\(\int \frac {(a+b \text {arctanh}(c x^2))^3}{x^5} \, dx\) [81]

Optimal result

Integrand size = 16, antiderivative size = 139 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\frac {3}{4} b c^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {3 b c \left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{4 x^2}+\frac {1}{4} c^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{4 x^4}+\frac {3}{2} b^2 c^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \log \left (2-\frac {2}{1+c x^2}\right )-\frac {3}{4} b^3 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x^2}\right ) \] Output:

3/4*b*c^2*(a+b*arctanh(c*x^2))^2-3/4*b*c*(a+b*arctanh(c*x^2))^2/x^2+1/4*c^ 
2*(a+b*arctanh(c*x^2))^3-1/4*(a+b*arctanh(c*x^2))^3/x^4+3/2*b^2*c^2*(a+b*a 
rctanh(c*x^2))*ln(2-2/(c*x^2+1))-3/4*b^3*c^2*polylog(2,-1+2/(c*x^2+1))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\frac {6 b^2 \left (-1+c x^2\right ) \left (a+a c x^2+b c x^2\right ) \text {arctanh}\left (c x^2\right )^2+2 b^3 \left (-1+c^2 x^4\right ) \text {arctanh}\left (c x^2\right )^3-6 b \text {arctanh}\left (c x^2\right ) \left (a^2+2 a b c x^2-2 b^2 c^2 x^4 \log \left (1-e^{-2 \text {arctanh}\left (c x^2\right )}\right )\right )+a \left (-2 a^2-6 a b c x^2-3 a b c^2 x^4 \log \left (1-c x^2\right )+3 a b c^2 x^4 \log \left (1+c x^2\right )+12 b^2 c^2 x^4 \log \left (\frac {c x^2}{\sqrt {1-c^2 x^4}}\right )\right )-6 b^3 c^2 x^4 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c x^2\right )}\right )}{8 x^4} \] Input:

Integrate[(a + b*ArcTanh[c*x^2])^3/x^5,x]
 

Output:

(6*b^2*(-1 + c*x^2)*(a + a*c*x^2 + b*c*x^2)*ArcTanh[c*x^2]^2 + 2*b^3*(-1 + 
 c^2*x^4)*ArcTanh[c*x^2]^3 - 6*b*ArcTanh[c*x^2]*(a^2 + 2*a*b*c*x^2 - 2*b^2 
*c^2*x^4*Log[1 - E^(-2*ArcTanh[c*x^2])]) + a*(-2*a^2 - 6*a*b*c*x^2 - 3*a*b 
*c^2*x^4*Log[1 - c*x^2] + 3*a*b*c^2*x^4*Log[1 + c*x^2] + 12*b^2*c^2*x^4*Lo 
g[(c*x^2)/Sqrt[1 - c^2*x^4]]) - 6*b^3*c^2*x^4*PolyLog[2, E^(-2*ArcTanh[c*x 
^2])])/(8*x^4)
 

Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6454, 6452, 6544, 6452, 6510, 6550, 6494, 2897}

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.

\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx\)

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6510

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

\(\Big \downarrow \) 6550

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

\(\Big \downarrow \) 6494

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

\(\Big \downarrow \) 2897

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

Input:

Int[(a + b*ArcTanh[c*x^2])^3/x^5,x]
 

Output:

(-1/2*(a + b*ArcTanh[c*x^2])^3/x^4 + (3*b*c*(-((a + b*ArcTanh[c*x^2])^2/x^ 
2) + (c*(a + b*ArcTanh[c*x^2])^3)/(3*b) + 2*b*c*((a + b*ArcTanh[c*x^2])^2/ 
(2*b) + (a + b*ArcTanh[c*x^2])*Log[2 - 2/(1 + c*x^2)] - (b*PolyLog[2, -1 + 
 2/(1 + c*x^2)])/2)))/2)/2
 

Defintions of rubi rules used

rule 2897

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ 
D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && 
PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

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 6494

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

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]
 

rule 6550

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

Maple [F]

\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (c \,x^{2}\right )\right )}^{3}}{x^{5}}d x\]

Input:

int((a+b*arctanh(c*x^2))^3/x^5,x)
 

Output:

int((a+b*arctanh(c*x^2))^3/x^5,x)
 

Fricas [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^2))^3/x^5,x, algorithm="fricas")
 

Output:

integral((b^3*arctanh(c*x^2)^3 + 3*a*b^2*arctanh(c*x^2)^2 + 3*a^2*b*arctan 
h(c*x^2) + a^3)/x^5, x)
 

