\(\int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx\) [1]

Optimal result

Integrand size = 16, antiderivative size = 149 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {b d e \left (2 c^2 d^2+e^2\right ) x}{c^3}+\frac {b e^2 \left (10 c^2 d^2+e^2\right ) x^2}{10 c^3}+\frac {b d e^3 x^3}{3 c}+\frac {b e^4 x^4}{20 c}+\frac {(d+e x)^5 (a+b \text {arctanh}(c x))}{5 e}+\frac {b (c d+e)^5 \log (1-c x)}{10 c^5 e}-\frac {b (c d-e)^5 \log (1+c x)}{10 c^5 e} \] Output:

b*d*e*(2*c^2*d^2+e^2)*x/c^3+1/10*b*e^2*(10*c^2*d^2+e^2)*x^2/c^3+1/3*b*d*e^ 
3*x^3/c+1/20*b*e^4*x^4/c+1/5*(e*x+d)^5*(a+b*arctanh(c*x))/e+1/10*b*(c*d+e) 
^5*ln(-c*x+1)/c^5/e-1/10*b*(c*d-e)^5*ln(c*x+1)/c^5/e
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.84 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {60 c^2 d \left (a c^3 d^3+b e \left (2 c^2 d^2+e^2\right )\right ) x+6 c^2 e \left (20 a c^3 d^3+b e \left (10 c^2 d^2+e^2\right )\right ) x^2+20 c^4 d e^2 (6 a c d+b e) x^3+3 c^4 e^3 (20 a c d+b e) x^4+12 a c^5 e^4 x^5+12 b c^5 x \left (5 d^4+10 d^3 e x+10 d^2 e^2 x^2+5 d e^3 x^3+e^4 x^4\right ) \text {arctanh}(c x)+6 b \left (5 c^4 d^4+10 c^3 d^3 e+10 c^2 d^2 e^2+5 c d e^3+e^4\right ) \log (1-c x)+6 b \left (5 c^4 d^4-10 c^3 d^3 e+10 c^2 d^2 e^2-5 c d e^3+e^4\right ) \log (1+c x)}{60 c^5} \] Input:

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

Output:

(60*c^2*d*(a*c^3*d^3 + b*e*(2*c^2*d^2 + e^2))*x + 6*c^2*e*(20*a*c^3*d^3 + 
b*e*(10*c^2*d^2 + e^2))*x^2 + 20*c^4*d*e^2*(6*a*c*d + b*e)*x^3 + 3*c^4*e^3 
*(20*a*c*d + b*e)*x^4 + 12*a*c^5*e^4*x^5 + 12*b*c^5*x*(5*d^4 + 10*d^3*e*x 
+ 10*d^2*e^2*x^2 + 5*d*e^3*x^3 + e^4*x^4)*ArcTanh[c*x] + 6*b*(5*c^4*d^4 + 
10*c^3*d^3*e + 10*c^2*d^2*e^2 + 5*c*d*e^3 + e^4)*Log[1 - c*x] + 6*b*(5*c^4 
*d^4 - 10*c^3*d^3*e + 10*c^2*d^2*e^2 - 5*c*d*e^3 + e^4)*Log[1 + c*x])/(60* 
c^5)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6478, 477, 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.

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

\(\Big \downarrow \) 6478

\(\displaystyle \frac {(d+e x)^5 (a+b \text {arctanh}(c x))}{5 e}-\frac {b c \int \frac {(d+e x)^5}{1-c^2 x^2}dx}{5 e}\)

