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

Optimal. Leaf size=159 \[ \frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x)^2}{12 d}+\frac {b^2 e^3 \log \left (1-(c+d x)^2\right )}{3 d}+\frac {b^2 e^3 (c+d x) \tanh ^{-1}(c+d x)}{2 d} \]

[Out]

1/2*a*b*e^3*x+1/12*b^2*e^3*(d*x+c)^2/d+1/2*b^2*e^3*(d*x+c)*arctanh(d*x+c)/d+1/6*b*e^3*(d*x+c)^3*(a+b*arctanh(d
*x+c))/d-1/4*e^3*(a+b*arctanh(d*x+c))^2/d+1/4*e^3*(d*x+c)^4*(a+b*arctanh(d*x+c))^2/d+1/3*b^2*e^3*ln(1-(d*x+c)^
2)/d

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Rubi [A]  time = 0.25, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {6107, 12, 5916, 5980, 266, 43, 5910, 260, 5948} \[ \frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x)^2}{12 d}+\frac {b^2 e^3 \log \left (1-(c+d x)^2\right )}{3 d}+\frac {b^2 e^3 (c+d x) \tanh ^{-1}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*e^3*x)/2 + (b^2*e^3*(c + d*x)^2)/(12*d) + (b^2*e^3*(c + d*x)*ArcTanh[c + d*x])/(2*d) + (b*e^3*(c + d*x)^3
*(a + b*ArcTanh[c + d*x]))/(6*d) - (e^3*(a + b*ArcTanh[c + d*x])^2)/(4*d) + (e^3*(c + d*x)^4*(a + b*ArcTanh[c
+ d*x])^2)/(4*d) + (b^2*e^3*Log[1 - (c + d*x)^2])/(3*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 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 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 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]

Rubi steps

\begin {align*} \int (c e+d e x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^3 x^3 \left (a+b \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int x^3 \left (a+b \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int x^2 \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{2 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}+\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{2 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{1-x^2} \, dx,x,c+d x\right )}{2 d}-\frac {\left (b^2 e^3\right ) \operatorname {Subst}\left (\int \frac {x^3}{1-x^2} \, dx,x,c+d x\right )}{6 d}\\ &=\frac {1}{2} a b e^3 x+\frac {b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}-\frac {\left (b^2 e^3\right ) \operatorname {Subst}\left (\int \frac {x}{1-x} \, dx,x,(c+d x)^2\right )}{12 d}+\frac {\left (b^2 e^3\right ) \operatorname {Subst}\left (\int \tanh ^{-1}(x) \, dx,x,c+d x\right )}{2 d}\\ &=\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x) \tanh ^{-1}(c+d x)}{2 d}+\frac {b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}-\frac {\left (b^2 e^3\right ) \operatorname {Subst}\left (\int \left (-1+\frac {1}{1-x}\right ) \, dx,x,(c+d x)^2\right )}{12 d}-\frac {\left (b^2 e^3\right ) \operatorname {Subst}\left (\int \frac {x}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x)^2}{12 d}+\frac {b^2 e^3 (c+d x) \tanh ^{-1}(c+d x)}{2 d}+\frac {b e^3 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d}+\frac {b^2 