\(\int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx\) [26]

Optimal result

Integrand size = 23, antiderivative size = 143 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\frac {(a+b \text {arctanh}(c+d x))^3}{d e^2}-\frac {(a+b \text {arctanh}(c+d x))^3}{d e^2 (c+d x)}+\frac {3 b (a+b \text {arctanh}(c+d x))^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c+d x}\right )}{d e^2}-\frac {3 b^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+c+d x}\right )}{2 d e^2} \] Output:

(a+b*arctanh(d*x+c))^3/d/e^2-(a+b*arctanh(d*x+c))^3/d/e^2/(d*x+c)+3*b*(a+b 
*arctanh(d*x+c))^2*ln(2-2/(d*x+c+1))/d/e^2-3*b^2*(a+b*arctanh(d*x+c))*poly 
log(2,-1+2/(d*x+c+1))/d/e^2-3/2*b^3*polylog(3,-1+2/(d*x+c+1))/d/e^2
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx=\frac {-\frac {2 a^3}{c+d x}-\frac {6 a^2 b \text {arctanh}(c+d x)}{c+d x}+6 a^2 b \log (c+d x)-3 a^2 b \log \left (1-c^2-2 c d x-d^2 x^2\right )+6 a b^2 \left (\text {arctanh}(c+d x) \left (\left (1-\frac {1}{c+d x}\right ) \text {arctanh}(c+d x)+2 \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )\right )-\operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )\right )+2 b^3 \left (\text {arctanh}(c+d x)^2 \left (\left (1-\frac {1}{c+d x}\right ) \text {arctanh}(c+d x)+3 \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )\right )-3 \text {arctanh}(c+d x) \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )-\frac {3}{2} \operatorname {PolyLog}\left (3,e^{-2 \text {arctanh}(c+d x)}\right )\right )}{2 d e^2} \] Input:

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

Output:

((-2*a^3)/(c + d*x) - (6*a^2*b*ArcTanh[c + d*x])/(c + d*x) + 6*a^2*b*Log[c 
 + d*x] - 3*a^2*b*Log[1 - c^2 - 2*c*d*x - d^2*x^2] + 6*a*b^2*(ArcTanh[c + 
d*x]*((1 - (c + d*x)^(-1))*ArcTanh[c + d*x] + 2*Log[1 - E^(-2*ArcTanh[c + 
d*x])]) - PolyLog[2, E^(-2*ArcTanh[c + d*x])]) + 2*b^3*(ArcTanh[c + d*x]^2 
*((1 - (c + d*x)^(-1))*ArcTanh[c + d*x] + 3*Log[1 - E^(-2*ArcTanh[c + d*x] 
)]) - 3*ArcTanh[c + d*x]*PolyLog[2, E^(-2*ArcTanh[c + d*x])] - (3*PolyLog[ 
3, E^(-2*ArcTanh[c + d*x])])/2))/(2*d*e^2)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6657, 27, 6452, 6550, 6494, 6618, 7164}

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.

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

\(\Big \downarrow \) 6657

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6550

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

\(\Big \downarrow \) 6494

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

\(\Big \downarrow \) 6618

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

\(\Big \downarrow \) 7164

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

Input:

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

Output:

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

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 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

rule 6494

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x 
_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - 
Simp[b*c*(p/d)   Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))] 
/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c 
^2*d^2 - e^2, 0]
 

rule 6550

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), 
 x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*d*(p + 1)), x] + Simp[1/ 
d   Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
 

rule 6618

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^ 
2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x 
] - Simp[b*(p/2)   Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/(d + 
 e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + 
e, 0] && EqQ[(1 - u)^2 - (1 - 2/(1 + c*x))^2, 0]
 

rule 6657

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( 
m_.), x_Symbol] :> Simp[1/d   Subst[Int[(f*(x/d))^m*(a + b*ArcTanh[x])^p, x 
], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] 
&& IGtQ[p, 0]
 

rule 7164

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, 
x]}, Simp[w*PolyLog[n + 1, v], x] /;  !FalseQ[w]] /; FreeQ[n, x]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.49 (sec) , antiderivative size = 1466, normalized size of antiderivative = 10.25

