\(\int \frac {(a+b \text {arctanh}(c x^2))^2}{x^5} \, dx\) [70]

Optimal result

Integrand size = 16, antiderivative size = 88 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^5} \, dx=-\frac {b c \left (a+b \text {arctanh}\left (c x^2\right )\right )}{2 x^2}+\frac {1}{4} c^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{4 x^4}+b^2 c^2 \log (x)-\frac {1}{4} b^2 c^2 \log \left (1-c^2 x^4\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^5} \, dx=\frac {1}{4} \left (-\frac {a^2}{x^4}-\frac {2 a b c}{x^2}-\frac {2 b \left (a+b c x^2\right ) \text {arctanh}\left (c x^2\right )}{x^4}+\frac {b^2 \left (-1+c^2 x^4\right ) \text {arctanh}\left (c x^2\right )^2}{x^4}+4 b^2 c^2 \log (x)-b (a+b) c^2 \log \left (1-c x^2\right )+(a-b) b c^2 \log \left (1+c x^2\right )\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, 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.

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

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 6510

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

Input:

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

Output:

(-1/2*(a + b*ArcTanh[c*x^2])^2/x^4 + b*c*(-((a + b*ArcTanh[c*x^2])/x^2) + 
(c*(a + b*ArcTanh[c*x^2])^2)/(2*b) + (b*c*(Log[x^4] - Log[1 - c^2*x^4]))/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 = 0.53 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.73

method	result	size
parallelrisch	\(\frac {b^{2} \operatorname {arctanh}\left (c \,x^{2}\right )^{2} x^{4} c^{2}+4 b^{2} c^{2} \ln \left (x \right ) x^{4}-2 \ln \left (c \,x^{2}-1\right ) b^{2} c^{2} x^{4}+2 x^{4} \operatorname {arctanh}\left (c \,x^{2}\right ) a b \,c^{2}-2 \,\operatorname {arctanh}\left (c \,x^{2}\right ) x^{4} b^{2} c^{2}-a^{2} c^{2} x^{4}-2 b^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{2} c -2 a b c \,x^{2}-b^{2} \operatorname {arctanh}\left (c \,x^{2}\right )^{2}-2 \,\operatorname {arctanh}\left (c \,x^{2}\right ) a b -a^{2}}{4 x^{4}}\)	\(152\)
risch	\(\frac {b^{2} \left (c^{2} x^{4}-1\right ) \ln \left (c \,x^{2}+1\right )^{2}}{16 x^{4}}-\frac {b \left (b \,c^{2} \ln \left (-c \,x^{2}+1\right ) x^{4}+2 b c \,x^{2}-b \ln \left (-c \,x^{2}+1\right )+2 a \right ) \ln \left (c \,x^{2}+1\right )}{8 x^{4}}+\frac {b^{2} c^{2} x^{4} \ln \left (-c \,x^{2}+1\right )^{2}+4 b \,c^{2} \ln \left (c \,x^{2}+1\right ) x^{4} a -4 \ln \left (c \,x^{2}+1\right ) b^{2} c^{2} x^{4}-4 b \,c^{2} \ln \left (c \,x^{2}-1\right ) x^{4} a -4 \ln \left (c \,x^{2}-1\right ) b^{2} c^{2} x^{4}+16 b^{2} c^{2} \ln \left (x \right ) x^{4}+4 b^{2} c \,x^{2} \ln \left (-c \,x^{2}+1\right )-8 a b c \,x^{2}-b^{2} \ln \left (-c \,x^{2}+1\right )^{2}+4 b \ln \left (-c \,x^{2}+1\right ) a -4 a^{2}}{16 x^{4}}\)	\(257\)
default	\(\text {Expression too large to display}\)	\(758\)
parts	\(\text {Expression too large to display}\)	\(758\)

Input:

int((a+b*arctanh(c*x^2))^2/x^5,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.72 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^5} \, dx=\frac {16 \, b^{2} c^{2} x^{4} \log \left (x\right ) + 4 \, {\left (a b - b^{2}\right )} c^{2} x^{4} \log \left (c x^{2} + 1\right ) - 4 \, {\left (a b + b^{2}\right )} c^{2} x^{4} \log \left (c x^{2} - 1\right ) - 8 \, a b c x^{2} + {\left (b^{2} c^{2} x^{4} - b^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2} - 4 \, a^{2} - 4 \, {\left (b^{2} c x^{2} + a b\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{16 \, x^{4}} \] Input:

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

Output:

1/16*(16*b^2*c^2*x^4*log(x) + 4*(a*b - b^2)*c^2*x^4*log(c*x^2 + 1) - 4*(a* 
b + b^2)*c^2*x^4*log(c*x^2 - 1) - 8*a*b*c*x^2 + (b^2*c^2*x^4 - b^2)*log(-( 
c*x^2 + 1)/(c*x^2 - 1))^2 - 4*a^2 - 4*(b^2*c*x^2 + a*b)*log(-(c*x^2 + 1)/( 
c*x^2 - 1)))/x^4
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (80) = 160\).

