3.26 \(\int \frac{(a+b \tanh ^{-1}(c+d x))^3}{(c e+d e x)^2} \, dx\)

Optimal. Leaf size=143 \[ -\frac{3 b^2 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2}-\frac{3 b^3 \text{PolyLog}\left (3,\frac{2}{c+d x+1}-1\right )}{2 d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}+\frac{3 b \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2} \]

[Out]

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

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Rubi [A]  time = 0.304231, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {6107, 12, 5916, 5988, 5932, 5948, 6056, 6610} \[ -\frac{3 b^2 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2}-\frac{3 b^3 \text{PolyLog}\left (3,\frac{2}{c+d x+1}-1\right )}{2 d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}+\frac{3 b \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6107

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5916

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

Rule 5988

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] + Dist[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 5932

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] - Dist[(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 5948

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

Rule 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(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 6610

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{x (1+x)} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1+c+d x}\right )}{d e^2}-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right ) \log \left (2-\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1+c+d x}\right )}{d e^2}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c+d x}\right )}{d e^2}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-1+\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1+c+d x}\right )}{d e^2}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c+d x}\right )}{d e^2}-\frac{3 b^3 \text{Li}_3\left (-1+\frac{2}{1+c+d x}\right )}{2 d e^2}\\ \end{align*}

Mathematica [C]  time = 0.597521, size = 248, normalized size = 1.73 \[ \frac{6 a b^2 \left (\tanh ^{-1}(c+d x) \left (\left (1-\frac{1}{c+d x}\right ) \tanh ^{-1}(c+d x)+2 \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+2 b^3 \left (3 \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c+d x)}\right )-\frac{3}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c+d x)}\right )-\frac{\tanh ^{-1}(c+d x)^3}{c+d x}-\tanh ^{-1}(c+d x)^3+3 \tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+\frac{i \pi ^3}{8}\right )-3 a^2 b \log \left (-c^2-2 c d x-d^2 x^2+1\right )+6 a^2 b \log (c+d x)-\frac{6 a^2 b \tanh ^{-1}(c+d x)}{c+d x}-\frac{2 a^3}{c+d x}}{2 d e^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-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*((I/8)*Pi^3 - ArcTanh[c + d*x]^3 - ArcTanh[c + d*x]^3
/(c + d*x) + 3*ArcTanh[c + d*x]^2*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)

