3.6.24 \(\int x^2 (a+b \tanh ^{-1}(c x)) (d+e \log (1-c^2 x^2)) \, dx\) [524]

Optimal. Leaf size=247 \[ -\frac {2 a e x}{3 c^2}-\frac {5 b e x^2}{18 c}-\frac {2}{9} a e x^3-\frac {2 b e x \tanh ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \tanh ^{-1}(c x)+\frac {b e \tanh ^{-1}(c x)^2}{3 c^3}-\frac {(2 a+b) e \log (1-c x)}{6 c^3}+\frac {(2 a-b) e \log (1+c x)}{6 c^3}-\frac {4 b e \log \left (1-c^2 x^2\right )}{9 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3} \]

[Out]

-2/3*a*e*x/c^2-5/18*b*e*x^2/c-2/9*a*e*x^3-2/3*b*e*x*arctanh(c*x)/c^2-2/9*b*e*x^3*arctanh(c*x)+1/3*b*e*arctanh(
c*x)^2/c^3-1/6*(2*a+b)*e*ln(-c*x+1)/c^3+1/6*(2*a-b)*e*ln(c*x+1)/c^3-4/9*b*e*ln(-c^2*x^2+1)/c^3-1/12*b*e*ln(-c^
2*x^2+1)^2/c^3+1/6*b*x^2*(d+e*ln(-c^2*x^2+1))/c+1/3*x^3*(a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))+1/6*b*ln(-c^2*
x^2+1)*(d+e*ln(-c^2*x^2+1))/c^3

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Rubi [A]
time = 0.43, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 15, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6037, 272, 45, 6232, 6857, 815, 647, 31, 6127, 6021, 266, 6095, 2525, 2437, 2338} \begin {gather*} -\frac {e (2 a+b) \log (1-c x)}{6 c^3}+\frac {e (2 a-b) \log (c x+1)}{6 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {2 a e x}{3 c^2}-\frac {2}{9} a e x^3+\frac {b e \tanh ^{-1}(c x)^2}{3 c^3}+\frac {b x^2 \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c}-\frac {2 b e x \tanh ^{-1}(c x)}{3 c^2}+\frac {b \log \left (1-c^2 x^2\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}-\frac {4 b e \log \left (1-c^2 x^2\right )}{9 c^3}-\frac {2}{9} b e x^3 \tanh ^{-1}(c x)-\frac {5 b e x^2}{18 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*e*x)/(3*c^2) - (5*b*e*x^2)/(18*c) - (2*a*e*x^3)/9 - (2*b*e*x*ArcTanh[c*x])/(3*c^2) - (2*b*e*x^3*ArcTanh[
c*x])/9 + (b*e*ArcTanh[c*x]^2)/(3*c^3) - ((2*a + b)*e*Log[1 - c*x])/(6*c^3) + ((2*a - b)*e*Log[1 + c*x])/(6*c^
3) - (4*b*e*Log[1 - c^2*x^2])/(9*c^3) - (b*e*Log[1 - c^2*x^2]^2)/(12*c^3) + (b*x^2*(d + e*Log[1 - c^2*x^2]))/(
6*c) + (x^3*(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/3 + (b*Log[1 - c^2*x^2]*(d + e*Log[1 - c^2*x^2]))/(
6*c^3)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 45

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 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 272

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 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 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 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 6232

