3.23 \(\int (c e+d e x)^2 (a+b \tanh ^{-1}(c+d x))^3 \, dx\)
Optimal. Leaf size=263 \[ -\frac{b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac{b^3 e^2 \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d}+a b^2 e^2 x-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{b e^2 \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d}+\frac{b^3 e^2 \log \left (1-(c+d x)^2\right )}{2 d}+\frac{b^3 e^2 (c+d x) \tanh ^{-1}(c+d x)}{d} \]
[Out]
a*b^2*e^2*x + (b^3*e^2*(c + d*x)*ArcTanh[c + d*x])/d - (b*e^2*(a + b*ArcTanh[c + d*x])^2)/(2*d) + (b*e^2*(c +
d*x)^2*(a + b*ArcTanh[c + d*x])^2)/(2*d) + (e^2*(a + b*ArcTanh[c + d*x])^3)/(3*d) + (e^2*(c + d*x)^3*(a + b*Ar
cTanh[c + d*x])^3)/(3*d) - (b*e^2*(a + b*ArcTanh[c + d*x])^2*Log[2/(1 - c - d*x)])/d + (b^3*e^2*Log[1 - (c + d
*x)^2])/(2*d) - (b^2*e^2*(a + b*ArcTanh[c + d*x])*PolyLog[2, 1 - 2/(1 - c - d*x)])/d + (b^3*e^2*PolyLog[3, 1 -
2/(1 - c - d*x)])/(2*d)
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Rubi [A] time = 0.46688, antiderivative size = 263, normalized size of antiderivative =
1., number of steps used = 14, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) =
0.478, Rules used = {6107, 12, 5916, 5980, 5910, 260, 5948, 5984, 5918, 6058, 6610}
\[ -\frac{b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac{b^3 e^2 \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d}+a b^2 e^2 x-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{b e^2 \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d}+\frac{b^3 e^2 \log \left (1-(c+d x)^2\right )}{2 d}+\frac{b^3 e^2 (c+d x) \tanh ^{-1}(c+d x)}{d} \]
Antiderivative was successfully verified.
[In]
Int[(c*e + d*e*x)^2*(a + b*ArcTanh[c + d*x])^3,x]
[Out]
a*b^2*e^2*x + (b^3*e^2*(c + d*x)*ArcTanh[c + d*x])/d - (b*e^2*(a + b*ArcTanh[c + d*x])^2)/(2*d) + (b*e^2*(c +
d*x)^2*(a + b*ArcTanh[c + d*x])^2)/(2*d) + (e^2*(a + b*ArcTanh[c + d*x])^3)/(3*d) + (e^2*(c + d*x)^3*(a + b*Ar
cTanh[c + d*x])^3)/(3*d) - (b*e^2*(a + b*ArcTanh[c + d*x])^2*Log[2/(1 - c - d*x)])/d + (b^3*e^2*Log[1 - (c + d
*x)^2])/(2*d) - (b^2*e^2*(a + b*ArcTanh[c + d*x])*PolyLog[2, 1 - 2/(1 - c - d*x)])/d + (b^3*e^2*PolyLog[3, 1 -
2/(1 - c - d*x)])/(2*d)
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 5980
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTanh[c*x])
^p)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Rule 5910
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]
Rule 260
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]
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 5984
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]
Rule 5918
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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 6058