\(\Big \downarrow \) 477

\(\displaystyle \frac {(d+e x)^5 (a+b \text {arctanh}(c x))}{5 e}-\frac {b c \int \left (\frac {(c d-e)^5}{2 c^5 (c x+1)}-\frac {e^5 x^3}{c^2}-\frac {5 d e^4 x^2}{c^2}-\frac {5 d e^2 \left (2 c^2 d^2+e^2\right )}{c^4}-\frac {e^3 \left (10 c^2 d^2+e^2\right ) x}{c^4}+\frac {(c d+e)^5}{2 c^5 (1-c x)}\right )dx}{5 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(d+e x)^5 (a+b \text {arctanh}(c x))}{5 e}-\frac {b c \left (\frac {(c d-e)^5 \log (c x+1)}{2 c^6}-\frac {(c d+e)^5 \log (1-c x)}{2 c^6}-\frac {5 d e^4 x^3}{3 c^2}-\frac {e^5 x^4}{4 c^2}-\frac {5 d e^2 x \left (2 c^2 d^2+e^2\right )}{c^4}-\frac {e^3 x^2 \left (10 c^2 d^2+e^2\right )}{2 c^4}\right )}{5 e}\)

Input:

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

Output:

((d + e*x)^5*(a + b*ArcTanh[c*x]))/(5*e) - (b*c*((-5*d*e^2*(2*c^2*d^2 + e^ 
2)*x)/c^4 - (e^3*(10*c^2*d^2 + e^2)*x^2)/(2*c^4) - (5*d*e^4*x^3)/(3*c^2) - 
 (e^5*x^4)/(4*c^2) - ((c*d + e)^5*Log[1 - c*x])/(2*c^6) + ((c*d - e)^5*Log 
[1 + c*x])/(2*c^6)))/(5*e)
 

Defintions of rubi rules used

rule 477

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
a^p   Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 
]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & 
& NiceSqrtQ[-b/a] &&  !FractionalPowerFactorQ[Rt[-b/a, 2]]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(137)=274\).

