\(\int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx\) [51]

Optimal result

Integrand size = 20, antiderivative size = 181 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=-\frac {2 a x}{c^3 d^2}+\frac {b x}{2 c^3 d^2}+\frac {b}{2 c^4 d^2 (1+c x)}-\frac {b \text {arctanh}(c x)}{c^4 d^2}-\frac {2 b x \text {arctanh}(c x)}{c^3 d^2}+\frac {x^2 (a+b \text {arctanh}(c x))}{2 c^2 d^2}+\frac {a+b \text {arctanh}(c x)}{c^4 d^2 (1+c x)}-\frac {3 (a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{c^4 d^2}-\frac {b \log \left (1-c^2 x^2\right )}{c^4 d^2}+\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c^4 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.78 \[ \int \frac {x^3 (a+b \text {arctanh}(c x))}{(d+c d x)^2} \, dx=\frac {-8 a c x+2 a c^2 x^2+\frac {4 a}{1+c x}+12 a \log (1+c x)+b \left (2 c x+\cosh (2 \text {arctanh}(c x))-4 \log \left (1-c^2 x^2\right )+6 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )+2 \text {arctanh}(c x) \left (-1-4 c x+c^2 x^2+\cosh (2 \text {arctanh}(c x))-6 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-\sinh (2 \text {arctanh}(c x))\right )-\sinh (2 \text {arctanh}(c x))\right )}{4 c^4 d^2} \] Input:

Integrate[(x^3*(a + b*ArcTanh[c*x]))/(d + c*d*x)^2,x]
 

Output:

(-8*a*c*x + 2*a*c^2*x^2 + (4*a)/(1 + c*x) + 12*a*Log[1 + c*x] + b*(2*c*x + 
 Cosh[2*ArcTanh[c*x]] - 4*Log[1 - c^2*x^2] + 6*PolyLog[2, -E^(-2*ArcTanh[c 
*x])] + 2*ArcTanh[c*x]*(-1 - 4*c*x + c^2*x^2 + Cosh[2*ArcTanh[c*x]] - 6*Lo 
g[1 + E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]]) - Sinh[2*ArcTanh[c*x]]) 
)/(4*c^4*d^2)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 181, 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.

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

\(\Big \downarrow \) 6502

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

\(\Big \downarrow \) 2009

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

Input:

Int[(x^3*(a + b*ArcTanh[c*x]))/(d + c*d*x)^2,x]
 

Output:

(-2*a*x)/(c^3*d^2) + (b*x)/(2*c^3*d^2) + b/(2*c^4*d^2*(1 + c*x)) - (b*ArcT 
anh[c*x])/(c^4*d^2) - (2*b*x*ArcTanh[c*x])/(c^3*d^2) + (x^2*(a + b*ArcTanh 
[c*x]))/(2*c^2*d^2) + (a + b*ArcTanh[c*x])/(c^4*d^2*(1 + c*x)) - (3*(a + b 
*ArcTanh[c*x])*Log[2/(1 + c*x)])/(c^4*d^2) - (b*Log[1 - c^2*x^2])/(c^4*d^2 
) + (3*b*PolyLog[2, 1 - 2/(1 + c*x)])/(2*c^4*d^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 = 0.35 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {\frac {a \left (\frac {c^{2} x^{2}}{2}-2 c x +\frac {1}{c x +1}+3 \ln \left (c x +1\right )\right )}{d^{2}}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}-2 \,\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}+3 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {3 \ln \left (c x +1\right )^{2}}{4}+\frac {3 \left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {3 \operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {c x}{2}+\frac {1}{2}+\frac {1}{2 c x +2}-\frac {3 \ln \left (c x +1\right )}{2}-\frac {\ln \left (c x -1\right )}{2}\right )}{d^{2}}}{c^{4}}\)	\(163\)
default	\(\frac {\frac {a \left (\frac {c^{2} x^{2}}{2}-2 c x +\frac {1}{c x +1}+3 \ln \left (c x +1\right )\right )}{d^{2}}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}-2 \,\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}+3 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {3 \ln \left (c x +1\right )^{2}}{4}+\frac {3 \left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {3 \operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {c x}{2}+\frac {1}{2}+\frac {1}{2 c x +2}-\frac {3 \ln \left (c x +1\right )}{2}-\frac {\ln \left (c x -1\right )}{2}\right )}{d^{2}}}{c^{4}}\)	\(163\)
parts	\(\frac {a \left (\frac {\frac {1}{2} c \,x^{2}-2 x}{c^{3}}+\frac {1}{c^{4} \left (c x +1\right )}+\frac {3 \ln \left (c x +1\right )}{c^{4}}\right )}{d^{2}}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{2} x^{2}}{2}-2 \,\operatorname {arctanh}\left (c x \right ) c x +\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}+3 \,\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {3 \ln \left (c x +1\right )^{2}}{4}+\frac {3 \left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {3 \operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {c x}{2}+\frac {1}{2}+\frac {1}{2 c x +2}-\frac {3 \ln \left (c x +1\right )}{2}-\frac {\ln \left (c x -1\right )}{2}\right )}{d^{2} c^{4}}\)	\(171\)
risch	\(\frac {3 b \ln \left (c x +1\right )^{2}}{4 c^{4} d^{2}}+\left (\frac {b \left (\frac {1}{2} c \,x^{2}-2 x \right )}{2 c^{3} d^{2}}+\frac {b}{2 c^{4} d^{2} \left (c x +1\right )}\right ) \ln \left (c x +1\right )+\frac {b x}{2 c^{3} d^{2}}+\frac {b}{2 c^{4} d^{2} \left (c x +1\right )}-\frac {5 b \ln \left (c x +1\right )}{4 c^{4} d^{2}}+\frac {a \,x^{2}}{2 d^{2} c^{2}}-\frac {2 a x}{c^{3} d^{2}}+\frac {3 a}{2 d^{2} c^{4}}-\frac {a}{d^{2} c^{4} \left (-c x -1\right )}+\frac {3 a \ln \left (-c x -1\right )}{d^{2} c^{4}}-\frac {b \ln \left (-c x +1\right ) x^{2}}{4 d^{2} c^{2}}+\frac {b \ln \left (-c x +1\right ) x}{d^{2} c^{3}}-\frac {3 b \ln \left (-c x +1\right )}{4 d^{2} c^{4}}+\frac {5 b}{8 c^{4} d^{2}}-\frac {b \ln \left (-c x -1\right )}{4 d^{2} c^{4}}-\frac {b \ln \left (-c x +1\right ) x}{4 d^{2} c^{3} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right )}{4 d^{2} c^{4} \left (-c x -1\right )}+\frac {3 b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{2} c^{4}}-\frac {3 b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d^{2} c^{4}}+\frac {3 b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{2} c^{4}}\)	\(354\)

