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

Optimal. Leaf size=185 \[ -\frac{b^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c^4}+\frac{b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}-\frac{2 b^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4}+\frac{3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}+\frac{b^3 x}{4 c^3}-\frac{b^3 \tanh ^{-1}(c x)}{4 c^4} \]

[Out]

(b^3*x)/(4*c^3) - (b^3*ArcTanh[c*x])/(4*c^4) + (b^2*x^2*(a + b*ArcTanh[c*x]))/(4*c^2) + (b*(a + b*ArcTanh[c*x]
)^2)/c^4 + (3*b*x*(a + b*ArcTanh[c*x])^2)/(4*c^3) + (b*x^3*(a + b*ArcTanh[c*x])^2)/(4*c) - (a + b*ArcTanh[c*x]
)^3/(4*c^4) + (x^4*(a + b*ArcTanh[c*x])^3)/4 - (2*b^2*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/c^4 - (b^3*PolyLo
g[2, 1 - 2/(1 - c*x)])/c^4

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Rubi [A]  time = 0.569391, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {5916, 5980, 321, 206, 5984, 5918, 2402, 2315, 5910, 5948} \[ -\frac{b^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c^4}+\frac{b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}-\frac{2 b^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4}+\frac{3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}+\frac{b^3 x}{4 c^3}-\frac{b^3 \tanh ^{-1}(c x)}{4 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x])^3,x]

[Out]

(b^3*x)/(4*c^3) - (b^3*ArcTanh[c*x])/(4*c^4) + (b^2*x^2*(a + b*ArcTanh[c*x]))/(4*c^2) + (b*(a + b*ArcTanh[c*x]
)^2)/c^4 + (3*b*x*(a + b*ArcTanh[c*x])^2)/(4*c^3) + (b*x^3*(a + b*ArcTanh[c*x])^2)/(4*c) - (a + b*ArcTanh[c*x]
)^3/(4*c^4) + (x^4*(a + b*ArcTanh[c*x])^3)/4 - (2*b^2*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/c^4 - (b^3*PolyLo
g[2, 1 - 2/(1 - c*x)])/c^4

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 321

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 2402

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 2315

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

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 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]

Rubi steps

\begin{align*} \int x^3 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{1}{4} (3 b c) \int \frac{x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{(3 b) \int x^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{4 c}-\frac{(3 b) \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{4 c}\\ &=\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{1}{2} b^2 \int \frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac{(3 b) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{4 c^3}-\frac{(3 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{4 c^3}\\ &=\frac{3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{b^2 \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c^2}-\frac{b^2 \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{2 c^2}-\frac{\left (3 b^2\right ) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{2 c^2}\\ &=\frac{b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{b^2 \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{2 c^3}-\frac{\left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{2 c^3}-\frac{b^3 \int \frac{x^2}{1-c^2 x^2} \, dx}{4 c}\\ &=\frac{b^3 x}{4 c^3}+\frac{b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{2 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^4}-\frac{b^3 \int \frac{1}{1-c^2 x^2} \, dx}{4 c^3}+\frac{b^3 \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{2 c^3}+\frac{\left (3 b^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{2 c^3}\\ &=\frac{b^3 x}{4 c^3}-\frac{b^3 \tanh ^{-1}(c x)}{4 c^4}+\frac{b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{2 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^4}-\frac{b^3 \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{2 c^4}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{2 c^4}\\ &=\frac{b^3 x}{4 c^3}-\frac{b^3 \tanh ^{-1}(c x)}{4 c^4}+\frac{b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac{3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{2 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c^4}-\frac{b^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c^4}\\ \end{align*}

Mathematica [A]  time = 0.49171, size = 245, normalized size = 1.32 \[ \frac{8 b^3 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+2 b \tanh ^{-1}(c x) \left (3 a^2 c^4 x^4+2 a b c x \left (c^2 x^2+3\right )+b^2 \left (c^2 x^2-1\right )-8 b^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+2 a^2 b c^3 x^3+6 a^2 b c x+3 a^2 b \log (1-c x)-3 a^2 b \log (c x+1)+2 a^3 c^4 x^4+2 a b^2 c^2 x^2+8 a b^2 \log \left (1-c^2 x^2\right )+2 b^2 \tanh ^{-1}(c x)^2 \left (3 a \left (c^4 x^4-1\right )+b \left (c^3 x^3+3 c x-4\right )\right )-2 a b^2+2 b^3 \left (c^4 x^4-1\right ) \tanh ^{-1}(c x)^3+2 b^3 c x}{8 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*a*b^2 + 6*a^2*b*c*x + 2*b^3*c*x + 2*a*b^2*c^2*x^2 + 2*a^2*b*c^3*x^3 + 2*a^3*c^4*x^4 + 2*b^2*(b*(-4 + 3*c*x
+ c^3*x^3) + 3*a*(-1 + c^4*x^4))*ArcTanh[c*x]^2 + 2*b^3*(-1 + c^4*x^4)*ArcTanh[c*x]^3 + 2*b*ArcTanh[c*x]*(3*a
^2*c^4*x^4 + b^2*(-1 + c^2*x^2) + 2*a*b*c*x*(3 + c^2*x^2) - 8*b^2*Log[1 + E^(-2*ArcTanh[c*x])]) + 3*a^2*b*Log[
1 - c*x] - 3*a^2*b*Log[1 + c*x] + 8*a*b^2*Log[1 - c^2*x^2] + 8*b^3*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(8*c^4)