Time = 6.56 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.99 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^5} \, dx=\begin {cases} - \frac {a^{2}}{4 x^{4}} + \frac {a b c^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2} - \frac {a b c}{2 x^{2}} - \frac {a b \operatorname {atanh}{\left (c x^{2} \right )}}{2 x^{4}} + b^{2} c^{2} \log {\left (x \right )} - \frac {b^{2} c^{2} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{2} - \frac {b^{2} c^{2} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{2} + \frac {b^{2} c^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4} + \frac {b^{2} c^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2} - \frac {b^{2} c \operatorname {atanh}{\left (c x^{2} \right )}}{2 x^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a**2/(4*x**4) + a*b*c**2*atanh(c*x**2)/2 - a*b*c/(2*x**2) - a* 
b*atanh(c*x**2)/(2*x**4) + b**2*c**2*log(x) - b**2*c**2*log(x - sqrt(-1/c) 
)/2 - b**2*c**2*log(x + sqrt(-1/c))/2 + b**2*c**2*atanh(c*x**2)**2/4 + b** 
2*c**2*atanh(c*x**2)/2 - b**2*c*atanh(c*x**2)/(2*x**2) - b**2*atanh(c*x**2 
)**2/(4*x**4), Ne(c, 0)), (-a**2/(4*x**4), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (80) = 160\).

Time = 0.03 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.99 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^5} \, dx=\frac {1}{4} \, {\left ({\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac {2}{x^{2}}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{4}}\right )} a b + \frac {1}{16} \, {\left ({\left (2 \, {\left (\log \left (c x^{2} - 1\right ) - 2\right )} \log \left (c x^{2} + 1\right ) - \log \left (c x^{2} + 1\right )^{2} - \log \left (c x^{2} - 1\right )^{2} - 4 \, \log \left (c x^{2} - 1\right ) + 16 \, \log \left (x\right )\right )} c^{2} + 4 \, {\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac {2}{x^{2}}\right )} c \operatorname {artanh}\left (c x^{2}\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x^{2}\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.96 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.16 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^5} \, dx=\frac {b^2\,c^2\,{\ln \left (c\,x^2+1\right )}^2}{16}-\frac {b^2\,c^2\,\ln \left (c\,x^2-1\right )}{4}-\frac {b^2\,c^2\,\ln \left (c\,x^2+1\right )}{4}-\frac {a^2}{4\,x^4}+\frac {b^2\,c^2\,{\ln \left (1-c\,x^2\right )}^2}{16}-\frac {b^2\,{\ln \left (c\,x^2+1\right )}^2}{16\,x^4}-\frac {b^2\,{\ln \left (1-c\,x^2\right )}^2}{16\,x^4}+b^2\,c^2\,\ln \left (x\right )-\frac {a\,b\,c^2\,\ln \left (c\,x^2-1\right )}{4}+\frac {a\,b\,c^2\,\ln \left (c\,x^2+1\right )}{4}-\frac {a\,b\,c}{2\,x^2}-\frac {a\,b\,\ln \left (c\,x^2+1\right )}{4\,x^4}+\frac {a\,b\,\ln \left (1-c\,x^2\right )}{4\,x^4}-\frac {b^2\,c^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8}-\frac {b^2\,c\,\ln \left (c\,x^2+1\right )}{4\,x^2}+\frac {b^2\,c\,\ln \left (1-c\,x^2\right )}{4\,x^2}+\frac {b^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8\,x^4} \] Input:

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

Output:

(b^2*c^2*log(c*x^2 + 1)^2)/16 - (b^2*c^2*log(c*x^2 - 1))/4 - (b^2*c^2*log( 
c*x^2 + 1))/4 - a^2/(4*x^4) + (b^2*c^2*log(1 - c*x^2)^2)/16 - (b^2*log(c*x 
^2 + 1)^2)/(16*x^4) - (b^2*log(1 - c*x^2)^2)/(16*x^4) + b^2*c^2*log(x) - ( 
a*b*c^2*log(c*x^2 - 1))/4 + (a*b*c^2*log(c*x^2 + 1))/4 - (a*b*c)/(2*x^2) - 
 (a*b*log(c*x^2 + 1))/(4*x^4) + (a*b*log(1 - c*x^2))/(4*x^4) - (b^2*c^2*lo 
g(c*x^2 + 1)*log(1 - c*x^2))/8 - (b^2*c*log(c*x^2 + 1))/(4*x^2) + (b^2*c*l 
og(1 - c*x^2))/(4*x^2) + (b^2*log(c*x^2 + 1)*log(1 - c*x^2))/(8*x^4)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^5} \, dx=\frac {\mathit {atanh} \left (c \,x^{2}\right )^{2} b^{2} c^{2} x^{4}-\mathit {atanh} \left (c \,x^{2}\right )^{2} b^{2}+2 \mathit {atanh} \left (c \,x^{2}\right ) a b \,c^{2} x^{4}-2 \mathit {atanh} \left (c \,x^{2}\right ) a b +2 \mathit {atanh} \left (c \,x^{2}\right ) b^{2} c^{2} x^{4}-2 \mathit {atanh} \left (c \,x^{2}\right ) b^{2} c \,x^{2}-2 \,\mathrm {log}\left (c \,x^{2}+1\right ) b^{2} c^{2} x^{4}+4 \,\mathrm {log}\left (x \right ) b^{2} c^{2} x^{4}-a^{2}-2 a b c \,x^{2}}{4 x^{4}} \] Input:

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

Output:

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