\(\int x (a+b \text {arctanh}(\frac {c}{x}))^2 \, dx\) [145]

Optimal result

Integrand size = 14, antiderivative size = 83 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=b c x \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )-\frac {1}{2} c^2 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{2} x^2 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{2} b^2 c^2 \log \left (1-\frac {c^2}{x^2}\right )+b^2 c^2 \log (x) \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.11 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{2} \left (2 a b c x+a^2 x^2+2 b x (b c+a x) \text {arctanh}\left (\frac {c}{x}\right )+b^2 \left (-c^2+x^2\right ) \text {arctanh}\left (\frac {c}{x}\right )^2+b (a+b) c^2 \log (-c+x)-a b c^2 \log (c+x)+b^2 c^2 \log (c+x)\right ) \] Input:

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

Output:

(2*a*b*c*x + a^2*x^2 + 2*b*x*(b*c + a*x)*ArcTanh[c/x] + b^2*(-c^2 + x^2)*A 
rcTanh[c/x]^2 + b*(a + b)*c^2*Log[-c + x] - a*b*c^2*Log[c + x] + b^2*c^2*L 
og[c + x])/2
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6454, 6452, 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.

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

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 6510

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

Input:

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

Output:

(x^2*(a + b*ArcTanh[c/x])^2)/2 - b*c*(-(x*(a + b*ArcTanh[c/x])) + (c*(a + 
b*ArcTanh[c/x])^2)/(2*b) + (b*c*(-Log[1 - c^2/x^2] + Log[x^(-2)]))/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 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 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 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 = 1.51 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.42

method	result	size
parallelrisch	\(\frac {b^{2} x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right )^{2} b^{2} c^{2}}{2}+b^{2} c^{2} \ln \left (x -c \right )+x^{2} a b \,\operatorname {arctanh}\left (\frac {c}{x}\right )+x \,\operatorname {arctanh}\left (\frac {c}{x}\right ) b^{2} c -\operatorname {arctanh}\left (\frac {c}{x}\right ) a b \,c^{2}+\operatorname {arctanh}\left (\frac {c}{x}\right ) b^{2} c^{2}+\frac {a^{2} x^{2}}{2}+a b c x +\frac {a^{2} c^{2}}{2}\)	\(118\)
parts	\(\frac {a^{2} x^{2}}{2}-b^{2} c^{2} \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{2 c^{2}}-\frac {x \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{c}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{2}+\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{4}-\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{8}-\frac {\ln \left (1+\frac {c}{x}\right )}{2}-\frac {\ln \left (\frac {c}{x}-1\right )}{2}+\ln \left (\frac {c}{x}\right )-\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{8}+\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{4}\right )-2 a b \,c^{2} \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c^{2}}-\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {x}{2 c}+\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )\)	\(227\)
derivativedivides	\(-c^{2} \left (-\frac {a^{2} x^{2}}{2 c^{2}}+b^{2} \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{2 c^{2}}-\frac {x \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{c}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{2}+\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{4}-\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{8}-\frac {\ln \left (1+\frac {c}{x}\right )}{2}-\frac {\ln \left (\frac {c}{x}-1\right )}{2}+\ln \left (\frac {c}{x}\right )-\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{8}+\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{4}\right )+2 a b \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c^{2}}-\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {x}{2 c}+\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )\right )\)	\(228\)
default	\(-c^{2} \left (-\frac {a^{2} x^{2}}{2 c^{2}}+b^{2} \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{2 c^{2}}-\frac {x \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{c}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{2}+\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{4}-\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{8}-\frac {\ln \left (1+\frac {c}{x}\right )}{2}-\frac {\ln \left (\frac {c}{x}-1\right )}{2}+\ln \left (\frac {c}{x}\right )-\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{8}+\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{4}\right )+2 a b \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c^{2}}-\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {x}{2 c}+\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )\right )\)	\(228\)
risch	\(\text {Expression too large to display}\)	\(14545\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.34 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=a b c x + \frac {1}{2} \, a^{2} x^{2} - \frac {1}{2} \, {\left (a b - b^{2}\right )} c^{2} \log \left (c + x\right ) + \frac {1}{2} \, {\left (a b + b^{2}\right )} c^{2} \log \left (-c + x\right ) - \frac {1}{8} \, {\left (b^{2} c^{2} - b^{2} x^{2}\right )} \log \left (-\frac {c + x}{c - x}\right )^{2} + \frac {1}{2} \, {\left (b^{2} c x + a b x^{2}\right )} \log \left (-\frac {c + x}{c - x}\right ) \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.25 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^{2} x^{2}}{2} - a b c^{2} \operatorname {atanh}{\left (\frac {c}{x} \right )} + a b c x + a b x^{2} \operatorname {atanh}{\left (\frac {c}{x} \right )} + b^{2} c^{2} \log {\left (- c + x \right )} - \frac {b^{2} c^{2} \operatorname {atanh}^{2}{\left (\frac {c}{x} \right )}}{2} + b^{2} c^{2} \operatorname {atanh}{\left (\frac {c}{x} \right )} + b^{2} c x \operatorname {atanh}{\left (\frac {c}{x} \right )} + \frac {b^{2} x^{2} \operatorname {atanh}^{2}{\left (\frac {c}{x} \right )}}{2} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.64 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{2} \, b^{2} x^{2} \operatorname {artanh}\left (\frac {c}{x}\right )^{2} + \frac {1}{2} \, a^{2} x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (\frac {c}{x}\right ) - {\left (c \log \left (c + x\right ) - c \log \left (-c + x\right ) - 2 \, x\right )} c\right )} a b + \frac {1}{8} \, {\left ({\left (\log \left (c + x\right )^{2} - 2 \, {\left (\log \left (c + x\right ) - 2\right )} \log \left (-c + x\right ) + \log \left (-c + x\right )^{2} + 4 \, \log \left (c + x\right )\right )} c^{2} - 4 \, {\left (c \log \left (c + x\right ) - c \log \left (-c + x\right ) - 2 \, x\right )} c \operatorname {artanh}\left (\frac {c}{x}\right )\right )} b^{2} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (77) = 154\).