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Wit
h[{u = IntHide[x^m*(a + b*ArcTanh[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegra
nd[x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx &=\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}+\left (2 c^2 e\right ) \int \left (-\frac {x^3 \left (b+2 a c x+2 b c x \tanh ^{-1}(c x)\right )}{6 c \left (-1+c^2 x^2\right )}-\frac {b x \log \left (1-c^2 x^2\right )}{6 c^3 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}-\frac {(b e) \int \frac {x \log \left (1-c^2 x^2\right )}{-1+c^2 x^2} \, dx}{3 c}-\frac {1}{3} (c e) \int \frac {x^3 \left (b+2 a c x+2 b c x \tanh ^{-1}(c x)\right )}{-1+c^2 x^2} \, dx\\ &=\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}-\frac {(b e) \text {Subst}\left (\int \frac {\log \left (1-c^2 x\right )}{-1+c^2 x} \, dx,x,x^2\right )}{6 c}-\frac {1}{3} (c e) \int \left (\frac {x^3 (b+2 a c x)}{-1+c^2 x^2}+\frac {2 b c x^4 \tanh ^{-1}(c x)}{-1+c^2 x^2}\right ) \, dx\\ &=\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}-\frac {(b e) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c^2 x^2\right )}{6 c^3}-\frac {1}{3} (c e) \int \frac {x^3 (b+2 a c x)}{-1+c^2 x^2} \, dx-\frac {1}{3} \left (2 b c^2 e\right ) \int \frac {x^4 \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx\\ &=-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}-\frac {1}{3} (2 b e) \int x^2 \tanh ^{-1}(c x) \, dx-\frac {1}{3} (2 b e) \int \frac {x^2 \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx-\frac {1}{3} (c e) \int \left (\frac {2 a}{c^3}+\frac {b x}{c^2}+\frac {2 a x^2}{c}+\frac {2 a+b c x}{c^3 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {2 a e x}{3 c^2}-\frac {b e x^2}{6 c}-\frac {2}{9} a e x^3-\frac {2}{9} b e x^3 \tanh ^{-1}(c x)-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}-\frac {e \int \frac {2 a+b c x}{-1+c^2 x^2} \, dx}{3 c^2}-\frac {(2 b e) \int \tanh ^{-1}(c x) \, dx}{3 c^2}-\frac {(2 b e) \int \frac {\tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{3 c^2}+\frac {1}{9} (2 b c e) \int \frac {x^3}{1-c^2 x^2} \, dx\\ &=-\frac {2 a e x}{3 c^2}-\frac {b e x^2}{6 c}-\frac {2}{9} a e x^3-\frac {2 b e x \tanh ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \tanh ^{-1}(c x)+\frac {b e \tanh ^{-1}(c x)^2}{3 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}+\frac {((2 a-b) e) \int \frac {1}{c+c^2 x} \, dx}{6 c}+\frac {(2 b e) \int \frac {x}{1-c^2 x^2} \, dx}{3 c}-\frac {((2 a+b) e) \int \frac {1}{-c+c^2 x} \, dx}{6 c}+\frac {1}{9} (b c e) \text {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {2 a e x}{3 c^2}-\frac {b e x^2}{6 c}-\frac {2}{9} a e x^3-\frac {2 b e x \tanh ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \tanh ^{-1}(c x)+\frac {b e \tanh ^{-1}(c x)^2}{3 c^3}-\frac {(2 a+b) e \log (1-c x)}{6 c^3}+\frac {(2 a-b) e \log (1+c x)}{6 c^3}-\frac {b e \log \left (1-c^2 x^2\right )}{3 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}+\frac {1}{9} (b c e) \text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {2 a e x}{3 c^2}-\frac {5 b e x^2}{18 c}-\frac {2}{9} a e x^3-\frac {2 b e x \tanh ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \tanh ^{-1}(c x)+\frac {b e \tanh ^{-1}(c x)^2}{3 c^3}-\frac {(2 a+b) e \log (1-c x)}{6 c^3}+\frac {(2 a-b) e \log (1+c x)}{6 c^3}-\frac {4 b e \log \left (1-c^2 x^2\right )}{9 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 183, normalized size = 0.74 \begin {gather*} \frac {-24 a c e x+2 b c^2 (3 d-5 e) x^2+4 a c^3 (3 d-2 e) x^3+4 b c x \left (3 c^2 d x^2-2 e \left (3+c^2 x^2\right )\right ) \tanh ^{-1}(c x)+12 b e \tanh ^{-1}(c x)^2+2 (3 b d-6 a e-11 b e) \log (1-c x)+2 (3 b d+6 a e-11 b e) \log (1+c x)+6 c^2 e x^2 \left (b+2 a c x+2 b c x \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )+3 b e \log ^2\left (1-c^2 x^2\right )}{36 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a*c*e*x + 2*b*c^2*(3*d - 5*e)*x^2 + 4*a*c^3*(3*d - 2*e)*x^3 + 4*b*c*x*(3*c^2*d*x^2 - 2*e*(3 + c^2*x^2))*A
rcTanh[c*x] + 12*b*e*ArcTanh[c*x]^2 + 2*(3*b*d - 6*a*e - 11*b*e)*Log[1 - c*x] + 2*(3*b*d + 6*a*e - 11*b*e)*Log
[1 + c*x] + 6*c^2*e*x^2*(b + 2*a*c*x + 2*b*c*x*ArcTanh[c*x])*Log[1 - c^2*x^2] + 3*b*e*Log[1 - c^2*x^2]^2)/(36*
c^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 11.62, size = 3994, normalized size = 16.17

