3.1.23 \(\int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx\) [23]

3.1.23.1 Optimal result
3.1.23.2 Mathematica [A] (verified)
3.1.23.3 Rubi [A] (verified)
3.1.23.4 Maple [A] (verified)
3.1.23.5 Fricas [A] (verification not implemented)
3.1.23.6 Sympy [B] (verification not implemented)
3.1.23.7 Maxima [B] (verification not implemented)
3.1.23.8 Giac [B] (verification not implemented)
3.1.23.9 Mupad [B] (verification not implemented)

3.1.23.1 Optimal result

Integrand size = 17, antiderivative size = 84 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=b d^3 x+\frac {b d^3 (1+c x)^2}{4 c}+\frac {b d^3 (1+c x)^3}{12 c}+\frac {d^3 (1+c x)^4 (a+b \text {arctanh}(c x))}{4 c}+\frac {2 b d^3 \log (1-c x)}{c} \]

output
b*d^3*x+1/4*b*d^3*(c*x+1)^2/c+1/12*b*d^3*(c*x+1)^3/c+1/4*d^3*(c*x+1)^4*(a+ 
b*arctanh(c*x))/c+2*b*d^3*ln(-c*x+1)/c
 
3.1.23.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.37 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^3 \left (24 a c x+42 b c x+36 a c^2 x^2+12 b c^2 x^2+24 a c^3 x^3+2 b c^3 x^3+6 a c^4 x^4+6 b c x \left (4+6 c x+4 c^2 x^2+c^3 x^3\right ) \text {arctanh}(c x)+45 b \log (1-c x)+3 b \log (1+c x)\right )}{24 c} \]

input
Integrate[(d + c*d*x)^3*(a + b*ArcTanh[c*x]),x]
 
output
(d^3*(24*a*c*x + 42*b*c*x + 36*a*c^2*x^2 + 12*b*c^2*x^2 + 24*a*c^3*x^3 + 2 
*b*c^3*x^3 + 6*a*c^4*x^4 + 6*b*c*x*(4 + 6*c*x + 4*c^2*x^2 + c^3*x^3)*ArcTa 
nh[c*x] + 45*b*Log[1 - c*x] + 3*b*Log[1 + c*x]))/(24*c)
 
3.1.23.3 Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6478, 27, 456, 49, 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.

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

\(\Big \downarrow \) 6478

\(\displaystyle \frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c}-\frac {b \int \frac {d^4 (c x+1)^4}{1-c^2 x^2}dx}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c}-\frac {1}{4} b d^3 \int \frac {(c x+1)^4}{1-c^2 x^2}dx\)

\(\Big \downarrow \) 456

\(\displaystyle \frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c}-\frac {1}{4} b d^3 \int \frac {(c x+1)^3}{1-c x}dx\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {d^3 (c x+1)^4 (a+b \text {arctanh}(c x))}{4 c}-\frac {1}{4} b d^3 \int \left (-(c x+1)^2-2 (c x+1)+\frac {8}{1-c x}-4\right )dx\)

\(\Big \downarrow \) 2009

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

input
Int[(d + c*d*x)^3*(a + b*ArcTanh[c*x]),x]
 
output
(d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*c) - (b*d^3*(-4*x - (1 + c*x)^2/ 
c - (1 + c*x)^3/(3*c) - (8*Log[1 - c*x])/c))/4
 

3.1.23.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 456
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && 
EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] &&  !Integ 
erQ[n]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6478
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTanh[c*x])/(e*(q + 1))), x] - Simp[b 
*(c/(e*(q + 1)))   Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ[{a, 
 b, c, d, e, q}, x] && NeQ[q, -1]
 