Time = 0.35 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.98

method	result	size
parts	\(\frac {a \left (e x +d \right )^{5}}{5 e}+\frac {b \left (\frac {c \,e^{4} \operatorname {arctanh}\left (c x \right ) x^{5}}{5}+c \,e^{3} \operatorname {arctanh}\left (c x \right ) x^{4} d +2 c \,e^{2} \operatorname {arctanh}\left (c x \right ) x^{3} d^{2}+2 c e \,\operatorname {arctanh}\left (c x \right ) x^{2} d^{3}+\operatorname {arctanh}\left (c x \right ) c x \,d^{4}+\frac {c \,\operatorname {arctanh}\left (c x \right ) d^{5}}{5 e}-\frac {-\frac {e^{5} c^{4} x^{4}}{4}-5 c^{4} d^{2} e^{3} x^{2}-\frac {5 c^{4} d \,e^{4} x^{3}}{3}-10 c^{4} d^{3} e^{2} x -5 c^{2} d \,e^{4} x +\frac {\left (c^{5} d^{5}-5 c^{4} d^{4} e +10 c^{3} d^{3} e^{2}-10 c^{2} d^{2} e^{3}+5 c d \,e^{4}-e^{5}\right ) \ln \left (c x +1\right )}{2}-\frac {\left (c^{5} d^{5}+5 c^{4} d^{4} e +10 c^{3} d^{3} e^{2}+10 c^{2} d^{2} e^{3}+5 c d \,e^{4}+e^{5}\right ) \ln \left (c x -1\right )}{2}-\frac {e^{5} c^{2} x^{2}}{2}}{5 c^{4} e}\right )}{c}\)	\(295\)
derivativedivides	\(\frac {\frac {a \left (c e x +c d \right )^{5}}{5 c^{4} e}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{5} d^{5}}{5 e}+\operatorname {arctanh}\left (c x \right ) c^{5} d^{4} x +2 e \,\operatorname {arctanh}\left (c x \right ) c^{5} d^{3} x^{2}+2 e^{2} \operatorname {arctanh}\left (c x \right ) c^{5} d^{2} x^{3}+e^{3} \operatorname {arctanh}\left (c x \right ) c^{5} d \,x^{4}+\frac {e^{4} \operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}-\frac {-\frac {e^{5} c^{4} x^{4}}{4}-5 c^{4} d^{2} e^{3} x^{2}-\frac {5 c^{4} d \,e^{4} x^{3}}{3}-10 c^{4} d^{3} e^{2} x -5 c^{2} d \,e^{4} x +\frac {\left (c^{5} d^{5}-5 c^{4} d^{4} e +10 c^{3} d^{3} e^{2}-10 c^{2} d^{2} e^{3}+5 c d \,e^{4}-e^{5}\right ) \ln \left (c x +1\right )}{2}-\frac {\left (c^{5} d^{5}+5 c^{4} d^{4} e +10 c^{3} d^{3} e^{2}+10 c^{2} d^{2} e^{3}+5 c d \,e^{4}+e^{5}\right ) \ln \left (c x -1\right )}{2}-\frac {e^{5} c^{2} x^{2}}{2}}{5 e}\right )}{c^{4}}}{c}\)	\(314\)
default	\(\frac {\frac {a \left (c e x +c d \right )^{5}}{5 c^{4} e}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{5} d^{5}}{5 e}+\operatorname {arctanh}\left (c x \right ) c^{5} d^{4} x +2 e \,\operatorname {arctanh}\left (c x \right ) c^{5} d^{3} x^{2}+2 e^{2} \operatorname {arctanh}\left (c x \right ) c^{5} d^{2} x^{3}+e^{3} \operatorname {arctanh}\left (c x \right ) c^{5} d \,x^{4}+\frac {e^{4} \operatorname {arctanh}\left (c x \right ) c^{5} x^{5}}{5}-\frac {-\frac {e^{5} c^{4} x^{4}}{4}-5 c^{4} d^{2} e^{3} x^{2}-\frac {5 c^{4} d \,e^{4} x^{3}}{3}-10 c^{4} d^{3} e^{2} x -5 c^{2} d \,e^{4} x +\frac {\left (c^{5} d^{5}-5 c^{4} d^{4} e +10 c^{3} d^{3} e^{2}-10 c^{2} d^{2} e^{3}+5 c d \,e^{4}-e^{5}\right ) \ln \left (c x +1\right )}{2}-\frac {\left (c^{5} d^{5}+5 c^{4} d^{4} e +10 c^{3} d^{3} e^{2}+10 c^{2} d^{2} e^{3}+5 c d \,e^{4}+e^{5}\right ) \ln \left (c x -1\right )}{2}-\frac {e^{5} c^{2} x^{2}}{2}}{5 e}\right )}{c^{4}}}{c}\)	\(314\)
