\(\int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x} \, dx\) [14]

Optimal result

Integrand size = 20, antiderivative size = 114 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x} \, dx=2 a c d^2 x+\frac {1}{2} b c d^2 x-\frac {1}{2} b d^2 \text {arctanh}(c x)+2 b c d^2 x \text {arctanh}(c x)+\frac {1}{2} c^2 d^2 x^2 (a+b \text {arctanh}(c x))+a d^2 \log (x)+b d^2 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^2 \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b d^2 \operatorname {PolyLog}(2,c x) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x} \, dx=\frac {1}{4} d^2 \left (8 a c x+2 b c x+2 a c^2 x^2+8 b c x \text {arctanh}(c x)+2 b c^2 x^2 \text {arctanh}(c x)+4 a \log (x)+b \log (1-c x)-b \log (1+c x)+4 b \log \left (1-c^2 x^2\right )-2 b \operatorname {PolyLog}(2,-c x)+2 b \operatorname {PolyLog}(2,c x)\right ) \] Input:

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

Output:

(d^2*(8*a*c*x + 2*b*c*x + 2*a*c^2*x^2 + 8*b*c*x*ArcTanh[c*x] + 2*b*c^2*x^2 
*ArcTanh[c*x] + 4*a*Log[x] + b*Log[1 - c*x] - b*Log[1 + c*x] + 4*b*Log[1 - 
 c^2*x^2] - 2*b*PolyLog[2, -(c*x)] + 2*b*PolyLog[2, c*x]))/4
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6502, 2009}

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.

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

\(\Big \downarrow \) 6502

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 6502

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e 
_.)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p, ( 
f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] 
 && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90

method	result	size
parts	\(d^{2} a \left (\frac {c^{2} x^{2}}{2}+2 c x +\ln \left (x \right )\right )+d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+2 \,\operatorname {arctanh}\left (c x \right ) c x +\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {c x}{2}+\frac {5 \ln \left (c x -1\right )}{4}+\frac {3 \ln \left (c x +1\right )}{4}\right )\)	\(103\)
derivativedivides	\(d^{2} a \left (\frac {c^{2} x^{2}}{2}+2 c x +\ln \left (c x \right )\right )+d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+2 \,\operatorname {arctanh}\left (c x \right ) c x +\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {c x}{2}+\frac {5 \ln \left (c x -1\right )}{4}+\frac {3 \ln \left (c x +1\right )}{4}\right )\)	\(105\)
default	\(d^{2} a \left (\frac {c^{2} x^{2}}{2}+2 c x +\ln \left (c x \right )\right )+d^{2} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}+2 \,\operatorname {arctanh}\left (c x \right ) c x +\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {c x}{2}+\frac {5 \ln \left (c x -1\right )}{4}+\frac {3 \ln \left (c x +1\right )}{4}\right )\)	\(105\)
risch	\(\frac {a \,c^{2} d^{2} x^{2}}{2}+2 d^{2} a c x +d^{2} a \ln \left (-c x \right )-\frac {5 d^{2} a}{2}-\frac {d^{2} b \ln \left (-c x +1\right ) c^{2} x^{2}}{4}-d^{2} b \ln \left (-c x +1\right ) c x +\frac {5 d^{2} b \ln \left (-c x +1\right )}{4}+\frac {b c \,d^{2} x}{2}-2 d^{2} b +\frac {d^{2} b \operatorname {dilog}\left (-c x +1\right )}{2}+\frac {d^{2} b \ln \left (c x +1\right ) c^{2} x^{2}}{4}+d^{2} b \ln \left (c x +1\right ) c x +\frac {3 d^{2} b \ln \left (c x +1\right )}{4}-\frac {d^{2} b \operatorname {dilog}\left (c x +1\right )}{2}\)	\(167\)

Input:

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

Output:

d^2*a*(1/2*c^2*x^2+2*c*x+ln(x))+d^2*b*(1/2*arctanh(c*x)*c^2*x^2+2*arctanh( 
c*x)*c*x+arctanh(c*x)*ln(c*x)-1/2*dilog(c*x)-1/2*dilog(c*x+1)-1/2*ln(c*x)* 
ln(c*x+1)+1/2*c*x+5/4*ln(c*x-1)+3/4*ln(c*x+1))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

d**2*(Integral(2*a*c, x) + Integral(a/x, x) + Integral(a*c**2*x, x) + Inte 
gral(2*b*c*atanh(c*x), x) + Integral(b*atanh(c*x)/x, x) + Integral(b*c**2* 
x*atanh(c*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.52 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x} \, dx=\frac {1}{4} \, b c^{2} d^{2} x^{2} \log \left (c x + 1\right ) - \frac {1}{4} \, b c^{2} d^{2} x^{2} \log \left (-c x + 1\right ) + \frac {1}{2} \, a c^{2} d^{2} x^{2} + 2 \, a c d^{2} x + \frac {1}{2} \, b c d^{2} x + {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{2} - \frac {1}{2} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b d^{2} + \frac {1}{2} \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b d^{2} - \frac {1}{4} \, b d^{2} \log \left (c x + 1\right ) + \frac {1}{4} \, b d^{2} \log \left (c x - 1\right ) + a d^{2} \log \left (x\right ) \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(d**2*(atanh(c*x)*b*c**2*x**2 + 4*atanh(c*x)*b*c*x + 3*atanh(c*x)*b + 2*in 
t(atanh(c*x)/x,x)*b + 4*log(c**2*x - c)*b + 2*log(x)*a + a*c**2*x**2 + 4*a 
*c*x + b*c*x))/2