3.1.77 \(\int x^3 (a+b \tanh ^{-1}(c x^2))^3 \, dx\) [77]

Optimal. Leaf size=141 \[ \frac {3 b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 c^2}+\frac {3 b x^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 c}-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {3 b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \log \left (\frac {2}{1-c x^2}\right )}{2 c^2}-\frac {3 b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right )}{4 c^2} \]

[Out]

3/4*b*(a+b*arctanh(c*x^2))^2/c^2+3/4*b*x^2*(a+b*arctanh(c*x^2))^2/c-1/4*(a+b*arctanh(c*x^2))^3/c^2+1/4*x^4*(a+
b*arctanh(c*x^2))^3-3/2*b^2*(a+b*arctanh(c*x^2))*ln(2/(-c*x^2+1))/c^2-3/4*b^3*polylog(2,1-2/(-c*x^2+1))/c^2

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Rubi [A]
time = 0.22, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {6039, 6037, 6127, 6021, 6131, 6055, 2449, 2352, 6095} \begin {gather*} -\frac {3 b^2 \log \left (\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{2 c^2}-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{4 c^2}+\frac {3 b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 c^2}+\frac {3 b x^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 c}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x^2])^3,x]

[Out]

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

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 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 6039

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

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]

Rubi steps

\begin {align*} \int x^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \, dx &=\int \left (\frac {1}{8} x^3 \left (2 a-b \log \left (1-c x^2\right )\right )^3+\frac {3}{8} b x^3 \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )-\frac {3}{8} b^2 x^3 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {1}{8} b^3 x^3 \log ^3\left (1+c x^2\right )\right ) \, dx\\ &=\frac {1}{8} \int x^3 \left (2 a-b \log \left (1-c x^2\right )\right )^3 \, dx+\frac {1}{8} (3 b) \int x^3 \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right ) \, dx-\frac {1}{8} \left (3 b^2\right ) \int x^3 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right ) \, dx+\frac {1}{8} b^3 \int x^3 \log ^3\left (1+c x^2\right ) \, dx\\ &=\frac {1}{16} \text {Subst}\left (\int x (2 a-b \log (1-c x))^3 \, dx,x,x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int x (-2 a+b \log (1-c x))^2 \log (1+c x) \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int x (-2 a+b \log (1-c x)) \log ^2(1+c x) \, dx,x,x^2\right )+\frac {1}{16} b^3 \text {Subst}\left (\int x \log ^3(1+c x) \, dx,x,x^2\right )\\ &=\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {1}{16} \text {Subst}\left (\int \left (\frac {(2 a-b \log (1-c x))^3}{c}-\frac {(1-c x) (2 a-b \log (1-c x))^3}{c}\right ) \, dx,x,x^2\right )+\frac {1}{16} b^3 \text {Subst}\left (\int \left (-\frac {\log ^3(1+c x)}{c}+\frac {(1+c x) \log ^3(1+c x)}{c}\right ) \, dx,x,x^2\right )-\frac {1}{32} (3 b c) \text {Subst}\left (\int \frac {x^2 (-2 a+b \log (1-c x))^2}{1+c x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^2 c\right ) \text {Subst}\left (\int \frac {x^2 (-2 a+b \log (1-c x)) \log (1+c x)}{1-c x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^2 c\right ) \text {Subst}\left (\int \frac {x^2 (-2 a+b \log (1-c x)) \log (1+c x)}{1+c x} \, dx,x,x^2\right )-\frac {1}{32} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {x^2 \log ^2(1+c x)}{1-c x} \, dx,x,x^2\right )\\ &=\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {\text {Subst}\left (\int (2 a-b \log (1-c x))^3 \, dx,x,x^2\right )}{16 c}-\frac {\text {Subst}\left (\int (1-c x) (2 a-b \log (1-c x))^3 \, dx,x,x^2\right )}{16 c}-\frac {b^3 \text {Subst}\left (\int \log ^3(1+c x) \, dx,x,x^2\right )}{16 c}+\frac {b^3 \text {Subst}\left (\int (1+c x) \log ^3(1+c x) \, dx,x,x^2\right )}{16 c}-\frac {1}{32} (3 b c) \text {Subst}\left (\int \left (-\frac {(-2 a+b \log (1-c x))^2}{c^2}+\frac {x (-2 a+b \log (1-c x))^2}{c}+\frac {(-2 a+b \log (1-c x))^2}{c^2 (1+c x)}\right ) \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^2 c\right ) \text {Subst}\left (\int \left (\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c^2}+\frac {x (2 a-b \log (1-c x)) \log (1+c x)}{c}+\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c^2 (-1+c x)}\right ) \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^2 c\right ) \text {Subst}\left (\int \left (\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c^2}-\frac {x (2 a-b \log (1-c x)) \log (1+c x)}{c}-\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c^2 (1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{32} \left (3 b^3 c\right ) \text {Subst}\left (\int \left (-\frac {\log ^2(1+c x)}{c^2}-\frac {x \log ^2(1+c x)}{c}-\frac {\log ^2(1+c x)}{c^2 (-1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {1}{32} (3 b) \text {Subst}\left (\int x (-2 a+b \log (1-c x))^2 \, dx,x,x^2\right )+\frac {1}{32} \left (3 b^3\right ) \text {Subst}\left (\int x \log ^2(1+c x) \, dx,x,x^2\right )-\frac {\text {Subst}\left (\int (2 a-b \log (x))^3 \, dx,x,1-c x^2\right )}{16 c^2}+\frac {\text {Subst}\left (\int x (2 a-b \log (x))^3 \, dx,x,1-c x^2\right )}{16 c^2}-\frac {b^3 \text {Subst}\left (\int \log ^3(x) \, dx,x,1+c x^2\right )}{16 c^2}+\frac {b^3 \text {Subst}\left (\int x \log ^3(x) \, dx,x,1+c x^2\right )}{16 c^2}+\frac {(3 b) \text {Subst}\left (\int (-2 a+b \log (1-c x))^2 \, dx,x,x^2\right )}{32 c}-\frac {(3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2}{1+c x} \, dx,x,x^2\right )}{32 c}+2 \frac {\left (3 b^2\right ) \text {Subst}\left (\int (2 a-b \log (1-c x)) \log (1+c x) \, dx,x,x^2\right )}{16 c}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x)) \log (1+c x)}{-1+c x} \, dx,x,x^2\right )}{16 c}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x)) \log (1+c x)}{1+c x} \, dx,x,x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log ^2(1+c x) \, dx,x,x^2\right )}{32 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log ^2(1+c x)}{-1+c x} \, dx,x,x^2\right )}{32 c}\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 c^2}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{32 c^2}+\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c^2}+\frac {b^3 \left (1+c x^2\right )^2 \log ^3\left (1+c x^2\right )}{32 c^2}-\frac {1}{32} (3 b) \text {Subst}\left (\int \left (\frac {(-2 a+b \log (1-c x))^2}{c}-\frac {(1-c x) (-2 a+b \log (1-c x))^2}{c}\right ) \, dx,x,x^2\right )+\frac {1}{32} \left (3 b^3\right ) \text {Subst}\left (\int \left (-\frac {\log ^2(1+c x)}{c}+\frac {(1+c x) \log ^2(1+c x)}{c}\right ) \, dx,x,x^2\right )+2 \left (\frac {3 b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c}-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {x (2 a-b \log (1-c x))}{1+c x} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^3\right ) \text {Subst}\left (\int \frac {x \log (1+c x)}{1-c x} \, dx,x,x^2\right )\right )+\frac {(3 b) \text {Subst}\left (\int x (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{32 c^2}-\frac {(3 b) \text {Subst}\left (\int (-2 a+b \log (x))^2 \, dx,x,1-c x^2\right )}{32 c^2}-\frac {(3 b) \text {Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{16 c^2}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (2-x)) \log (x)}{x} \, dx,x,1+c x^2\right )}{16 c^2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\log (2-x) (2 a-b \log (x))}{x} \, dx,x,1-c x^2\right )}{16 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{32 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int x \log ^2(x) \, dx,x,1+c x^2\right )}{32 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{16 c^2}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )}{16 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right ) \log (1+c x)}{1+c x} \, dx,x,x^2\right )}{16 c}\\ &=-\frac {9 b \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{32 c^2}+\frac {3 b \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{64 c^2}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 c^2}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{32 c^2}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {9 b^3 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{32 c^2}-\frac {3 b^3 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{64 