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[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 (c e+d e x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \tanh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \tanh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int x \left (a+b \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{1-x} \, dx,x,c+d x\right )}{d}-\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d}+\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}-\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{1-x^2} \, dx,x,c+d x\right )}{d}+\frac{\left (2 b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right ) \log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=a b^2 e^2 x-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d}-\frac{b^2 e^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d}+\frac{\left (b^3 e^2\right ) \operatorname{Subst}\left (\int \tanh ^{-1}(x) \, dx,x,c+d x\right )}{d}+\frac{\left (b^3 e^2\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}\\ &=a b^2 e^2 x+\frac{b^3 e^2 (c+d x) \tanh ^{-1}(c+d x)}{d}-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d}-\frac{b^2 e^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d}+\frac{b^3 e^2 \text{Li}_3\left (1-\frac{2}{1-c-d x}\right )}{2 d}-\frac{\left (b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=a b^2 e^2 x+\frac{b^3 e^2 (c+d x) \tanh ^{-1}(c+d x)}{d}-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac{e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac{b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d}+\frac{b^3 e^2 \log \left (1-(c+d x)^2\right )}{2 d}-\frac{b^2 e^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d}+\frac{b^3 e^2 \text{Li}_3\left (1-\frac{2}{1-c-d x}\right )}{2 d}\\ \end{align*}
Mathematica [A] time = 0.720965, size = 336, normalized size = 1.28 \[ \frac{e^2 \left (6 a b^2 \left (\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )+(c+d x)^3 \tanh ^{-1}(c+d x)^2+(c+d x)^2 \tanh ^{-1}(c+d x)-\tanh ^{-1}(c+d x)^2-\tanh ^{-1}(c+d x)-2 \tanh ^{-1}(c+d x) \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )+c+d x\right )+b^3 \left (6 \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )+3 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c+d x)}\right )-6 \log \left (\frac{1}{\sqrt{1-(c+d x)^2}}\right )+2 (c+d x) \tanh ^{-1}(c+d x)^3-2 (c+d x) \left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^3-2 \tanh ^{-1}(c+d x)^3-3 \left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^2+6 (c+d x) \tanh ^{-1}(c+d x)-6 \tanh ^{-1}(c+d x)^2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )+3 a^2 b (c+d x)^2+3 a^2 b \log \left (1-(c+d x)^2\right )+6 a^2 b (c+d x)^3 \tanh ^{-1}(c+d x)+2 a^3 (c+d x)^3\right )}{6 d} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c*e + d*e*x)^2*(a + b*ArcTanh[c + d*x])^3,x]
[Out]
(e^2*(3*a^2*b*(c + d*x)^2 + 2*a^3*(c + d*x)^3 + 6*a^2*b*(c + d*x)^3*ArcTanh[c + d*x] + 3*a^2*b*Log[1 - (c + d*
x)^2] + 6*a*b^2*(c + d*x - ArcTanh[c + d*x] + (c + d*x)^2*ArcTanh[c + d*x] - ArcTanh[c + d*x]^2 + (c + d*x)^3*
ArcTanh[c + d*x]^2 - 2*ArcTanh[c + d*x]*Log[1 + E^(-2*ArcTanh[c + d*x])] + PolyLog[2, -E^(-2*ArcTanh[c + d*x])
]) + b^3*(6*(c + d*x)*ArcTanh[c + d*x] - 3*(1 - (c + d*x)^2)*ArcTanh[c + d*x]^2 - 2*ArcTanh[c + d*x]^3 + 2*(c