parallelrisch	\(\frac {6 b \,e^{4}+60 b \,c^{2} d^{2} e^{2}+120 a \,c^{3} d^{3} e +12 \,\operatorname {arctanh}\left (c x \right ) b \,e^{4}+12 \ln \left (c x -1\right ) b \,e^{4}+60 \ln \left (c x -1\right ) b \,c^{4} d^{4}+60 x a \,c^{5} d^{4}+3 x^{4} b \,c^{4} e^{4}+12 x^{5} a \,c^{5} e^{4}+60 \,\operatorname {arctanh}\left (c x \right ) b \,c^{4} d^{4}+6 x^{2} b \,c^{2} e^{4}-120 \,\operatorname {arctanh}\left (c x \right ) b \,c^{3} d^{3} e +120 \,\operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{2} e^{2}-60 \,\operatorname {arctanh}\left (c x \right ) b c d \,e^{3}+120 x^{2} a \,c^{5} d^{3} e +120 \ln \left (c x -1\right ) b \,c^{2} d^{2} e^{2}+60 b \,d^{4} \operatorname {arctanh}\left (c x \right ) x \,c^{5}+12 x^{5} \operatorname {arctanh}\left (c x \right ) b \,c^{5} e^{4}+120 x b \,c^{4} d^{3} e +60 x b \,c^{2} d \,e^{3}+120 x^{3} a \,c^{5} d^{2} e^{2}+20 x^{3} b \,c^{4} d \,e^{3}+60 x^{4} a \,c^{5} d \,e^{3}+60 x^{2} b \,c^{4} d^{2} e^{2}+60 x^{4} \operatorname {arctanh}\left (c x \right ) b \,c^{5} d \,e^{3}+120 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{5} d^{3} e +120 x^{3} \operatorname {arctanh}\left (c x \right ) b \,c^{5} d^{2} e^{2}}{60 c^{5}}\)	\(365\)
risch	\(\frac {b \,e^{4} x^{4}}{20 c}+\frac {b d \,e^{3} x^{3}}{3 c}+e^{3} a d \,x^{4}+2 e^{2} a \,d^{2} x^{3}+2 e a \,d^{3} x^{2}+a \,d^{4} x +\frac {e^{2} b \,d^{2} x^{2}}{c}+\frac {2 e b \,d^{3} x}{c}+\frac {e^{3} b d x}{c^{3}}-\frac {e^{3} b d \,x^{4} \ln \left (-c x +1\right )}{2}-e^{2} b \,d^{2} x^{3} \ln \left (-c x +1\right )-e b \,d^{3} x^{2} \ln \left (-c x +1\right )+\frac {e \ln \left (-c x +1\right ) b \,d^{3}}{c^{2}}-\frac {e \ln \left (c x +1\right ) b \,d^{3}}{c^{2}}+\frac {e^{2} \ln \left (-c x +1\right ) b \,d^{2}}{c^{3}}+\frac {e^{2} \ln \left (c x +1\right ) b \,d^{2}}{c^{3}}+\frac {e^{3} \ln \left (-c x +1\right ) b d}{2 c^{4}}-\frac {e^{3} \ln \left (c x +1\right ) b d}{2 c^{4}}+\frac {e^{4} a \,x^{5}}{5}+\frac {e^{4} b \,x^{2}}{10 c^{3}}-\frac {b \,d^{4} x \ln \left (-c x +1\right )}{2}-\frac {e^{4} b \,x^{5} \ln \left (-c x +1\right )}{10}-\frac {\ln \left (c x +1\right ) b \,d^{5}}{10 e}+\frac {\ln \left (-c x +1\right ) b \,d^{4}}{2 c}+\frac {\ln \left (c x +1\right ) b \,d^{4}}{2 c}+\frac {e^{4} \ln \left (-c x +1\right ) b}{10 c^{5}}+\frac {e^{4} \ln \left (c x +1\right ) b}{10 c^{5}}+\frac {\left (e x +d \right )^{5} b \ln \left (c x +1\right )}{10 e}\)	\(399\)

