\(\int x^2 (a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2)) \, dx\) [273]

Optimal result

Integrand size = 27, antiderivative size = 247 \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {2 a e x}{3 c^2}-\frac {5 b e x^2}{18 c}-\frac {2}{9} a e x^3-\frac {2 b e x \coth ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \coth ^{-1}(c x)+\frac {b e \coth ^{-1}(c x)^2}{3 c^3}-\frac {(2 a+b) e \log (1-c x)}{6 c^3}+\frac {(2 a-b) e \log (1+c x)}{6 c^3}-\frac {4 b e \log \left (1-c^2 x^2\right )}{9 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3} \]

[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*arccoth(c*x)/c^2-2/9*b*e*x^3*arccoth(c*x)+1/3*b*e*arccoth(
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*arccoth(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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6038, 272, 45, 6233, 6857, 815, 647, 31, 6128, 6022, 266, 6096, 2525, 2437, 2338} \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {e (2 a+b) \log (1-c x)}{6 c^3}+\frac {e (2 a-b) \log (c x+1)}{6 c^3}+\frac {1}{3} x^3 \left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {2 a e x}{3 c^2}-\frac {2}{9} a e x^3+\frac {b e \coth ^{-1}(c x)^2}{3 c^3}+\frac {b x^2 \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c}-\frac {2 b e x \coth ^{-1}(c x)}{3 c^2}+\frac {b \log \left (1-c^2 x^2\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}-\frac {4 b e \log \left (1-c^2 x^2\right )}{9 c^3}-\frac {2}{9} b e x^3 \coth ^{-1}(c x)-\frac {5 b e x^2}{18 c} \]

[In]

Int[x^2*(a + b*ArcCoth[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*ArcCoth[c*x])/(3*c^2) - (2*b*e*x^3*ArcCoth[
c*x])/9 + (b*e*ArcCoth[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*ArcCoth[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 6022

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcCoth[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 6038

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCoth[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcCoth[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 6096

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[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 6128

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcCoth[c*x])^p, x], x] - Dist[d*(f^2/e), Int[(f*x)^(m - 2)*((a + b*ArcCoth[c*x])
^p/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 6233

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Wit
h[{u = IntHide[x^m*(a + b*ArcCoth[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*} \text {integral}& = \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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-1}(c x) \, dx-\frac {1}{3} (2 b e) \int \frac {x^2 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-1}(c x) \, dx}{3 c^2}-\frac {(2 b e) \int \frac {\coth ^{-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 \coth ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \coth ^{-1}(c x)+\frac {b e \coth ^{-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 \coth ^{-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 \coth ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \coth ^{-1}(c x)+\frac {b e \coth ^{-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 \coth ^{-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 \coth ^{-1}(c x)}{3 c^2}-\frac {2}{9} b e x^3 \coth ^{-1}(c x)+\frac {b e \coth ^{-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 \coth ^{-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*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.74 \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {-24 a c e x+2 b c^2 (3 d-5 e) x^2+4 a c^3 (3 d-2 e) x^3+4 b c x \left (3 c^2 d x^2-2 e \left (3+c^2 x^2\right )\right ) \coth ^{-1}(c x)+12 b e \coth ^{-1}(c x)^2+2 (3 b d-6 a e-11 b e) \log (1-c x)+2 (3 b d+6 a e-11 b e) \log (1+c x)+6 c^2 e x^2 \left (b+2 a c x+2 b c x \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )+3 b e \log ^2\left (1-c^2 x^2\right )}{36 c^3} \]

[In]

Integrate[x^2*(a + b*ArcCoth[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
rcCoth[c*x] + 12*b*e*ArcCoth[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*ArcCoth[c*x])*Log[1 - c^2*x^2] + 3*b*e*Log[1 - c^2*x^2]^2)/(36*
c^3)

