\(\int \frac {(a+b \text {arctanh}(c x^3))^2}{x^4} \, dx\) [121]

Optimal result

Integrand size = 16, antiderivative size = 90 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\frac {1}{3} c \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^3}+\frac {2}{3} b c \left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1+c x^3}\right )-\frac {1}{3} b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x^3}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\frac {b^2 \left (-1+c x^3\right ) \text {arctanh}\left (c x^3\right )^2+2 b \text {arctanh}\left (c x^3\right ) \left (-a+b c x^3 \log \left (1-e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )-a \left (a-2 b c x^3 \log \left (c x^3\right )+b c x^3 \log \left (1-c^2 x^6\right )\right )-b^2 c x^3 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c x^3\right )}\right )}{3 x^3} \] Input:

Integrate[(a + b*ArcTanh[c*x^3])^2/x^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6454, 6452, 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.

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

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6550

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

\(\Big \downarrow \) 6494

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

\(\Big \downarrow \) 2897

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

Input:

Int[(a + b*ArcTanh[c*x^3])^2/x^4,x]
 

Output:

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

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 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 = 0.87 (sec) , antiderivative size = 2993, normalized size of antiderivative = 33.26

method	result	size
default	\(\text {Expression too large to display}\)	\(2993\)
parts	\(\text {Expression too large to display}\)	\(2993\)

Input:

int((a+b*arctanh(c*x^3))^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*a^2/x^3+b^2*(-1/3/x^3*arctanh(c*x^3)^2+2*c*(arctanh(c*x^3)*ln(x)-1/6* 
arctanh(c*x^3)*ln(c*x^3-1)-1/6*arctanh(c*x^3)*ln(c*x^3+1)-1/2*c*(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)/Roo 
tOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+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*_alpha+3*_alpha 
^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*_alpha+6*ln((Root 
Of(_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)/RootOf(_Z^2+3*_Z 
*_alpha+3*_alpha^2,index=1))*_alpha^2+2*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alp 
ha^2,index=2)-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*RootO 
f(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2, 
index=2)+6*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)-x+_alpha)/RootO 
f(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2 
,index=1)*_alpha+3*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)-x+_alph 
a)/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*_alpha^2,index=2) 
-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*_alpha^2)/(3*_a...
 

Fricas [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^3))^2/x^4,x, algorithm="fricas")
 

Output:

integral((b^2*arctanh(c*x^3)^2 + 2*a*b*arctanh(c*x^3) + a^2)/x^4, x)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\text {Timed out} \] Input:

integrate((a+b*atanh(c*x**3))**2/x**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^3))^2/x^4,x, algorithm="maxima")
 

Output:

-1/3*(c*(log(c^2*x^6 - 1) - log(x^6)) + 2*arctanh(c*x^3)/x^3)*a*b - 1/12*b 
^2*(log(-c*x^3 + 1)^2/x^3 + 3*integrate(-((c*x^3 - 1)*log(c*x^3 + 1)^2 + 2 
*(c*x^3 - (c*x^3 - 1)*log(c*x^3 + 1))*log(-c*x^3 + 1))/(c*x^7 - x^4), x)) 
- 1/3*a^2/x^3
 

Giac [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^3))^2/x^4,x, algorithm="giac")
 

Output:

integrate((b*arctanh(c*x^3) + a)^2/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^2}{x^4} \,d x \] Input:

int((a + b*atanh(c*x^3))^2/x^4,x)
 

Output:

int((a + b*atanh(c*x^3))^2/x^4, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\frac {-\mathit {atanh} \left (c \,x^{3}\right )^{2} b^{2}+2 \mathit {atanh} \left (c \,x^{3}\right ) a b c \,x^{3}-2 \mathit {atanh} \left (c \,x^{3}\right ) a b -6 \left (\int \frac {\mathit {atanh} \left (c \,x^{3}\right )}{c^{2} x^{7}-x}d x \right ) b^{2} c \,x^{3}-2 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) a b c \,x^{3}-2 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) a b c \,x^{3}+6 \,\mathrm {log}\left (x \right ) a b c \,x^{3}-a^{2}}{3 x^{3}} \] Input:

int((a+b*atanh(c*x^3))^2/x^4,x)
 

Output:

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