Input:

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

Output:

1/5*a*(e*x+d)^5/e+b/c*(1/5*c*e^4*arctanh(c*x)*x^5+c*e^3*arctanh(c*x)*x^4*d 
+2*c*e^2*arctanh(c*x)*x^3*d^2+2*c*e*arctanh(c*x)*x^2*d^3+arctanh(c*x)*c*x* 
d^4+1/5*c/e*arctanh(c*x)*d^5-1/5/c^4/e*(-1/4*e^5*c^4*x^4-5*c^4*d^2*e^3*x^2 
-5/3*c^4*d*e^4*x^3-10*c^4*d^3*e^2*x-5*c^2*d*e^4*x+1/2*(c^5*d^5-5*c^4*d^4*e 
+10*c^3*d^3*e^2-10*c^2*d^2*e^3+5*c*d*e^4-e^5)*ln(c*x+1)-1/2*(c^5*d^5+5*c^4 
*d^4*e+10*c^3*d^3*e^2+10*c^2*d^2*e^3+5*c*d*e^4+e^5)*ln(c*x-1)-1/2*e^5*c^2* 
x^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (137) = 274\).

Time = 0.09 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.16 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {12 \, a c^{5} e^{4} x^{5} + 3 \, {\left (20 \, a c^{5} d e^{3} + b c^{4} e^{4}\right )} x^{4} + 20 \, {\left (6 \, a c^{5} d^{2} e^{2} + b c^{4} d e^{3}\right )} x^{3} + 6 \, {\left (20 \, a c^{5} d^{3} e + 10 \, b c^{4} d^{2} e^{2} + b c^{2} e^{4}\right )} x^{2} + 60 \, {\left (a c^{5} d^{4} + 2 \, b c^{4} d^{3} e + b c^{2} d e^{3}\right )} x + 6 \, {\left (5 \, b c^{4} d^{4} - 10 \, b c^{3} d^{3} e + 10 \, b c^{2} d^{2} e^{2} - 5 \, b c d e^{3} + b e^{4}\right )} \log \left (c x + 1\right ) + 6 \, {\left (5 \, b c^{4} d^{4} + 10 \, b c^{3} d^{3} e + 10 \, b c^{2} d^{2} e^{2} + 5 \, b c d e^{3} + b e^{4}\right )} \log \left (c x - 1\right ) + 6 \, {\left (b c^{5} e^{4} x^{5} + 5 \, b c^{5} d e^{3} x^{4} + 10 \, b c^{5} d^{2} e^{2} x^{3} + 10 \, b c^{5} d^{3} e x^{2} + 5 \, b c^{5} d^{4} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, c^{5}} \] Input:

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

Output:

1/60*(12*a*c^5*e^4*x^5 + 3*(20*a*c^5*d*e^3 + b*c^4*e^4)*x^4 + 20*(6*a*c^5* 
d^2*e^2 + b*c^4*d*e^3)*x^3 + 6*(20*a*c^5*d^3*e + 10*b*c^4*d^2*e^2 + b*c^2* 
e^4)*x^2 + 60*(a*c^5*d^4 + 2*b*c^4*d^3*e + b*c^2*d*e^3)*x + 6*(5*b*c^4*d^4 
 - 10*b*c^3*d^3*e + 10*b*c^2*d^2*e^2 - 5*b*c*d*e^3 + b*e^4)*log(c*x + 1) + 
 6*(5*b*c^4*d^4 + 10*b*c^3*d^3*e + 10*b*c^2*d^2*e^2 + 5*b*c*d*e^3 + b*e^4) 
*log(c*x - 1) + 6*(b*c^5*e^4*x^5 + 5*b*c^5*d*e^3*x^4 + 10*b*c^5*d^2*e^2*x^ 
3 + 10*b*c^5*d^3*e*x^2 + 5*b*c^5*d^4*x)*log(-(c*x + 1)/(c*x - 1)))/c^5
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (134) = 268\).