c^2}+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{32 c^2}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c^2}+\frac {b^3 \left (1+c x^2\right )^2 \log ^3\left (1+c x^2\right )}{32 c^2}+2 \left (\frac {3 b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c}-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \left (\frac {2 a-b \log (1-c x)}{c}-\frac {2 a-b \log (1-c x)}{c (1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^3\right ) \text {Subst}\left (\int \left (-\frac {\log (1+c x)}{c}-\frac {\log (1+c x)}{c (-1+c x)}\right ) \, dx,x,x^2\right )\right )-\frac {(3 b) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{2-x} \, dx,x,1-c x^2\right )}{32 c^2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int x (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{32 c^2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int (-2 a+b \log (x)) \, dx,x,1-c x^2\right )}{16 c^2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {2-x}{2}\right ) (-2 a+b \log (x))}{x} \, dx,x,1-c x^2\right )}{16 c^2}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{8 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int x \log (x) \, dx,x,1+c x^2\right )}{32 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{2-x} \, dx,x,1+c x^2\right )}{32 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {2-x}{2}\right ) \log (x)}{x} \, dx,x,1+c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{8 c^2}-\frac {(3 b) \text {Subst}\left (\int (-2 a+b \log (1-c x))^2 \, dx,x,x^2\right )}{32 c}+\frac {(3 b) \text {Subst}\left (\int (1-c x) (-2 a+b \log (1-c x))^2 \, dx,x,x^2\right )}{32 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log ^2(1+c x) \, dx,x,x^2\right )}{32 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int (1+c x) \log ^2(1+c x) \, dx,x,x^2\right )}{32 c}\\ &=\frac {9 a b^2 x^2}{8 c}+\frac {9 b^3 x^2}{16 c}+\frac {3 b^3 \left (1-c x^2\right )^2}{128 c^2}-\frac {3 b^3 \left (1+c x^2\right )^2}{128 c^2}+\frac {3 b^2 \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )}{64 c^2}-\frac {9 b \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{32 c^2}+\frac {3 b \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{64 c^2}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 c^2}-\frac {9 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{16 c^2}+\frac {3 b^3 \left (1+c x^2\right )^2 \log \left (1+c x^2\right )}{64 c^2}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {9 b^3 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{32 c^2}-\frac {3 b^3 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{64 c^2}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c^2}+\frac {b^3 \left (1+c x^2\right )^2 \log ^3\left (1+c x^2\right )}{32 c^2}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{16 c^2}+\frac {3 b^3 \log \left (1+c x^2\right ) \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^2}+\frac {(3 b) \text {Subst}\left (\int (-2 a+b \log (x))^2 \, dx,x,1-c x^2\right )}{32 c^2}-\frac {(3 b) \text {Subst}\left (\int x (-2 a+b \log (x))^2 \, dx,x,1-c x^2\right )}{32 c^2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right ) (2 a-b \log (x))}{x} \, dx,x,1-c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{32 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int x \log ^2(x) \, dx,x,1+c x^2\right )}{32 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{16 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right ) \log (x)}{x} \, dx,x,1+c x^2\right )}{16 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{16 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{8 c^2}+2 \left (\frac {3 b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int (2 a-b \log (1-c x)) \, dx,x,x^2\right )}{16 c}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {2 a-b \log (1-c x)}{1+c x} \, dx,x,x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (1+c x) \, dx,x,x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^2\right )}{16 c}\right )\\ &=\frac {9 a b^2 x^2}{8 c}+\frac {9 b^3 x^2}{8 c}+\frac {3 b^3 \left (1-c x^2\right )^2}{128 c^2}-\frac {3 b^3 \left (1+c x^2\right )^2}{128 c^2}+\frac {9 b^3 