\(\int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^2} \, dx\) [36]

Optimal result

Integrand size = 20, antiderivative size = 178 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^2} \, dx=6 a c^2 d^4 x+2 b c^2 d^4 x+\frac {1}{6} b c^3 d^4 x^2-2 b c d^4 \text {arctanh}(c x)+6 b c^2 d^4 x \text {arctanh}(c x)-\frac {d^4 (a+b \text {arctanh}(c x))}{x}+2 c^3 d^4 x^2 (a+b \text {arctanh}(c x))+\frac {1}{3} c^4 d^4 x^3 (a+b \text {arctanh}(c x))+4 a c d^4 \log (x)+b c d^4 \log (x)+\frac {8}{3} b c d^4 \log \left (1-c^2 x^2\right )-2 b c d^4 \operatorname {PolyLog}(2,-c x)+2 b c d^4 \operatorname {PolyLog}(2,c x) \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.09 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^2} \, dx=\frac {d^4 \left (-6 a+36 a c^2 x^2+12 b c^2 x^2+12 a c^3 x^3+b c^3 x^3+2 a c^4 x^4-6 b \text {arctanh}(c x)+36 b c^2 x^2 \text {arctanh}(c x)+12 b c^3 x^3 \text {arctanh}(c x)+2 b c^4 x^4 \text {arctanh}(c x)+24 a c x \log (x)+6 b c x \log (c x)+6 b c x \log (1-c x)-6 b c x \log (1+c x)+15 b c x \log \left (1-c^2 x^2\right )+b c x \log \left (-1+c^2 x^2\right )-12 b c x \operatorname {PolyLog}(2,-c x)+12 b c x \operatorname {PolyLog}(2,c x)\right )}{6 x} \] Input:

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

Output:

(d^4*(-6*a + 36*a*c^2*x^2 + 12*b*c^2*x^2 + 12*a*c^3*x^3 + b*c^3*x^3 + 2*a* 
c^4*x^4 - 6*b*ArcTanh[c*x] + 36*b*c^2*x^2*ArcTanh[c*x] + 12*b*c^3*x^3*ArcT 
anh[c*x] + 2*b*c^4*x^4*ArcTanh[c*x] + 24*a*c*x*Log[x] + 6*b*c*x*Log[c*x] + 
 6*b*c*x*Log[1 - c*x] - 6*b*c*x*Log[1 + c*x] + 15*b*c*x*Log[1 - c^2*x^2] + 
 b*c*x*Log[-1 + c^2*x^2] - 12*b*c*x*PolyLog[2, -(c*x)] + 12*b*c*x*PolyLog[ 
2, c*x]))/(6*x)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 178, 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.

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

\(\Big \downarrow \) 6502

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

6*a*c^2*d^4*x + 2*b*c^2*d^4*x + (b*c^3*d^4*x^2)/6 - 2*b*c*d^4*ArcTanh[c*x] 
 + 6*b*c^2*d^4*x*ArcTanh[c*x] - (d^4*(a + b*ArcTanh[c*x]))/x + 2*c^3*d^4*x 
^2*(a + b*ArcTanh[c*x]) + (c^4*d^4*x^3*(a + b*ArcTanh[c*x]))/3 + 4*a*c*d^4 
*Log[x] + b*c*d^4*Log[x] + (8*b*c*d^4*Log[1 - c^2*x^2])/3 - 2*b*c*d^4*Poly 
Log[2, -(c*x)] + 2*b*c*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 = 0.58 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.89

