\(\int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^4} \, dx\) [38]

Optimal result

Integrand size = 20, antiderivative size = 189 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^4} \, dx=-\frac {b c d^4}{6 x^2}-\frac {2 b c^2 d^4}{x}+a c^4 d^4 x+2 b c^3 d^4 \text {arctanh}(c x)+b c^4 d^4 x \text {arctanh}(c x)-\frac {d^4 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {2 c d^4 (a+b \text {arctanh}(c x))}{x^2}-\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{x}+4 a c^3 d^4 \log (x)+\frac {19}{3} b c^3 d^4 \log (x)-\frac {8}{3} b c^3 d^4 \log \left (1-c^2 x^2\right )-2 b c^3 d^4 \operatorname {PolyLog}(2,-c x)+2 b c^3 d^4 \operatorname {PolyLog}(2,c x) \] Output:

-1/6*b*c*d^4/x^2-2*b*c^2*d^4/x+a*c^4*d^4*x+2*b*c^3*d^4*arctanh(c*x)+b*c^4* 
d^4*x*arctanh(c*x)-1/3*d^4*(a+b*arctanh(c*x))/x^3-2*c*d^4*(a+b*arctanh(c*x 
))/x^2-6*c^2*d^4*(a+b*arctanh(c*x))/x+4*a*c^3*d^4*ln(x)+19/3*b*c^3*d^4*ln( 
x)-8/3*b*c^3*d^4*ln(-c^2*x^2+1)-2*b*c^3*d^4*polylog(2,-c*x)+2*b*c^3*d^4*po 
lylog(2,c*x)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.04 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {d^4 \left (-2 a-12 a c x-b c x-36 a c^2 x^2-12 b c^2 x^2+6 a c^4 x^4-2 b \text {arctanh}(c x)-12 b c x \text {arctanh}(c x)-36 b c^2 x^2 \text {arctanh}(c x)+6 b c^4 x^4 \text {arctanh}(c x)+24 a c^3 x^3 \log (x)+38 b c^3 x^3 \log (c x)-6 b c^3 x^3 \log (1-c x)+6 b c^3 x^3 \log (1+c x)-16 b c^3 x^3 \log \left (1-c^2 x^2\right )-12 b c^3 x^3 \operatorname {PolyLog}(2,-c x)+12 b c^3 x^3 \operatorname {PolyLog}(2,c x)\right )}{6 x^3} \] Input:

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

Output:

(d^4*(-2*a - 12*a*c*x - b*c*x - 36*a*c^2*x^2 - 12*b*c^2*x^2 + 6*a*c^4*x^4 
- 2*b*ArcTanh[c*x] - 12*b*c*x*ArcTanh[c*x] - 36*b*c^2*x^2*ArcTanh[c*x] + 6 
*b*c^4*x^4*ArcTanh[c*x] + 24*a*c^3*x^3*Log[x] + 38*b*c^3*x^3*Log[c*x] - 6* 
b*c^3*x^3*Log[1 - c*x] + 6*b*c^3*x^3*Log[1 + c*x] - 16*b*c^3*x^3*Log[1 - c 
^2*x^2] - 12*b*c^3*x^3*PolyLog[2, -(c*x)] + 12*b*c^3*x^3*PolyLog[2, c*x])) 
/(6*x^3)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 189, 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.

\(\displaystyle \int \frac {(c d x+d)^4 (a+b \text {arctanh}(c x))}{x^4} \, dx\)

\(\Big \downarrow \) 6502

\(\displaystyle \int \left (c^4 d^4 (a+b \text {arctanh}(c x))+\frac {4 c^3 d^4 (a+b \text {arctanh}(c x))}{x}+\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{x^2}+\frac {d^4 (a+b \text {arctanh}(c x))}{x^4}+\frac {4 c d^4 (a+b \text {arctanh}(c x))}{x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {6 c^2 d^4 (a+b \text {arctanh}(c x))}{x}-\frac {d^4 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {2 c d^4 (a+b \text {arctanh}(c x))}{x^2}+a c^4 d^4 x+4 a c^3 d^4 \log (x)+b c^4 d^4 x \text {arctanh}(c x)+2 b c^3 d^4 \text {arctanh}(c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,-c x)+2 b c^3 d^4 \operatorname {PolyLog}(2,c x)+\frac {19}{3} b c^3 d^4 \log (x)-\frac {2 b c^2 d^4}{x}-\frac {8}{3} b c^3 d^4 \log \left (1-c^2 x^2\right )-\frac {b c d^4}{6 x^2}\)

