\(\int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^4} \, dx\) [27]

Optimal result

Integrand size = 20, antiderivative size = 176 \[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^4} \, dx=-\frac {b c d^3}{6 x^2}-\frac {3 b c^2 d^3}{2 x}+\frac {3}{2} b c^3 d^3 \text {arctanh}(c x)-\frac {d^3 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {3 c d^3 (a+b \text {arctanh}(c x))}{2 x^2}-\frac {3 c^2 d^3 (a+b \text {arctanh}(c x))}{x}+a c^3 d^3 \log (x)+\frac {10}{3} b c^3 d^3 \log (x)-\frac {5}{3} b c^3 d^3 \log \left (1-c^2 x^2\right )-\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,c x) \] Output:

-1/6*b*c*d^3/x^2-3/2*b*c^2*d^3/x+3/2*b*c^3*d^3*arctanh(c*x)-1/3*d^3*(a+b*a 
rctanh(c*x))/x^3-3/2*c*d^3*(a+b*arctanh(c*x))/x^2-3*c^2*d^3*(a+b*arctanh(c 
*x))/x+a*c^3*d^3*ln(x)+10/3*b*c^3*d^3*ln(x)-5/3*b*c^3*d^3*ln(-c^2*x^2+1)-1 
/2*b*c^3*d^3*polylog(2,-c*x)+1/2*b*c^3*d^3*polylog(2,c*x)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.99 \[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {d^3 \left (-4 a-18 a c x-2 b c x-36 a c^2 x^2-18 b c^2 x^2-4 b \text {arctanh}(c x)-18 b c x \text {arctanh}(c x)-36 b c^2 x^2 \text {arctanh}(c x)+12 a c^3 x^3 \log (x)+40 b c^3 x^3 \log (c x)-9 b c^3 x^3 \log (1-c x)+9 b c^3 x^3 \log (1+c x)-20 b c^3 x^3 \log \left (1-c^2 x^2\right )-6 b c^3 x^3 \operatorname {PolyLog}(2,-c x)+6 b c^3 x^3 \operatorname {PolyLog}(2,c x)\right )}{12 x^3} \] Input:

Integrate[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^4,x]
 

Output:

(d^3*(-4*a - 18*a*c*x - 2*b*c*x - 36*a*c^2*x^2 - 18*b*c^2*x^2 - 4*b*ArcTan 
h[c*x] - 18*b*c*x*ArcTanh[c*x] - 36*b*c^2*x^2*ArcTanh[c*x] + 12*a*c^3*x^3* 
Log[x] + 40*b*c^3*x^3*Log[c*x] - 9*b*c^3*x^3*Log[1 - c*x] + 9*b*c^3*x^3*Lo 
g[1 + c*x] - 20*b*c^3*x^3*Log[1 - c^2*x^2] - 6*b*c^3*x^3*PolyLog[2, -(c*x) 
] + 6*b*c^3*x^3*PolyLog[2, c*x]))/(12*x^3)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 176, 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.

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

\(\Big \downarrow \) 6502

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 c^2 d^3 (a+b \text {arctanh}(c x))}{x}-\frac {d^3 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {3 c d^3 (a+b \text {arctanh}(c x))}{2 x^2}+a c^3 d^3 \log (x)+\frac {3}{2} b c^3 d^3 \text {arctanh}(c x)-\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b c^3 d^3 \operatorname {PolyLog}(2,c x)+\frac {10}{3} b c^3 d^3 \log (x)-\frac {3 b c^2 d^3}{2 x}-\frac {5}{3} b c^3 d^3 \log \left (1-c^2 x^2\right )-\frac {b c d^3}{6 x^2}\)

Input:

Int[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^4,x]
 

Output:

-1/6*(b*c*d^3)/x^2 - (3*b*c^2*d^3)/(2*x) + (3*b*c^3*d^3*ArcTanh[c*x])/2 - 
(d^3*(a + b*ArcTanh[c*x]))/(3*x^3) - (3*c*d^3*(a + b*ArcTanh[c*x]))/(2*x^2 
) - (3*c^2*d^3*(a + b*ArcTanh[c*x]))/x + a*c^3*d^3*Log[x] + (10*b*c^3*d^3* 
Log[x])/3 - (5*b*c^3*d^3*Log[1 - c^2*x^2])/3 - (b*c^3*d^3*PolyLog[2, -(c*x 
)])/2 + (b*c^3*d^3*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 = 0.53 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.86