e^3 \log \left (1-(c+d x)^2\right )}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 148, normalized size = 0.93 \[ \frac {e^3 \left (3 a^2 (c+d x)^4+2 a b (c+d x)^3+6 a b (c+d x)+b (3 a+4 b) \log (-c-d x+1)+b (4 b-3 a) \log (c+d x+1)+2 b (c+d x) \tanh ^{-1}(c+d x) \left (3 a (c+d x)^3+b (c+d x)^2+3 b\right )+b^2 (c+d x)^2+3 b^2 \left ((c+d x)^4-1\right ) \tanh ^{-1}(c+d x)^2\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.74, size = 383, normalized size = 2.41 \[ \frac {12 \, a^{2} d^{4} e^{3} x^{4} + 8 \, {\left (6 \, a^{2} c + a b\right )} d^{3} e^{3} x^{3} + 4 \, {\left (18 \, a^{2} c^{2} + 6 \, a b c + b^{2}\right )} d^{2} e^{3} x^{2} + 8 \, {\left (6 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c + 3 \, a b\right )} d e^{3} x + 4 \, {\left (3 \, a b c^{4} + b^{2} c^{3} + 3 \, b^{2} c - 3 \, a b + 4 \, b^{2}\right )} e^{3} \log \left (d x + c + 1\right ) - 4 \, {\left (3 \, a b c^{4} + b^{2} c^{3} + 3 \, b^{2} c - 3 \, a b - 4 \, b^{2}\right )} e^{3} \log \left (d x + c - 1\right ) + 3 \, {\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x + {\left (b^{2} c^{4} - b^{2}\right )} e^{3}\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )^{2} + 4 \, {\left (3 \, a b d^{4} e^{3} x^{4} + {\left (12 \, a b c + b^{2}\right )} d^{3} e^{3} x^{3} + 3 \, {\left (6 \, a b c^{2} + b^{2} c\right )} d^{2} e^{3} x^{2} + 3 \, {\left (4 \, a b c^{3} + b^{2} c^{2} + b^{2}\right )} d e^{3} x\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(12*a^2*d^4*e^3*x^4 + 8*(6*a^2*c + a*b)*d^3*e^3*x^3 + 4*(18*a^2*c^2 + 6*a*b*c + b^2)*d^2*e^3*x^2 + 8*(6*a
^2*c^3 + 3*a*b*c^2 + b^2*c + 3*a*b)*d*e^3*x + 4*(3*a*b*c^4 + b^2*c^3 + 3*b^2*c - 3*a*b + 4*b^2)*e^3*log(d*x +
c + 1) - 4*(3*a*b*c^4 + b^2*c^3 + 3*b^2*c - 3*a*b - 4*b^2)*e^3*log(d*x + c - 1) + 3*(b^2*d^4*e^3*x^4 + 4*b^2*c
*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (b^2*c^4 - b^2)*e^3)*log(-(d*x + c + 1)/(d*x + c -
1))^2 + 4*(3*a*b*d^4*e^3*x^4 + (12*a*b*c + b^2)*d^3*e^3*x^3 + 3*(6*a*b*c^2 + b^2*c)*d^2*e^3*x^2 + 3*(4*a*b*c^3
+ b^2*c^2 + b^2)*d*e^3*x)*log(-(d*x + c + 1)/(d*x + c - 1)))/d

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giac [B]  time = 0.25, size = 791, normalized size = 4.97 \[ \frac {{\left (\frac {3 \, {\left (d x + c + 1\right )}^{3} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )^{2}}{{\left (d x + c - 1\right )}^{3}} + \frac {3 \, {\left (d x + c + 1\right )} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )^{2}}{d x + c - 1} - \frac {4 \, {\left (d x + c + 1\right )}^{4} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{{\left (d x + c - 1\right )}^{4}} + \frac {16 \, {\left (d x + c + 1\right )}^{3} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{{\left (d x + c - 1\right )}^{3}} - \frac {24 \, {\left (d x + c + 1\right )}^{2} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{{\left (d x + c - 1\right )}^{2}} + \frac {16 \, {\left (d x + c + 1\right )} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{d x + c - 1} - 4 \, b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right ) + \frac {12 \, {\left (d x + c + 1\right )}^{3} a b e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{{\left (d x + c - 1\right )}^{3}} + \frac {12 \, {\left (d x + c + 1\right )} a b e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + \frac {4 \, {\left (d x + c + 1\right )}^{4} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{{\left (d x + c - 1\right )}^{4}} - \frac {10 \, {\left (d x + c + 1\right )}^{3} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{{\left (d x + c - 1\right )}^{3}} + \frac {12 \, {\left (d x + c + 1\right )}^{2} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{{\left (d x + c - 1\right )}^{2}} - \frac {6 \, {\left (d x + c + 1\right )} b^{2} e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + \frac {12 \, {\left (d x + c + 1\right )}^{3} a^{2} e^{3}}{{\left (d x + c - 1\right )}^{3}} + \frac {12 \, {\left (d x + c + 1\right )} a^{2} e^{3}}{d x + c - 1} + \frac {12 \, {\left (d x + c + 1\right )}^{3} a b e^{3}}{{\left (d x + c - 1\right )}^{3}} - \frac {24 \, {\left (d x + c + 1\right )}^{2} a b e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {20 \, {\left (d x + c + 1\right )} a b e^{3}}{d x + c - 1} - 8 \, a b e^{3} + \frac {2 \, {\left (d x + c + 1\right )}^{3} b^{2} e^{3}}{{\left (d x + c - 1\right )}^{3}} - \frac {4 \, {\left (d x + c + 1\right )}^{2} b^{2} e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {2 \, {\left (d x + c + 1\right )} b^{2} e^{3}}{d x + c - 1}\right )} {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )}}{12 \, {\left (\frac {{\left (d x + c + 1\right )}^{4} d^{2}}{{\left (d x + c - 1\right )}^{4}} - \frac {4 \, {\left (d x + c + 1\right )}^{3} d^{2}}{{\left (d x + c - 1\right )}^{3}} + \frac {6 \, {\left (d x + c + 1\right )}^{2} d^{2}}{{\left (d x + c - 1\right )}^{2}} - \frac {4 \, {\left (d x + c + 1\right )} d^{2}}{d x + c - 1} + d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(d*x + c + 1)^3*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1))^2/(d*x + c - 1)^3 + 3*(d*x + c + 1)*b^2*e^3*
log(-(d*x + c + 1)/(d*x + c - 1))^2/(d*x + c - 1) - 4*(d*x + c + 1)^4*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1)
+ 1)/(d*x + c - 1)^4 + 16*(d*x + c + 1)^3*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1) + 1)/(d*x + c - 1)^3 - 24*
(d*x + c + 1)^2*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1) + 1)/(d*x + c - 1)^2 + 16*(d*x + c + 1)*b^2*e^3*log(-
(d*x + c + 1)/(d*x + c - 1) + 1)/(d*x + c - 1) - 4*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1) + 1) + 12*(d*x + c
+ 1)^3*a*b*e^3*log(-(d*x + c + 1)/(d*x + c - 1))/(d*x + c - 1)^3 + 12*(d*x + c + 1)*a*b*e^3*log(-(d*x + c + 1
