\(\int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx\) [59]

Optimal result

Integrand size = 20, antiderivative size = 194 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {a x}{c^3 d^3}+\frac {b}{8 c^4 d^3 (1+c x)^2}-\frac {11 b}{8 c^4 d^3 (1+c x)}+\frac {11 b \text {arctanh}(c x)}{8 c^4 d^3}+\frac {b x \text {arctanh}(c x)}{c^3 d^3}+\frac {a+b \text {arctanh}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac {3 (a+b \text {arctanh}(c x))}{c^4 d^3 (1+c x)}+\frac {3 (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c^4 d^3} \] Output:

a*x/c^3/d^3+1/8*b/c^4/d^3/(c*x+1)^2-11/8*b/c^4/d^3/(c*x+1)+11/8*b*arctanh( 
c*x)/c^4/d^3+b*x*arctanh(c*x)/c^3/d^3+1/2*(a+b*arctanh(c*x))/c^4/d^3/(c*x+ 
1)^2-3*(a+b*arctanh(c*x))/c^4/d^3/(c*x+1)+3*(a+b*arctanh(c*x))*ln(2/(c*x+1 
))/c^4/d^3+1/2*b*ln(-c^2*x^2+1)/c^4/d^3-3/2*b*polylog(2,1-2/(c*x+1))/c^4/d 
^3
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {32 a c x+\frac {16 a}{(1+c x)^2}-\frac {96 a}{1+c x}-96 a \log (1+c x)+b \left (-20 \cosh (2 \text {arctanh}(c x))+\cosh (4 \text {arctanh}(c x))+16 \log \left (1-c^2 x^2\right )-48 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+20 \sinh (2 \text {arctanh}(c x))+4 \text {arctanh}(c x) \left (8 c x-10 \cosh (2 \text {arctanh}(c x))+\cosh (4 \text {arctanh}(c x))+24 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )+10 \sinh (2 \text {arctanh}(c x))-\sinh (4 \text {arctanh}(c x))\right )-\sinh (4 \text {arctanh}(c x))\right )}{32 c^4 d^3} \] Input:

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

Output:

(32*a*c*x + (16*a)/(1 + c*x)^2 - (96*a)/(1 + c*x) - 96*a*Log[1 + c*x] + b* 
(-20*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 16*Log[1 - c^2*x^2] - 4 
8*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 20*Sinh[2*ArcTanh[c*x]] + 4*ArcTanh[c 
*x]*(8*c*x - 10*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 24*Log[1 + E 
^(-2*ArcTanh[c*x])] + 10*Sinh[2*ArcTanh[c*x]] - Sinh[4*ArcTanh[c*x]]) - Si 
nh[4*ArcTanh[c*x]]))/(32*c^4*d^3)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 194, 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.

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

\(\Big \downarrow \) 6502

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 (a+b \text {arctanh}(c x))}{c^4 d^3 (c x+1)}+\frac {a+b \text {arctanh}(c x)}{2 c^4 d^3 (c x+1)^2}+\frac {3 \log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c^4 d^3}+\frac {a x}{c^3 d^3}+\frac {11 b \text {arctanh}(c x)}{8 c^4 d^3}+\frac {b x \text {arctanh}(c x)}{c^3 d^3}-\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^4 d^3}-\frac {11 b}{8 c^4 d^3 (c x+1)}+\frac {b}{8 c^4 d^3 (c x+1)^2}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}\)

Input:

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

Output:

(a*x)/(c^3*d^3) + b/(8*c^4*d^3*(1 + c*x)^2) - (11*b)/(8*c^4*d^3*(1 + c*x)) 
 + (11*b*ArcTanh[c*x])/(8*c^4*d^3) + (b*x*ArcTanh[c*x])/(c^3*d^3) + (a + b 
*ArcTanh[c*x])/(2*c^4*d^3*(1 + c*x)^2) - (3*(a + b*ArcTanh[c*x]))/(c^4*d^3 
*(1 + c*x)) + (3*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(c^4*d^3) + (b*Log 
[1 - c^2*x^2])/(2*c^4*d^3) - (3*b*PolyLog[2, 1 - 2/(1 + c*x)])/(2*c^4*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 = 170, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {a \left (c x +\frac {1}{2 \left (c x +1\right )^{2}}-\frac {3}{c x +1}-3 \ln \left (c x +1\right )\right )}{d^{3}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x +1}-3 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {3 \ln \left (c x +1\right )^{2}}{4}-\frac {3 \left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {3 \operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {1}{8 \left (c x +1\right )^{2}}-\frac {11}{8 \left (c x +1\right )}+\frac {19 \ln \left (c x +1\right )}{16}-\frac {3 \ln \left (c x -1\right )}{16}\right )}{d^{3}}}{c^{4}}\)	\(170\)
default	\(\frac {\frac {a \left (c x +\frac {1}{2 \left (c x +1\right )^{2}}-\frac {3}{c x +1}-3 \ln \left (c x +1\right )\right )}{d^{3}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x +1}-3 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {3 \ln \left (c x +1\right )^{2}}{4}-\frac {3 \left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {3 \operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {1}{8 \left (c x +1\right )^{2}}-\frac {11}{8 \left (c x +1\right )}+\frac {19 \ln \left (c x +1\right )}{16}-\frac {3 \ln \left (c x -1\right )}{16}\right )}{d^{3}}}{c^{4}}\)	\(170\)
parts	\(\frac {a \left (\frac {x}{c^{3}}-\frac {3}{c^{4} \left (c x +1\right )}-\frac {3 \ln \left (c x +1\right )}{c^{4}}+\frac {1}{2 c^{4} \left (c x +1\right )^{2}}\right )}{d^{3}}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x +1}-3 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+\frac {3 \ln \left (c x +1\right )^{2}}{4}-\frac {3 \left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {3 \operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {1}{8 \left (c x +1\right )^{2}}-\frac {11}{8 \left (c x +1\right )}+\frac {19 \ln \left (c x +1\right )}{16}-\frac {3 \ln \left (c x -1\right )}{16}\right )}{d^{3} c^{4}}\)	\(180\)
risch	\(\frac {a}{2 d^{3} c^{4} \left (-c x -1\right )^{2}}+\frac {b \ln \left (-c x +1\right )}{2 d^{3} c^{4}}+\frac {11 b \ln \left (-c x -1\right )}{16 d^{3} c^{4}}-\frac {3 b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3} c^{4}}-\frac {3 a \ln \left (-c x -1\right )}{d^{3} c^{4}}+\frac {b \ln \left (-c x +1\right ) x}{8 d^{3} c^{3} \left (-c x -1\right )^{2}}+\frac {3 b \ln \left (-c x +1\right ) x}{4 d^{3} c^{3} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right ) x^{2}}{16 d^{3} c^{2} \left (-c x -1\right )^{2}}-\frac {3 b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3} c^{4}}+\frac {3 b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d^{3} c^{4}}-\frac {3 b \ln \left (-c x +1\right )}{16 d^{3} c^{4} \left (-c x -1\right )^{2}}-\frac {b \ln \left (-c x +1\right ) x}{2 d^{3} c^{3}}-\frac {3 b \ln \left (-c x +1\right )}{4 d^{3} c^{4} \left (-c x -1\right )}+\frac {b \ln \left (c x +1\right )}{2 c^{4} d^{3}}-\frac {b}{2 c^{4} d^{3}}+\frac {a x}{c^{3} d^{3}}-\frac {3 b \ln \left (c x +1\right )^{2}}{4 c^{4} d^{3}}-\frac {b}{8 d^{3} c^{4} \left (-c x -1\right )}+\frac {3 a}{d^{3} c^{4} \left (-c x -1\right )}+\left (\frac {b x}{2 c^{3} d^{3}}+\frac {-3 b \,d^{3} x -\frac {5 d^{3} b}{2 c}}{2 c^{3} d^{6} \left (c x +1\right )^{2}}\right ) \ln \left (c x +1\right )-\frac {a}{d^{3} c^{4}}+\frac {b}{8 c^{4} d^{3} \left (c x +1\right )^{2}}-\frac {3 b}{2 c^{4} d^{3} \left (c x +1\right )}\)	\(446\)

Input:

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

Output:

1/c^4*(a/d^3*(c*x+1/2/(c*x+1)^2-3/(c*x+1)-3*ln(c*x+1))+b/d^3*(arctanh(c*x) 
*c*x+1/2/(c*x+1)^2*arctanh(c*x)-3/(c*x+1)*arctanh(c*x)-3*arctanh(c*x)*ln(c 
*x+1)+3/4*ln(c*x+1)^2-3/2*(ln(c*x+1)-ln(1/2*c*x+1/2))*ln(-1/2*c*x+1/2)+3/2 
*dilog(1/2*c*x+1/2)+1/8/(c*x+1)^2-11/8/(c*x+1)+19/16*ln(c*x+1)-3/16*ln(c*x 
-1)))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {\int \frac {a x^{3}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b x^{3} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \] Input:

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

Output:

(Integral(a*x**3/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b*x* 
*3*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3
 

Maxima [F]

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

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

Output:

-1/32*(2*c^4*(2*(7*c*x + 6)/(c^10*d^3*x^2 + 2*c^9*d^3*x + c^8*d^3) - 8*x/( 
c^7*d^3) + 17*log(c*x + 1)/(c^8*d^3) - log(c*x - 1)/(c^8*d^3)) - 32*c^4*in 
tegrate(1/2*x^4*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - 
c^3*d^3), x) - 6*c^3*(2*(5*c*x + 4)/(c^9*d^3*x^2 + 2*c^8*d^3*x + c^7*d^3) 
+ 7*log(c*x + 1)/(c^7*d^3) + log(c*x - 1)/(c^7*d^3)) + 128*c^3*integrate(1 
/2*x^3*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), 
 x) + 288*c^2*integrate(1/2*x^2*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 
- 2*c^4*d^3*x - c^3*d^3), x) + 9*c*(2*x/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d 
^3) - log(c*x + 1)/(c^5*d^3) + log(c*x - 1)/(c^5*d^3)) + 288*c*integrate(1 
/2*x*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x 
) + 8*(2*c^3*x^3 + 4*c^2*x^2 - 4*c*x - 6*(c^2*x^2 + 2*c*x + 1)*log(c*x + 1 
) - 5)*log(-c*x + 1)/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) + 10*(c*x + 2)/ 
(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - 5*log(c*x + 1)/(c^4*d^3) + 5*log(c 
*x - 1)/(c^4*d^3) + 96*integrate(1/2*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3 
*x^3 - 2*c^4*d^3*x - c^3*d^3), x))*b - 1/2*a*((6*c*x + 5)/(c^6*d^3*x^2 + 2 
*c^5*d^3*x + c^4*d^3) - 2*x/(c^3*d^3) + 6*log(c*x + 1)/(c^4*d^3))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {2 \left (\int \frac {\mathit {atanh} \left (c x \right ) x^{3}}{c^{3} x^{3}+3 c^{2} x^{2}+3 c x +1}d x \right ) b \,c^{6} x^{2}+4 \left (\int \frac {\mathit {atanh} \left (c x \right ) x^{3}}{c^{3} x^{3}+3 c^{2} x^{2}+3 c x +1}d x \right ) b \,c^{5} x +2 \left (\int \frac {\mathit {atanh} \left (c x \right ) x^{3}}{c^{3} x^{3}+3 c^{2} x^{2}+3 c x +1}d x \right ) b \,c^{4}-6 \,\mathrm {log}\left (c x +1\right ) a \,c^{2} x^{2}-12 \,\mathrm {log}\left (c x +1\right ) a c x -6 \,\mathrm {log}\left (c x +1\right ) a +2 a \,c^{3} x^{3}+6 a \,c^{2} x^{2}-3 a}{2 c^{4} d^{3} \left (c^{2} x^{2}+2 c x +1\right )} \] Input:

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

Output:

(2*int((atanh(c*x)*x**3)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1),x)*b*c**6*x 
**2 + 4*int((atanh(c*x)*x**3)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1),x)*b*c 
**5*x + 2*int((atanh(c*x)*x**3)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1),x)*b 
*c**4 - 6*log(c*x + 1)*a*c**2*x**2 - 12*log(c*x + 1)*a*c*x - 6*log(c*x + 1 
)*a + 2*a*c**3*x**3 + 6*a*c**2*x**2 - 3*a)/(2*c**4*d**3*(c**2*x**2 + 2*c*x 
 + 1))