Sympy [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}}{x^{5}}\, dx \] Input:

integrate((a+b*atanh(c*x**2))**3/x**5,x)
 

Output:

Integral((a + b*atanh(c*x**2))**3/x**5, x)
 

Maxima [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^2))^3/x^5,x, algorithm="maxima")
 

Output:

3/8*((c*log(c*x^2 + 1) - c*log(c*x^2 - 1) - 2/x^2)*c - 2*arctanh(c*x^2)/x^ 
4)*a^2*b + 3/16*((2*(log(c*x^2 - 1) - 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 
 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1) + 16*log(x))*c^2 + 4*(c*log(c*x^2 + 
 1) - c*log(c*x^2 - 1) - 2/x^2)*c*arctanh(c*x^2))*a*b^2 - 1/32*b^3*(((c^2* 
x^4 - 1)*log(-c*x^2 + 1)^3 + 3*(2*c*x^2 - (c^2*x^4 - 1)*log(c*x^2 + 1))*lo 
g(-c*x^2 + 1)^2)/x^4 + 4*integrate(-((c*x^2 - 1)*log(c*x^2 + 1)^3 + 3*(2*c 
^2*x^4 - (c*x^2 - 1)*log(c*x^2 + 1)^2 - (c^3*x^6 - c*x^2)*log(c*x^2 + 1))* 
log(-c*x^2 + 1))/(c*x^7 - x^5), x)) - 3/4*a*b^2*arctanh(c*x^2)^2/x^4 - 1/4 
*a^3/x^4
 

Giac [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^2))^3/x^5,x, algorithm="giac")
 

Output:

integrate((b*arctanh(c*x^2) + a)^3/x^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3}{x^5} \,d x \] Input:

int((a + b*atanh(c*x^2))^3/x^5,x)
 

Output:

int((a + b*atanh(c*x^2))^3/x^5, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\frac {\mathit {atanh} \left (c \,x^{2}\right )^{3} b^{3} c^{2} x^{4}-\mathit {atanh} \left (c \,x^{2}\right )^{3} b^{3}+3 \mathit {atanh} \left (c \,x^{2}\right )^{2} a \,b^{2} c^{2} x^{4}-3 \mathit {atanh} \left (c \,x^{2}\right )^{2} a \,b^{2}-3 \mathit {atanh} \left (c \,x^{2}\right )^{2} b^{3} c \,x^{2}+3 \mathit {atanh} \left (c \,x^{2}\right ) a^{2} b \,c^{2} x^{4}-3 \mathit {atanh} \left (c \,x^{2}\right ) a^{2} b +6 \mathit {atanh} \left (c \,x^{2}\right ) a \,b^{2} c^{2} x^{4}-6 \mathit {atanh} \left (c \,x^{2}\right ) a \,b^{2} c \,x^{2}-3 \mathit {atanh} \left (c \,x^{2}\right ) b^{3} c^{2} x^{4}+3 \mathit {atanh} \left (c \,x^{2}\right ) b^{3}-12 \left (\int \frac {\mathit {atanh} \left (c \,x^{2}\right )}{c^{2} x^{9}-x^{5}}d x \right ) b^{3} x^{4}-6 \,\mathrm {log}\left (c \,x^{2}+1\right ) a \,b^{2} c^{2} x^{4}+12 \,\mathrm {log}\left (x \right ) a \,b^{2} c^{2} x^{4}-a^{3}-3 a^{2} b c \,x^{2}+3 b^{3} c \,x^{2}}{4 x^{4}} \] Input:

int((a+b*atanh(c*x^2))^3/x^5,x)
 

Output:

(atanh(c*x**2)**3*b**3*c**2*x**4 - atanh(c*x**2)**3*b**3 + 3*atanh(c*x**2) 
**2*a*b**2*c**2*x**4 - 3*atanh(c*x**2)**2*a*b**2 - 3*atanh(c*x**2)**2*b**3 
*c*x**2 + 3*atanh(c*x**2)*a**2*b*c**2*x**4 - 3*atanh(c*x**2)*a**2*b + 6*at 
anh(c*x**2)*a*b**2*c**2*x**4 - 6*atanh(c*x**2)*a*b**2*c*x**2 - 3*atanh(c*x 
**2)*b**3*c**2*x**4 + 3*atanh(c*x**2)*b**3 - 12*int(atanh(c*x**2)/(c**2*x* 
*9 - x**5),x)*b**3*x**4 - 6*log(c*x**2 + 1)*a*b**2*c**2*x**4 + 12*log(x)*a 
*b**2*c**2*x**4 - a**3 - 3*a**2*b*c*x**2 + 3*b**3*c*x**2)/(4*x**4)