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.93

method	result	size
parallelrisch	\(\frac {12 \,\operatorname {arccoth}\left (c x \right ) b d +6 b \,c^{2} d \,x^{2}-24 a e c x -10 b e \,x^{2} c^{2}+12 b e \ln \left (-c^{2} x^{2}+1\right ) \operatorname {arccoth}\left (c x \right ) x^{3} c^{3}+12 \ln \left (c x -1\right ) b d -44 \ln \left (c x -1\right ) b e +6 x^{2} \ln \left (-c^{2} x^{2}+1\right ) b \,c^{2} e -24 e b x \,\operatorname {arccoth}\left (c x \right ) c -8 e b \,\operatorname {arccoth}\left (c x \right ) x^{3} c^{3}+12 e b \operatorname {arccoth}\left (c x \right )^{2}+12 a e \,x^{3} \ln \left (-c^{2} x^{2}+1\right ) c^{3}+12 b \,\operatorname {arccoth}\left (c x \right ) x^{3} c^{3} d -8 a e \,x^{3} c^{3}-44 \,\operatorname {arccoth}\left (c x \right ) b e +12 a d \,x^{3} c^{3}+3 e b \ln \left (-c^{2} x^{2}+1\right )^{2}+24 \,\operatorname {arccoth}\left (c x \right ) a e}{36 c^{3}}\)	\(229\)
risch	\(\text {Expression too large to display}\)	\(1459\)
default	\(\text {Expression too large to display}\)	\(3355\)
parts	\(\text {Expression too large to display}\)	\(3355\)

[In]

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

[Out]

1/36*(12*arccoth(c*x)*b*d+6*b*c^2*d*x^2-24*a*e*c*x-10*b*e*x^2*c^2+12*b*e*ln(-c^2*x^2+1)*arccoth(c*x)*x^3*c^3+1
2*ln(c*x-1)*b*d-44*ln(c*x-1)*b*e+6*x^2*ln(-c^2*x^2+1)*b*c^2*e-24*e*b*x*arccoth(c*x)*c-8*e*b*arccoth(c*x)*x^3*c
^3+12*e*b*arccoth(c*x)^2+12*a*e*x^3*ln(-c^2*x^2+1)*c^3+12*b*arccoth(c*x)*x^3*c^3*d-8*a*e*x^3*c^3-44*arccoth(c*
x)*b*e+12*a*d*x^3*c^3+3*e*b*ln(-c^2*x^2+1)^2+24*arccoth(c*x)*a*e)/c^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.80 \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {24 \, a c e x - 4 \, {\left (3 \, a c^{3} d - 2 \, a c^{3} e\right )} x^{3} - 3 \, b e \log \left (-c^{2} x^{2} + 1\right )^{2} - 3 \, b e \log \left (\frac {c x + 1}{c x - 1}\right )^{2} - 2 \, {\left (3 \, b c^{2} d - 5 \, b c^{2} e\right )} x^{2} - 2 \, {\left (6 \, a c^{3} e x^{3} + 3 \, b c^{2} e x^{2} + 3 \, b d - 11 \, b e\right )} \log \left (-c^{2} x^{2} + 1\right ) - 2 \, {\left (3 \, b c^{3} e x^{3} \log \left (-c^{2} x^{2} + 1\right ) - 6 \, b c e x + {\left (3 \, b c^{3} d - 2 \, b c^{3} e\right )} x^{3} + 6 \, a e\right )} \log \left (\frac {c x + 1}{c x - 1}\right )}{36 \, c^{3}} \]

[In]

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

[Out]