+ d*x)*ArcTanh[c + d*x]^3 - 2*(c + d*x)*(1 - (c + d*x)^2)*ArcTanh[c + d*x]^3 - 6*ArcTanh[c + d*x]^2*Log[1 + E^
(-2*ArcTanh[c + d*x])] - 6*Log[1/Sqrt[1 - (c + d*x)^2]] + 6*ArcTanh[c + d*x]*PolyLog[2, -E^(-2*ArcTanh[c + d*x
])] + 3*PolyLog[3, -E^(-2*ArcTanh[c + d*x])])))/(6*d)
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Maple [C] time = 0.468, size = 1768, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*e*x+c*e)^2*(a+b*arctanh(d*x+c))^3,x)
[Out]
1/3/d*a^3*c^3*e^2+a*b^2*e^2*x+1/4*I/d*e^2*b^3*arctanh(d*x+c)^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))^2-1/2*I/d*e^2*b^3*arctanh(d*x+c)^2*Pi*csgn(I*(d*x+c+1)^2
/((d*x+c)^2-1))^2*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/4*I/d*e^2*b^3*arctanh(d*x+c)^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))^2-1/4*I/d*e^2*b^3*arctanh(
d*x+c)^2*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))^2+d*arctanh(d*x+c)^3*x^2*b
^3*c*e^2+d^2*arctanh(d*x+c)^2*x^3*a*b^2*e^2+d*arctanh(d*x+c)*x^2*a*b^2*e^2+d^2*arctanh(d*x+c)*x^3*a^2*b*e^2+1/
d*arctanh(d*x+c)^2*a*b^2*c^3*e^2+1/2/d*e^2*b^3*arctanh(d*x+c)^2*ln(d*x+c-1)+1/d*arctanh(d*x+c)*a*b^2*c^2*e^2+1
/d*arctanh(d*x+c)*a^2*b*c^3*e^2+1/d*e^2*a*b^2*arctanh(d*x+c)*ln(d*x+c-1)+1/d*e^2*a*b^2*arctanh(d*x+c)*ln(d*x+c
+1)-1/2/d*e^2*a*b^2*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/2*c)+1/2/d*e^2*a*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)-1/2/d
*e^2*a*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+1/2*c)+3*arctanh(d*x+c)^2*x*a*b^2*c^2*e^2+2*arctanh(d*x+c)*x*
a*b^2*c*e^2+3*arctanh(d*x+c)*x*a^2*b*c^2*e^2-1/2*I/d*e^2*b^3*arctanh(d*x+c)^2*Pi+1/3/d*e^2*b^3*arctanh(d*x+c)^
3+1/3*d^2*x^3*a^3*e^2+1/d*e^2*b^3*arctanh(d*x+c)-1/d*e^2*b^3*ln((d*x+c+1)^2/(1-(d*x+c)^2)+1)+1/2/d*a^2*b*c^2*e
^2+1/2/d*e^2*b^3*polylog(3,-(d*x+c+1)^2/(1-(d*x+c)^2))-1/2/d*e^2*a*b^2*ln(d*x+c+1)+1/2/d*e^2*a*b^2*ln(d*x+c-1)
+1/d*a*b^2*c*e^2+x*a^3*c^2*e^2+1/2/d*e^2*b^3*arctanh(d*x+c)^2*ln(d*x+c+1)+1/2/d*e^2*a^2*b*ln(d*x+c-1)+1/2/d*e^
2*a^2*b*ln(d*x+c+1)-1/d*e^2*b^3*arctanh(d*x+c)^2*ln(2)-1/2/d*e^2*b^3*arctanh(d*x+c)^2+arctanh(d*x+c)*x*b^3*e^2
+x*a^2*b*c*e^2+1/2*d*x^2*a^2*b*e^2+d*x^2*a^3*c*e^2+1/d*arctanh(d*x+c)*b^3*c*e^2+1/3/d*arctanh(d*x+c)^3*b^3*c^3
*e^2+1/2/d*arctanh(d*x+c)^2*b^3*c^2*e^2+1/4*I/d*e^2*b^3*arctanh(d*x+c)^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))-1/4*I/d*
e^2*b^3*arctanh(d*x+c)^2*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3+1/2*I/d*e^2*b^3*
arctanh(d*x+c)^2*Pi*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2+3*d*arctanh(d*x+c)^2*x^2*a*b^2*c*e^2+3*d*arctanh(d
*x+c)*x^2*a^2*b*c*e^2-1/2*I/d*e^2*b^3*arctanh(d*x+c)^2*Pi*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3-1/4*I/d*e^2*
b^3*arctanh(d*x+c)^2*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))^3+arctanh(d*x+c)^3*x*b^3*c^2*e^2+arctanh(d*x+c)^2*x*
b^3*c*e^2-1/d*e^2*b^3*arctanh(d*x+c)*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))-1/d*e^2*b^3*arctanh(d*x+c)^2*ln((d*
x+c+1)/(1-(d*x+c)^2)^(1/2))+1/2*d*arctanh(d*x+c)^2*x^2*b^3*e^2+1/3*d^2*arctanh(d*x+c)^3*x^3*b^3*e^2+1/4/d*e^2*