)/(d*x + c - 1))/(d*x + c - 1) + 4*(d*x + c + 1)^4*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1))/(d*x + c - 1)^4 -
10*(d*x + c + 1)^3*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1))/(d*x + c - 1)^3 + 12*(d*x + c + 1)^2*b^2*e^3*log
(-(d*x + c + 1)/(d*x + c - 1))/(d*x + c - 1)^2 - 6*(d*x + c + 1)*b^2*e^3*log(-(d*x + c + 1)/(d*x + c - 1))/(d*
x + c - 1) + 12*(d*x + c + 1)^3*a^2*e^3/(d*x + c - 1)^3 + 12*(d*x + c + 1)*a^2*e^3/(d*x + c - 1) + 12*(d*x + c
+ 1)^3*a*b*e^3/(d*x + c - 1)^3 - 24*(d*x + c + 1)^2*a*b*e^3/(d*x + c - 1)^2 + 20*(d*x + c + 1)*a*b*e^3/(d*x +
c - 1) - 8*a*b*e^3 + 2*(d*x + c + 1)^3*b^2*e^3/(d*x + c - 1)^3 - 4*(d*x + c + 1)^2*b^2*e^3/(d*x + c - 1)^2 +
2*(d*x + c + 1)*b^2*e^3/(d*x + c - 1))*((c + 1)*d - (c - 1)*d)/((d*x + c + 1)^4*d^2/(d*x + c - 1)^4 - 4*(d*x +
c + 1)^3*d^2/(d*x + c - 1)^3 + 6*(d*x + c + 1)^2*d^2/(d*x + c - 1)^2 - 4*(d*x + c + 1)*d^2/(d*x + c - 1) + d^
2)

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maple [B]  time = 0.07, size = 732, normalized size = 4.60 \[ \frac {a b \,e^{3} x}{2}+\frac {d^{3} x^{4} a^{2} e^{3}}{4}+\frac {\arctanh \left (d x +c \right ) x \,b^{2} e^{3}}{2}+2 d^{2} \arctanh \left (d x +c \right ) x^{3} a b c \,e^{3}+3 d \arctanh \left (d x +c \right ) x^{2} a b \,c^{2} e^{3}+\frac {e^{3} b^{2} \ln \left (d x +c -1\right )^{2}}{16 d}+\frac {e^{3} b^{2} \ln \left (d x +c +1\right )}{3 d}+\frac {e^{3} b^{2} \ln \left (d x +c -1\right )}{3 d}+\frac {e^{3} b^{2} \ln \left (d x +c +1\right )^{2}}{16 d}+\frac {e^{3} b^{2} c^{2}}{12 d}+\frac {a^{2} c^{4} e^{3}}{4 d}+2 \arctanh \left (d x +c \right ) x a b \,c^{3} e^{3}+\frac {d \arctanh \left (d x +c \right ) x^{2} b^{2} c \,e^{3}}{2}+\frac {3 d \arctanh \left (d x +c \right )^{2} x^{2} b^{2} c^{2} e^{3}}{2}+\frac {\arctanh \left (d x +c \right ) a b \,c^{4} e^{3}}{2 d}+d^{2} \arctanh \left (d x +c \right )^{2} x^{3} b^{2} c \,e^{3}+\frac {d \,x^{2} a b c \,e^{3}}{2}+\frac {e^{3} b^{2} x c}{6}+x \,a^{2} c^{3} e^{3}+\frac {d \,e^{3} b^{2} x^{2}}{12}+\frac {x a b \,c^{2} e^{3}}{2}+\frac {3 d \,x^{2} a^{2} c^{2} e^{3}}{2}+d^{2} x^{3} a^{2} c \,e^{3}+\frac {e^{3} a b \ln \left (d x +c -1\right )}{4 d}-\frac {e^{3} a b \ln \left (d x +c +1\right )}{4 d}-\frac {e^{3} b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {e^{3} b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {e^{3} b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{8 d}+\frac {d^{2} x^{3} a b \,e^{3}}{6}+\frac {\arctanh \left (d x +c \right ) x \,b^{2} c^{2} e^{3}}{2}+\arctanh \left (d x +c \right )^{2} x \,b^{2} c^{3} e^{3}+\frac {e^{3} b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{4 d}+\frac {\arctanh \left (d x +c \right ) b^{2} c^{3} e^{3}}{6 d}+\frac {\arctanh \left (d x +c \right ) b^{2} c \,e^{3}}{2 d}+\frac {d^{2} \arctanh \left (d x +c \right ) x^{3} b^{2} e^{3}}{6}+\frac {d^{3} \arctanh \left (d x +c \right )^{2} x^{4} b^{2} e^{3}}{4}-\frac {e^{3} b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{4 d}+\frac {\arctanh \left (d x +c \right )^{2} b^{2} c^{4} e^{3}}{4 d}+\frac {a b \,c^{3} e^{3}}{6 d}+\frac {d^{3} \arctanh \left (d x +c \right ) x^{4} a b \,e^{3}}{2}+\frac {a b c \,e^{3}}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*b*e^3*x+1/4*d^3*x^4*a^2*e^3+3*d*arctanh(d*x+c)*x^2*a*b*c^2*e^3+2*d^2*arctanh(d*x+c)*x^3*a*b*c*e^3+1/16/d
*e^3*b^2*ln(d*x+c-1)^2+1/3/d*e^3*b^2*ln(d*x+c+1)+1/3/d*e^3*b^2*ln(d*x+c-1)+1/16/d*e^3*b^2*ln(d*x+c+1)^2+1/2*ar
ctanh(d*x+c)*x*b^2*e^3+1/12/d*e^3*b^2*c^2+1/4/d*a^2*c^4*e^3+2*arctanh(d*x+c)*x*a*b*c^3*e^3+1/2*d*arctanh(d*x+c
)*x^2*b^2*c*e^3+1/2*d^3*arctanh(d*x+c)*x^4*a*b*e^3+3/2*d*arctanh(d*x+c)^2*x^2*b^2*c^2*e^3+1/2/d*arctanh(d*x+c)
*a*b*c^4*e^3+d^2*arctanh(d*x+c)^2*x^3*b^2*c*e^3+1/2*d*x^2*a*b*c*e^3+1/6*e^3*b^2*x*c+1/2*arctanh(d*x+c)*x*b^2*c
^2*e^3+arctanh(d*x+c)^2*x*b^2*c^3*e^3+1/4/d*e^3*b^2*arctanh(d*x+c)*ln(d*x+c-1)+x*a^2*c^3*e^3+1/12*d*e^3*b^2*x^
2+1/2*x*a*b*c^2*e^3+3/2*d*x^2*a^2*c^2*e^3+d^2*x^3*a^2*c*e^3+1/4/d*e^3*a*b*ln(d*x+c-1)-1/4/d*e^3*a*b*ln(d*x+c+1
)+1/6/d*arctanh(d*x+c)*b^2*c^3*e^3+1/2/d*arctanh(d*x+c)*b^2*c*e^3+1/6*d^2*arctanh(d*x+c)*x^3*b^2*e^3+1/4*d^3*a
rctanh(d*x+c)^2*x^4*b^2*e^3-1/4/d*e^3*b^2*arctanh(d*x+c)*ln(d*x+c+1)+1/4/d*arctanh(d*x+c)^2*b^2*c^4*e^3-1/8/d*
e^3*b^2*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/2*c)+1/8/d*e^3*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+1/2*c)-1/8/d*e^3
*b^2*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)+1/6*d^2*x^3*a*b*e^3+1/6/d*a*b*c^3*e^3+1/2/d*a*b*c*e^3

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maxima [B]  time = 0.62, size = 827, normalized size = 5.20 \[ \frac {1}{4} \, a^{2} d^{3} e^{3} x^{4} + a^{2} c d^{2} e^{3} x^{3} + \frac {3}{2} \, a^{2} c^{2} d e^{3} x^{2} + \frac {3}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} a b c^{2} d e^{3} + {\left (2 \, x^{3} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} + \frac {{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac {{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} a b c d^{2} e^{3} + \frac {1}{12} \, {\left (6 \, x^{4} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, {\left (d^{2} x^{3} - 3 \, c d x^{2} + 3 \, {\left (3 \, c^{2} + 1\right )} x\right )}}{d^{4}} - \frac {3 \, {\left (c^{4} + 4 \, c^{3} + 6 \, c^{2} + 4 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{5}} + \frac {3 \, {\left (c^{4} - 4 \, c^{3} + 6 \, c^{2} - 4 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{5}}\right )}\right )} a b d^{3} e^{3} + a^{2} c^{3} e^{3} x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a b c^{3} e^{3}}{d} + \frac {4 \, b^{2} d^{2} e^{3} x^{2} + 8 \, b^{2} c d e^{3} x + 3 \, {\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x + {\left (c^{4} e^{3} - e^{3}\right )} b^{2}\right )} \log \left (d x + c + 1\right )^{2} + 3 \, {\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x + {\left (c^{4} e^{3} - e^{3}\right )} b^{2}\right )} \log \left (-d x - c + 1\right )^{2} + 4 \, {\left (b^{2} d^{3} e^{3} x^{3} + 3 \, b^{2} c d^{2} e^{3} x^{2} + 3 \, {\left (c^{2} d e^{3} + d e^{3}\right )} b^{2} x + {\left (c^{3} e^{3} + 3 \, c e^{3} + 4 \, e^{3}\right )} b^{2}\right )} \log \left (d x + c + 1\right ) - 2 \, {\left (2 \, b^{2} d^{3} e^{3} x^{3} + 6 \, b^{2} c d^{2} e^{3} x^{2} + 6 \, {\left (c^{2} d e^{3} + d e^{3}\right )} b^{2} x + 2 \, {\left (c^{3} e^{3} + 3 \, c e^{3} - 4 \, e^{3}\right )} b^{2} + 3 \, {\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x + {\left (c^{4} e^{3} - e^{3}\right )} b^{2}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*d^3*e^3*x^4 + a^2*c*d^2*e^3*x^3 + 3/2*a^2*c^2*d*e^3*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*b*c^2*d*e^3 + (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*b*c*d^2*e^3 + 1/12*(6*x^4*arctanh(d*x + c) + d*(2*(d^2*x^3 - 3*c*d*x^2 + 3*(3*c^2 + 1)*x)/
d^4 - 3*(c^4 + 4*c^3 + 6*c^2 + 4*c + 1)*log(d*x + c + 1)/d^5 + 3*(c^4 - 4*c^3 + 6*c^2 - 4*c + 1)*log(d*x + c -
1)/d^5))*a*b*d^3*e^3 + a^2*c^3*e^3*x + (2*(d*x + c)*arctanh(d*x + c) + log(-(d*x + c)^2 + 1))*a*b*c^3*e^3/d +
1/48*(4*b^2*d^2*e^3*x^2 + 8*b^2*c*d*e^3*x + 3*(b^2*d^4*e^3*x^4 + 4*b^2*c*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2
+ 4*b^2*c^3*d*e^3*x + (c^4*e^3 - e^3)*b^2)*log(d*x + c + 1)^2 + 3*(b^2*d^4*e^3*x^4 + 4*b^2*c*d^3*e^3*x^3 + 6*b
^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (c^4*e^3 - e^3)*b^2)*log(-d*x - c + 1)^2 + 4*(b^2*d^3*e^3*x^3 + 3*b^2
*c*d^2*e^3*x^2 + 3*(c^2*d*e^3 + d*e^3)*b^2*x + (c^3*e^3 + 3*c*e^3 + 4*e^3)*b^2)*log(d*x + c + 1) - 2*(2*b^2*d^
3*e^3*x^3 + 6*b^2*c*d^2*e^3*x^2 + 6*(c^2*d*e^3 + d*e^3)*b^2*x + 2*(c^3*e^3 + 3*c*e^3 - 4*e^3)*b^2 + 3*(b^2*d^4
*e^3*x^4 + 4*b^2*c*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (c^4*e^3 - e^3)*b^2)*log(d*x + c
+ 1))*log(-d*x - c + 1))/d

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mupad [B]  time = 2.12, size = 1730, normalized size = 10.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1 - d*x - c)^2*((b^2*c^3*e^3*x)/4 - (b^2*e^3 - b^2*c^4*e^3)/(16*d) + (b^2*d^3*e^3*x^4)/16 + (3*b^2*c^2*d*e
^3*x^2)/8 + (b^2*c*d^2*e^3*x^3)/4) + x*((c*e^3*(b^2 - 6*a^2 + 20*a^2*c^2 + 6*a*b*c))/2 + ((6*c^2 - 6)*(2*a^2*c
*d^2*e^3 - (a*d^2*e^3*(b + 10*a*c))/2))/(6*d^2) - (2*c*((2*c*(2*a^2*c*d^2*e^3 - (a*d^2*e^3*(b + 10*a*c))/2))/d
+ (d*e^3*(b^2 - 6*a^2 + 60*a^2*c^2 + 12*a*b*c))/6 - (a^2*d*e^3*(6*c^2 - 6))/6))/d) - log(1 - d*x - c)*(log(c
+ d*x + 1)*((b^2*c^3*e^3*x)/2 - ((b^2*e^3)/2 - (b^2*c^4*e^3)/2)/(4*d) + (b^2*d^3*e^3*x^4)/8 + (3*b^2*c^2*d*e^3