Input:

int(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

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

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^2,x, algorithm="maxima")
 

Output:

1/16*(c^4*(2/(c^9*d^2*x + c^8*d^2) + 2*(c*x^2 - 2*x)/(c^7*d^2) + 7*log(c*x 
 + 1)/(c^8*d^2) + log(c*x - 1)/(c^8*d^2)) + 16*c^4*integrate(1/2*x^4*log(c 
*x + 1)/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2*x - c^3*d^2), x) + 2*c^3*(2/( 
c^8*d^2*x + c^7*d^2) - 4*x/(c^6*d^2) + 5*log(c*x + 1)/(c^7*d^2) - log(c*x 
- 1)/(c^7*d^2)) - 16*c^3*integrate(1/2*x^3*log(c*x + 1)/(c^6*d^2*x^3 + c^5 
*d^2*x^2 - c^4*d^2*x - c^3*d^2), x) - 7*c^2*(2/(c^7*d^2*x + c^6*d^2) + 3*l 
og(c*x + 1)/(c^6*d^2) + log(c*x - 1)/(c^6*d^2)) + 48*c^2*integrate(1/2*x^2 
*log(c*x + 1)/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2*x - c^3*d^2), x) + 2*c* 
(2/(c^6*d^2*x + c^5*d^2) + log(c*x + 1)/(c^5*d^2) - log(c*x - 1)/(c^5*d^2) 
) + 96*c*integrate(1/2*x*log(c*x + 1)/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2 
*x - c^3*d^2), x) - 4*(c^3*x^3 - 3*c^2*x^2 - 4*c*x + 6*(c*x + 1)*log(c*x + 
 1) + 2)*log(-c*x + 1)/(c^5*d^2*x + c^4*d^2) + 4/(c^5*d^2*x + c^4*d^2) - 2 
*log(c*x + 1)/(c^4*d^2) + 2*log(c*x - 1)/(c^4*d^2) + 48*integrate(1/2*log( 
c*x + 1)/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2*x - c^3*d^2), x))*b + 1/2*a* 
(2/(c^5*d^2*x + c^4*d^2) + (c*x^2 - 4*x)/(c^3*d^2) + 6*log(c*x + 1)/(c^4*d 
^2))
 

Giac [F]

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

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((x^3*(a + b*atanh(c*x)))/(d + c*d*x)^2,x)
 

Output:

int((x^3*(a + b*atanh(c*x)))/(d + c*d*x)^2, x)
 

Reduce [F]

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

int(x^3*(a+b*atanh(c*x))/(c*d*x+d)^2,x)
 

Output:

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