\(\int x^3 (a+b \text {arctanh}(\frac {c}{x}))^2 \, dx\) [143]

Optimal result

Integrand size = 16, antiderivative size = 123 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{12} b^2 c^2 x^2+\frac {1}{2} b c^3 x \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )+\frac {1}{6} b c x^3 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )-\frac {1}{4} c^4 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{4} x^4 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{3} b^2 c^4 \log \left (1-\frac {c^2}{x^2}\right )+\frac {2}{3} b^2 c^4 \log (x) \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{12} \left (6 a b c^3 x+b^2 c^2 x^2+2 a b c x^3+3 a^2 x^4+2 b x \left (3 a x^3+b c \left (3 c^2+x^2\right )\right ) \text {arctanh}\left (\frac {c}{x}\right )+3 b^2 \left (-c^4+x^4\right ) \text {arctanh}\left (\frac {c}{x}\right )^2+b (3 a+4 b) c^4 \log (-c+x)-3 a b c^4 \log (c+x)+4 b^2 c^4 \log (c+x)\right ) \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.01 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.14, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {6454, 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.

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

\(\Big \downarrow \) 6454

\(\displaystyle -\int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2d\frac {1}{x}\)

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 54

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 6510

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

Input:

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

Output:

(x^4*(a + b*ArcTanh[c/x])^2)/4 - (b*c*(-1/3*(x^3*(a + b*ArcTanh[c/x])) + ( 
b*c*(-x - c^2*Log[1 - c^2/x^2] + c^2*Log[x^(-2)]))/6 + c^2*(-(x*(a + b*Arc 
Tanh[c/x])) + (c*(a + b*ArcTanh[c/x])^2)/(2*b) + (b*c*(-Log[1 - c^2/x^2] + 
 Log[x^(-2)]))/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 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 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 = 1.83 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.31

method	result	size
parallelrisch	\(\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2} b^{2}}{4}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right )^{2} b^{2} c^{4}}{4}+\frac {2 b^{2} c^{4} \ln \left (x -c \right )}{3}+\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right ) a b}{2}+\frac {x^{3} \operatorname {arctanh}\left (\frac {c}{x}\right ) b^{2} c}{6}+\frac {x \,\operatorname {arctanh}\left (\frac {c}{x}\right ) b^{2} c^{3}}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) a b \,c^{4}}{2}+\frac {2 \,\operatorname {arctanh}\left (\frac {c}{x}\right ) b^{2} c^{4}}{3}+\frac {a^{2} x^{4}}{4}+\frac {a b c \,x^{3}}{6}+\frac {b^{2} c^{2} x^{2}}{12}+\frac {c^{3} a b x}{2}+\frac {b^{2} c^{4}}{12}\)	\(161\)
parts	\(\frac {a^{2} x^{4}}{4}-b^{2} c^{4} \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{4 c^{4}}-\frac {x^{3} \operatorname {arctanh}\left (\frac {c}{x}\right )}{6 c^{3}}-\frac {x \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{4}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{16}+\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{8}+\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{8}-\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{16}-\frac {x^{2}}{12 c^{2}}+\frac {2 \ln \left (\frac {c}{x}\right )}{3}-\frac {\ln \left (1+\frac {c}{x}\right )}{3}-\frac {\ln \left (\frac {c}{x}-1\right )}{3}\right )-2 a b \,c^{4} \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\ln \left (1+\frac {c}{x}\right )}{8}-\frac {\ln \left (\frac {c}{x}-1\right )}{8}-\frac {x^{3}}{12 c^{3}}-\frac {x}{4 c}\right )\)	\(259\)
derivativedivides	\(-c^{4} \left (-\frac {a^{2} x^{4}}{4 c^{4}}+b^{2} \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{4 c^{4}}-\frac {x^{3} \operatorname {arctanh}\left (\frac {c}{x}\right )}{6 c^{3}}-\frac {x \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{4}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{16}+\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{8}+\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{8}-\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{16}-\frac {x^{2}}{12 c^{2}}+\frac {2 \ln \left (\frac {c}{x}\right )}{3}-\frac {\ln \left (1+\frac {c}{x}\right )}{3}-\frac {\ln \left (\frac {c}{x}-1\right )}{3}\right )+2 a b \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\ln \left (1+\frac {c}{x}\right )}{8}-\frac {\ln \left (\frac {c}{x}-1\right )}{8}-\frac {x^{3}}{12 c^{3}}-\frac {x}{4 c}\right )\right )\)	\(260\)
default	\(-c^{4} \left (-\frac {a^{2} x^{4}}{4 c^{4}}+b^{2} \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )^{2}}{4 c^{4}}-\frac {x^{3} \operatorname {arctanh}\left (\frac {c}{x}\right )}{6 c^{3}}-\frac {x \,\operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c}+\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (1+\frac {c}{x}\right )}{4}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) \ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (\frac {c}{x}-1\right )^{2}}{16}+\frac {\ln \left (\frac {c}{x}-1\right ) \ln \left (\frac {c}{2 x}+\frac {1}{2}\right )}{8}+\frac {\left (\ln \left (1+\frac {c}{x}\right )-\ln \left (\frac {c}{2 x}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c}{2 x}+\frac {1}{2}\right )}{8}-\frac {\ln \left (1+\frac {c}{x}\right )^{2}}{16}-\frac {x^{2}}{12 c^{2}}+\frac {2 \ln \left (\frac {c}{x}\right )}{3}-\frac {\ln \left (1+\frac {c}{x}\right )}{3}-\frac {\ln \left (\frac {c}{x}-1\right )}{3}\right )+2 a b \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\ln \left (1+\frac {c}{x}\right )}{8}-\frac {\ln \left (\frac {c}{x}-1\right )}{8}-\frac {x^{3}}{12 c^{3}}-\frac {x}{4 c}\right )\right )\)	\(260\)
risch	\(\text {Expression too large to display}\)	\(14899\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.21 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{2} \, a b c^{3} x + \frac {1}{12} \, b^{2} c^{2} x^{2} + \frac {1}{6} \, a b c x^{3} + \frac {1}{4} \, a^{2} x^{4} - \frac {1}{12} \, {\left (3 \, a b - 4 \, b^{2}\right )} c^{4} \log \left (c + x\right ) + \frac {1}{12} \, {\left (3 \, a b + 4 \, b^{2}\right )} c^{4} \log \left (-c + x\right ) - \frac {1}{16} \, {\left (b^{2} c^{4} - b^{2} x^{4}\right )} \log \left (-\frac {c + x}{c - x}\right )^{2} + \frac {1}{12} \, {\left (3 \, b^{2} c^{3} x + b^{2} c x^{3} + 3 \, a b x^{4}\right )} \log \left (-\frac {c + x}{c - x}\right ) \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.28 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^{2} x^{4}}{4} - \frac {a b c^{4} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{2} + \frac {a b c^{3} x}{2} + \frac {a b c x^{3}}{6} + \frac {a b x^{4} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{2} + \frac {2 b^{2} c^{4} \log {\left (- c + x \right )}}{3} - \frac {b^{2} c^{4} \operatorname {atanh}^{2}{\left (\frac {c}{x} \right )}}{4} + \frac {2 b^{2} c^{4} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{3} + \frac {b^{2} c^{3} x \operatorname {atanh}{\left (\frac {c}{x} \right )}}{2} + \frac {b^{2} c^{2} x^{2}}{12} + \frac {b^{2} c x^{3} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{6} + \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (\frac {c}{x} \right )}}{4} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.54 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (\frac {c}{x}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{12} \, {\left (6 \, x^{4} \operatorname {artanh}\left (\frac {c}{x}\right ) - {\left (3 \, c^{3} \log \left (c + x\right ) - 3 \, c^{3} \log \left (-c + x\right ) - 6 \, c^{2} x - 2 \, x^{3}\right )} c\right )} a b + \frac {1}{48} \, {\left ({\left (3 \, c^{2} \log \left (c + x\right )^{2} + 3 \, c^{2} \log \left (-c + x\right )^{2} + 16 \, c^{2} \log \left (c + x\right ) + 4 \, x^{2} - 2 \, {\left (3 \, c^{2} \log \left (c + x\right ) - 8 \, c^{2}\right )} \log \left (-c + x\right )\right )} c^{2} - 4 \, {\left (3 \, c^{3} \log \left (c + x\right ) - 3 \, c^{3} \log \left (-c + x\right ) - 6 \, c^{2} x - 2 \, x^{3}\right )} c \operatorname {artanh}\left (\frac {c}{x}\right )\right )} b^{2} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 552 vs. \(2 (109) = 218\).