Time = 0.50 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.56 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} a d^{4} x + 2 a d^{3} e x^{2} + 2 a d^{2} e^{2} x^{3} + a d e^{3} x^{4} + \frac {a e^{4} x^{5}}{5} + b d^{4} x \operatorname {atanh}{\left (c x \right )} + 2 b d^{3} e x^{2} \operatorname {atanh}{\left (c x \right )} + 2 b d^{2} e^{2} x^{3} \operatorname {atanh}{\left (c x \right )} + b d e^{3} x^{4} \operatorname {atanh}{\left (c x \right )} + \frac {b e^{4} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b d^{4} \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{c} + \frac {2 b d^{3} e x}{c} + \frac {b d^{2} e^{2} x^{2}}{c} + \frac {b d e^{3} x^{3}}{3 c} + \frac {b e^{4} x^{4}}{20 c} - \frac {2 b d^{3} e \operatorname {atanh}{\left (c x \right )}}{c^{2}} + \frac {2 b d^{2} e^{2} \log {\left (x - \frac {1}{c} \right )}}{c^{3}} + \frac {2 b d^{2} e^{2} \operatorname {atanh}{\left (c x \right )}}{c^{3}} + \frac {b d e^{3} x}{c^{3}} + \frac {b e^{4} x^{2}}{10 c^{3}} - \frac {b d e^{3} \operatorname {atanh}{\left (c x \right )}}{c^{4}} + \frac {b e^{4} \log {\left (x - \frac {1}{c} \right )}}{5 c^{5}} + \frac {b e^{4} \operatorname {atanh}{\left (c x \right )}}{5 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{4} x + 2 d^{3} e x^{2} + 2 d^{2} e^{2} x^{3} + d e^{3} x^{4} + \frac {e^{4} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a*d**4*x + 2*a*d**3*e*x**2 + 2*a*d**2*e**2*x**3 + a*d*e**3*x**4 
 + a*e**4*x**5/5 + b*d**4*x*atanh(c*x) + 2*b*d**3*e*x**2*atanh(c*x) + 2*b* 
d**2*e**2*x**3*atanh(c*x) + b*d*e**3*x**4*atanh(c*x) + b*e**4*x**5*atanh(c 
*x)/5 + b*d**4*log(x - 1/c)/c + b*d**4*atanh(c*x)/c + 2*b*d**3*e*x/c + b*d 
**2*e**2*x**2/c + b*d*e**3*x**3/(3*c) + b*e**4*x**4/(20*c) - 2*b*d**3*e*at 
anh(c*x)/c**2 + 2*b*d**2*e**2*log(x - 1/c)/c**3 + 2*b*d**2*e**2*atanh(c*x) 
/c**3 + b*d*e**3*x/c**3 + b*e**4*x**2/(10*c**3) - b*d*e**3*atanh(c*x)/c**4 
 + b*e**4*log(x - 1/c)/(5*c**5) + b*e**4*atanh(c*x)/(5*c**5), Ne(c, 0)), ( 
a*(d**4*x + 2*d**3*e*x**2 + 2*d**2*e**2*x**3 + d*e**3*x**4 + e**4*x**5/5), 
 True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.83 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{5} \, a e^{4} x^{5} + a d e^{3} x^{4} + 2 \, a d^{2} e^{2} x^{3} + 2 \, a d^{3} e x^{2} + {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{3} e + {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d^{2} e^{2} + \frac {1}{6} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b d e^{3} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b e^{4} + a d^{4} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \] Input:

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

Output:

1/5*a*e^4*x^5 + a*d*e^3*x^4 + 2*a*d^2*e^2*x^3 + 2*a*d^3*e*x^2 + (2*x^2*arc 
tanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*d^3*e + ( 
2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*d^2*e^2 + 1/6*( 
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*d*e^3 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/ 
c^4 + 2*log(c^2*x^2 - 1)/c^6))*b*e^4 + a*d^4*x + 1/2*(2*c*x*arctanh(c*x) + 
 log(-c^2*x^2 + 1))*b*d^4/c
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1576 vs. \(2 (137) = 274\).