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{16 c^2}+\frac {3 b^2 \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )}{64 c^2}-\frac {3 b \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 c^2}-\frac {9 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{16 c^2}+\frac {3 b^3 \left (1+c x^2\right )^2 \log \left (1+c x^2\right )}{64 c^2}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{16 c^2}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c^2}+\frac {b^3 \left (1+c x^2\right )^2 \log ^3\left (1+c x^2\right )}{32 c^2}+\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1-c x^2\right )\right )}{16 c^2}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^2}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int x (-2 a+b \log (x)) \, dx,x,1-c x^2\right )}{32 c^2}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int (-2 a+b \log (x)) \, dx,x,1-c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int x \log (x) \, dx,x,1+c x^2\right )}{32 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{16 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{16 c^2}+2 \left (-\frac {3 a b^2 x^2}{8 c}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^2}+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c^2}+\frac {3 b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (1-c x) \, dx,x,x^2\right )}{16 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )}{16 c}\right )\\ &=\frac {3 a b^2 x^2}{4 c}+\frac {15 b^3 x^2}{16 c}+\frac {9 b^3 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{16 c^2}-\frac {3 b \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 c^2}-\frac {3 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c^2}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{16 c^2}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c^2}+\frac {b^3 \left (1+c x^2\right )^2 \log ^3\left (1+c x^2\right )}{32 c^2}+2 \left (-\frac {3 a b^2 x^2}{8 c}-\frac {3 b^3 x^2}{16 c}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^2}+\frac {3 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{16 c^2}+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c^2}+\frac {3 b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{16 c^2}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{16 c^2}\right )-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{16 c^2}\\ &=\frac {3 a b^2 x^2}{4 c}+\frac {3 b^3 x^2}{4 c}+\frac {3 b^3 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{8 c^2}-\frac {3 b \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c^2}+\frac {\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^3}{32 c^2}-\frac {3 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c^2}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{16 c^2}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{32 c^2}+\frac {3}{32} b^2 x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c^2}+\frac {b^3 \left (1+c x^2\right )^2 \log ^3\left (1+c x^2\right )}{32 c^2}+2 \left (-\frac {3 a b^2 x^2}{8 c}-\frac {3 b^3 x^2}{8 c}-\frac {3 b^3 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{16 c^2}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^2}+\frac {3 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{16 c^2}+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c^2}+\frac {3 b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c}-\frac {3 b^3 \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{16 c^2}+\frac {3 b^3 \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 185, normalized size = 1.31 \begin {gather*} \frac {6 b^2 \left (-1+c x^2\right ) \left (a+b+a c x^2\right ) \tanh ^{-1}\left (c x^2\right )^2+2 b^3 \left (-1+c^2 x^4\right ) \tanh ^{-1}\left (c x^2\right )^3+6 b \tanh ^{-1}\left (c x^2\right ) \left (a c x^2 \left (2 b+a c x^2\right )-2 b^2 \log \left (1+e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )+a \left (6 a b c x^2+2 a^2 c^2 x^4+3 a b \log \left (1-c x^2\right )-3 a b \log \left (1+c x^2\right )+6 b^2 \log \left (1-c^2 x^4\right )\right )+6 b^3 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )}{8 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.32, size = 798, normalized size = 5.66