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Maple [C]  time = 0.602, size = 1245, 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(x^3*(a+b*arctanh(c*x))^3,x)

[Out]

-1/4/c^4*b^3*arctanh(c*x)^3+1/c^4*b^3*arctanh(c*x)^2-2/c^4*b^3*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-2/c^4*b^3
*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+1/4*x^4*b^3*arctanh(c*x)^3-3/16*I/c^4*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^
2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2
+1/4*b^3*x/c^3-1/4*b^3*arctanh(c*x)/c^4+1/4*x^4*a^3-3/16*I/c^4*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*
x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+3/16*I/c^4*b^3*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^
(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*arctanh(c*x)^2+3/16*I/c^4*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csg
n(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+3/8*I/c^4*b^3*Pi*csgn(I*(c*x+1)/(-c^2*x
^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*arctanh(c*x)^2+1/2/c*a*b^2*arctanh(c*x)*x^3+3/2/c^3*a*b^2*arctanh
(c*x)*x+3/4/c^4*a*b^2*arctanh(c*x)*ln(c*x-1)-3/4/c^4*a*b^2*arctanh(c*x)*ln(c*x+1)-3/8/c^4*a*b^2*ln(c*x-1)*ln(1
/2+1/2*c*x)-3/8/c^4*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+3/8/c^4*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-3/8*I/c^4*
b^3*Pi*arctanh(c*x)^2+1/4/c^2*x^2*a*b^2+3/4/c^3*x*a^2*b+1/4/c*a^2*b*x^3+3/4*a*b^2*x^4*arctanh(c*x)^2+3/4*x^4*a
^2*b*arctanh(c*x)+3/8/c^4*b^3*arctanh(c*x)^2*ln(c*x-1)-3/8/c^4*b^3*arctanh(c*x)^2*ln(c*x+1)-2/c^4*b^3*arctanh(
c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-2/c^4*b^3*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+3/4/c^4*b^3*
arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+1/c^4*a*b^2*ln(c*x-1)+1/c^4*a*b^2*ln(c*x+1)+3/8/c^4*a^2*b*ln(c*x
-1)-3/8/c^4*a^2*b*ln(c*x+1)+3/16/c^4*a*b^2*ln(c*x-1)^2+3/16/c^4*a*b^2*ln(c*x+1)^2+1/4/c^2*b^3*arctanh(c*x)*x^2
+1/4/c*b^3*arctanh(c*x)^2*x^3+3/4/c^3*b^3*arctanh(c*x)^2*x-1/4/c^4*b^3+3/8*I/c^4*b^3*Pi*csgn(I/((c*x+1)^2/(-c^
2*x^2+1)+1))^2*arctanh(c*x)^2+3/16*I/c^4*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arc
tanh(c*x)^2+3/16*I/c^4*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2-3/8*I/c^4*b^3*Pi*csgn(I/((c*x+1)^
2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^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(x^3*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

3/4*a*b^2*x^4*arctanh(c*x)^2 + 1/4*a^3*x^4 + 1/8*(6*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x +
1)/c^5 + 3*log(c*x - 1)/c^5))*a^2*b + 1/16*(4*c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c
^5)*arctanh(c*x) + (4*c^2*x^2 - 2*(3*log(c*x - 1) - 8)*log(c*x + 1) + 3*log(c*x + 1)^2 + 3*log(c*x - 1)^2 + 16
*log(c*x - 1))/c^4)*a*b^2 - 1/9216*(27*c^5*((c^2*x^4 + 2*x^2)/c^7 + 2*log(c^2*x^2 - 1)/c^9) + 74*c^4*(2*(c^2*x
^3 + 3*x)/c^7 - 3*log(c*x + 1)/c^8 + 3*log(c*x - 1)/c^8) + 60*c^3*(x^2/c^5 + log(c^2*x^2 - 1)/c^7) - 221184*c^
3*integrate(1/96*x^3*log(c*x + 1)/(c^5*x^2 - c^3), x) + 1692*c^2*(2*x/c^5 - log(c*x + 1)/c^6 + log(c*x - 1)/c^
6) - 1105920*c*integrate(1/96*x*log(c*x + 1)/(c^5*x^2 - c^3), x) + (9*(32*log(-c*x + 1)^3 - 24*log(-c*x + 1)^2
+ 12*log(-c*x + 1) - 3)*(c*x - 1)^4 + 128*(9*log(-c*x + 1)^3 - 9*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 2)*(c*x
- 1)^3 + 432*(4*log(-c*x + 1)^3 - 6*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 3)*(c*x - 1)^2 + 1152*(log(-c*x + 1)^3
- 3*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 6)*(c*x - 1))/c^4 - 12*(24*(c^4*x^4 - 1)*log(c*x + 1)^3 + 48*(c^3*x^3
+ 3*c*x)*log(c*x + 1)^2 - 6*(3*c^4*x^4 - 4*c^3*x^3 + 6*c^2*x^2 - 12*c*x - 12*(c^4*x^4 - 1)*log(c*x + 1) + 7)*
log(-c*x + 1)^2 + (9*c^4*x^4 + 28*c^3*x^3 - 18*c^2*x^2 - 72*(c^4*x^4 - 1)*log(c*x + 1)^2 + 300*c*x - 96*(c^3*x
^3 + 3*c*x + 4)*log(c*x + 1))*log(-c*x + 1))/c^4 + 1800*log(96*c^5*x^2 - 96*c^3)/c^4 - 442368*integrate(1/96*l
og(c*x + 1)/(c^5*x^2 - c^3), x))*b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*atanh(c*x))**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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