\(\int \frac {(a+b \text {arctanh}(c x^3))^2}{x^{10}} \, dx\) [123]

Optimal result

Integrand size = 16, antiderivative size = 144 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=-\frac {b^2 c^2}{9 x^3}+\frac {1}{9} b^2 c^3 \text {arctanh}\left (c x^3\right )-\frac {b c \left (a+b \text {arctanh}\left (c x^3\right )\right )}{9 x^6}+\frac {1}{9} c^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{9 x^9}+\frac {2}{9} b c^3 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1+c x^3}\right )-\frac {1}{9} b^2 c^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x^3}\right ) \] Output:

-1/9*b^2*c^2/x^3+1/9*b^2*c^3*arctanh(c*x^3)-1/9*b*c*(a+b*arctanh(c*x^3))/x 
^6+1/9*c^3*(a+b*arctanh(c*x^3))^2-1/9*(a+b*arctanh(c*x^3))^2/x^9+2/9*b*c^3 
*(a+b*arctanh(c*x^3))*ln(2-2/(c*x^3+1))-1/9*b^2*c^3*polylog(2,-1+2/(c*x^3+ 
1))
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=-\frac {a^2+a b c x^3+b^2 c^2 x^6+b^2 \left (1-c^3 x^9\right ) \text {arctanh}\left (c x^3\right )^2+b \text {arctanh}\left (c x^3\right ) \left (2 a+b c x^3-b c^3 x^9-2 b c^3 x^9 \log \left (1-e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )-2 a b c^3 x^9 \log \left (c x^3\right )+a b c^3 x^9 \log \left (1-c^2 x^6\right )+b^2 c^3 x^9 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c x^3\right )}\right )}{9 x^9} \] Input:

Integrate[(a + b*ArcTanh[c*x^3])^2/x^10,x]
 

Output:

-1/9*(a^2 + a*b*c*x^3 + b^2*c^2*x^6 + b^2*(1 - c^3*x^9)*ArcTanh[c*x^3]^2 + 
 b*ArcTanh[c*x^3]*(2*a + b*c*x^3 - b*c^3*x^9 - 2*b*c^3*x^9*Log[1 - E^(-2*A 
rcTanh[c*x^3])]) - 2*a*b*c^3*x^9*Log[c*x^3] + a*b*c^3*x^9*Log[1 - c^2*x^6] 
 + b^2*c^3*x^9*PolyLog[2, E^(-2*ArcTanh[c*x^3])])/x^9
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {6454, 6452, 6544, 6452, 264, 219, 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.

\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx\)

\(\Big \downarrow \) 6454

\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{12}}dx^3\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^9 \left (1-c^2 x^6\right )}dx^3-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 6544

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (1-c^2 x^6\right )}dx^3+\int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^9}dx^3\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (1-c^2 x^6\right )}dx^3+\frac {1}{2} b c \int \frac {1}{x^6 \left (1-c^2 x^6\right )}dx^3-\frac {a+b \text {arctanh}\left (c x^3\right )}{2 x^6}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (1-c^2 x^6\right )}dx^3+\frac {1}{2} b c \left (c^2 \int \frac {1}{1-c^2 x^6}dx^3-\frac {1}{x^3}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{2 x^6}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (1-c^2 x^6\right )}dx^3-\frac {a+b \text {arctanh}\left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (c \text {arctanh}\left (c x^3\right )-\frac {1}{x^3}\right )\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 6550

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \left (\int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (c x^3+1\right )}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (c \text {arctanh}\left (c x^3\right )-\frac {1}{x^3}\right )\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 6494

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \left (-b c \int \frac {\log \left (2-\frac {2}{c x^3+1}\right )}{1-c^2 x^6}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (c \text {arctanh}\left (c x^3\right )-\frac {1}{x^3}\right )\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

\(\Big \downarrow \) 2897

\(\displaystyle \frac {1}{3} \left (\frac {2}{3} b c \left (c^2 \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x^3+1}-1\right )\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{2 x^6}+\frac {1}{2} b c \left (c \text {arctanh}\left (c x^3\right )-\frac {1}{x^3}\right )\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^9}\right )\)

Input:

Int[(a + b*ArcTanh[c*x^3])^2/x^10,x]
 

Output:

(-1/3*(a + b*ArcTanh[c*x^3])^2/x^9 + (2*b*c*(-1/2*(a + b*ArcTanh[c*x^3])/x 
^6 + (b*c*(-x^(-3) + c*ArcTanh[c*x^3]))/2 + c^2*((a + b*ArcTanh[c*x^3])^2/ 
(2*b) + (a + b*ArcTanh[c*x^3])*Log[2 - 2/(1 + c*x^3)] - (b*PolyLog[2, -1 + 
 2/(1 + c*x^3)])/2)))/3)/3
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

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 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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.24 (sec) , antiderivative size = 3062, normalized size of antiderivative = 21.26

method	result	size
default	\(\text {Expression too large to display}\)	\(3062\)
parts	\(\text {Expression too large to display}\)	\(3062\)

