\(\int \frac {(a+b \text {arctanh}(\frac {c}{x}))^2}{x^3} \, dx\) [149]

Optimal result

Integrand size = 16, antiderivative size = 87 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=-\frac {a b}{c x}-\frac {b^2 \coth ^{-1}\left (\frac {x}{c}\right )}{c x}+\frac {\left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2}{2 c^2}-\frac {\left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2}{2 x^2}-\frac {b^2 \log \left (1-\frac {c^2}{x^2}\right )}{2 c^2} \] Output:

-a*b/c/x-b^2*arccoth(x/c)/c/x+1/2*(a+b*arccoth(x/c))^2/c^2-1/2*(a+b*arccot 
h(x/c))^2/x^2-1/2*b^2*ln(1-c^2/x^2)/c^2
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=-\frac {a^2 c^2+2 a b c x+2 b c (a c+b x) \text {arctanh}\left (\frac {c}{x}\right )+b^2 \left (c^2-x^2\right ) \text {arctanh}\left (\frac {c}{x}\right )^2-2 b^2 x^2 \log (x)+a b x^2 \log (-c+x)+b^2 x^2 \log (-c+x)-a b x^2 \log (c+x)+b^2 x^2 \log (c+x)}{2 c^2 x^2} \] Input:

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

Output:

-1/2*(a^2*c^2 + 2*a*b*c*x + 2*b*c*(a*c + b*x)*ArcTanh[c/x] + b^2*(c^2 - x^ 
2)*ArcTanh[c/x]^2 - 2*b^2*x^2*Log[x] + a*b*x^2*Log[-c + x] + b^2*x^2*Log[- 
c + x] - a*b*x^2*Log[c + x] + b^2*x^2*Log[c + x])/(c^2*x^2)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6454, 6452, 6542, 2009, 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.

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

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6542

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6510

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

Input:

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

Output:

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

Defintions of rubi rules used

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 6454

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
 Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x 
], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simpl 
ify[(m + 1)/n]]
 

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 6542

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

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.52

method	result	size
parallelrisch	\(\frac {b^{2} x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}-\operatorname {arctanh}\left (\frac {c}{x}\right )^{2} b^{2} c^{2}+2 b^{2} x^{2} \ln \left (x \right )-2 \ln \left (x -c \right ) x^{2} b^{2}+2 x^{2} a b \,\operatorname {arctanh}\left (\frac {c}{x}\right )-2 x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right ) b^{2}-2 x \,\operatorname {arctanh}\left (\frac {c}{x}\right ) b^{2} c -2 \,\operatorname {arctanh}\left (\frac {c}{x}\right ) a b \,c^{2}-2 a b c x -a^{2} c^{2}}{2 x^{2} c^{2}}\)	\(132\)
parts	\(-\frac {a^{2}}{2 x^{2}}-\frac {b^{2} \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{2 x^{2}}+\frac {c \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{x}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{2}-\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{4}+\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{8}+\frac {\ln \left (\frac {c}{x}-1\right )}{2}+\frac {\ln \left (1+\frac {c}{x}\right )}{2}+\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{8}-\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{4}\right )}{c^{2}}-\frac {2 a b \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 x^{2}}+\frac {c}{2 x}+\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )}{c^{2}}\)	\(220\)
derivativedivides	\(-\frac {\frac {a^{2} c^{2}}{2 x^{2}}+b^{2} \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{2 x^{2}}+\frac {c \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{x}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{2}-\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{4}+\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{8}+\frac {\ln \left (\frac {c}{x}-1\right )}{2}+\frac {\ln \left (1+\frac {c}{x}\right )}{2}+\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{8}-\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{4}\right )+2 a b \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 x^{2}}+\frac {c}{2 x}+\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )}{c^{2}}\)	\(221\)
default	\(-\frac {\frac {a^{2} c^{2}}{2 x^{2}}+b^{2} \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{2 x^{2}}+\frac {c \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{x}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{2}-\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{4}+\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{8}+\frac {\ln \left (\frac {c}{x}-1\right )}{2}+\frac {\ln \left (1+\frac {c}{x}\right )}{2}+\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{8}-\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{4}\right )+2 a b \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 x^{2}}+\frac {c}{2 x}+\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )}{c^{2}}\)	\(221\)
risch	\(\text {Expression too large to display}\)	\(117644\)