Time = 0.13 (sec) , antiderivative size = 552, normalized size of antiderivative = 4.49 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.61 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^2\,x^4}{4}-\frac {b^2\,c^4\,{\mathrm {atanh}\left (\frac {c}{x}\right )}^2}{4}+\frac {b^2\,x^4\,{\mathrm {atanh}\left (\frac {c}{x}\right )}^2}{4}+\frac {b^2\,c^4\,\ln \left (x^2-c^2\right )}{3}+\frac {b^2\,c^2\,x^2}{12}+\frac {b^2\,c\,x^3\,\mathrm {atanh}\left (\frac {c}{x}\right )}{6}+\frac {b^2\,c^3\,x\,\mathrm {atanh}\left (\frac {c}{x}\right )}{2}+\frac {a\,b\,c\,x^3}{6}+\frac {a\,b\,c^3\,x}{2}-\frac {a\,b\,c^4\,\mathrm {atanh}\left (\frac {c}{x}\right )}{2}+\frac {a\,b\,x^4\,\mathrm {atanh}\left (\frac {c}{x}\right )}{2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.25 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )^2 \, dx=-\frac {\mathit {atanh} \left (\frac {c}{x}\right )^{2} b^{2} c^{4}}{4}+\frac {\mathit {atanh} \left (\frac {c}{x}\right )^{2} b^{2} x^{4}}{4}-\frac {\mathit {atanh} \left (\frac {c}{x}\right ) a b \,c^{4}}{2}+\frac {\mathit {atanh} \left (\frac {c}{x}\right ) a b \,x^{4}}{2}-\frac {2 \mathit {atanh} \left (\frac {c}{x}\right ) b^{2} c^{4}}{3}+\frac {\mathit {atanh} \left (\frac {c}{x}\right ) b^{2} c^{3} x}{2}+\frac {\mathit {atanh} \left (\frac {c}{x}\right ) b^{2} c \,x^{3}}{6}+\frac {2 \,\mathrm {log}\left (-c -x \right ) b^{2} c^{4}}{3}+\frac {a^{2} x^{4}}{4}+\frac {a b \,c^{3} x}{2}+\frac {a b c \,x^{3}}{6}+\frac {b^{2} c^{2} x^{2}}{12} \] Input:

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

Output:

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