-1/36*(24*a*c*e*x - 4*(3*a*c^3*d - 2*a*c^3*e)*x^3 - 3*b*e*log(-c^2*x^2 + 1)^2 - 3*b*e*log((c*x + 1)/(c*x - 1))
^2 - 2*(3*b*c^2*d - 5*b*c^2*e)*x^2 - 2*(6*a*c^3*e*x^3 + 3*b*c^2*e*x^2 + 3*b*d - 11*b*e)*log(-c^2*x^2 + 1) - 2*
(3*b*c^3*e*x^3*log(-c^2*x^2 + 1) - 6*b*c*e*x + (3*b*c^3*d - 2*b*c^3*e)*x^3 + 6*a*e)*log((c*x + 1)/(c*x - 1)))/
c^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.07 \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{3} \log {\left (- c^{2} x^{2} + 1 \right )}}{3} - \frac {2 a e x^{3}}{9} - \frac {2 a e x}{3 c^{2}} + \frac {2 a e \operatorname {acoth}{\left (c x \right )}}{3 c^{3}} + \frac {b d x^{3} \operatorname {acoth}{\left (c x \right )}}{3} + \frac {b e x^{3} \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {acoth}{\left (c x \right )}}{3} - \frac {2 b e x^{3} \operatorname {acoth}{\left (c x \right )}}{9} + \frac {b d x^{2}}{6 c} + \frac {b e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )}}{6 c} - \frac {5 b e x^{2}}{18 c} - \frac {2 b e x \operatorname {acoth}{\left (c x \right )}}{3 c^{2}} + \frac {b d \log {\left (- c^{2} x^{2} + 1 \right )}}{6 c^{3}} + \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}^{2}}{12 c^{3}} - \frac {11 b e \log {\left (- c^{2} x^{2} + 1 \right )}}{18 c^{3}} + \frac {b e \operatorname {acoth}^{2}{\left (c x \right )}}{3 c^{3}} & \text {for}\: c \neq 0 \\\frac {d x^{3} \left (a + \frac {i \pi b}{2}\right )}{3} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(a+b*acoth(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*acoth(c*x)/(3
*c**3) + b*d*x**3*acoth(c*x)/3 + b*e*x**3*log(-c**2*x**2 + 1)*acoth(c*x)/3 - 2*b*e*x**3*acoth(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*acoth(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*acoth(
c*x)**2/(3*c**3), Ne(c, 0)), (d*x**3*(a + I*pi*b/2)/3, True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.02 \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b e \operatorname {arcoth}\left (c x\right ) + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {arcoth}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} a e + \frac {{\left ({\left (3 i \, \pi c^{2} - 5 \, c^{2}\right )} x^{2} + {\left (3 i \, \pi + 3 \, c^{2} x^{2} + 6 \, \log \left (c x - 1\right ) - 11\right )} \log \left (c x + 1\right ) + {\left (3 i \, \pi + 3 \, c^{2} x^{2} - 11\right )} \log \left (c x - 1\right )\right )} b e}{18 \, c^{3}} \]

[In]

