3.1.23 \(\int (c e+d e x)^2 (a+b \tanh ^{-1}(c+d x))^3 \, dx\) [23]

Optimal. Leaf size=263 \[ a b^2 e^2 x+\frac {b^3 e^2 (c+d x) \tanh ^{-1}(c+d x)}{d}-\frac {b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+\frac {e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}+\frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d}-\frac {b e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d}+\frac {b^3 e^2 \log \left (1-(c+d x)^2\right )}{2 d}-\frac {b^2 e^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c-d x}\right )}{d}+\frac {b^3 e^2 \text {PolyLog}\left (3,1-\frac {2}{1-c-d x}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.34, antiderivative size = 263, normalized size of antiderivative = 1.00, 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 = {6242, 12, 6037, 6127, 6021, 266, 6095, 6131, 6055, 6205, 6745} \begin {gather*} -\frac {b^2 e^2 \text {Li}_2\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(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 (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 \text {Li}_3\left (1-\frac {2}{-c-d x+1}\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} \end {gather*}

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 12

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

Rule 266

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 6021

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

Rule 6037

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] - Dist[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 6055

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 6095

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 6127

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 6131

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 6205

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 6242

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 6745

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 {\text {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 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}-\frac {\left (b^2 e^2\right ) \text {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 ) \text {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 ) \text {Subst}\left (\int \tanh ^{-1}(x) \, dx,x,c+d x\right )}{d}+\frac {\left (b^3 e^2\right ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.44, size = 336, normalized size = 1.28 \begin {gather*} \frac {e^2 \left (3 a^2 b (c+d x)^2+2 a^3 (c+d x)^3+6 a^2 b (c+d x)^3 \tanh ^{-1}(c+d x)+3 a^2 b \log \left (1-(c+d x)^2\right )+6 a b^2 \left (c+d x-\tanh ^{-1}(c+d x)+(c+d x)^2 \tanh ^{-1}(c+d x)-\tanh ^{-1}(c+d x)^2+(c+d x)^3 \tanh ^{-1}(c+d x)^2-2 \tanh ^{-1}(c+d x) \log \left (1+e^{-2 \tanh ^{-1}(c+d x)}\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+b^3 \left (6 (c+d x) \tanh ^{-1}(c+d x)-3 \left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^2-2 \tanh ^{-1}(c+d x)^3+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)^3-6 \tanh ^{-1}(c+d x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c+d x)}\right )-6 \log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+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 )\right )\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 15.65, size = 1345, normalized size = 5.11

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1345\)
default	\(\text {Expression too large to display}\)	\(1345\)

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,method=_RETURNVERBOSE)

[Out]