Time = 0.12 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.23 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=-\frac {2 \, b^{2} c^{3} \log \left (-\frac {c + x}{c - x} - 1\right ) - 2 \, b^{2} c^{3} \log \left (-\frac {c + x}{c - x}\right ) + \frac {b^{2} {\left (c + x\right )} c^{3} \log \left (-\frac {c + x}{c - x}\right )^{2}}{{\left (c - x\right )} {\left (\frac {{\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {2 \, {\left (c + x\right )}}{c - x} + 1\right )}} + \frac {2 \, {\left (b^{2} c^{3} + \frac {2 \, a b {\left (c + x\right )} c^{3}}{c - x} + \frac {b^{2} {\left (c + x\right )} c^{3}}{c - x}\right )} \log \left (-\frac {c + x}{c - x}\right )}{\frac {{\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {2 \, {\left (c + x\right )}}{c - x} + 1} + \frac {4 \, {\left (a b c^{3} + \frac {a^{2} {\left (c + x\right )} c^{3}}{c - x} + \frac {a b {\left (c + x\right )} c^{3}}{c - x}\right )}}{\frac {{\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {2 \, {\left (c + x\right )}}{c - x} + 1}}{2 \, c} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.47 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.22 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^2\,x^2}{2}-\frac {b^2\,c^2\,{\mathrm {atanh}\left (\frac {c}{x}\right )}^2}{2}+\frac {b^2\,x^2\,{\mathrm {atanh}\left (\frac {c}{x}\right )}^2}{2}+\frac {b^2\,c^2\,\ln \left (x^2-c^2\right )}{2}-a\,b\,c^2\,\mathrm {atanh}\left (\frac {c}{x}\right )+a\,b\,x^2\,\mathrm {atanh}\left (\frac {c}{x}\right )+b^2\,c\,x\,\mathrm {atanh}\left (\frac {c}{x}\right )+a\,b\,c\,x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.35 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=-\frac {\mathit {atanh} \left (\frac {c}{x}\right )^{2} b^{2} c^{2}}{2}+\frac {\mathit {atanh} \left (\frac {c}{x}\right )^{2} b^{2} x^{2}}{2}-\mathit {atanh} \left (\frac {c}{x}\right ) a b \,c^{2}+\mathit {atanh} \left (\frac {c}{x}\right ) a b \,x^{2}-\mathit {atanh} \left (\frac {c}{x}\right ) b^{2} c^{2}+\mathit {atanh} \left (\frac {c}{x}\right ) b^{2} c x +\mathrm {log}\left (-c -x \right ) b^{2} c^{2}+\frac {a^{2} x^{2}}{2}+a b c x \] Input:

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