3.1.17 \(\int (d+e x) (a+b \tanh ^{-1}(c x))^3 \, dx\) [17]

Optimal. Leaf size=244 \[ \frac {3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2+\frac {e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}-\frac {3 b d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^3 e \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c^2}-\frac {3 b^2 d \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 d \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 c} \]

[Out]

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

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Rubi [A]
time = 0.43, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6065, 6021, 6131, 6055, 2449, 2352, 6195, 6095, 6205, 6745} \begin {gather*} -\frac {3 b^2 e \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2}-\frac {3 b^2 d \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac {\left (\frac {e^2}{c^2}+d^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac {3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac {3 b d \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}-\frac {3 b^3 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 c^2}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 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 6065

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

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 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 6195

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 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 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 (d+e x) \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b c) \int \left (-\frac {e^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2}+\frac {\left (c^2 d^2+e^2+2 c^2 d e x\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 \left (1-c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b) \int \frac {\left (c^2 d^2+e^2+2 c^2 d e x\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c e}+\frac {(3 b e) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{2 c}\\ &=\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b) \int \left (\frac {c^2 d^2 \left (1+\frac {e^2}{c^2 d^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}+\frac {2 c^2 d e x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}\right ) \, dx}{2 c e}-\left (3 b^2 e\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-(3 b c d) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx-\frac {\left (3 b^2 e\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c}-\frac {\left (3 b \left (c^2 d^2+e^2\right )\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c e}\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2+\frac {e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}-(3 b d) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx+\frac {\left (3 b^3 e\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c}\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2+\frac {e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}-\frac {3 b d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}+\left (6 b^2 d\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\frac {\left (3 b^3 e\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c^2}\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2+\frac {e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}-\frac {3 b d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^3 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 c^2}-\frac {3 b^2 d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\left (3 b^3 d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac {3 b e \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2}+\frac {3 b e x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2+\frac {e^2}{c^2}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2}-\frac {3 b d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^3 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 c^2}-\frac {3 b^2 d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 331, normalized size = 1.36 \begin {gather*} \frac {2 a^2 c (2 a c d+3 b e) x+2 a^3 c^2 e x^2+6 a^2 b c^2 x (2 d+e x) \tanh ^{-1}(c x)+3 a^2 b (2 c d+e) \log (1-c x)+3 a^2 b (2 c d-e) \log (1+c x)+6 a b^2 e \left (2 c x \tanh ^{-1}(c x)+\left (-1+c^2 x^2\right ) \tanh ^{-1}(c x)^2+\log \left (1-c^2 x^2\right )\right )-2 b^3 e \left (\tanh ^{-1}(c x) \left ((3-3 c x) \tanh ^{-1}(c x)+\left (1-c^2 x^2\right ) \tanh ^{-1}(c x)^2+6 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )-3 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )+12 a b^2 c d \left (\tanh ^{-1}(c x) \left ((-1+c x) \tanh ^{-1}(c x)-2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )+4 b^3 c d \left (\tanh ^{-1}(c x)^2 \left ((-1+c x) \tanh ^{-1}(c x)-3 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+3 \tanh ^{-1}(c x) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+\frac {3}{2} \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 6.66, size = 12134, normalized size = 49.73

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(12134\)
default	\(\text {Expression too large to display}\)	\(12134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

1/2*a^3*x^2*e + a^3*d*x + 3/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a^2*b*e
+ 3/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a^2*b*d/c - 1/16*((b^3*c^2*x^2*e + 2*b^3*c^2*d*x - 2*b^3*c*d -
b^3*e)*log(-c*x + 1)^3 - 3*(2*a*b^2*c^2*x^2*e + 2*(2*a*b^2*c^2*d + b^3*c*e)*x + (b^3*c^2*x^2*e + 2*b^3*c^2*d*
x + 2*b^3*c*d - b^3*e)*log(c*x + 1))*log(-c*x + 1)^2)/c^2 - integrate(-1/8*((b^3*c^2*x^2*e - b^3*c*d + (b^3*c^
2*d - b^3*c*e)*x)*log(c*x + 1)^3 + 6*(a*b^2*c^2*x^2*e - a*b^2*c*d + (a*b^2*c^2*d - a*b^2*c*e)*x)*log(c*x + 1)^
2 - 3*(2*a*b^2*c^2*x^2*e + (b^3*c^2*x^2*e - b^3*c*d + (b^3*c^2*d - b^3*c*e)*x)*log(c*x + 1)^2 + 2*(2*a*b^2*c^2
*d + b^3*c*e)*x - (4*a*b^2*c*d - 2*b^3*c*d + b^3*e - (4*a*b^2*c^2 + b^3*c^2)*x^2*e - 2*(2*a*b^2*c^2*d + b^3*c^
2*d - 2*a*b^2*c*e)*x)*log(c*x + 1))*log(-c*x + 1))/(c^2*x - c), 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((e*x+d)*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^3*(d + e*x),x)

[Out]

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

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