method	result	size
default	\(\text {Expression too large to display}\)	\(3994\)
risch	\(\text {Expression too large to display}\)	\(4015\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1)),x,method=_RETURNVERBOSE)

[Out]

-2/3*b*e*x*arctanh(c*x)/c^2+1/6*b*d*x^2/c-2/9*a*e*x^3-2/3*a*e*x/c^2-5/18*b*e*x^2/c-2/9*b*e*x^3*arctanh(c*x)+1/
3*a*d*x^3+1/3*a*e*x^3*ln(-c^2*x^2+1)+5/18*e*b/c^3-1/12*I*b/c^3*e*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(
I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2+1/6*I*b/c^3*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))*cs
gn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2-1/12*I*b/c^3*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))^2*csgn(I*(1+(c*x+1)^
2/(-c^2*x^2+1))^2)+1/12*I*b/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*x^2*e+1/12*I*b/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2
-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3*x^2*e+1/12*I*b/c*Pi*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3*x^2*e+1/6*I*b/c
^3*arctanh(c*x)*Pi*e*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3+1/6*I*b/c^3*arctanh(c*x)*Pi*e*csgn(I*(c*x+1)^2/(c^2*x^2-1
)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3+1/6*I*b/c^3*arctanh(c*x)*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3-1/6*I*b
/c^3*Pi*ln(1+(c*x+1)^2/(-c^2*x^2+1))*e*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3+1/6*I*b*Pi*arctanh(c*x)*csgn(I*(1+(c*x+
1)^2/(-c^2*x^2+1))^2)^3*x^3*e+1/6*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*x^3*e+1/6*I*b*Pi*arctanh
(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3*x^3*e-1/6*I*b/c^3*Pi*ln(1+(c*x+1)^2/(-c^2*x
^2+1))*e*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3-1/6*I*b/c^3*Pi*ln(1+(c*x+1)^2/(-c^2*x^2+
1))*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3-1/6*I*b/c^3*e*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+1/12*I*b/c^3*e*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(
-c^2*x^2+1))^2)^2-1/12*I*b/c^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*Pi*e-1/3*I*b
*Pi*arctanh(c*x)*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))*x^3*e+1/6*I*b*Pi*ar
ctanh(c*x)*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))^2*x^3*e+1/6*I*b/c*Pi*csgn(I
*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*x^2*e-1/12*I*b/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))
*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*x^2*e+1/12*I*b/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)
)*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*x^2*e+1/12*I*b/c*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(c*x+1)