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1466\)
default	\(\text {Expression too large to display}\)	\(1466\)
parts	\(\text {Expression too large to display}\)	\(1474\)

Input:

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

Output:

1/d*(-a^3/e^2/(d*x+c)+b^3/e^2*(-1/(d*x+c)*arctanh(d*x+c)^3-3/2*arctanh(d*x 
+c)^2*ln(d*x+c-1)+3*ln(d*x+c)*arctanh(d*x+c)^2-3/2*arctanh(d*x+c)^2*ln(d*x 
+c+1)+3*arctanh(d*x+c)^2*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2))-arctanh(d*x+c)^ 
3-3*arctanh(d*x+c)^2*ln((d*x+c+1)^2/(1-(d*x+c)^2)-1)+3*arctanh(d*x+c)^2*ln 
(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+6*arctanh(d*x+c)*polylog(2,-(d*x+c+1)/(1 
-(d*x+c)^2)^(1/2))-6*polylog(3,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3*arctanh(d 
*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+6*arctanh(d*x+c)*polylog(2,(d* 
x+c+1)/(1-(d*x+c)^2)^(1/2))-6*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/4 
*(-I*Pi*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))*csgn(I*(d*x+c+1)^2/((d*x+c)^ 
2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))+I*Pi 
*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^3+2*I*Pi+ 
2*I*Pi*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^3+2*I*Pi*csgn(I*(-(d*x+c+1)^2 
/((d*x+c)^2-1)-1))*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))*csgn(I*(-(d*x+c+1 
)^2/((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))-2*I*Pi*csgn(I/(1-(d*x+ 
c+1)^2/((d*x+c)^2-1)))^2+2*I*Pi*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))*csgn 
(I*(d*x+c+1)^2/((d*x+c)^2-1))^2+I*Pi*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1))) 
*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^2+I*Pi*cs 
gn(I*(d*x+c+1)^2/((d*x+c)^2-1))^3+I*Pi*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2 
))^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))-2*I*Pi*csgn(I*(-(d*x+c+1)^2/((d*x+c 
)^2-1)-1))*csgn(I*(-(d*x+c+1)^2/((d*x+c)^2-1)-1)/(1-(d*x+c+1)^2/((d*x+c...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-3/2*(d*(log(d*x + c + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2) + log(d*x + 
 c - 1)/(d^2*e^2)) + 2*arctanh(d*x + c)/(d^2*e^2*x + c*d*e^2))*a^2*b - a^3 
/(d^2*e^2*x + c*d*e^2) - 1/8*((b^3*d*x + b^3*(c - 1))*log(-d*x - c + 1)^3 
+ 3*(2*a*b^2 + (b^3*d*x + b^3*(c + 1))*log(d*x + c + 1))*log(-d*x - c + 1) 
^2)/(d^2*e^2*x + c*d*e^2) - integrate(-1/8*((b^3*d*x + b^3*(c - 1))*log(d* 
x + c + 1)^3 + 6*(a*b^2*d*x + a*b^2*(c - 1))*log(d*x + c + 1)^2 + 3*(4*a*b 
^2*d*x + 4*a*b^2*c - (b^3*d*x + b^3*(c - 1))*log(d*x + c + 1)^2 + 2*(b^3*d 
^2*x^2 + (c^2 + c)*b^3 - 2*a*b^2*(c - 1) + ((2*c*d + d)*b^3 - 2*a*b^2*d)*x 
)*log(d*x + c + 1))*log(-d*x - c + 1))/(d^3*e^2*x^3 + c^3*e^2 - c^2*e^2 + 
(3*c*d^2*e^2 - d^2*e^2)*x^2 + (3*c^2*d*e^2 - 2*c*d*e^2)*x), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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