3.1.23.4 Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20

method result size
derivativedivides \(\frac {\frac {d^{3} a \left (c x +1\right )^{4}}{4}+d^{3} b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+c^{3} x^{3} \operatorname {arctanh}\left (c x \right )+\frac {3 c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c^{2} x^{2}}{2}+\frac {7 c x}{4}+2 \ln \left (c x -1\right )\right )}{c}\) \(101\)
default \(\frac {\frac {d^{3} a \left (c x +1\right )^{4}}{4}+d^{3} b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+c^{3} x^{3} \operatorname {arctanh}\left (c x \right )+\frac {3 c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c^{2} x^{2}}{2}+\frac {7 c x}{4}+2 \ln \left (c x -1\right )\right )}{c}\) \(101\)
parts \(\frac {d^{3} a \left (c x +1\right )^{4}}{4 c}+\frac {d^{3} b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+c^{3} x^{3} \operatorname {arctanh}\left (c x \right )+\frac {3 c^{2} x^{2} \operatorname {arctanh}\left (c x \right )}{2}+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c^{2} x^{2}}{2}+\frac {7 c x}{4}+2 \ln \left (c x -1\right )\right )}{c}\) \(103\)
parallelrisch \(\frac {3 x^{4} \operatorname {arctanh}\left (c x \right ) b \,c^{4} d^{3}+3 a \,c^{4} d^{3} x^{4}+12 x^{3} \operatorname {arctanh}\left (c x \right ) b \,d^{3} c^{3}+12 a \,c^{3} d^{3} x^{3}+b \,c^{3} d^{3} x^{3}+18 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{3}+18 a \,c^{2} d^{3} x^{2}+6 b \,c^{2} d^{3} x^{2}+12 b c \,d^{3} x \,\operatorname {arctanh}\left (c x \right )+12 d^{3} a c x +21 b c \,d^{3} x +24 \ln \left (c x -1\right ) b \,d^{3}+3 b \,d^{3} \operatorname {arctanh}\left (c x \right )}{12 c}\) \(164\)
risch \(\frac {d^{3} \left (c x +1\right )^{4} b \ln \left (c x +1\right )}{8 c}-\frac {d^{3} c^{3} b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {a \,c^{3} d^{3} x^{4}}{4}-\frac {d^{3} c^{2} b \,x^{3} \ln \left (-c x +1\right )}{2}+a \,c^{2} d^{3} x^{3}+\frac {b \,c^{2} d^{3} x^{3}}{12}-\frac {3 d^{3} c b \,x^{2} \ln \left (-c x +1\right )}{4}+\frac {3 a c \,d^{3} x^{2}}{2}+\frac {b c \,d^{3} x^{2}}{2}-\frac {b \,d^{3} x \ln \left (-c x +1\right )}{2}+a \,d^{3} x +\frac {7 b \,d^{3} x}{4}-\frac {\ln \left (-c x +1\right ) b \,d^{3}}{8 c}+\frac {2 d^{3} b \ln \left (c x -1\right )}{c}\) \(192\)

input
int((c*d*x+d)^3*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
 
output
1/c*(1/4*d^3*a*(c*x+1)^4+d^3*b*(1/4*c^4*x^4*arctanh(c*x)+c^3*x^3*arctanh(c 
*x)+3/2*c^2*x^2*arctanh(c*x)+c*x*arctanh(c*x)+1/4*arctanh(c*x)+1/12*c^3*x^ 
3+1/2*c^2*x^2+7/4*c*x+2*ln(c*x-1)))
 
3.1.23.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.77 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {6 \, a c^{4} d^{3} x^{4} + 2 \, {\left (12 \, a + b\right )} c^{3} d^{3} x^{3} + 12 \, {\left (3 \, a + b\right )} c^{2} d^{3} x^{2} + 6 \, {\left (4 \, a + 7 \, b\right )} c d^{3} x + 3 \, b d^{3} \log \left (c x + 1\right ) + 45 \, b d^{3} \log \left (c x - 1\right ) + 3 \, {\left (b c^{4} d^{3} x^{4} + 4 \, b c^{3} d^{3} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c d^{3} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c} \]

input
integrate((c*d*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="fricas")
 
output
1/24*(6*a*c^4*d^3*x^4 + 2*(12*a + b)*c^3*d^3*x^3 + 12*(3*a + b)*c^2*d^3*x^ 
2 + 6*(4*a + 7*b)*c*d^3*x + 3*b*d^3*log(c*x + 1) + 45*b*d^3*log(c*x - 1) + 
 3*(b*c^4*d^3*x^4 + 4*b*c^3*d^3*x^3 + 6*b*c^2*d^3*x^2 + 4*b*c*d^3*x)*log(- 
(c*x + 1)/(c*x - 1)))/c
 
3.1.23.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (73) = 146\).

Time = 0.31 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.17 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{3} d^{3} x^{4}}{4} + a c^{2} d^{3} x^{3} + \frac {3 a c d^{3} x^{2}}{2} + a d^{3} x + \frac {b c^{3} d^{3} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + b c^{2} d^{3} x^{3} \operatorname {atanh}{\left (c x \right )} + \frac {b c^{2} d^{3} x^{3}}{12} + \frac {3 b c d^{3} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b c d^{3} x^{2}}{2} + b d^{3} x \operatorname {atanh}{\left (c x \right )} + \frac {7 b d^{3} x}{4} + \frac {2 b d^{3} \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b d^{3} \operatorname {atanh}{\left (c x \right )}}{4 c} & \text {for}\: c \neq 0 \\a d^{3} x & \text {otherwise} \end {cases} \]

input
integrate((c*d*x+d)**3*(a+b*atanh(c*x)),x)
 
output
Piecewise((a*c**3*d**3*x**4/4 + a*c**2*d**3*x**3 + 3*a*c*d**3*x**2/2 + a*d 
**3*x + b*c**3*d**3*x**4*atanh(c*x)/4 + b*c**2*d**3*x**3*atanh(c*x) + b*c* 
*2*d**3*x**3/12 + 3*b*c*d**3*x**2*atanh(c*x)/2 + b*c*d**3*x**2/2 + b*d**3* 
x*atanh(c*x) + 7*b*d**3*x/4 + 2*b*d**3*log(x - 1/c)/c + b*d**3*atanh(c*x)/ 
(4*c), Ne(c, 0)), (a*d**3*x, True))
 
3.1.23.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (78) = 156\).