^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*x^2*e-1/6*I*b/c*Pi*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(
I*(1+(c*x+1)^2/(-c^2*x^2+1)))*x^2*e+1/12*I*b/c*Pi*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(1+(c*x+1)^2/(-c
^2*x^2+1)))^2*x^2*e+1/3*I*b/c^3*arctanh(c*x)*Pi*e*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2
-1))^2-1/6*I*b/c^3*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*
Pi*e*arctanh(c*x)+1/6*I*b/c^3*arctanh(c*x)*Pi*e*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2
-1))+1/6*I*b/c^3*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)
)^2)*Pi*e*arctanh(c*x)-1/3*I*b/c^3*arctanh(c*x)*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(1+(c*x+1)^2/(-
c^2*x^2+1))^2)^2+1/6*I*b/c^3*arctanh(c*x)*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))^2*csgn(I*(1+(c*x+1)^2/(-c^2*
x^2+1))^2)-1/3*I*b/c^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*e*ln(1+(c*x+1)^2/(-c
^2*x^2+1))*Pi+1/6*I*b/c^3*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(I*(c*x+1)^2/(c^2*x
^2-1))*e*ln(1+(c*x+1)^2/(-c^2*x^2+1))*Pi-1/6*I*b/c^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)
^(1/2))^2*e*ln(1+(c*x+1)^2/(-c^2*x^2+1))*Pi-1/6*I*b/c^3*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)
)^2)^2*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*e*ln(1+(c*x+1)^2/(-c^2*x^2+1))*Pi+1/3*I*b/c^3*csgn(I*(1+(c*x+1)^2/
(-c^2*x^2+1))^2)^2*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))*e*ln(1+(c*x+1)^2/(-c^2*x^2+1))*Pi-1/6*I*b/c^3*csgn(I*(1+
(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))^2*e*ln(1+(c*x+1)^2/(-c^2*x^2+1))*Pi+1/12*I*b/c^3
*e*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+
1)^2/(-c^2*x^2+1))^2)+1/3*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2
))*x^3*e-1/6*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2
*x^2+1))^2)^2*x^3*e+1/6*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2
*x^3*e+1/6*I*b*Pi*arctanh(c*x)*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/
(-c^2*x^2+1))^2)^2*x^3*e+1/3*b/c*ln(2)*x^2*e-1/3*b/c*ln(1+(c*x+1)^2/(-c^2*x^2+1))*x^2*e+2/3*b/c^3*arctanh(c*x)
*ln(2)*e-2/3*b/c^3*ln(2)*ln(1+(c*x+1)^2/(-c^2*x^2+1))*e+2/3*b*ln(2)*arctanh(c*x)*x^3*e-2/3*b*arctanh(c*x)*ln(1
+(c*x+1)^2/(-c^2*x^2+1))*x^3*e+1/3*b/c^3*e*(2*arctanh(c*x)*x^3*c^3+c^2*x^2+2*arctanh(c*x)-2*ln(1+(c*x+1)^2/(-c
^2*x^2+1))-1)*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-1/12*I*b/c^3*Pi*e*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-1/12*I*b/c^3*e*
Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3-1/12*I*b/c^3*e*Pi*csgn(I*(1+(c*x+1)^2/(-c^2*x^
2+1))^2)^3-1/3*a*e/c^3*ln(c*x-1)+1/3*a*e/c^3*ln...