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Maple [C]  time = 0.338, size = 2001, normalized size = 14. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2/d*a*b^2/e^2*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+1/2*c)-3/d*a*b^2/e^2*ln(d*x+c)*ln(d*x+c+1)-3/d*a^2*b/e^2
/(d*x+c)*arctanh(d*x+c)-3/d*a*b^2/e^2/(d*x+c)*arctanh(d*x+c)^2-3/d*a*b^2/e^2*arctanh(d*x+c)*ln(d*x+c-1)+6/d*a*
b^2/e^2*arctanh(d*x+c)*ln(d*x+c)+3/2*I/d*b^3/e^2*Pi*arctanh(d*x+c)^2-3/d*a*b^2/e^2*arctanh(d*x+c)*ln(d*x+c+1)+
3/2/d*a*b^2/e^2*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/2*c)-3/2/d*a*b^2/e^2*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)+3/d*a*b^2
/e^2*dilog(1/2+1/2*d*x+1/2*c)+3/4/d*a*b^2/e^2*ln(d*x+c+1)^2-3/d*a*b^2/e^2*dilog(d*x+c)+3/4*I/d*b^3/e^2*Pi*csgn
(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2*arctanh(d*
x+c)^2-3/2*I/d*b^3/e^2*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/((d*x+c+1
)^2/(1-(d*x+c)^2)+1))^2*arctanh(d*x+c)^2-3/2*I/d*b^3/e^2*Pi*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))*csgn(I*((d*x
+c+1)^2/(1-(d*x+c)^2)-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2*arctanh(d*x+c)^2+3/4*I/d*b^3/e^2*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))*arctanh(d*x+c)^2+3/2*I/d*b^3/e^2*Pi*csgn(I*((d*x+c+1
)^2/(1-(d*x+c)^2)-1))*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/((d*x+c+1)^2/
(1-(d*x+c)^2)+1))*arctanh(d*x+c)^2-3/4/d*a*b^2/e^2*ln(d*x+c-1)^2-3/4*I/d*b^3/e^2*Pi*csgn(I/((d*x+c+1)^2/(1-(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)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))*
arctanh(d*x+c)^2+3/2*I/d*b^3/e^2*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*
arctanh(d*x+c)^2-3/4*I/d*b^3/e^2*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/((d*x+c
+1)^2/(1-(d*x+c)^2)+1))^2*arctanh(d*x+c)^2-1/d*a^3/e^2/(d*x+c)-1/d*b^3/e^2*arctanh(d*x+c)^3-6/d*b^3/e^2*polylo
g(3,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-6/d*b^3/e^2*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/2*I/d*b^3/e^2*Pi*cs
gn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3*arctanh(d*x+c)^2+3/4*I/d*b^3/e^2*Pi*csgn(I
*(d*x+c+1)^2/((d*x+c)^2-1))^3*arctanh(d*x+c)^2-3/2*I/d*b^3/e^2*Pi*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2*arct
anh(d*x+c)^2+3/2*I/d*b^3/e^2*Pi*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3*arctanh(d*x+c)^2+3/4*I/d*b^3/e^2*Pi*cs
gn(I*(d*x+c+1)^2/((d*x+c)^2-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3*arctanh(d*x+c)^2-3/d*b^3/e^2*arctanh(d*x+c)^2*
ln((d*x+c+1)^2/(1-(d*x+c)^2)-1)-3/2/d*b^3/e^2*arctanh(d*x+c)^2*ln(d*x+c-1)-3/2/d*b^3/e^2*arctanh(d*x+c)^2*ln(d
*x+c+1)+3/d*b^3/e^2*ln(d*x+c)*arctanh(d*x+c)^2+3/d*b^3/e^2*arctanh(d*x+c)^2*ln(2)-3/2/d*a^2*b/e^2*ln(d*x+c-1)+
3/d*a^2*b/e^2*ln(d*x+c)-3/2/d*a^2*b/e^2*ln(d*x+c+1)-3/d*a*b^2/e^2*dilog(d*x+c+1)+3/d*b^3/e^2*arctanh(d*x+c)^2*
ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+6/d*b^3/e^2*arctanh(d*x+c)*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/d*b
^3/e^2*arctanh(d*x+c)^2*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/d*b^3/e^2/(d*x+c)*arctanh(d*x+c)^3+6/d*b^3/e^2*arc
tanh(d*x+c)*polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/d*b^3/e^2*arctanh(d*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)
^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{3}{2} \,{\left (d{\left (\frac{\log \left (d x + c + 1\right )}{d^{2} e^{2}} - \frac{2 \, \log \left (d x + c\right )}{d^{2} e^{2}} + \frac{\log \left (d x + c - 1\right )}{d^{2} e^{2}}\right )} + \frac{2 \, \operatorname{artanh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} a^{2} b - \frac{a^{3}}{d^{2} e^{2} x + c d e^{2}} - \frac{{\left (b^{3} d x + b^{3}{\left (c - 1\right )}\right )} \log \left (-d x - c + 1\right )^{3} + 3 \,{\left (2 \, a b^{2} +{\left (b^{3} d x + b^{3}{\left (c + 1\right )}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )^{2}}{8 \,{\left (d^{2} e^{2} x + c d e^{2}\right )}} - \int -\frac{{\left (b^{3} d x + b^{3}{\left (c - 1\right )}\right )} \log \left (d x + c + 1\right )^{3} + 6 \,{\left (a b^{2} d x + a b^{2}{\left (c - 1\right )}\right )} \log \left (d x + c + 1\right )^{2} + 3 \,{\left (4 \, a b^{2} d x + 4 \, a b^{2} c -{\left (b^{3} d x + b^{3}{\left (c - 1\right )}\right )} \log \left (d x + c + 1\right )^{2} + 2 \,{\left (b^{3} d^{2} x^{2} +{\left (c^{2} + c\right )} b^{3} - 2 \, a b^{2}{\left (c - 1\right )} +{\left ({\left (2 \, c d + d\right )} b^{3} - 2 \, a b^{2} d\right )} x\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{8 \,{\left (d^{3} e^{2} x^{3} + c^{3} e^{2} - c^{2} e^{2} +{\left (3 \, c d^{2} e^{2} - d^{2} e^{2}\right )} x^{2} +{\left (3 \, c^{2} d e^{2} - 2 \, c d e^{2}\right )} x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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) - integr
ate(-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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \operatorname{artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (d x + c\right ) + a^{3}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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