\(\int (a+b \text {arctanh}(c x)) (d+e \log (1-c^2 x^2)) \, dx\) [525]

Optimal result

Integrand size = 24, antiderivative size = 104 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-2 a e x-2 b e x \text {arctanh}(c x)+\frac {e (a+b \text {arctanh}(c x))^2}{b c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+x (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.38 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=a d x-2 a e x+\frac {2 a e \text {arctanh}(c x)}{c}+b d x \text {arctanh}(c x)-2 b e x \text {arctanh}(c x)+\frac {b e \text {arctanh}(c x)^2}{c}+\frac {b d \log \left (1-c^2 x^2\right )}{2 c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+a e x \log \left (1-c^2 x^2\right )+b e x \text {arctanh}(c x) \log \left (1-c^2 x^2\right )+\frac {b e \log ^2\left (1-c^2 x^2\right )}{4 c} \] Input:

Integrate[(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {6635, 2925, 2837, 2738, 6542, 2009, 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.

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

\(\Big \downarrow \) 6635

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

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2837

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

\(\Big \downarrow \) 2738

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

\(\Big \downarrow \) 6542

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6510

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

Input:

Int[(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]),x]
 

Output:

x*(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]) + (b*(d + e*Log[1 - c^2*x^ 
2])^2)/(4*c*e) + 2*c^2*e*((a + b*ArcTanh[c*x])^2/(2*b*c^3) - (a*x + b*x*Ar 
cTanh[c*x] + (b*Log[1 - c^2*x^2])/(2*c))/c^2)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2738

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Lo 
g[c*x^n])^2/(2*b*n), x] /; FreeQ[{a, b, c, n}, x]
 

rule 2837

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Simp[1/e   Subst[Int[(f*(x/d))^q*(a + b*Log[c*x 
^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x] && 
EqQ[e*f - d*g, 0]
 

rule 2925

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Si 
mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], 
x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer 
Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 
] || IGtQ[q, 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 6542

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( 
e_.)*(x_)^2), x_Symbol] :> Simp[f^2/e   Int[(f*x)^(m - 2)*(a + b*ArcTanh[c* 
x])^p, x], x] - Simp[d*(f^2/e)   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] && GtQ[m, 
 1]
 

rule 6635

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]* 
(e_.)), x_Symbol] :> Simp[x*(d + e*Log[f + g*x^2])*(a + b*ArcTanh[c*x]), x] 
 + (-Simp[b*c   Int[x*((d + e*Log[f + g*x^2])/(1 - c^2*x^2)), x], x] - Simp 
[2*e*g   Int[x^2*((a + b*ArcTanh[c*x])/(f + g*x^2)), x], x]) /; FreeQ[{a, b 
, c, d, e, f, g}, x]
 

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\frac {4 x \,\operatorname {arctanh}\left (c x \right ) \ln \left (-c^{2} x^{2}+1\right ) b e c +4 b c d x \,\operatorname {arctanh}\left (c x \right )-8 e b \,\operatorname {arctanh}\left (c x \right ) x c +4 x \ln \left (-c^{2} x^{2}+1\right ) a e c +4 a d x c -8 a c e x +4 e b \operatorname {arctanh}\left (c x \right )^{2}+e b \ln \left (-c^{2} x^{2}+1\right )^{2}+8 \,\operatorname {arctanh}\left (c x \right ) a e +2 b d \ln \left (-c^{2} x^{2}+1\right )-4 \ln \left (-c^{2} x^{2}+1\right ) b e}{4 c}\)	\(141\)
default	\(\text {Expression too large to display}\)	\(2431\)
parts	\(\text {Expression too large to display}\)	\(2431\)
risch	\(\text {Expression too large to display}\)	\(3552\)

Input:

int((a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1)),x,method=_RETURNVERBOSE)
 

Output:

1/4*(4*x*arctanh(c*x)*ln(-c^2*x^2+1)*b*e*c+4*b*c*d*x*arctanh(c*x)-8*e*b*ar 
ctanh(c*x)*x*c+4*x*ln(-c^2*x^2+1)*a*e*c+4*a*d*x*c-8*a*c*e*x+4*e*b*arctanh( 
c*x)^2+e*b*ln(-c^2*x^2+1)^2+8*arctanh(c*x)*a*e+2*b*d*ln(-c^2*x^2+1)-4*ln(- 
c^2*x^2+1)*b*e)/c
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.27 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {b e \log \left (-c^{2} x^{2} + 1\right )^{2} + b e \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + 4 \, {\left (a c d - 2 \, a c e\right )} x + 2 \, {\left (2 \, a c e x + b d - 2 \, b e\right )} \log \left (-c^{2} x^{2} + 1\right ) + 2 \, {\left (b c e x \log \left (-c^{2} x^{2} + 1\right ) + 2 \, a e + {\left (b c d - 2 \, b c e\right )} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, c} \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="fricas")
 