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Maxima [C] Result contains complex when optimal does not.
time = 0.28, size = 255, normalized size = 1.03 \begin {gather*} \frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b \operatorname {artanh}\left (c x\right ) e + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} a e + \frac {{\left ({\left (3 i \, \pi c^{2} - 5 \, c^{2}\right )} x^{2} + {\left (3 i \, \pi + 3 \, c^{2} x^{2} + 6 \, \log \left (c x - 1\right ) - 11\right )} \log \left (c x + 1\right ) + {\left (3 i \, \pi + 3 \, c^{2} x^{2} - 11\right )} \log \left (c x - 1\right )\right )} b e}{18 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="maxima")

[Out]

1/3*a*d*x^3 + 1/9*(3*x^3*log(-c^2*x^2 + 1) - c^2*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/
c^5))*b*arctanh(c*x)*e + 1/6*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*d + 1/9*(3*x^3*log(-c
^2*x^2 + 1) - c^2*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*a*e + 1/18*((3*I*pi*c^2 -
5*c^2)*x^2 + (3*I*pi + 3*c^2*x^2 + 6*log(c*x - 1) - 11)*log(c*x + 1) + (3*I*pi + 3*c^2*x^2 - 11)*log(c*x - 1)
)*b*e/c^3

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Fricas [A]
time = 0.37, size = 306, normalized size = 1.24 \begin {gather*} \frac {12 \, a c^{3} d x^{3} + 6 \, b c^{2} d x^{2} + 3 \, {\left (b \cosh \left (1\right ) + b \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right )^{2} + 3 \, {\left (b \cosh \left (1\right ) + b \sinh \left (1\right )\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 2 \, {\left (4 \, a c^{3} x^{3} + 5 \, b c^{2} x^{2} + 12 \, a c x\right )} \cosh \left (1\right ) + 2 \, {\left (3 \, b d + {\left (6 \, a c^{3} x^{3} + 3 \, b c^{2} x^{2} - 11 \, b\right )} \cosh \left (1\right ) + {\left (6 \, a c^{3} x^{3} + 3 \, b c^{2} x^{2} - 11 \, b\right )} \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right ) + 2 \, {\left (3 \, b c^{3} d x^{3} - 2 \, {\left (b c^{3} x^{3} + 3 \, b c x - 3 \, a\right )} \cosh \left (1\right ) + 3 \, {\left (b c^{3} x^{3} \cosh \left (1\right ) + b c^{3} x^{3} \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right ) - 2 \, {\left (b c^{3} x^{3} + 3 \, b c x - 3 \, a\right )} \sinh \left (1\right )\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) - 2 \, {\left (4 \, a c^{3} x^{3} + 5 \, b c^{2} x^{2} + 12 \, a c x\right )} \sinh \left (1\right )}{36 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="fricas")

[Out]

1/36*(12*a*c^3*d*x^3 + 6*b*c^2*d*x^2 + 3*(b*cosh(1) + b*sinh(1))*log(-c^2*x^2 + 1)^2 + 3*(b*cosh(1) + b*sinh(1
))*log(-(c*x + 1)/(c*x - 1))^2 - 2*(4*a*c^3*x^3 + 5*b*c^2*x^2 + 12*a*c*x)*cosh(1) + 2*(3*b*d + (6*a*c^3*x^3 +
3*b*c^2*x^2 - 11*b)*cosh(1) + (6*a*c^3*x^3 + 3*b*c^2*x^2 - 11*b)*sinh(1))*log(-c^2*x^2 + 1) + 2*(3*b*c^3*d*x^3
- 2*(b*c^3*x^3 + 3*b*c*x - 3*a)*cosh(1) + 3*(b*c^3*x^3*cosh(1) + b*c^3*x^3*sinh(1))*log(-c^2*x^2 + 1) - 2*(b*
c^3*x^3 + 3*b*c*x - 3*a)*sinh(1))*log(-(c*x + 1)/(c*x - 1)) - 2*(4*a*c^3*x^3 + 5*b*c^2*x^2 + 12*a*c*x)*sinh(1)
)/c^3

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Sympy [A]
time = 0.99, size = 258, normalized size = 1.04 \begin {gather*} \begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{3} \log {\left (- c^{2} x^{2} + 1 \right )}}{3} - \frac {2 a e x^{3}}{9} - \frac {2 a e x}{3 c^{2}} + \frac {2 a e \operatorname {atanh}{\left (c x \right )}}{3 c^{3}} + \frac {b d x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b e x^{3} \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{3} - \frac {2 b e x^{3} \operatorname {atanh}{\left (c x \right )}}{9} + \frac {b d x^{2}}{6 c} + \frac {b e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )}}{6 c} - \frac {5 b e x^{2}}{18 c} - \frac {2 b e x \operatorname {atanh}{\left (c x \right )}}{3 c^{2}} + \frac {b d \log {\left (- c^{2} x^{2} + 1 \right )}}{6 c^{3}} + \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}^{2}}{12 c^{3}} - \frac {11 b e \log {\left (- c^{2} x^{2} + 1 \right )}}{18 c^{3}} + \frac {b e \operatorname {atanh}^{2}{\left (c x \right )}}{3 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d x^{3}}{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atanh(c*x))*(d+e*ln(-c**2*x**2+1)),x)

[Out]