integrate(x^2*(a+b*arccoth(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*e*arccoth(c*x) + 1/6*(2*x^3*arccoth(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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.14 \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {1}{6} \, b e x^{3} \log \left (-c x + 1\right )^{2} - \frac {1}{18} \, {\left (-3 i \, \pi b d + 2 i \, \pi b e - 6 \, a d + 4 \, a e\right )} x^{3} + \frac {1}{6} \, {\left (b e x^{3} + \frac {b e}{c^{3}}\right )} \log \left (c x + 1\right )^{2} + \frac {{\left (3 \, b d - 5 \, b e\right )} x^{2}}{18 \, c} - \frac {1}{18} \, {\left ({\left (-3 i \, \pi b e - 3 \, b d - 6 \, a e + 2 \, b e\right )} x^{3} - \frac {3 \, b e x^{2}}{c} + \frac {6 \, b e x}{c^{2}}\right )} \log \left (c x + 1\right ) - \frac {1}{18} \, {\left ({\left (-3 i \, \pi b e + 3 \, b d - 6 \, a e - 2 \, b e\right )} x^{3} - \frac {3 \, b e x^{2}}{c} - \frac {6 \, b e x}{c^{2}} - \frac {6 \, b e \log \left (c x - 1\right )}{c^{3}}\right )} \log \left (-c x + 1\right ) - \frac {b e \log \left (c x - 1\right )^{2}}{6 \, c^{3}} - \frac {{\left (i \, \pi b e + 2 \, a e\right )} x}{3 \, c^{2}} + \frac {{\left (3 i \, \pi b e + 3 \, b d + 6 \, a e - 11 \, b e\right )} \log \left (c x + 1\right )}{18 \, c^{3}} + \frac {{\left (-3 i \, \pi b e + 3 \, b d - 6 \, a e - 11 \, b e\right )} \log \left (c x - 1\right )}{18 \, c^{3}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.68 \[ \int x^2 \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\ln \left (\frac {1}{c\,x}+1\right )\,\left (\frac {b\,d\,x^3}{6}-\frac {\frac {2\,b\,e\,c^3\,x^3}{3}+2\,b\,e\,c\,x}{6\,c^3}+\frac {b\,e\,x^3\,\ln \left (1-c^2\,x^2\right )}{6}\right )+x\,\left (\frac {a\,\left (3\,d-2\,e\right )}{3\,c^2}-\frac {a\,d}{c^2}\right )+\ln \left (1-\frac {1}{c\,x}\right )\,\left (\frac {\frac {4\,b\,e\,x^4}{9}-\frac {2\,b\,e\,x^2}{3\,c^2}+\frac {2\,b\,c^2\,e\,x^6}{9}}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}+\frac {\frac {b\,d\,x^4}{3}-\frac {b\,c^2\,d\,x^6}{3}}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}+\frac {\ln \left (1-c^2\,x^2\right )\,\left (\frac {b\,e\,x^4}{3}-\frac {b\,c^2\,e\,x^6}{3}\right )}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}-\frac {b\,e\,\ln \left (\frac {1}{c\,x}+1\right )}{6\,c^3}\right )+\frac {a\,x^3\,\left (3\,d-2\,e\right )}{9}+c^2\,\ln \left (1-c^2\,x^2\right )\,\left (\frac {a\,e\,x^3}{3\,c^2}+\frac {b\,e\,x^2}{6\,c^3}\right )-\frac {\ln \left (c\,x-1\right )\,\left (6\,a\,e-3\,b\,d+11\,b\,e\right )}{18\,c^3}+\frac {\ln \left (c\,x+1\right )\,\left (6\,a\,e+3\,b\,d-11\,b\,e\right )}{18\,c^3}+\frac {b\,e\,{\ln \left (\frac {1}{c\,x}+1\right )}^2}{12\,c^3}+\frac {b\,e\,{\ln \left (1-\frac {1}{c\,x}\right )}^2}{12\,c^3}+\frac {b\,e\,{\ln \left (1-c^2\,x^2\right )}^2}{12\,c^3}+\frac {b\,x^2\,\left (3\,d-5\,e\right )}{18\,c} \]

[In]

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

[Out]

log(1/(c*x) + 1)*((b*d*x^3)/6 - (2*b*c*e*x + (2*b*c^3*e*x^3)/3)/(6*c^3) + (b*e*x^3*log(1 - c^2*x^2))/6) + x*((
a*(3*d - 2*e))/(3*c^2) - (a*d)/c^2) + log(1 - 1/(c*x))*(((4*b*e*x^4)/9 - (2*b*e*x^2)/(3*c^2) + (2*b*c^2*e*x^6)
/9)/(2*(x + c*x^2)*(c*x - 1)) + ((b*d*x^4)/3 - (b*c^2*d*x^6)/3)/(2*(x + c*x^2)*(c*x - 1)) + (log(1 - c^2*x^2)*
((b*e*x^4)/3 - (b*c^2*e*x^6)/3))/(2*(x + c*x^2)*(c*x - 1)) - (b*e*log(1/(c*x) + 1))/(6*c^3)) + (a*x^3*(3*d - 2
*e))/9 + c^2*log(1 - c^2*x^2)*((a*e*x^3)/(3*c^2) + (b*e*x^2)/(6*c^3)) - (log(c*x - 1)*(6*a*e - 3*b*d + 11*b*e)
)/(18*c^3) + (log(c*x + 1)*(6*a*e + 3*b*d - 11*b*e))/(18*c^3) + (b*e*log(1/(c*x) + 1)^2)/(12*c^3) + (b*e*log(1
- 1/(c*x))^2)/(12*c^3) + (b*e*log(1 - c^2*x^2)^2)/(12*c^3) + (b*x^2*(3*d - 5*e))/(18*c)