Output:

1/4*(b*e*log(-c^2*x^2 + 1)^2 + b*e*log(-(c*x + 1)/(c*x - 1))^2 + 4*(a*c*d 
- 2*a*c*e)*x + 2*(2*a*c*e*x + b*d - 2*b*e)*log(-c^2*x^2 + 1) + 2*(b*c*e*x* 
log(-c^2*x^2 + 1) + 2*a*e + (b*c*d - 2*b*c*e)*x)*log(-(c*x + 1)/(c*x - 1)) 
)/c
 

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.42 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\begin {cases} a d x + a e x \log {\left (- c^{2} x^{2} + 1 \right )} - 2 a e x + \frac {2 a e \operatorname {atanh}{\left (c x \right )}}{c} + b d x \operatorname {atanh}{\left (c x \right )} + b e x \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )} - 2 b e x \operatorname {atanh}{\left (c x \right )} + \frac {b d \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c} + \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}^{2}}{4 c} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}}{c} + \frac {b e \operatorname {atanh}^{2}{\left (c x \right )}}{c} & \text {for}\: c \neq 0 \\a d x & \text {otherwise} \end {cases} \] Input:

integrate((a+b*atanh(c*x))*(d+e*ln(-c**2*x**2+1)),x)
 

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.04 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.71 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-{\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} b e \operatorname {artanh}\left (c x\right ) - {\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} a e + a d x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} + \frac {{\left ({\left (i \, \pi + 2 \, \log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) + {\left (i \, \pi - 2\right )} \log \left (c x - 1\right )\right )} b e}{2 \, c} \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="maxima")
 

Output:

-(c^2*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3) - x*log(-c^2*x^2 + 1 
))*b*e*arctanh(c*x) - (c^2*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3) 
 - x*log(-c^2*x^2 + 1))*a*e + a*d*x + 1/2*(2*c*x*arctanh(c*x) + log(-c^2*x 
^2 + 1))*b*d/c + 1/2*((I*pi + 2*log(c*x - 1) - 2)*log(c*x + 1) + (I*pi - 2 
)*log(c*x - 1))*b*e/c
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.59 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {1}{2} \, b e x \log \left (-c x + 1\right )^{2} + \frac {1}{2} \, {\left (b d + 2 \, a e - 2 \, b e\right )} x \log \left (c x + 1\right ) + \frac {1}{2} \, {\left (b e x + \frac {b e}{c}\right )} \log \left (c x + 1\right )^{2} - \frac {b e \log \left (c x - 1\right )^{2}}{2 \, c} + {\left (a d - 2 \, a e\right )} x - \frac {1}{2} \, {\left ({\left (b d - 2 \, a e - 2 \, b e\right )} x - \frac {2 \, b e \log \left (c x - 1\right )}{c}\right )} \log \left (-c x + 1\right ) + \frac {{\left (b d + 2 \, a e - 2 \, b e\right )} \log \left (c x + 1\right )}{2 \, c} + \frac {{\left (b d - 2 \, a e - 2 \, b e\right )} \log \left (c x - 1\right )}{2 \, c} \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="giac")
 

Output:

-1/2*b*e*x*log(-c*x + 1)^2 + 1/2*(b*d + 2*a*e - 2*b*e)*x*log(c*x + 1) + 1/ 
2*(b*e*x + b*e/c)*log(c*x + 1)^2 - 1/2*b*e*log(c*x - 1)^2/c + (a*d - 2*a*e 
)*x - 1/2*((b*d - 2*a*e - 2*b*e)*x - 2*b*e*log(c*x - 1)/c)*log(-c*x + 1) + 
 1/2*(b*d + 2*a*e - 2*b*e)*log(c*x + 1)/c + 1/2*(b*d - 2*a*e - 2*b*e)*log( 
c*x - 1)/c
 

Mupad [B] (verification not implemented)