Piecewise((a*d*x**3/3 + a*e*x**3*log(-c**2*x**2 + 1)/3 - 2*a*e*x**3/9 - 2*a*e*x/(3*c**2) + 2*a*e*atanh(c*x)/(3
*c**3) + b*d*x**3*atanh(c*x)/3 + b*e*x**3*log(-c**2*x**2 + 1)*atanh(c*x)/3 - 2*b*e*x**3*atanh(c*x)/9 + b*d*x**
2/(6*c) + b*e*x**2*log(-c**2*x**2 + 1)/(6*c) - 5*b*e*x**2/(18*c) - 2*b*e*x*atanh(c*x)/(3*c**2) + b*d*log(-c**2
*x**2 + 1)/(6*c**3) + b*e*log(-c**2*x**2 + 1)**2/(12*c**3) - 11*b*e*log(-c**2*x**2 + 1)/(18*c**3) + b*e*atanh(
c*x)**2/(3*c**3), Ne(c, 0)), (a*d*x**3/3, True))

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Giac [A]
time = 0.54, size = 244, normalized size = 0.99 \begin {gather*} -\frac {1}{6} \, b e x^{3} \log \left (-c x + 1\right )^{2} + \frac {1}{9} \, {\left (3 \, a d - 2 \, a e\right )} x^{3} + \frac {1}{6} \, {\left (b e x^{3} + \frac {b e}{c^{3}}\right )} \log \left (c x + 1\right )^{2} + \frac {{\left (3 \, b d - 5 \, b e\right )} x^{2}}{18 \, c} + \frac {1}{18} \, {\left ({\left (3 \, b d + 6 \, a e - 2 \, b e\right )} x^{3} + \frac {3 \, b e x^{2}}{c} - \frac {6 \, b e x}{c^{2}}\right )} \log \left (c x + 1\right ) - \frac {1}{18} \, {\left ({\left (3 \, b d - 6 \, a e - 2 \, b e\right )} x^{3} - \frac {3 \, b e x^{2}}{c} - \frac {6 \, b e x}{c^{2}} - \frac {6 \, b e \log \left (c x - 1\right )}{c^{3}}\right )} \log \left (-c x + 1\right ) - \frac {2 \, a e x}{3 \, c^{2}} - \frac {b e \log \left (c x - 1\right )^{2}}{6 \, c^{3}} + \frac {{\left (3 \, b d + 6 \, a e - 11 \, b e\right )} \log \left (c x + 1\right )}{18 \, c^{3}} + \frac {{\left (3 \, b d - 6 \, a e - 11 \, b e\right )} \log \left (c x - 1\right )}{18 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="giac")

[Out]

-1/6*b*e*x^3*log(-c*x + 1)^2 + 1/9*(3*a*d - 2*a*e)*x^3 + 1/6*(b*e*x^3 + b*e/c^3)*log(c*x + 1)^2 + 1/18*(3*b*d
- 5*b*e)*x^2/c + 1/18*((3*b*d + 6*a*e - 2*b*e)*x^3 + 3*b*e*x^2/c - 6*b*e*x/c^2)*log(c*x + 1) - 1/18*((3*b*d -
6*a*e - 2*b*e)*x^3 - 3*b*e*x^2/c - 6*b*e*x/c^2 - 6*b*e*log(c*x - 1)/c^3)*log(-c*x + 1) - 2/3*a*e*x/c^2 - 1/6*b
*e*log(c*x - 1)^2/c^3 + 1/18*(3*b*d + 6*a*e - 11*b*e)*log(c*x + 1)/c^3 + 1/18*(3*b*d - 6*a*e - 11*b*e)*log(c*x
- 1)/c^3