Time = 0.15 (sec) , antiderivative size = 1576, normalized size of antiderivative = 10.58 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/15*c*(3*(5*(c*x + 1)^4*b*c^4*d^4/(c*x - 1)^4 - 20*(c*x + 1)^3*b*c^4*d^4/ 
(c*x - 1)^3 + 30*(c*x + 1)^2*b*c^4*d^4/(c*x - 1)^2 - 20*(c*x + 1)*b*c^4*d^ 
4/(c*x - 1) + 5*b*c^4*d^4 + 20*(c*x + 1)^4*b*c^3*d^3*e/(c*x - 1)^4 - 60*(c 
*x + 1)^3*b*c^3*d^3*e/(c*x - 1)^3 + 60*(c*x + 1)^2*b*c^3*d^3*e/(c*x - 1)^2 
 - 20*(c*x + 1)*b*c^3*d^3*e/(c*x - 1) + 30*(c*x + 1)^4*b*c^2*d^2*e^2/(c*x 
- 1)^4 - 60*(c*x + 1)^3*b*c^2*d^2*e^2/(c*x - 1)^3 + 40*(c*x + 1)^2*b*c^2*d 
^2*e^2/(c*x - 1)^2 - 20*(c*x + 1)*b*c^2*d^2*e^2/(c*x - 1) + 10*b*c^2*d^2*e 
^2 + 20*(c*x + 1)^4*b*c*d*e^3/(c*x - 1)^4 - 20*(c*x + 1)^3*b*c*d*e^3/(c*x 
- 1)^3 + 20*(c*x + 1)^2*b*c*d*e^3/(c*x - 1)^2 - 20*(c*x + 1)*b*c*d*e^3/(c* 
x - 1) + 5*(c*x + 1)^4*b*e^4/(c*x - 1)^4 + 10*(c*x + 1)^2*b*e^4/(c*x - 1)^ 
2 + b*e^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^6/(c*x - 1)^5 - 5*(c*x 
 + 1)^4*c^6/(c*x - 1)^4 + 10*(c*x + 1)^3*c^6/(c*x - 1)^3 - 10*(c*x + 1)^2* 
c^6/(c*x - 1)^2 + 5*(c*x + 1)*c^6/(c*x - 1) - c^6) + 2*(15*(c*x + 1)^4*a*c 
^4*d^4/(c*x - 1)^4 - 60*(c*x + 1)^3*a*c^4*d^4/(c*x - 1)^3 + 90*(c*x + 1)^2 
*a*c^4*d^4/(c*x - 1)^2 - 60*(c*x + 1)*a*c^4*d^4/(c*x - 1) + 15*a*c^4*d^4 + 
 60*(c*x + 1)^4*a*c^3*d^3*e/(c*x - 1)^4 - 180*(c*x + 1)^3*a*c^3*d^3*e/(c*x 
 - 1)^3 + 180*(c*x + 1)^2*a*c^3*d^3*e/(c*x - 1)^2 - 60*(c*x + 1)*a*c^3*d^3 
*e/(c*x - 1) + 30*(c*x + 1)^4*b*c^3*d^3*e/(c*x - 1)^4 - 120*(c*x + 1)^3*b* 
c^3*d^3*e/(c*x - 1)^3 + 180*(c*x + 1)^2*b*c^3*d^3*e/(c*x - 1)^2 - 120*(c*x 
 + 1)*b*c^3*d^3*e/(c*x - 1) + 30*b*c^3*d^3*e + 90*(c*x + 1)^4*a*c^2*d^2...
 

Mupad [B] (verification not implemented)

Time = 4.21 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.83 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {a\,e^4\,x^5}{5}+a\,d^4\,x+\frac {b\,d^4\,\ln \left (c^2\,x^2-1\right )}{2\,c}+\frac {b\,e^4\,\ln \left (c^2\,x^2-1\right )}{10\,c^5}+2\,a\,d^2\,e^2\,x^3+\frac {b\,e^4\,x^4}{20\,c}+\frac {b\,e^4\,x^2}{10\,c^3}+b\,d^4\,x\,\mathrm {atanh}\left (c\,x\right )+2\,a\,d^3\,e\,x^2+a\,d\,e^3\,x^4+\frac {b\,e^4\,x^5\,\mathrm {atanh}\left (c\,x\right )}{5}+\frac {2\,b\,d^3\,e\,x}{c}+\frac {b\,d\,e^3\,x}{c^3}-\frac {2\,b\,d^3\,e\,\mathrm {atanh}\left (c\,x\right )}{c^2}-\frac {b\,d\,e^3\,\mathrm {atanh}\left (c\,x\right )}{c^4}+2\,b\,d^3\,e\,x^2\,\mathrm {atanh}\left (c\,x\right )+b\,d\,e^3\,x^4\,\mathrm {atanh}\left (c\,x\right )+\frac {b\,d\,e^3\,x^3}{3\,c}+2\,b\,d^2\,e^2\,x^3\,\mathrm {atanh}\left (c\,x\right )+\frac {b\,d^2\,e^2\,\ln \left (c^2\,x^2-1\right )}{c^3}+\frac {b\,d^2\,e^2\,x^2}{c} \] Input:

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

Output:

(a*e^4*x^5)/5 + a*d^4*x + (b*d^4*log(c^2*x^2 - 1))/(2*c) + (b*e^4*log(c^2* 
x^2 - 1))/(10*c^5) + 2*a*d^2*e^2*x^3 + (b*e^4*x^4)/(20*c) + (b*e^4*x^2)/(1 
0*c^3) + b*d^4*x*atanh(c*x) + 2*a*d^3*e*x^2 + a*d*e^3*x^4 + (b*e^4*x^5*ata 
nh(c*x))/5 + (2*b*d^3*e*x)/c + (b*d*e^3*x)/c^3 - (2*b*d^3*e*atanh(c*x))/c^ 
2 - (b*d*e^3*atanh(c*x))/c^4 + 2*b*d^3*e*x^2*atanh(c*x) + b*d*e^3*x^4*atan 
h(c*x) + (b*d*e^3*x^3)/(3*c) + 2*b*d^2*e^2*x^3*atanh(c*x) + (b*d^2*e^2*log 
(c^2*x^2 - 1))/c^3 + (b*d^2*e^2*x^2)/c
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.34 \[ \int (d+e x)^4 (a+b \text {arctanh}(c x)) \, dx=\frac {60 \mathit {atanh} \left (c x \right ) b \,c^{5} d^{4} x +120 \mathit {atanh} \left (c x \right ) b \,c^{5} d^{3} e \,x^{2}+120 \mathit {atanh} \left (c x \right ) b \,c^{5} d^{2} e^{2} x^{3}+60 \mathit {atanh} \left (c x \right ) b \,c^{5} d \,e^{3} x^{4}+12 \mathit {atanh} \left (c x \right ) b \,c^{5} e^{4} x^{5}+60 \mathit {atanh} \left (c x \right ) b \,c^{4} d^{4}-120 \mathit {atanh} \left (c x \right ) b \,c^{3} d^{3} e +120 \mathit {atanh} \left (c x \right ) b \,c^{2} d^{2} e^{2}-60 \mathit {atanh} \left (c x \right ) b c d \,e^{3}+12 \mathit {atanh} \left (c x \right ) b \,e^{4}+60 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{4} d^{4}+120 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{2} d^{2} e^{2}+12 \,\mathrm {log}\left (c^{2} x -c \right ) b \,e^{4}+60 a \,c^{5} d^{4} x +120 a \,c^{5} d^{3} e \,x^{2}+120 a \,c^{5} d^{2} e^{2} x^{3}+60 a \,c^{5} d \,e^{3} x^{4}+12 a \,c^{5} e^{4} x^{5}+120 b \,c^{4} d^{3} e x +60 b \,c^{4} d^{2} e^{2} x^{2}+20 b \,c^{4} d \,e^{3} x^{3}+3 b \,c^{4} e^{4} x^{4}+60 b \,c^{2} d \,e^{3} x +6 b \,c^{2} e^{4} x^{2}}{60 c^{5}} \] Input:

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

Output:

(60*atanh(c*x)*b*c**5*d**4*x + 120*atanh(c*x)*b*c**5*d**3*e*x**2 + 120*ata 
nh(c*x)*b*c**5*d**2*e**2*x**3 + 60*atanh(c*x)*b*c**5*d*e**3*x**4 + 12*atan 
h(c*x)*b*c**5*e**4*x**5 + 60*atanh(c*x)*b*c**4*d**4 - 120*atanh(c*x)*b*c** 
3*d**3*e + 120*atanh(c*x)*b*c**2*d**2*e**2 - 60*atanh(c*x)*b*c*d*e**3 + 12 
*atanh(c*x)*b*e**4 + 60*log(c**2*x - c)*b*c**4*d**4 + 120*log(c**2*x - c)* 
b*c**2*d**2*e**2 + 12*log(c**2*x - c)*b*e**4 + 60*a*c**5*d**4*x + 120*a*c* 
*5*d**3*e*x**2 + 120*a*c**5*d**2*e**2*x**3 + 60*a*c**5*d*e**3*x**4 + 12*a* 
c**5*e**4*x**5 + 120*b*c**4*d**3*e*x + 60*b*c**4*d**2*e**2*x**2 + 20*b*c** 
4*d*e**3*x**3 + 3*b*c**4*e**4*x**4 + 60*b*c**2*d*e**3*x + 6*b*c**2*e**4*x* 
*2)/(60*c**5)