method	result	size
parts	\(d^{4} a \left (\frac {x^{3} c^{4}}{3}+2 c^{3} x^{2}+6 c^{2} x +4 c \ln \left (x \right )-\frac {1}{x}\right )+d^{4} b c \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+2 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}+6 \,\operatorname {arctanh}\left (c x \right ) c x +4 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-2 \operatorname {dilog}\left (c x \right )-2 \operatorname {dilog}\left (c x +1\right )-2 \ln \left (c x \right ) \ln \left (c x +1\right )+\frac {c^{2} x^{2}}{6}+2 c x +\frac {11 \ln \left (c x -1\right )}{3}+\ln \left (c x \right )+\frac {5 \ln \left (c x +1\right )}{3}\right )\)	\(159\)
derivativedivides	\(c \left (d^{4} a \left (\frac {x^{3} c^{3}}{3}+2 c^{2} x^{2}+6 c x +4 \ln \left (c x \right )-\frac {1}{c x}\right )+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+2 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}+6 \,\operatorname {arctanh}\left (c x \right ) c x +4 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-2 \operatorname {dilog}\left (c x \right )-2 \operatorname {dilog}\left (c x +1\right )-2 \ln \left (c x \right ) \ln \left (c x +1\right )+\frac {c^{2} x^{2}}{6}+2 c x +\frac {11 \ln \left (c x -1\right )}{3}+\ln \left (c x \right )+\frac {5 \ln \left (c x +1\right )}{3}\right )\right )\)	\(162\)
default	\(c \left (d^{4} a \left (\frac {x^{3} c^{3}}{3}+2 c^{2} x^{2}+6 c x +4 \ln \left (c x \right )-\frac {1}{c x}\right )+d^{4} b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+2 \,\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}+6 \,\operatorname {arctanh}\left (c x \right ) c x +4 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-2 \operatorname {dilog}\left (c x \right )-2 \operatorname {dilog}\left (c x +1\right )-2 \ln \left (c x \right ) \ln \left (c x +1\right )+\frac {c^{2} x^{2}}{6}+2 c x +\frac {11 \ln \left (c x -1\right )}{3}+\ln \left (c x \right )+\frac {5 \ln \left (c x +1\right )}{3}\right )\right )\)	\(162\)
risch	\(-\frac {119 b c \,d^{4}}{18}-\frac {25 c \,d^{4} a}{3}+\frac {b \,c^{4} d^{4} \ln \left (c x +1\right ) x^{3}}{6}+b \,c^{3} d^{4} \ln \left (c x +1\right ) x^{2}+3 b \,c^{2} d^{4} \ln \left (c x +1\right ) x +\frac {5 b c \,d^{4} \ln \left (c x +1\right )}{3}-2 b c \,d^{4} \operatorname {dilog}\left (c x +1\right )+\frac {b c \,d^{4} \ln \left (c x \right )}{2}-\frac {b \,d^{4} \ln \left (c x +1\right )}{2 x}+\frac {11 c \,d^{4} b \ln \left (-c x +1\right )}{3}+2 c \,d^{4} b \operatorname {dilog}\left (-c x +1\right )+\frac {c \,d^{4} b \ln \left (-c x \right )}{2}+\frac {d^{4} b \ln \left (-c x +1\right )}{2 x}+\frac {c^{4} d^{4} a \,x^{3}}{3}+2 c^{3} d^{4} a \,x^{2}-\frac {d^{4} a}{x}+4 c \,d^{4} a \ln \left (-c x \right )-\frac {c^{4} d^{4} b \ln \left (-c x +1\right ) x^{3}}{6}-c^{3} d^{4} b \ln \left (-c x +1\right ) x^{2}-3 c^{2} d^{4} b \ln \left (-c x +1\right ) x +6 a \,c^{2} d^{4} x +2 b \,c^{2} d^{4} x +\frac {b \,c^{3} d^{4} x^{2}}{6}\)	\(307\)

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

d**4*(Integral(6*a*c**2, x) + Integral(a/x**2, x) + Integral(4*a*c/x, x) + 
 Integral(4*a*c**3*x, x) + Integral(a*c**4*x**2, x) + Integral(6*b*c**2*at 
anh(c*x), x) + Integral(b*atanh(c*x)/x**2, x) + Integral(4*b*c*atanh(c*x)/ 
x, x) + Integral(4*b*c**3*x*atanh(c*x), x) + Integral(b*c**4*x**2*atanh(c* 
x), x))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.58 \[ \int \frac {(d+c d x)^4 (a+b \text {arctanh}(c x))}{x^2} \, dx=\frac {1}{3} \, a c^{4} d^{4} x^{3} + 2 \, a c^{3} d^{4} x^{2} + \frac {1}{6} \, b c^{3} d^{4} x^{2} + 6 \, a c^{2} d^{4} x + 2 \, b c^{2} d^{4} x + 3 \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b c d^{4} - 2 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b c d^{4} + 2 \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b c d^{4} - \frac {5}{6} \, b c d^{4} \log \left (c x + 1\right ) + \frac {7}{6} \, b c d^{4} \log \left (c x - 1\right ) + 4 \, a c d^{4} \log \left (x\right ) - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} b d^{4} - \frac {a d^{4}}{x} + \frac {1}{6} \, {\left (b c^{4} d^{4} x^{3} + 6 \, b c^{3} d^{4} x^{2}\right )} \log \left (c x + 1\right ) - \frac {1}{6} \, {\left (b c^{4} d^{4} x^{3} + 6 \, b c^{3} d^{4} x^{2}\right )} \log \left (-c x + 1\right ) \] Input:

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

Output:

1/3*a*c^4*d^4*x^3 + 2*a*c^3*d^4*x^2 + 1/6*b*c^3*d^4*x^2 + 6*a*c^2*d^4*x + 
2*b*c^2*d^4*x + 3*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*c*d^4 - 2*(lo 
g(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*b*c*d^4 + 2*(log(c*x + 1)*log(-c*x 
) + dilog(c*x + 1))*b*c*d^4 - 5/6*b*c*d^4*log(c*x + 1) + 7/6*b*c*d^4*log(c 
*x - 1) + 4*a*c*d^4*log(x) - 1/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arct 
anh(c*x)/x)*b*d^4 - a*d^4/x + 1/6*(b*c^4*d^4*x^3 + 6*b*c^3*d^4*x^2)*log(c* 
x + 1) - 1/6*(b*c^4*d^4*x^3 + 6*b*c^3*d^4*x^2)*log(-c*x + 1)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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