Time = 0.19 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.61 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{4} \, a c^{3} d^{3} x^{4} + a c^{2} d^{3} x^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{3} d^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{3} + \frac {3}{2} \, a c d^{3} x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b c d^{3} + a d^{3} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{3}}{2 \, c} \]

input
integrate((c*d*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="maxima")
 
output
1/4*a*c^3*d^3*x^4 + a*c^2*d^3*x^3 + 1/24*(6*x^4*arctanh(c*x) + c*(2*(c^2*x 
^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*b*c^3*d^3 + 1/2* 
(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*c^2*d^3 + 3/2* 
a*c*d^3*x^2 + 3/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + lo 
g(c*x - 1)/c^3))*b*c*d^3 + a*d^3*x + 1/2*(2*c*x*arctanh(c*x) + log(-c^2*x^ 
2 + 1))*b*d^3/c
 
3.1.23.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (78) = 156\).

Time = 0.28 (sec) , antiderivative size = 425, normalized size of antiderivative = 5.06 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=-\frac {1}{3} \, {\left (\frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} - \frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {6 \, {\left (\frac {4 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} - b d^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}} - \frac {\frac {48 \, {\left (c x + 1\right )}^{3} a d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {72 \, {\left (c x + 1\right )}^{2} a d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {48 \, {\left (c x + 1\right )} a d^{3}}{c x - 1} - 12 \, a d^{3} + \frac {18 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {45 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {38 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} - 11 \, b d^{3}}{\frac {{\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}}\right )} c \]

input
integrate((c*d*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="giac")
 
output
-1/3*(6*b*d^3*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 - 6*b*d^3*log(-(c*x + 1)/( 
c*x - 1))/c^2 - 6*(4*(c*x + 1)^3*b*d^3/(c*x - 1)^3 - 6*(c*x + 1)^2*b*d^3/( 
c*x - 1)^2 + 4*(c*x + 1)*b*d^3/(c*x - 1) - b*d^3)*log(-(c*x + 1)/(c*x - 1) 
)/((c*x + 1)^4*c^2/(c*x - 1)^4 - 4*(c*x + 1)^3*c^2/(c*x - 1)^3 + 6*(c*x + 
1)^2*c^2/(c*x - 1)^2 - 4*(c*x + 1)*c^2/(c*x - 1) + c^2) - (48*(c*x + 1)^3* 
a*d^3/(c*x - 1)^3 - 72*(c*x + 1)^2*a*d^3/(c*x - 1)^2 + 48*(c*x + 1)*a*d^3/ 
(c*x - 1) - 12*a*d^3 + 18*(c*x + 1)^3*b*d^3/(c*x - 1)^3 - 45*(c*x + 1)^2*b 
*d^3/(c*x - 1)^2 + 38*(c*x + 1)*b*d^3/(c*x - 1) - 11*b*d^3)/((c*x + 1)^4*c 
^2/(c*x - 1)^4 - 4*(c*x + 1)^3*c^2/(c*x - 1)^3 + 6*(c*x + 1)^2*c^2/(c*x - 
1)^2 - 4*(c*x + 1)*c^2/(c*x - 1) + c^2))*c
 
3.1.23.9 Mupad [B] (verification not implemented)

Time = 3.47 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.62 \[ \int (d+c d x)^3 (a+b \text {arctanh}(c x)) \, dx=\frac {d^3\,\left (12\,a\,x+21\,b\,x+12\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c^3\,d^3\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{12}-\frac {d^3\,\left (21\,b\,\mathrm {atanh}\left (c\,x\right )-12\,b\,\ln \left (c^2\,x^2-1\right )\right )}{12\,c}+\frac {c\,d^3\,\left (18\,a\,x^2+6\,b\,x^2+18\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c^2\,d^3\,\left (12\,a\,x^3+b\,x^3+12\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{12} \]

input
int((a + b*atanh(c*x))*(d + c*d*x)^3,x)
 
output
(d^3*(12*a*x + 21*b*x + 12*b*x*atanh(c*x)))/12 + (c^3*d^3*(3*a*x^4 + 3*b*x 
^4*atanh(c*x)))/12 - (d^3*(21*b*atanh(c*x) - 12*b*log(c^2*x^2 - 1)))/(12*c 
) + (c*d^3*(18*a*x^2 + 6*b*x^2 + 18*b*x^2*atanh(c*x)))/12 + (c^2*d^3*(12*a 
*x^3 + b*x^3 + 12*b*x^3*atanh(c*x)))/12