method	result	size
parts	\(d^{3} a \left (c^{3} \ln \left (x \right )-\frac {3 c^{2}}{x}-\frac {3 c}{2 x^{2}}-\frac {1}{3 x^{3}}\right )+d^{3} b \,c^{3} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}+\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {29 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}-\frac {3}{2 c x}+\frac {10 \ln \left (c x \right )}{3}-\frac {11 \ln \left (c x +1\right )}{12}-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\)	\(151\)
derivativedivides	\(c^{3} \left (d^{3} a \left (-\frac {1}{3 c^{3} x^{3}}-\frac {3}{2 c^{2} x^{2}}+\ln \left (c x \right )-\frac {3}{c x}\right )+d^{3} b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}+\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {29 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}-\frac {3}{2 c x}+\frac {10 \ln \left (c x \right )}{3}-\frac {11 \ln \left (c x +1\right )}{12}-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\right )\)	\(155\)
default	\(c^{3} \left (d^{3} a \left (-\frac {1}{3 c^{3} x^{3}}-\frac {3}{2 c^{2} x^{2}}+\ln \left (c x \right )-\frac {3}{c x}\right )+d^{3} b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}+\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {29 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}-\frac {3}{2 c x}+\frac {10 \ln \left (c x \right )}{3}-\frac {11 \ln \left (c x +1\right )}{12}-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\right )\)	\(155\)
risch	\(\frac {29 c^{3} d^{3} b \ln \left (-c x \right )}{12}-\frac {3 b \,c^{2} d^{3}}{2 x}-\frac {29 \ln \left (-c x +1\right ) b \,c^{3} d^{3}}{12}+\frac {3 c \,d^{3} b \ln \left (-c x +1\right )}{4 x^{2}}+\frac {c^{3} d^{3} b \operatorname {dilog}\left (-c x +1\right )}{2}-\frac {b c \,d^{3}}{6 x^{2}}+\frac {d^{3} b \ln \left (-c x +1\right )}{6 x^{3}}+\frac {3 c^{2} d^{3} b \ln \left (-c x +1\right )}{2 x}-\frac {3 c \,d^{3} a}{2 x^{2}}+c^{3} d^{3} a \ln \left (-c x \right )-\frac {d^{3} a}{3 x^{3}}-\frac {3 c^{2} d^{3} a}{x}+\frac {11 b \,c^{3} d^{3} \ln \left (c x \right )}{12}-\frac {11 \ln \left (c x +1\right ) b \,c^{3} d^{3}}{12}-\frac {3 b c \,d^{3} \ln \left (c x +1\right )}{4 x^{2}}-\frac {b \,c^{3} d^{3} \operatorname {dilog}\left (c x +1\right )}{2}-\frac {b \,d^{3} \ln \left (c x +1\right )}{6 x^{3}}-\frac {3 b \,c^{2} d^{3} \ln \left (c x +1\right )}{2 x}\)	\(258\)

Input:

int((c*d*x+d)^3*(a+b*arctanh(c*x))/x^4,x,method=_RETURNVERBOSE)
 

Output:

d^3*a*(c^3*ln(x)-3*c^2/x-3/2*c/x^2-1/3/x^3)+d^3*b*c^3*(-1/3*arctanh(c*x)/c 
^3/x^3-3/2*arctanh(c*x)/c^2/x^2+arctanh(c*x)*ln(c*x)-3*arctanh(c*x)/c/x-29 
/12*ln(c*x-1)-1/6/c^2/x^2-3/2/c/x+10/3*ln(c*x)-11/12*ln(c*x+1)-1/2*dilog(c 
*x)-1/2*dilog(c*x+1)-1/2*ln(c*x)*ln(c*x+1))
 

Fricas [F]

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

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^4,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((c*d*x+d)**3*(a+b*atanh(c*x))/x**4,x)
 

Output:

d**3*(Integral(a/x**4, x) + Integral(3*a*c/x**3, x) + Integral(3*a*c**2/x* 
*2, x) + Integral(a*c**3/x, x) + Integral(b*atanh(c*x)/x**4, x) + Integral 
(3*b*c*atanh(c*x)/x**3, x) + Integral(3*b*c**2*atanh(c*x)/x**2, x) + Integ 
ral(b*c**3*atanh(c*x)/x, x))
 

Maxima [F]

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

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^4,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^4,x, algorithm="giac")
 

Output:

integrate((c*d*x + d)^3*(b*arctanh(c*x) + a)/x^4, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

int(((a + b*atanh(c*x))*(d + c*d*x)^3)/x^4,x)
 

Output:

int(((a + b*atanh(c*x))*(d + c*d*x)^3)/x^4, x)
 

Reduce [F]

\[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {d^{3} \left (-11 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}-18 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-9 \mathit {atanh} \left (c x \right ) b c x -2 \mathit {atanh} \left (c x \right ) b +6 \left (\int \frac {\mathit {atanh} \left (c x \right )}{x}d x \right ) b \,c^{3} x^{3}-20 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{3} x^{3}+6 \,\mathrm {log}\left (x \right ) a \,c^{3} x^{3}+20 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{3}-18 a \,c^{2} x^{2}-9 a c x -2 a -9 b \,c^{2} x^{2}-b c x \right )}{6 x^{3}} \] Input:

int((c*d*x+d)^3*(a+b*atanh(c*x))/x^4,x)
 

Output:

(d**3*( - 11*atanh(c*x)*b*c**3*x**3 - 18*atanh(c*x)*b*c**2*x**2 - 9*atanh( 
c*x)*b*c*x - 2*atanh(c*x)*b + 6*int(atanh(c*x)/x,x)*b*c**3*x**3 - 20*log(c 
**2*x - c)*b*c**3*x**3 + 6*log(x)*a*c**3*x**3 + 20*log(x)*b*c**3*x**3 - 18 
*a*c**2*x**2 - 9*a*c*x - 2*a - 9*b*c**2*x**2 - b*c*x))/(6*x**3)