*x^2)/4 + (b^2*c*d^2*e^3*x^3)/2) + (x^2*(((d*(c - 1) + d*(c + 1))*(32*b^2*c*d^4*e^3 - 8*b^2*d^3*e^3*(d*(c - 1)
+ d*(c + 1)) + 8*b^2*d^4*e^3*(c - 1)))/d^2 - 48*b^2*c^2*d^3*e^3 + 8*b^2*d^3*e^3*(c - 1)*(c + 1) - 32*b^2*c*d^
3*e^3*(c - 1)))/(128*d^2) - (x^2*(((d*(c - 1) + d*(c + 1))*(32*b*d^4*e^3*(8*a*c - 2*a + b*c) - 8*b*d^3*e^3*(d*
(c - 1) + d*(c + 1))*(8*a + b) + 8*b*d^4*e^3*(8*a + b)*(c + 1)))/d^2 - 48*b*c*d^3*e^3*(8*a*c - 4*a + b*c) - 32
*b*d^3*e^3*(c + 1)*(8*a*c - 2*a + b*c) + 8*b*d^3*e^3*(8*a + b)*(c - 1)*(c + 1)))/(128*d^2) + (x^3*(32*b*d^4*e^
3*(8*a*c - 2*a + b*c) - 8*b*d^3*e^3*(d*(c - 1) + d*(c + 1))*(8*a + b) + 8*b*d^4*e^3*(8*a + b)*(c + 1)))/(192*d
^2) - (x^3*(32*b^2*c*d^4*e^3 - 8*b^2*d^3*e^3*(d*(c - 1) + d*(c + 1)) + 8*b^2*d^4*e^3*(c - 1)))/(192*d^2) + (x*
(((d*(c - 1) + d*(c + 1))*(((d*(c - 1) + d*(c + 1))*(32*b*d^4*e^3*(8*a*c - 2*a + b*c) - 8*b*d^3*e^3*(d*(c - 1)
+ d*(c + 1))*(8*a + b) + 8*b*d^4*e^3*(8*a + b)*(c + 1)))/d^2 - 48*b*c*d^3*e^3*(8*a*c - 4*a + b*c) - 32*b*d^3*
e^3*(c + 1)*(8*a*c - 2*a + b*c) + 8*b*d^3*e^3*(8*a + b)*(c - 1)*(c + 1)))/d^2 - ((c - 1)*(c + 1)*(32*b*d^4*e^3
*(8*a*c - 2*a + b*c) - 8*b*d^3*e^3*(d*(c - 1) + d*(c + 1))*(8*a + b) + 8*b*d^4*e^3*(8*a + b)*(c + 1)))/d^2 + 3
2*b*c^2*d^2*e^3*(8*a*c - 6*a + b*c) + 48*b*c*d^2*e^3*(c + 1)*(8*a*c - 4*a + b*c)))/(64*d^2) - (x*(((d*(c - 1)
+ d*(c + 1))*(((d*(c - 1) + d*(c + 1))*(32*b^2*c*d^4*e^3 - 8*b^2*d^3*e^3*(d*(c - 1) + d*(c + 1)) + 8*b^2*d^4*e
^3*(c - 1)))/d^2 - 48*b^2*c^2*d^3*e^3 + 8*b^2*d^3*e^3*(c - 1)*(c + 1) - 32*b^2*c*d^3*e^3*(c - 1)))/d^2 + 32*b^
2*c^3*d^2*e^3 - ((c - 1)*(c + 1)*(32*b^2*c*d^4*e^3 - 8*b^2*d^3*e^3*(d*(c - 1) + d*(c + 1)) + 8*b^2*d^4*e^3*(c
- 1)))/d^2 + 48*b^2*c^2*d^2*e^3*(c - 1)))/(64*d^2) - (b^2*d^3*e^3*x^4)/32 + (b*d^3*e^3*x^4*(8*a + b))/32) + x^
2*((c*(2*a^2*c*d^2*e^3 - (a*d^2*e^3*(b + 10*a*c))/2))/d + (d*e^3*(b^2 - 6*a^2 + 60*a^2*c^2 + 12*a*b*c))/12 - (
a^2*d*e^3*(6*c^2 - 6))/12) - x^3*((2*a^2*c*d^2*e^3)/3 - (a*d^2*e^3*(b + 10*a*c))/6) + log(c + d*x + 1)^2*((b^2
*c^3*e^3*x)/4 - (b^2*e^3 - b^2*c^4*e^3)/(16*d) + (b^2*d^3*e^3*x^4)/16 + (3*b^2*c^2*d*e^3*x^2)/8 + (b^2*c*d^2*e
^3*x^3)/4) - (log(c + d*x - 1)*(3*b^2*c*e^3 - 4*b^2*e^3 + b^2*c^3*e^3 - 3*a*b*e^3 + 3*a*b*c^4*e^3))/(12*d) + (
log(c + d*x + 1)*(4*b^2*e^3 + 3*b^2*c*e^3 + b^2*c^3*e^3 - 3*a*b*e^3 + 3*a*b*c^4*e^3))/(12*d) + d*log(c + d*x +
1)*(x^2*((b^2*c*e^3)/4 + (3*a*b*c^2*e^3)/2) + x^3*((b^2*d*e^3)/12 + a*b*c*d*e^3) + (x*(b^2*e^3 + b^2*c^2*e^3
+ 4*a*b*c^3*e^3))/(4*d) + (a*b*d^2*e^3*x^4)/4) + (a^2*d^3*e^3*x^4)/4