a*b^2*ln(d*x+c-1)^2-1/d*e^2*a*b^2*dilog(1/2+1/2*d*x+1/2*c)-1/4/d*e^2*a*b^2*ln(d*x+c+1)^2
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)^2*(a+b*arctanh(d*x+c))^3,x, algorithm="maxima")
[Out]
1/3*a^3*d^2*e^2*x^3 + a^3*c*d*e^2*x^2 + 3/2*(2*x^2*arctanh(d*x + c) + d*(2*x/d^2 - (c^2 + 2*c + 1)*log(d*x + c
+ 1)/d^3 + (c^2 - 2*c + 1)*log(d*x + c - 1)/d^3))*a^2*b*c*d*e^2 + 1/2*(2*x^3*arctanh(d*x + c) + d*((d*x^2 - 4
*c*x)/d^3 + (c^3 + 3*c^2 + 3*c + 1)*log(d*x + c + 1)/d^4 - (c^3 - 3*c^2 + 3*c - 1)*log(d*x + c - 1)/d^4))*a^2*
b*d^2*e^2 + a^3*c^2*e^2*x + 3/2*(2*(d*x + c)*arctanh(d*x + c) + log(-(d*x + c)^2 + 1))*a^2*b*c^2*e^2/d - 1/24*
((b^3*d^3*e^2*x^3 + 3*b^3*c*d^2*e^2*x^2 + 3*b^3*c^2*d*e^2*x + (c^3*e^2 - e^2)*b^3)*log(-d*x - c + 1)^3 - 3*(2*
a*b^2*d^3*e^2*x^3 + (6*a*b^2*c*d^2*e^2 + b^3*d^2*e^2)*x^2 + 2*(3*a*b^2*c^2*d*e^2 + b^3*c*d*e^2)*x + (b^3*d^3*e
^2*x^3 + 3*b^3*c*d^2*e^2*x^2 + 3*b^3*c^2*d*e^2*x + (c^3*e^2 + e^2)*b^3)*log(d*x + c + 1))*log(-d*x - c + 1)^2)
/d - integrate(-1/8*((b^3*d^3*e^2*x^3 + (3*c*d^2*e^2 - d^2*e^2)*b^3*x^2 + (3*c^2*d*e^2 - 2*c*d*e^2)*b^3*x + (c
^3*e^2 - c^2*e^2)*b^3)*log(d*x + c + 1)^3 + 6*(a*b^2*d^3*e^2*x^3 + (3*c*d^2*e^2 - d^2*e^2)*a*b^2*x^2 + (3*c^2*
d*e^2 - 2*c*d*e^2)*a*b^2*x + (c^3*e^2 - c^2*e^2)*a*b^2)*log(d*x + c + 1)^2 - (4*a*b^2*d^3*e^2*x^3 + 2*(6*a*b^2
*c*d^2*e^2 + b^3*d^2*e^2)*x^2 + 3*(b^3*d^3*e^2*x^3 + (3*c*d^2*e^2 - d^2*e^2)*b^3*x^2 + (3*c^2*d*e^2 - 2*c*d*e^
2)*b^3*x + (c^3*e^2 - c^2*e^2)*b^3)*log(d*x + c + 1)^2 + 4*(3*a*b^2*c^2*d*e^2 + b^3*c*d*e^2)*x + 2*(6*(c^3*e^2
- c^2*e^2)*a*b^2 + (c^3*e^2 + e^2)*b^3 + (6*a*b^2*d^3*e^2 + b^3*d^3*e^2)*x^3 + 3*(b^3*c*d^2*e^2 + 2*(3*c*d^2*
e^2 - d^2*e^2)*a*b^2)*x^2 + 3*(b^3*c^2*d*e^2 + 2*(3*c^2*d*e^2 - 2*c*d*e^2)*a*b^2)*x)*log(d*x + c + 1))*log(-d*
x - c + 1))/(d*x + c - 1), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{3} d^{2} e^{2} x^{2} + 2 \, a^{3} c d e^{2} x + a^{3} c^{2} e^{2} +{\left (b^{3} d^{2} e^{2} x^{2} + 2 \, b^{3} c d e^{2} x + b^{3} c^{2} e^{2}\right )} \operatorname{artanh}\left (d x + c\right )^{3} + 3 \,{\left (a b^{2} d^{2} e^{2} x^{2} + 2 \, a b^{2} c d e^{2} x + a b^{2} c^{2} e^{2}\right )} \operatorname{artanh}\left (d x + c\right )^{2} + 3 \,{\left (a^{2} b d^{2} e^{2} x^{2} + 2 \, a^{2} b c d e^{2} x + a^{2} b c^{2} e^{2}\right )} \operatorname{artanh}\left (d x + c\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)^2*(a+b*arctanh(d*x+c))^3,x, algorithm="fricas")
[Out]
integral(a^3*d^2*e^2*x^2 + 2*a^3*c*d*e^2*x + a^3*c^2*e^2 + (b^3*d^2*e^2*x^2 + 2*b^3*c*d*e^2*x + b^3*c^2*e^2)*a
rctanh(d*x + c)^3 + 3*(a*b^2*d^2*e^2*x^2 + 2*a*b^2*c*d*e^2*x + a*b^2*c^2*e^2)*arctanh(d*x + c)^2 + 3*(a^2*b*d^
2*e^2*x^2 + 2*a^2*b*c*d*e^2*x + a^2*b*c^2*e^2)*arctanh(d*x + c), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)**2*(a+b*atanh(d*x+c))**3,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{2}{\left (b \operatorname{artanh}\left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*e*x+c*e)^2*(a+b*arctanh(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((d*e*x + c*e)^2*(b*arctanh(d*x + c) + a)^3, x)