Time = 4.49 (sec) , antiderivative size = 385, normalized size of antiderivative = 3.70 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=a\,d\,x-2\,a\,e\,x+\frac {b\,e\,{\ln \left (c\,x+1\right )}^2}{2\,c}+\frac {b\,e\,{\ln \left (1-c\,x\right )}^2}{2\,c}+a\,e\,x\,\ln \left (1-c^2\,x^2\right )+\frac {b\,d\,x\,\ln \left (c\,x+1\right )}{2}-\frac {b\,d\,x\,\ln \left (1-c\,x\right )}{2}-b\,e\,x\,\ln \left (c\,x+1\right )+b\,e\,x\,\ln \left (1-c\,x\right )-\frac {a\,e\,\ln \left (c\,x-1\right )}{c}+\frac {a\,e\,\ln \left (c\,x+1\right )}{c}+\frac {b\,d\,\ln \left (c\,x-1\right )}{2\,c}+\frac {b\,d\,\ln \left (c\,x+1\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x-1\right )}{c}-\frac {b\,e\,\ln \left (c\,x+1\right )}{c}+\frac {b\,e\,x\,\ln \left (c\,x+1\right )\,\ln \left (1-c^2\,x^2\right )}{2}-\frac {b\,e\,x\,\ln \left (1-c\,x\right )\,\ln \left (1-c^2\,x^2\right )}{2}+\frac {b\,e\,\ln \left (1-c^2\,x^2\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}+\frac {b\,e\,\ln \left (1-c^2\,x^2\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c} \] Input:

int((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)),x)
 

Output:

a*d*x - 2*a*e*x + (b*e*log(c*x + 1)^2)/(2*c) + (b*e*log(1 - c*x)^2)/(2*c) 
+ a*e*x*log(1 - c^2*x^2) + (b*d*x*log(c*x + 1))/2 - (b*d*x*log(1 - c*x))/2 
 - b*e*x*log(c*x + 1) + b*e*x*log(1 - c*x) - (a*e*log(c*x - 1))/c + (a*e*l 
og(c*x + 1))/c + (b*d*log(c*x - 1))/(2*c) + (b*d*log(c*x + 1))/(2*c) - (b* 
e*log(c*x - 1))/c - (b*e*log(c*x + 1))/c + (b*e*x*log(c*x + 1)*log(1 - c^2 
*x^2))/2 - (b*e*x*log(1 - c*x)*log(1 - c^2*x^2))/2 + (b*e*log(1 - c^2*x^2) 
*log(- 2*a*e - 2*a*c*e*x))/(2*c) + (b*e*log(1 - c^2*x^2)*log(2*a*c*e*x - 2 
*a*e))/(2*c) - (b*e*log(c*x + 1)*log(- 2*a*e - 2*a*c*e*x))/(2*c) - (b*e*lo 
g(c*x + 1)*log(2*a*c*e*x - 2*a*e))/(2*c) - (b*e*log(1 - c*x)*log(- 2*a*e - 
 2*a*c*e*x))/(2*c) - (b*e*log(1 - c*x)*log(2*a*c*e*x - 2*a*e))/(2*c)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.55 \[ \int (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {4 \mathit {atanh} \left (c x \right )^{2} b e +4 \mathit {atanh} \left (c x \right ) \mathrm {log}\left (-c^{2} x^{2}+1\right ) b c e x +4 \mathit {atanh} \left (c x \right ) b c d x -8 \mathit {atanh} \left (c x \right ) b c e x +\mathrm {log}\left (-c^{2} x^{2}+1\right )^{2} b e +4 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) a c e x +4 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) a e +2 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) b d -4 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) b e -8 \,\mathrm {log}\left (c^{2} x -c \right ) a e +4 a c d x -8 a c e x}{4 c} \] Input:

int((a+b*atanh(c*x))*(d+e*log(-c^2*x^2+1)),x)
 

Output:

(4*atanh(c*x)**2*b*e + 4*atanh(c*x)*log( - c**2*x**2 + 1)*b*c*e*x + 4*atan 
h(c*x)*b*c*d*x - 8*atanh(c*x)*b*c*e*x + log( - c**2*x**2 + 1)**2*b*e + 4*l 
og( - c**2*x**2 + 1)*a*c*e*x + 4*log( - c**2*x**2 + 1)*a*e + 2*log( - c**2 
*x**2 + 1)*b*d - 4*log( - c**2*x**2 + 1)*b*e - 8*log(c**2*x - c)*a*e + 4*a 
*c*d*x - 8*a*c*e*x)/(4*c)