = 206\).

Time = 0.13 (sec) , antiderivative size = 612, normalized size of antiderivative = 5.23 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*(4*b^2*c^3*log(-(c*x + 1)/(c*x - 1) - 1) - 4*b^2*c^3*log(-(c*x + 1)/(c 
*x - 1)) + 3*((c*x + 1)^3*b^2*c^3/(c*x - 1)^3 + (c*x + 1)*b^2*c^3/(c*x - 1 
))*log(-(c*x + 1)/(c*x - 1))^2/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c 
*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1) + 2*(6* 
(c*x + 1)^3*a*b*c^3/(c*x - 1)^3 + 6*(c*x + 1)*a*b*c^3/(c*x - 1) + 3*(c*x + 
 1)^3*b^2*c^3/(c*x - 1)^3 + 6*(c*x + 1)^2*b^2*c^3/(c*x - 1)^2 + 5*(c*x + 1 
)*b^2*c^3/(c*x - 1) + 2*b^2*c^3)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^4/(c 
*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x 
 + 1)/(c*x - 1) + 1) + 2*(6*(c*x + 1)^3*a^2*c^3/(c*x - 1)^3 + 6*(c*x + 1)* 
a^2*c^3/(c*x - 1) + 6*(c*x + 1)^3*a*b*c^3/(c*x - 1)^3 + 12*(c*x + 1)^2*a*b 
*c^3/(c*x - 1)^2 + 10*(c*x + 1)*a*b*c^3/(c*x - 1) + 4*a*b*c^3 + (c*x + 1)^ 
3*b^2*c^3/(c*x - 1)^3 + 2*(c*x + 1)^2*b^2*c^3/(c*x - 1)^2 + (c*x + 1)*b^2* 
c^3/(c*x - 1))/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c 
*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1))*c
 

Mupad [B] (verification not implemented)

Time = 4.50 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.59 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx=\frac {b^2\,c^4\,{\ln \left (c\,x+1\right )}^2}{16}-\frac {a^2}{4\,x^4}+\frac {b^2\,c^4\,{\ln \left (1-c\,x\right )}^2}{16}-\frac {b^2\,{\ln \left (c\,x+1\right )}^2}{16\,x^4}-\frac {b^2\,{\ln \left (1-c\,x\right )}^2}{16\,x^4}-\frac {b^2\,c^2}{12\,x^2}+\frac {2\,b^2\,c^4\,\ln \left (x\right )}{3}-\frac {b^2\,c^4\,\ln \left (c\,x-1\right )}{3}-\frac {b^2\,c^4\,\ln \left (c\,x+1\right )}{3}-\frac {a\,b\,\ln \left (c\,x+1\right )}{4\,x^4}+\frac {a\,b\,\ln \left (1-c\,x\right )}{4\,x^4}+\frac {b^2\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )}{8\,x^4}-\frac {a\,b\,c}{6\,x^3}-\frac {b^2\,c\,\ln \left (c\,x+1\right )}{12\,x^3}+\frac {b^2\,c\,\ln \left (1-c\,x\right )}{12\,x^3}-\frac {a\,b\,c^3}{2\,x}-\frac {b^2\,c^3\,\ln \left (c\,x+1\right )}{4\,x}+\frac {b^2\,c^3\,\ln \left (1-c\,x\right )}{4\,x}-\frac {a\,b\,c^4\,\ln \left (c\,x-1\right )}{4}+\frac {a\,b\,c^4\,\ln \left (c\,x+1\right )}{4}-\frac {b^2\,c^4\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )}{8} \] Input:

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

Output:

(b^2*c^4*log(c*x + 1)^2)/16 - a^2/(4*x^4) + (b^2*c^4*log(1 - c*x)^2)/16 - 
(b^2*log(c*x + 1)^2)/(16*x^4) - (b^2*log(1 - c*x)^2)/(16*x^4) - (b^2*c^2)/ 
(12*x^2) + (2*b^2*c^4*log(x))/3 - (b^2*c^4*log(c*x - 1))/3 - (b^2*c^4*log( 
c*x + 1))/3 - (a*b*log(c*x + 1))/(4*x^4) + (a*b*log(1 - c*x))/(4*x^4) + (b 
^2*log(c*x + 1)*log(1 - c*x))/(8*x^4) - (a*b*c)/(6*x^3) - (b^2*c*log(c*x + 
 1))/(12*x^3) + (b^2*c*log(1 - c*x))/(12*x^3) - (a*b*c^3)/(2*x) - (b^2*c^3 
*log(c*x + 1))/(4*x) + (b^2*c^3*log(1 - c*x))/(4*x) - (a*b*c^4*log(c*x - 1 
))/4 + (a*b*c^4*log(c*x + 1))/4 - (b^2*c^4*log(c*x + 1)*log(1 - c*x))/8
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b \text {arctanh}(c x))^2}{x^5} \, dx=\frac {3 \mathit {atanh} \left (c x \right )^{2} b^{2} c^{4} x^{4}-3 \mathit {atanh} \left (c x \right )^{2} b^{2}+6 \mathit {atanh} \left (c x \right ) a b \,c^{4} x^{4}-6 \mathit {atanh} \left (c x \right ) a b -8 \mathit {atanh} \left (c x \right ) b^{2} c^{4} x^{4}-6 \mathit {atanh} \left (c x \right ) b^{2} c^{3} x^{3}-2 \mathit {atanh} \left (c x \right ) b^{2} c x -8 \,\mathrm {log}\left (c^{2} x -c \right ) b^{2} c^{4} x^{4}+8 \,\mathrm {log}\left (x \right ) b^{2} c^{4} x^{4}-3 a^{2}-6 a b \,c^{3} x^{3}-2 a b c x -b^{2} c^{2} x^{2}}{12 x^{4}} \] Input:

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

Output:

(3*atanh(c*x)**2*b**2*c**4*x**4 - 3*atanh(c*x)**2*b**2 + 6*atanh(c*x)*a*b* 
c**4*x**4 - 6*atanh(c*x)*a*b - 8*atanh(c*x)*b**2*c**4*x**4 - 6*atanh(c*x)* 
b**2*c**3*x**3 - 2*atanh(c*x)*b**2*c*x - 8*log(c**2*x - c)*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)