\(\int \frac {a+b \text {arctanh}(c x)}{x (d+c d x)^3} \, dx\) [63]

Optimal result

Integrand size = 20, antiderivative size = 161 \[ \int \frac {a+b \text {arctanh}(c x)}{x (d+c d x)^3} \, dx=\frac {b}{8 d^3 (1+c x)^2}+\frac {5 b}{8 d^3 (1+c x)}-\frac {5 b \text {arctanh}(c x)}{8 d^3}+\frac {a+b \text {arctanh}(c x)}{2 d^3 (1+c x)^2}+\frac {a+b \text {arctanh}(c x)}{d^3 (1+c x)}+\frac {a \log (x)}{d^3}+\frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b \operatorname {PolyLog}(2,-c x)}{2 d^3}+\frac {b \operatorname {PolyLog}(2,c x)}{2 d^3}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 d^3} \] Output:

1/8*b/d^3/(c*x+1)^2+5/8*b/d^3/(c*x+1)-5/8*b*arctanh(c*x)/d^3+1/2*(a+b*arct 
anh(c*x))/d^3/(c*x+1)^2+(a+b*arctanh(c*x))/d^3/(c*x+1)+a*ln(x)/d^3+(a+b*ar 
ctanh(c*x))*ln(2/(c*x+1))/d^3-1/2*b*polylog(2,-c*x)/d^3+1/2*b*polylog(2,c* 
x)/d^3-1/2*b*polylog(2,1-2/(c*x+1))/d^3
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \text {arctanh}(c x)}{x (d+c d x)^3} \, dx=\frac {\frac {16 a}{(1+c x)^2}+\frac {32 a}{1+c x}+32 a \log (x)-32 a \log (1+c x)+b \left (12 \cosh (2 \text {arctanh}(c x))+\cosh (4 \text {arctanh}(c x))-16 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c x)}\right )-12 \sinh (2 \text {arctanh}(c x))+4 \text {arctanh}(c x) \left (6 \cosh (2 \text {arctanh}(c x))+\cosh (4 \text {arctanh}(c x))+8 \log \left (1-e^{-2 \text {arctanh}(c x)}\right )-6 \sinh (2 \text {arctanh}(c x))-\sinh (4 \text {arctanh}(c x))\right )-\sinh (4 \text {arctanh}(c x))\right )}{32 d^3} \] Input:

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

Output:

((16*a)/(1 + c*x)^2 + (32*a)/(1 + c*x) + 32*a*Log[x] - 32*a*Log[1 + c*x] + 
 b*(12*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] - 16*PolyLog[2, E^(-2*A 
rcTanh[c*x])] - 12*Sinh[2*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(6*Cosh[2*ArcTanh 
[c*x]] + Cosh[4*ArcTanh[c*x]] + 8*Log[1 - E^(-2*ArcTanh[c*x])] - 6*Sinh[2* 
ArcTanh[c*x]] - Sinh[4*ArcTanh[c*x]]) - Sinh[4*ArcTanh[c*x]]))/(32*d^3)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 161, 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.

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

\(\Big \downarrow \) 6502

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a+b \text {arctanh}(c x)}{d^3 (c x+1)}+\frac {a+b \text {arctanh}(c x)}{2 d^3 (c x+1)^2}+\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{d^3}+\frac {a \log (x)}{d^3}-\frac {5 b \text {arctanh}(c x)}{8 d^3}-\frac {b \operatorname {PolyLog}(2,-c x)}{2 d^3}+\frac {b \operatorname {PolyLog}(2,c x)}{2 d^3}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 d^3}+\frac {5 b}{8 d^3 (c x+1)}+\frac {b}{8 d^3 (c x+1)^2}\)

Input:

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

Output:

b/(8*d^3*(1 + c*x)^2) + (5*b)/(8*d^3*(1 + c*x)) - (5*b*ArcTanh[c*x])/(8*d^ 
3) + (a + b*ArcTanh[c*x])/(2*d^3*(1 + c*x)^2) + (a + b*ArcTanh[c*x])/(d^3* 
(1 + c*x)) + (a*Log[x])/d^3 + ((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d^3 
- (b*PolyLog[2, -(c*x)])/(2*d^3) + (b*PolyLog[2, c*x])/(2*d^3) - (b*PolyLo 
g[2, 1 - 2/(1 + c*x)])/(2*d^3)
 

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 = 0.42 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.18

method	result	size
parts	\(\frac {a \left (\frac {1}{2 \left (c x +1\right )^{2}}+\frac {1}{c x +1}-\ln \left (c x +1\right )+\ln \left (x \right )\right )}{d^{3}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}+\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {5 \ln \left (c x -1\right )}{16}+\frac {1}{8 \left (c x +1\right )^{2}}+\frac {5}{8 \left (c x +1\right )}-\frac {5 \ln \left (c x +1\right )}{16}\right )}{d^{3}}\)	\(190\)
derivativedivides	\(\frac {a \left (\ln \left (c x \right )+\frac {1}{2 \left (c x +1\right )^{2}}+\frac {1}{c x +1}-\ln \left (c x +1\right )\right )}{d^{3}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}+\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {5 \ln \left (c x -1\right )}{16}+\frac {1}{8 \left (c x +1\right )^{2}}+\frac {5}{8 \left (c x +1\right )}-\frac {5 \ln \left (c x +1\right )}{16}\right )}{d^{3}}\)	\(192\)
default	\(\frac {a \left (\ln \left (c x \right )+\frac {1}{2 \left (c x +1\right )^{2}}+\frac {1}{c x +1}-\ln \left (c x +1\right )\right )}{d^{3}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}+\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\operatorname {dilog}\left (c x \right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {5 \ln \left (c x -1\right )}{16}+\frac {1}{8 \left (c x +1\right )^{2}}+\frac {5}{8 \left (c x +1\right )}-\frac {5 \ln \left (c x +1\right )}{16}\right )}{d^{3}}\)	\(192\)
risch	\(-\frac {5 b \ln \left (-c x -1\right )}{16 d^{3}}-\frac {b \ln \left (-c x +1\right ) c x}{4 d^{3} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right )}{4 d^{3} \left (-c x -1\right )}+\frac {b \operatorname {dilog}\left (-c x +1\right )}{2 d^{3}}-\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}+\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d^{3}}-\frac {b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}-\frac {b}{8 d^{3} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right ) c^{2} x^{2}}{16 d^{3} \left (-c x -1\right )^{2}}+\frac {b \ln \left (-c x +1\right ) c x}{8 d^{3} \left (-c x -1\right )^{2}}-\frac {3 b \ln \left (-c x +1\right )}{16 d^{3} \left (-c x -1\right )^{2}}+\frac {a \ln \left (-c x \right )}{d^{3}}+\frac {a}{2 d^{3} \left (-c x -1\right )^{2}}-\frac {a \ln \left (-c x -1\right )}{d^{3}}-\frac {a}{d^{3} \left (-c x -1\right )}-\frac {b \ln \left (c x +1\right )^{2}}{4 d^{3}}-\frac {b \operatorname {dilog}\left (c x +1\right )}{2 d^{3}}+\frac {b \ln \left (c x +1\right )}{2 d^{3} \left (c x +1\right )}+\frac {b}{2 d^{3} \left (c x +1\right )}+\frac {b \ln \left (c x +1\right )}{4 d^{3} \left (c x +1\right )^{2}}+\frac {b}{8 d^{3} \left (c x +1\right )^{2}}\)	\(351\)

Input:

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

Output:

a/d^3*(1/2/(c*x+1)^2+1/(c*x+1)-ln(c*x+1)+ln(x))+b/d^3*(arctanh(c*x)*ln(c*x 
)+1/2/(c*x+1)^2*arctanh(c*x)+1/(c*x+1)*arctanh(c*x)-arctanh(c*x)*ln(c*x+1) 
-1/2*dilog(c*x)-1/2*dilog(c*x+1)-1/2*ln(c*x)*ln(c*x+1)+1/4*ln(c*x+1)^2-1/2 
*(ln(c*x+1)-ln(1/2*c*x+1/2))*ln(-1/2*c*x+1/2)+1/2*dilog(1/2*c*x+1/2)+5/16* 
ln(c*x-1)+1/8/(c*x+1)^2+5/8/(c*x+1)-5/16*ln(c*x+1))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \text {arctanh}(c x)}{x (d+c d x)^3} \, dx=\frac {-4 \mathit {atanh} \left (c x \right )^{2} b \,c^{2} x^{2}-8 \mathit {atanh} \left (c x \right )^{2} b c x -4 \mathit {atanh} \left (c x \right )^{2} b +8 \mathit {atanh} \left (c x \right ) b c x +16 \mathit {atanh} \left (c x \right ) b -32 \left (\int \frac {\mathit {atanh} \left (c x \right )}{c^{4} x^{5}+2 c^{3} x^{4}-2 c \,x^{2}-x}d x \right ) b \,c^{2} x^{2}-64 \left (\int \frac {\mathit {atanh} \left (c x \right )}{c^{4} x^{5}+2 c^{3} x^{4}-2 c \,x^{2}-x}d x \right ) b c x -32 \left (\int \frac {\mathit {atanh} \left (c x \right )}{c^{4} x^{5}+2 c^{3} x^{4}-2 c \,x^{2}-x}d x \right ) b +3 \,\mathrm {log}\left (c x -1\right ) b \,c^{2} x^{2}+6 \,\mathrm {log}\left (c x -1\right ) b c x +3 \,\mathrm {log}\left (c x -1\right ) b -32 \,\mathrm {log}\left (c x +1\right ) a \,c^{2} x^{2}-64 \,\mathrm {log}\left (c x +1\right ) a c x -32 \,\mathrm {log}\left (c x +1\right ) a -3 \,\mathrm {log}\left (c x +1\right ) b \,c^{2} x^{2}-6 \,\mathrm {log}\left (c x +1\right ) b c x -3 \,\mathrm {log}\left (c x +1\right ) b +32 \,\mathrm {log}\left (x \right ) a \,c^{2} x^{2}+64 \,\mathrm {log}\left (x \right ) a c x +32 \,\mathrm {log}\left (x \right ) a -16 a \,c^{2} x^{2}+32 a -3 b \,c^{2} x^{2}+5 b}{32 d^{3} \left (c^{2} x^{2}+2 c x +1\right )} \] Input:

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

Output:

( - 4*atanh(c*x)**2*b*c**2*x**2 - 8*atanh(c*x)**2*b*c*x - 4*atanh(c*x)**2* 
b + 8*atanh(c*x)*b*c*x + 16*atanh(c*x)*b - 32*int(atanh(c*x)/(c**4*x**5 + 
2*c**3*x**4 - 2*c*x**2 - x),x)*b*c**2*x**2 - 64*int(atanh(c*x)/(c**4*x**5 
+ 2*c**3*x**4 - 2*c*x**2 - x),x)*b*c*x - 32*int(atanh(c*x)/(c**4*x**5 + 2* 
c**3*x**4 - 2*c*x**2 - x),x)*b + 3*log(c*x - 1)*b*c**2*x**2 + 6*log(c*x - 
1)*b*c*x + 3*log(c*x - 1)*b - 32*log(c*x + 1)*a*c**2*x**2 - 64*log(c*x + 1 
)*a*c*x - 32*log(c*x + 1)*a - 3*log(c*x + 1)*b*c**2*x**2 - 6*log(c*x + 1)* 
b*c*x - 3*log(c*x + 1)*b + 32*log(x)*a*c**2*x**2 + 64*log(x)*a*c*x + 32*lo 
g(x)*a - 16*a*c**2*x**2 + 32*a - 3*b*c**2*x**2 + 5*b)/(32*d**3*(c**2*x**2 
+ 2*c*x + 1))