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sympy [A]  time = 8.07, size = 581, normalized size = 3.65 \[ \begin {cases} a^{2} c^{3} e^{3} x + \frac {3 a^{2} c^{2} d e^{3} x^{2}}{2} + a^{2} c d^{2} e^{3} x^{3} + \frac {a^{2} d^{3} e^{3} x^{4}}{4} + \frac {a b c^{4} e^{3} \operatorname {atanh}{\left (c + d x \right )}}{2 d} + 2 a b c^{3} e^{3} x \operatorname {atanh}{\left (c + d x \right )} + 3 a b c^{2} d e^{3} x^{2} \operatorname {atanh}{\left (c + d x \right )} + \frac {a b c^{2} e^{3} x}{2} + 2 a b c d^{2} e^{3} x^{3} \operatorname {atanh}{\left (c + d x \right )} + \frac {a b c d e^{3} x^{2}}{2} + \frac {a b d^{3} e^{3} x^{4} \operatorname {atanh}{\left (c + d x \right )}}{2} + \frac {a b d^{2} e^{3} x^{3}}{6} + \frac {a b e^{3} x}{2} - \frac {a b e^{3} \operatorname {atanh}{\left (c + d x \right )}}{2 d} + \frac {b^{2} c^{4} e^{3} \operatorname {atanh}^{2}{\left (c + d x \right )}}{4 d} + b^{2} c^{3} e^{3} x \operatorname {atanh}^{2}{\left (c + d x \right )} + \frac {b^{2} c^{3} e^{3} \operatorname {atanh}{\left (c + d x \right )}}{6 d} + \frac {3 b^{2} c^{2} d e^{3} x^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{2} + \frac {b^{2} c^{2} e^{3} x \operatorname {atanh}{\left (c + d x \right )}}{2} + b^{2} c d^{2} e^{3} x^{3} \operatorname {atanh}^{2}{\left (c + d x \right )} + \frac {b^{2} c d e^{3} x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2} + \frac {b^{2} c e^{3} x}{6} + \frac {b^{2} c e^{3} \operatorname {atanh}{\left (c + d x \right )}}{2 d} + \frac {b^{2} d^{3} e^{3} x^{4} \operatorname {atanh}^{2}{\left (c + d x \right )}}{4} + \frac {b^{2} d^{2} e^{3} x^{3} \operatorname {atanh}{\left (c + d x \right )}}{6} + \frac {b^{2} d e^{3} x^{2}}{12} + \frac {b^{2} e^{3} x \operatorname {atanh}{\left (c + d x \right )}}{2} + \frac {2 b^{2} e^{3} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{3 d} - \frac {b^{2} e^{3} \operatorname {atanh}^{2}{\left (c + d x \right )}}{4 d} - \frac {2 b^{2} e^{3} \operatorname {atanh}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {atanh}{\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c**3*e**3*x + 3*a**2*c**2*d*e**3*x**2/2 + a**2*c*d**2*e**3*x**3 + a**2*d**3*e**3*x**4/4 + a*b*
c**4*e**3*atanh(c + d*x)/(2*d) + 2*a*b*c**3*e**3*x*atanh(c + d*x) + 3*a*b*c**2*d*e**3*x**2*atanh(c + d*x) + a*
b*c**2*e**3*x/2 + 2*a*b*c*d**2*e**3*x**3*atanh(c + d*x) + a*b*c*d*e**3*x**2/2 + a*b*d**3*e**3*x**4*atanh(c + d
*x)/2 + a*b*d**2*e**3*x**3/6 + a*b*e**3*x/2 - a*b*e**3*atanh(c + d*x)/(2*d) + b**2*c**4*e**3*atanh(c + d*x)**2
/(4*d) + b**2*c**3*e**3*x*atanh(c + d*x)**2 + b**2*c**3*e**3*atanh(c + d*x)/(6*d) + 3*b**2*c**2*d*e**3*x**2*at
anh(c + d*x)**2/2 + b**2*c**2*e**3*x*atanh(c + d*x)/2 + b**2*c*d**2*e**3*x**3*atanh(c + d*x)**2 + b**2*c*d*e**
3*x**2*atanh(c + d*x)/2 + b**2*c*e**3*x/6 + b**2*c*e**3*atanh(c + d*x)/(2*d) + b**2*d**3*e**3*x**4*atanh(c + d
*x)**2/4 + b**2*d**2*e**3*x**3*atanh(c + d*x)/6 + b**2*d*e**3*x**2/12 + b**2*e**3*x*atanh(c + d*x)/2 + 2*b**2*
e**3*log(c/d + x + 1/d)/(3*d) - b**2*e**3*atanh(c + d*x)**2/(4*d) - 2*b**2*e**3*atanh(c + d*x)/(3*d), Ne(d, 0)
), (c**3*e**3*x*(a + b*atanh(c))**2, True))

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