Input:

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

Output:

1/2*(b^2*x^2*arctanh(c/x)^2-arctanh(c/x)^2*b^2*c^2+2*b^2*x^2*ln(x)-2*ln(x- 
c)*x^2*b^2+2*x^2*a*b*arctanh(c/x)-2*x^2*arctanh(c/x)*b^2-2*x*arctanh(c/x)* 
b^2*c-2*arctanh(c/x)*a*b*c^2-2*a*b*c*x-a^2*c^2)/x^2/c^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=\frac {8 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a^{2} c^{2} - 8 \, a b c x + 4 \, {\left (a b - b^{2}\right )} x^{2} \log \left (c + x\right ) - 4 \, {\left (a b + b^{2}\right )} x^{2} \log \left (-c + x\right ) - {\left (b^{2} c^{2} - b^{2} x^{2}\right )} \log \left (-\frac {c + x}{c - x}\right )^{2} - 4 \, {\left (a b c^{2} + b^{2} c x\right )} \log \left (-\frac {c + x}{c - x}\right )}{8 \, c^{2} x^{2}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=\begin {cases} - \frac {a^{2}}{2 x^{2}} - \frac {a b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{x^{2}} - \frac {a b}{c x} + \frac {a b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{c^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\frac {c}{x} \right )}}{2 x^{2}} - \frac {b^{2} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{c x} + \frac {b^{2} \log {\left (x \right )}}{c^{2}} - \frac {b^{2} \log {\left (- c + x \right )}}{c^{2}} + \frac {b^{2} \operatorname {atanh}^{2}{\left (\frac {c}{x} \right )}}{2 c^{2}} - \frac {b^{2} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{c^{2}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{2 x^{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a**2/(2*x**2) - a*b*atanh(c/x)/x**2 - a*b/(c*x) + a*b*atanh(c/ 
x)/c**2 - b**2*atanh(c/x)**2/(2*x**2) - b**2*atanh(c/x)/(c*x) + b**2*log(x 
)/c**2 - b**2*log(-c + x)/c**2 + b**2*atanh(c/x)**2/(2*c**2) - b**2*atanh( 
c/x)/c**2, Ne(c, 0)), (-a**2/(2*x**2), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (81) = 162\).

Time = 0.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.90 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=\frac {1}{2} \, {\left (c {\left (\frac {\log \left (c + x\right )}{c^{3}} - \frac {\log \left (-c + x\right )}{c^{3}} - \frac {2}{c^{2} x}\right )} - \frac {2 \, \operatorname {artanh}\left (\frac {c}{x}\right )}{x^{2}}\right )} a b - \frac {1}{8} \, {\left (c^{2} {\left (\frac {\log \left (c + x\right )^{2} - 2 \, {\left (\log \left (c + x\right ) - 2\right )} \log \left (-c + x\right ) + \log \left (-c + x\right )^{2} + 4 \, \log \left (c + x\right )}{c^{4}} - \frac {8 \, \log \left (x\right )}{c^{4}}\right )} - 4 \, c {\left (\frac {\log \left (c + x\right )}{c^{3}} - \frac {\log \left (-c + x\right )}{c^{3}} - \frac {2}{c^{2} x}\right )} \operatorname {artanh}\left (\frac {c}{x}\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (\frac {c}{x}\right )^{2}}{2 \, x^{2}} - \frac {a^{2}}{2 \, x^{2}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (81) = 162\).

Time = 0.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.93 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=-\frac {\frac {b^{2} {\left (c + x\right )} \log \left (-\frac {c + x}{c - x}\right )^{2}}{{\left (\frac {{\left (c + x\right )}^{2} c}{{\left (c - x\right )}^{2}} - \frac {2 \, {\left (c + x\right )} c}{c - x} + c\right )} {\left (c - x\right )}} - \frac {2 \, b^{2} \log \left (-\frac {c + x}{c - x} + 1\right )}{c} + \frac {2 \, b^{2} \log \left (-\frac {c + x}{c - x}\right )}{c} - \frac {2 \, {\left (b^{2} - \frac {2 \, a b {\left (c + x\right )}}{c - x} - \frac {b^{2} {\left (c + x\right )}}{c - x}\right )} \log \left (-\frac {c + x}{c - x}\right )}{\frac {{\left (c + x\right )}^{2} c}{{\left (c - x\right )}^{2}} - \frac {2 \, {\left (c + x\right )} c}{c - x} + c} - \frac {4 \, {\left (a b - \frac {a^{2} {\left (c + x\right )}}{c - x} - \frac {a b {\left (c + x\right )}}{c - x}\right )}}{\frac {{\left (c + x\right )}^{2} c}{{\left (c - x\right )}^{2}} - \frac {2 \, {\left (c + x\right )} c}{c - x} + c}}{2 \, c} \] Input:

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

Output:

-1/2*(b^2*(c + x)*log(-(c + x)/(c - x))^2/(((c + x)^2*c/(c - x)^2 - 2*(c + 
 x)*c/(c - x) + c)*(c - x)) - 2*b^2*log(-(c + x)/(c - x) + 1)/c + 2*b^2*lo 
g(-(c + x)/(c - x))/c - 2*(b^2 - 2*a*b*(c + x)/(c - x) - b^2*(c + x)/(c - 
x))*log(-(c + x)/(c - x))/((c + x)^2*c/(c - x)^2 - 2*(c + x)*c/(c - x) + c 
) - 4*(a*b - a^2*(c + x)/(c - x) - a*b*(c + x)/(c - x))/((c + x)^2*c/(c - 
x)^2 - 2*(c + x)*c/(c - x) + c))/c
 

Mupad [B] (verification not implemented)

Time = 4.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.70 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=\ln \left (1-\frac {c}{x}\right )\,\left (\frac {a\,b}{2\,x^2}-\ln \left (\frac {c}{x}+1\right )\,\left (\frac {b^2}{4\,c^2}-\frac {b^2}{4\,x^2}\right )+\frac {b^2\,\left (2\,c\,x-c^2\right )}{8\,c^2\,x^2}+\frac {b^2\,\left (2\,c^2+4\,x\,c\right )}{16\,c^2\,x^2}\right )-\frac {\frac {a^2}{2}+\frac {a\,b\,x}{c}}{x^2}+{\ln \left (\frac {c}{x}+1\right )}^2\,\left (\frac {b^2}{8\,c^2}-\frac {b^2}{8\,x^2}\right )+{\ln \left (1-\frac {c}{x}\right )}^2\,\left (\frac {b^2}{8\,c^2}-\frac {b^2}{8\,x^2}\right )-\frac {\ln \left (x-c\right )\,\left (b^2+a\,b\right )}{2\,c^2}+\frac {\ln \left (c+x\right )\,\left (a\,b-b^2\right )}{2\,c^2}-\frac {\ln \left (\frac {c}{x}+1\right )\,\left (\frac {a\,b}{2}+\frac {b^2\,x}{2\,c}\right )}{x^2}+\frac {b^2\,\ln \left (x\right )}{c^2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2}{x^3} \, dx=\frac {-\mathit {atanh} \left (\frac {c}{x}\right )^{2} b^{2} c^{2}+\mathit {atanh} \left (\frac {c}{x}\right )^{2} b^{2} x^{2}-2 \mathit {atanh} \left (\frac {c}{x}\right ) a b \,c^{2}+2 \mathit {atanh} \left (\frac {c}{x}\right ) a b \,x^{2}-2 \mathit {atanh} \left (\frac {c}{x}\right ) b^{2} c x +2 \mathit {atanh} \left (\frac {c}{x}\right ) b^{2} x^{2}-2 \,\mathrm {log}\left (-c -x \right ) b^{2} x^{2}+2 \,\mathrm {log}\left (x \right ) b^{2} x^{2}-a^{2} c^{2}-2 a b c x}{2 c^{2} x^{2}} \] Input:

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

Output:

( - atanh(c/x)**2*b**2*c**2 + atanh(c/x)**2*b**2*x**2 - 2*atanh(c/x)*a*b*c 
**2 + 2*atanh(c/x)*a*b*x**2 - 2*atanh(c/x)*b**2*c*x + 2*atanh(c/x)*b**2*x* 
*2 - 2*log( - c - x)*b**2*x**2 + 2*log(x)*b**2*x**2 - a**2*c**2 - 2*a*b*c* 
x)/(2*c**2*x**2)