Input:

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

Output:

-1/6*(b*c*d^4)/x^2 - (2*b*c^2*d^4)/x + a*c^4*d^4*x + 2*b*c^3*d^4*ArcTanh[c 
*x] + b*c^4*d^4*x*ArcTanh[c*x] - (d^4*(a + b*ArcTanh[c*x]))/(3*x^3) - (2*c 
*d^4*(a + b*ArcTanh[c*x]))/x^2 - (6*c^2*d^4*(a + b*ArcTanh[c*x]))/x + 4*a* 
c^3*d^4*Log[x] + (19*b*c^3*d^4*Log[x])/3 - (8*b*c^3*d^4*Log[1 - c^2*x^2])/ 
3 - 2*b*c^3*d^4*PolyLog[2, -(c*x)] + 2*b*c^3*d^4*PolyLog[2, c*x]
 

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 = 0.78 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.87

method	result	size
parts	\(d^{4} a \left (c^{4} x +4 c^{3} \ln \left (x \right )-\frac {6 c^{2}}{x}-\frac {2 c}{x^{2}}-\frac {1}{3 x^{3}}\right )+d^{4} b \,c^{3} \left (\operatorname {arctanh}\left (c x \right ) c x -\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}+4 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {6 \,\operatorname {arctanh}\left (c x \right )}{c x}-2 \operatorname {dilog}\left (c x \right )-2 \operatorname {dilog}\left (c x +1\right )-2 \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {11 \ln \left (c x -1\right )}{3}-\frac {1}{6 c^{2} x^{2}}-\frac {2}{c x}+\frac {19 \ln \left (c x \right )}{3}-\frac {5 \ln \left (c x +1\right )}{3}\right )\)	\(165\)
derivativedivides	\(c^{3} \left (d^{4} a \left (c x -\frac {1}{3 c^{3} x^{3}}-\frac {2}{c^{2} x^{2}}+4 \ln \left (c x \right )-\frac {6}{c x}\right )+d^{4} b \left (\operatorname {arctanh}\left (c x \right ) c x -\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}+4 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {6 \,\operatorname {arctanh}\left (c x \right )}{c x}-2 \operatorname {dilog}\left (c x \right )-2 \operatorname {dilog}\left (c x +1\right )-2 \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {11 \ln \left (c x -1\right )}{3}-\frac {1}{6 c^{2} x^{2}}-\frac {2}{c x}+\frac {19 \ln \left (c x \right )}{3}-\frac {5 \ln \left (c x +1\right )}{3}\right )\right )\)	\(168\)
default	\(c^{3} \left (d^{4} a \left (c x -\frac {1}{3 c^{3} x^{3}}-\frac {2}{c^{2} x^{2}}+4 \ln \left (c x \right )-\frac {6}{c x}\right )+d^{4} b \left (\operatorname {arctanh}\left (c x \right ) c x -\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c^{2} x^{2}}+4 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {6 \,\operatorname {arctanh}\left (c x \right )}{c x}-2 \operatorname {dilog}\left (c x \right )-2 \operatorname {dilog}\left (c x +1\right )-2 \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {11 \ln \left (c x -1\right )}{3}-\frac {1}{6 c^{2} x^{2}}-\frac {2}{c x}+\frac {19 \ln \left (c x \right )}{3}-\frac {5 \ln \left (c x +1\right )}{3}\right )\right )\)	\(168\)
risch	\(-\frac {b c \,d^{4} \ln \left (c x +1\right )}{x^{2}}-\frac {3 b \,c^{2} d^{4} \ln \left (c x +1\right )}{x}-b \,c^{3} d^{4}-c^{3} d^{4} a -\frac {c^{4} d^{4} b \ln \left (-c x +1\right ) x}{2}+\frac {c \,d^{4} b \ln \left (-c x +1\right )}{x^{2}}+\frac {3 c^{2} d^{4} b \ln \left (-c x +1\right )}{x}-\frac {5 b \,c^{3} d^{4} \ln \left (c x +1\right )}{3}+\frac {13 b \,c^{3} d^{4} \ln \left (c x \right )}{6}-2 b \,c^{3} d^{4} \operatorname {dilog}\left (c x +1\right )-\frac {b \,d^{4} \ln \left (c x +1\right )}{6 x^{3}}-\frac {2 c \,d^{4} a}{x^{2}}-\frac {d^{4} a}{3 x^{3}}-\frac {6 c^{2} d^{4} a}{x}+4 c^{3} d^{4} a \ln \left (-c x \right )-\frac {11 c^{3} d^{4} b \ln \left (-c x +1\right )}{3}+\frac {25 c^{3} d^{4} b \ln \left (-c x \right )}{6}+2 c^{3} d^{4} b \operatorname {dilog}\left (-c x +1\right )+\frac {d^{4} b \ln \left (-c x +1\right )}{6 x^{3}}-\frac {b c \,d^{4}}{6 x^{2}}-\frac {2 b \,c^{2} d^{4}}{x}+a \,c^{4} d^{4} x +\frac {b \,c^{4} d^{4} \ln \left (c x +1\right ) x}{2}\)	\(318\)