Input:

int((a+b*arctanh(c*x^3))^2/x^10,x,method=_RETURNVERBOSE)
 

Output:

-1/9*a^2/x^9+b^2*(-1/9/x^9*arctanh(c*x^3)^2+2/3*c*(-1/6/x^6*arctanh(c*x^3) 
+arctanh(c*x^3)*c^2*ln(x)-1/6*arctanh(c*x^3)*c^2*ln(c*x^3-1)-1/6*arctanh(c 
*x^3)*c^2*ln(c*x^3+1)-1/2*c*(1/6*c*ln(c*x^3-1)+1/3/x^3-1/6*c*ln(c*x^3+1)+c 
^2*(Sum(1/6*(ln(x-_alpha)*ln(c*x^3-1)-3*c*(1/6/_alpha^2/c*ln(x-_alpha)^2-1 
/3*_alpha*ln(x-_alpha)*(2*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)- 
x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_a 
lpha+3*_alpha^2,index=1)*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)+3*ln( 
(RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+3*_Z*_a 
lpha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*_alp 
ha+6*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2 
+3*_Z*_alpha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index 
=2)*_alpha+9*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/Roo 
tOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1))*_alpha^2+2*ln((RootOf(_Z^2+3*_Z* 
_alpha+3*_alpha^2,index=2)-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,in 
dex=2))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*RootOf(_Z^2+3*_Z*_alph 
a+3*_alpha^2,index=2)+6*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)-x+ 
_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*RootOf(_Z^2+3*_Z*_alp 
ha+3*_alpha^2,index=1)*_alpha+3*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,ind 
ex=2)-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*RootOf(_Z^2+3 
*_Z*_alpha+3*_alpha^2,index=2)*_alpha+9*ln((RootOf(_Z^2+3*_Z*_alpha+3*_...
 

Fricas [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{10}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^3))^2/x^10,x, algorithm="fricas")
 

Output:

integral((b^2*arctanh(c*x^3)^2 + 2*a*b*arctanh(c*x^3) + a^2)/x^10, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=\text {Timed out} \] Input:

integrate((a+b*atanh(c*x**3))**2/x**10,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{10}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^3))^2/x^10,x, algorithm="maxima")
 

Output:

-1/9*((c^2*log(c^2*x^6 - 1) - c^2*log(x^6) + 1/x^6)*c + 2*arctanh(c*x^3)/x 
^9)*a*b - 1/36*b^2*(log(-c*x^3 + 1)^2/x^9 + 9*integrate(-1/3*(3*(c*x^3 - 1 
)*log(c*x^3 + 1)^2 + 2*(c*x^3 - 3*(c*x^3 - 1)*log(c*x^3 + 1))*log(-c*x^3 + 
 1))/(c*x^13 - x^10), x)) - 1/9*a^2/x^9
 

Giac [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{10}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^3))^2/x^10,x, algorithm="giac")
 

Output:

integrate((b*arctanh(c*x^3) + a)^2/x^10, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^2}{x^{10}} \,d x \] Input:

int((a + b*atanh(c*x^3))^2/x^10,x)
 

Output:

int((a + b*atanh(c*x^3))^2/x^10, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^{10}} \, dx=\frac {-\mathit {atanh} \left (c \,x^{3}\right )^{2} b^{2}+2 \mathit {atanh} \left (c \,x^{3}\right ) a b \,c^{3} x^{9}-2 \mathit {atanh} \left (c \,x^{3}\right ) a b +\mathit {atanh} \left (c \,x^{3}\right ) b^{2} c^{3} x^{9}-\mathit {atanh} \left (c \,x^{3}\right ) b^{2} c \,x^{3}-6 \left (\int \frac {\mathit {atanh} \left (c \,x^{3}\right )}{c^{2} x^{7}-x}d x \right ) b^{2} c^{3} x^{9}-2 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) a b \,c^{3} x^{9}-2 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) a b \,c^{3} x^{9}+6 \,\mathrm {log}\left (x \right ) a b \,c^{3} x^{9}-a^{2}-a b c \,x^{3}-b^{2} c^{2} x^{6}}{9 x^{9}} \] Input:

int((a+b*atanh(c*x^3))^2/x^10,x)
 

Output:

( - atanh(c*x**3)**2*b**2 + 2*atanh(c*x**3)*a*b*c**3*x**9 - 2*atanh(c*x**3 
)*a*b + atanh(c*x**3)*b**2*c**3*x**9 - atanh(c*x**3)*b**2*c*x**3 - 6*int(a 
tanh(c*x**3)/(c**2*x**7 - x),x)*b**2*c**3*x**9 - 2*log(c**(2/3)*x**2 - c** 
(1/3)*x + 1)*a*b*c**3*x**9 - 2*log(c**(2/3)*x + c**(1/3))*a*b*c**3*x**9 + 
6*log(x)*a*b*c**3*x**9 - a**2 - a*b*c*x**3 - b**2*c**2*x**6)/(9*x**9)