method	result	size
risch	\(\text {Expression too large to display}\)	\(798\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x^2))^3,x,method=_RETURNVERBOSE)

[Out]

1/32*b^3*(c^2*x^4-1)/c^2*ln(c*x^2+1)^3+3/32*b^2*(-b*c^2*ln(-c*x^2+1)*x^4+2*a*c^2*x^4+2*b*c*x^2+b*ln(-c*x^2+1)-
2*a+2*b)/c^2*ln(c*x^2+1)^2+(3/32*b^3*(c^2*x^4-1)/c^2*ln(-c*x^2+1)^2-3/32*b^2*(2*a*c*x^2+b)^2/c^2/a*ln(-c*x^2+1
)-3/32*b*(-4*a^3*c^2*x^4-8*a^2*b*c*x^2-4*ln(-c*x^2+1)*a^2*b-4*ln(-c*x^2+1)*a*b^2-ln(-c*x^2+1)*b^3-4*a*b^2)/c^2
/a)*ln(c*x^2+1)-3/4*a*b^2/c*x^2*ln(-c*x^2+1)+1/4*a^3*x^4+3/4*b^2/c^2*ln(c*x^2+1)*a+3/16*b^2/c^2*a*ln(c*x^2-1)-
3/8*b/c^2*ln(c*x^2+1)*a^2-1/32*b^3*x^4*ln(-c*x^2+1)^3-3/16*b^3/c^2*ln(-c*x^2+1)^2+1/32*b^3/c^2*ln(-c*x^2+1)^3-
3/8/c^2*b^3*ln(c*x^2-1)-3/8/c^2*b^3*ln(c*x^2+1)-3/16*b^3/c^2+3/8/c^2*b^3*ln(-c*x^2+1)+3/16*b^3/c*x^2*ln(-c*x^2
+1)^2-3/8*a^2*b*x^4*ln(-c*x^2+1)+3/4*a^2*b/c*x^2+3/8*a^2*b/c^2*ln(c*x^2-1)+3/16*a*b^2*x^4*ln(-c*x^2+1)^2+9/16*
a*b^2/c^2*ln(-c*x^2+1)-3/16*a*b^2/c^2*ln(-c*x^2+1)^2+3/4*b^2/c*Sum(-(ln(x-_alpha)*ln(-c*x^2+1)+2*c*(-1/2*ln(x-
_alpha)*(ln((RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1)-x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1))+ln((Root
Of(_Z^2*c+2*_Z*_alpha*c-2,index=2)-x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2)))/c-1/2*(dilog((RootOf(_Z^
2*c+2*_Z*_alpha*c-2,index=1)-x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1))+dilog((RootOf(_Z^2*c+2*_Z*_alph
a*c-2,index=2)-x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2)))/c))*b/c,_alpha=RootOf(_Z^2*c+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(x^3*(a+b*arctanh(c*x^2))^3,x, algorithm="maxima")

[Out]

3/4*a*b^2*x^4*arctanh(c*x^2)^2 + 1/4*a^3*x^4 + 3/8*(2*x^4*arctanh(c*x^2) + c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 +
log(c*x^2 - 1)/c^3))*a^2*b + 3/16*(4*c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3)*arctanh(c*x^2) -
(2*(log(c*x^2 - 1) - 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1))/c^2)*a*b^2 -
1/128*(4*x^4*log(-c*x^2 + 1)^3 + 3*c^3*(x^4/c^3 + log(c^2*x^4 - 1)/c^5) - 6*c*((c*x^4 + 2*x^2)/c^2 + 2*log(c*
x^2 - 1)/c^3)*log(-c*x^2 + 1)^2 + 21*c^2*(2*x^2/c^3 - log(c*x^2 + 1)/c^4 + log(c*x^2 - 1)/c^4) + c*(6*(c^2*x^4
+ 6*c*x^2 + 2*log(c*x^2 - 1)^2 + 6*log(c*x^2 - 1))*log(-c*x^2 + 1)/c^3 - (3*c^2*x^4 + 42*c*x^2 + 4*log(c*x^2
- 1)^3 + 18*log(c*x^2 - 1)^2 + 42*log(c*x^2 - 1))/c^3) - 1152*c*integrate(1/4*x^3*log(c*x^2 + 1)/(c^3*x^4 - c)
, x) - 2*(12*c*x^2*log(c*x^2 + 1)^2 + 2*(c^2*x^4 - 1)*log(c*x^2 + 1)^3 - 3*(c^2*x^4 - 2*c*x^2 - 2*(c^2*x^4 - 1
)*log(c*x^2 + 1) + 1)*log(-c*x^2 + 1)^2 + 3*(c^2*x^4 + 6*c*x^2 - 2*(c^2*x^4 - 1)*log(c*x^2 + 1)^2 - 8*(c*x^2 +
1)*log(c*x^2 + 1))*log(-c*x^2 + 1))/c^2 + 18*log(4*c^3*x^4 - 4*c)/c^2 - 384*integrate(1/4*x*log(c*x^2 + 1)/(c
^3*x^4 - c), x))*b^3

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*atanh(c*x**2))**3, 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(x^3*(a+b*arctanh(c*x^2))^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x^2))^3,x)

[Out]

int(x^3*(a + b*atanh(c*x^2))^3, x)

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