Input:

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

Output:

d^4*a*(c^4*x+4*c^3*ln(x)-6*c^2/x-2*c/x^2-1/3/x^3)+d^4*b*c^3*(arctanh(c*x)* 
c*x-1/3*arctanh(c*x)/c^3/x^3-2*arctanh(c*x)/c^2/x^2+4*arctanh(c*x)*ln(c*x) 
-6*arctanh(c*x)/c/x-2*dilog(c*x)-2*dilog(c*x+1)-2*ln(c*x)*ln(c*x+1)-11/3*l 
n(c*x-1)-1/6/c^2/x^2-2/c/x+19/3*ln(c*x)-5/3*ln(c*x+1))
 

Fricas [F]

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

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

Output:

integral((a*c^4*d^4*x^4 + 4*a*c^3*d^4*x^3 + 6*a*c^2*d^4*x^2 + 4*a*c*d^4*x 
+ a*d^4 + (b*c^4*d^4*x^4 + 4*b*c^3*d^4*x^3 + 6*b*c^2*d^4*x^2 + 4*b*c*d^4*x 
 + b*d^4)*arctanh(c*x))/x^4, x)
 

Sympy [F]

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

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

Output:

d**4*(Integral(a*c**4, x) + Integral(a/x**4, x) + Integral(4*a*c/x**3, x) 
+ Integral(6*a*c**2/x**2, x) + Integral(4*a*c**3/x, x) + Integral(b*c**4*a 
tanh(c*x), x) + Integral(b*atanh(c*x)/x**4, x) + Integral(4*b*c*atanh(c*x) 
/x**3, x) + Integral(6*b*c**2*atanh(c*x)/x**2, x) + Integral(4*b*c**3*atan 
h(c*x)/x, x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {d^{4} \left (6 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}-20 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}-36 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-12 \mathit {atanh} \left (c x \right ) b c x -2 \mathit {atanh} \left (c x \right ) b +24 \left (\int \frac {\mathit {atanh} \left (c x \right )}{x}d x \right ) b \,c^{3} x^{3}-32 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{3} x^{3}+24 \,\mathrm {log}\left (x \right ) a \,c^{3} x^{3}+38 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{3}+6 a \,c^{4} x^{4}-36 a \,c^{2} x^{2}-12 a c x -2 a -12 b \,c^{2} x^{2}-b c x \right )}{6 x^{3}} \] Input:

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

Output:

(d**4*(6*atanh(c*x)*b*c**4*x**4 - 20*atanh(c*x)*b*c**3*x**3 - 36*atanh(c*x 
)*b*c**2*x**2 - 12*atanh(c*x)*b*c*x - 2*atanh(c*x)*b + 24*int(atanh(c*x)/x 
,x)*b*c**3*x**3 - 32*log(c**2*x - c)*b*c**3*x**3 + 24*log(x)*a*c**3*x**3 + 
 38*log(x)*b*c**3*x**3 + 6*a*c**4*x**4 - 36*a*c**2*x**2 - 12*a*c*x - 2*a - 
 12*b*c**2*x**2 - b*c*x))/(6*x**3)