1/d*(-1/2*I*e^2*b^3*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3+1/3*e^2*(d*x+c)^3*a^3+1/4*I*e^
2*b^3*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^
2/(1-(d*x+c)^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))-1/4*I*e^2*b^3*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x
+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3-1/4*I*e^2*b^3*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))^
3+1/2*I*e^2*b^3*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2+e^2*a*b^2*(d*x+c)^3*arctanh(d*x+c)
^2+e^2*a*b^2*(d*x+c)^2*arctanh(d*x+c)+e^2*a*b^2*arctanh(d*x+c)*ln(d*x+c-1)+e^2*a*b^2*arctanh(d*x+c)*ln(d*x+c+1
)-1/2*e^2*a*b^2*ln(d*x+c-1)*ln(1/2*d*x+1/2*c+1/2)+1/2*e^2*a*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)-1/2*e^2*a*b
^2*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2*d*x+1/2*c+1/2)+e^2*a^2*b*(d*x+c)^3*arctanh(d*x+c)-1/2*I*e^2*b^3*Pi*arctanh(d*
x+c)^2+1/2*e^2*a^2*b*ln(d*x+c-1)+1/2*e^2*a^2*b*ln(d*x+c+1)+e^2*a*b^2*(d*x+c)+1/2*e^2*a*b^2*ln(d*x+c-1)-1/2*e^2
*a*b^2*ln(d*x+c+1)-e^2*a*b^2*dilog(1/2*d*x+1/2*c+1/2)+1/4*e^2*a*b^2*ln(d*x+c-1)^2-1/4*e^2*a*b^2*ln(d*x+c+1)^2+
1/3*e^2*b^3*(d*x+c)^3*arctanh(d*x+c)^3+1/2*e^2*b^3*(d*x+c)^2*arctanh(d*x+c)^2+1/2*e^2*b^3*arctanh(d*x+c)^2*ln(
d*x+c-1)+1/2*e^2*b^3*arctanh(d*x+c)^2*ln(d*x+c+1)-e^2*b^3*arctanh(d*x+c)^2*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2))-e
^2*b^3*arctanh(d*x+c)*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))-e^2*b^3*ln(2)*arctanh(d*x+c)^2+e^2*b^3*(d*x+c)*arc
tanh(d*x+c)+1/2*e^2*(d*x+c)^2*a^2*b+1/2*e^2*b^3*polylog(3,-(d*x+c+1)^2/(1-(d*x+c)^2))+1/3*e^2*b^3*arctanh(d*x+
c)^3-1/2*e^2*b^3*arctanh(d*x+c)^2+e^2*b^3*arctanh(d*x+c)-e^2*b^3*ln(1+(d*x+c+1)^2/(1-(d*x+c)^2))-1/2*I*e^2*b^3
*Pi*arctanh(d*x+c)^2*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*e^2*b^3*P
i*arctanh(d*x+c)^2*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))^2-1/4*I*e^2*b^3*Pi*
arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x
+c)^2)))^2+1/4*I*e^2*b^3*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2
*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*x^3*e^2 + a^3*c*d*x^2*e^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*x*e^2 + 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*x^3*e^2 + 3*b^3*c*d^2*x^2*e^2 + 3*b^3*c^2*d*x*e^2 + (c^3 - 1)*b^3*e^2)*log(-d*x - c + 1)^3 - 3*(2*a*
b^2*d^3*x^3*e^2 + (6*a*b^2*c*d^2 + b^3*d^2)*x^2*e^2 + 2*(3*a*b^2*c^2*d + b^3*c*d)*x*e^2 + (b^3*d^3*x^3*e^2 + 3
*b^3*c*d^2*x^2*e^2 + 3*b^3*c^2*d*x*e^2 + (c^3 + 1)*b^3*e^2)*log(d*x + c + 1))*log(-d*x - c + 1)^2)/d - integra
te(-1/8*((b^3*d^3*x^3*e^2 + (3*c*d^2 - d^2)*b^3*x^2*e^2 + (3*c^2*d - 2*c*d)*b^3*x*e^2 + (c^3 - c^2)*b^3*e^2)*l
og(d*x + c + 1)^3 + 6*(a*b^2*d^3*x^3*e^2 + (3*c*d^2 - d^2)*a*b^2*x^2*e^2 + (3*c^2*d - 2*c*d)*a*b^2*x*e^2 + (c^
3 - c^2)*a*b^2*e^2)*log(d*x + c + 1)^2 - (4*a*b^2*d^3*x^3*e^2 + 2*(6*a*b^2*c*d^2 + b^3*d^2)*x^2*e^2 + 4*(3*a*b
^2*c^2*d + b^3*c*d)*x*e^2 + 3*(b^3*d^3*x^3*e^2 + (3*c*d^2 - d^2)*b^3*x^2*e^2 + (3*c^2*d - 2*c*d)*b^3*x*e^2 + (
c^3 - c^2)*b^3*e^2)*log(d*x + c + 1)^2 + 2*((6*a*b^2*d^3 + b^3*d^3)*x^3*e^2 + 3*(b^3*c*d^2 + 2*(3*c*d^2 - d^2)
*a*b^2)*x^2*e^2 + 3*(b^3*c^2*d + 2*(3*c^2*d - 2*c*d)*a*b^2)*x*e^2 + (6*(c^3 - c^2)*a*b^2 + (c^3 + 1)*b^3)*e^2)
*log(d*x + c + 1))*log(-d*x - c + 1))/(d*x + c - 1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{2} \left (\int a^{3} c^{2}\, dx + \int a^{3} d^{2} x^{2}\, dx + \int b^{3} c^{2} \operatorname {atanh}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} c^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b c^{2} \operatorname {atanh}{\left (c + d x \right )}\, dx + \int 2 a^{3} c d x\, dx + \int b^{3} d^{2} x^{2} \operatorname {atanh}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} d^{2} x^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b d^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )}\, dx + \int 2 b^{3} c d x \operatorname {atanh}^{3}{\left (c + d x \right )}\, dx + \int 6 a b^{2} c d x \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 6 a^{2} b c d x \operatorname {atanh}{\left (c + d x \right )}\, dx\right ) \end {gather*}

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]

e**2*(Integral(a**3*c**2, x) + Integral(a**3*d**2*x**2, x) + Integral(b**3*c**2*atanh(c + d*x)**3, x) + Integr
al(3*a*b**2*c**2*atanh(c + d*x)**2, x) + Integral(3*a**2*b*c**2*atanh(c + d*x), x) + Integral(2*a**3*c*d*x, x)
+ Integral(b**3*d**2*x**2*atanh(c + d*x)**3, x) + Integral(3*a*b**2*d**2*x**2*atanh(c + d*x)**2, x) + Integra
l(3*a**2*b*d**2*x**2*atanh(c + d*x), x) + Integral(2*b**3*c*d*x*atanh(c + d*x)**3, x) + Integral(6*a*b**2*c*d*
x*atanh(c + d*x)**2, x) + Integral(6*a**2*b*c*d*x*atanh(c + d*x), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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