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Mupad [B]
time = 2.81, size = 515, normalized size = 2.09 \begin {gather*} \frac {a\,d\,x^3}{3}-\frac {2\,a\,e\,x^3}{9}+\frac {b\,d\,x^3\,\ln \left (c\,x+1\right )}{6}-\frac {b\,d\,x^3\,\ln \left (1-c\,x\right )}{6}-\frac {b\,e\,x^3\,\ln \left (c\,x+1\right )}{9}+\frac {b\,e\,x^3\,\ln \left (1-c\,x\right )}{9}+\frac {b\,e\,{\ln \left (c\,x+1\right )}^2}{6\,c^3}+\frac {b\,e\,{\ln \left (1-c\,x\right )}^2}{6\,c^3}-\frac {2\,a\,e\,x}{3\,c^2}+\frac {b\,d\,x^2}{6\,c}-\frac {5\,b\,e\,x^2}{18\,c}+\frac {a\,e\,x^3\,\ln \left (1-c^2\,x^2\right )}{3}-\frac {a\,e\,\ln \left (c\,x-1\right )}{3\,c^3}+\frac {a\,e\,\ln \left (c\,x+1\right )}{3\,c^3}+\frac {b\,d\,\ln \left (c\,x-1\right )}{6\,c^3}+\frac {b\,d\,\ln \left (c\,x+1\right )}{6\,c^3}-\frac {11\,b\,e\,\ln \left (c\,x-1\right )}{18\,c^3}-\frac {11\,b\,e\,\ln \left (c\,x+1\right )}{18\,c^3}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (-\frac {2\,a\,e-2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (-\frac {2\,a\,e+2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (-\frac {2\,a\,e-2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (-\frac {2\,a\,e+2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,x\,\ln \left (c\,x+1\right )}{3\,c^2}+\frac {b\,e\,x\,\ln \left (1-c\,x\right )}{3\,c^2}+\frac {b\,e\,x^2\,\ln \left (1-c^2\,x^2\right )}{6\,c}+\frac {b\,e\,\ln \left (-\frac {2\,a\,e-2\,a\,c\,e\,x}{3\,c^2}\right )\,\ln \left (1-c^2\,x^2\right )}{6\,c^3}+\frac {b\,e\,\ln \left (-\frac {2\,a\,e+2\,a\,c\,e\,x}{3\,c^2}\right )\,\ln \left (1-c^2\,x^2\right )}{6\,c^3}+\frac {b\,e\,x^3\,\ln \left (c\,x+1\right )\,\ln \left (1-c^2\,x^2\right )}{6}-\frac {b\,e\,x^3\,\ln \left (1-c\,x\right )\,\ln \left (1-c^2\,x^2\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)),x)

[Out]

(a*d*x^3)/3 - (2*a*e*x^3)/9 + (b*d*x^3*log(c*x + 1))/6 - (b*d*x^3*log(1 - c*x))/6 - (b*e*x^3*log(c*x + 1))/9 +
(b*e*x^3*log(1 - c*x))/9 + (b*e*log(c*x + 1)^2)/(6*c^3) + (b*e*log(1 - c*x)^2)/(6*c^3) - (2*a*e*x)/(3*c^2) +
(b*d*x^2)/(6*c) - (5*b*e*x^2)/(18*c) + (a*e*x^3*log(1 - c^2*x^2))/3 - (a*e*log(c*x - 1))/(3*c^3) + (a*e*log(c*
x + 1))/(3*c^3) + (b*d*log(c*x - 1))/(6*c^3) + (b*d*log(c*x + 1))/(6*c^3) - (11*b*e*log(c*x - 1))/(18*c^3) - (
11*b*e*log(c*x + 1))/(18*c^3) - (b*e*log(c*x + 1)*log(-(2*a*e - 2*a*c*e*x)/(3*c^2)))/(6*c^3) - (b*e*log(c*x +
1)*log(-(2*a*e + 2*a*c*e*x)/(3*c^2)))/(6*c^3) - (b*e*log(1 - c*x)*log(-(2*a*e - 2*a*c*e*x)/(3*c^2)))/(6*c^3) -
(b*e*log(1 - c*x)*log(-(2*a*e + 2*a*c*e*x)/(3*c^2)))/(6*c^3) - (b*e*x*log(c*x + 1))/(3*c^2) + (b*e*x*log(1 -
c*x))/(3*c^2) + (b*e*x^2*log(1 - c^2*x^2))/(6*c) + (b*e*log(-(2*a*e - 2*a*c*e*x)/(3*c^2))*log(1 - c^2*x^2))/(6
*c^3) + (b*e*log(-(2*a*e + 2*a*c*e*x)/(3*c^2))*log(1 - c^2*x^2))/(6*c^3) + (b*e*x^3*log(c*x + 1)*log(1 - c^2*x
^2))/6 - (b*e*x^3*log(1 - c*x)*log(1 - c^2*x^2))/6

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