3.77 \(\int x^2 (d+c d x)^2 (a+b \tanh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=312 \[ \frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^3}-\frac {16 b d^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{15 c^3}+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {a b d^2 x}{c^2}+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{10} b c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{3} b d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {8 b d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}-\frac {8 b^2 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{15 c^3}-\frac {19 b^2 d^2 \tanh ^{-1}(c x)}{30 c^3}+\frac {19 b^2 d^2 x}{30 c^2}+\frac {b^2 d^2 x \tanh ^{-1}(c x)}{c^2}+\frac {2 b^2 d^2 \log \left (1-c^2 x^2\right )}{3 c^3}+\frac {b^2 d^2 x^2}{6 c}+\frac {1}{30} b^2 d^2 x^3 \]

[Out]

a*b*d^2*x/c^2+19/30*b^2*d^2*x/c^2+1/6*b^2*d^2*x^2/c+1/30*b^2*d^2*x^3-19/30*b^2*d^2*arctanh(c*x)/c^3+b^2*d^2*x*
arctanh(c*x)/c^2+8/15*b*d^2*x^2*(a+b*arctanh(c*x))/c+1/3*b*d^2*x^3*(a+b*arctanh(c*x))+1/10*b*c*d^2*x^4*(a+b*ar
ctanh(c*x))+1/30*d^2*(a+b*arctanh(c*x))^2/c^3+1/3*d^2*x^3*(a+b*arctanh(c*x))^2+1/2*c*d^2*x^4*(a+b*arctanh(c*x)
)^2+1/5*c^2*d^2*x^5*(a+b*arctanh(c*x))^2-16/15*b*d^2*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^3+2/3*b^2*d^2*ln(-c^2
*x^2+1)/c^3-8/15*b^2*d^2*polylog(2,1-2/(-c*x+1))/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.88, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 36, number of rules used = 15, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.682, Rules used = {5940, 5916, 5980, 321, 206, 5984, 5918, 2402, 2315, 266, 43, 5910, 260, 5948, 302} \[ -\frac {8 b^2 d^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{15 c^3}+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {a b d^2 x}{c^2}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^3}-\frac {16 b d^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{15 c^3}+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{10} b c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{3} b d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {8 b d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {2 b^2 d^2 \log \left (1-c^2 x^2\right )}{3 c^3}+\frac {19 b^2 d^2 x}{30 c^2}+\frac {b^2 d^2 x \tanh ^{-1}(c x)}{c^2}-\frac {19 b^2 d^2 \tanh ^{-1}(c x)}{30 c^3}+\frac {b^2 d^2 x^2}{6 c}+\frac {1}{30} b^2 d^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*d^2*x)/c^2 + (19*b^2*d^2*x)/(30*c^2) + (b^2*d^2*x^2)/(6*c) + (b^2*d^2*x^3)/30 - (19*b^2*d^2*ArcTanh[c*x])
/(30*c^3) + (b^2*d^2*x*ArcTanh[c*x])/c^2 + (8*b*d^2*x^2*(a + b*ArcTanh[c*x]))/(15*c) + (b*d^2*x^3*(a + b*ArcTa
nh[c*x]))/3 + (b*c*d^2*x^4*(a + b*ArcTanh[c*x]))/10 + (d^2*(a + b*ArcTanh[c*x])^2)/(30*c^3) + (d^2*x^3*(a + b*
ArcTanh[c*x])^2)/3 + (c*d^2*x^4*(a + b*ArcTanh[c*x])^2)/2 + (c^2*d^2*x^5*(a + b*ArcTanh[c*x])^2)/5 - (16*b*d^2
*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(15*c^3) + (2*b^2*d^2*Log[1 - c^2*x^2])/(3*c^3) - (8*b^2*d^2*PolyLog[2
, 1 - 2/(1 - c*x)])/(15*c^3)

Rule 43

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 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 260

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 266

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 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 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 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 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 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 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 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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]

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 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]

Rubi steps

\begin {align*} \int x^2 (d+c d x)^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\int \left (d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+2 c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^2 \int x^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx+\left (2 c d^2\right ) \int x^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx+\left (c^2 d^2\right ) \int x^4 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{3} \left (2 b c d^2\right ) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\left (b c^2 d^2\right ) \int \frac {x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\frac {1}{5} \left (2 b c^3 d^2\right ) \int \frac {x^5 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\left (b d^2\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx-\left (b d^2\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac {\left (2 b d^2\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c}-\frac {\left (2 b d^2\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c}+\frac {1}{5} \left (2 b c d^2\right ) \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx-\frac {1}{5} \left (2 b c d^2\right ) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {b d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+\frac {1}{3} b d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{10} b c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^3}+\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{3} \left (b^2 d^2\right ) \int \frac {x^2}{1-c^2 x^2} \, dx-\frac {\left (2 b d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{3 c^2}+\frac {\left (b d^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^2}-\frac {\left (b d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{c^2}+\frac {\left (2 b d^2\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c}-\frac {\left (2 b d^2\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c}-\frac {1}{3} \left (b^2 c d^2\right ) \int \frac {x^3}{1-c^2 x^2} \, dx-\frac {1}{10} \left (b^2 c^2 d^2\right ) \int \frac {x^4}{1-c^2 x^2} \, dx\\ &=\frac {a b d^2 x}{c^2}+\frac {b^2 d^2 x}{3 c^2}+\frac {8 b d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {1}{3} b d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{10} b c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^3}+\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^3}-\frac {1}{5} \left (b^2 d^2\right ) \int \frac {x^2}{1-c^2 x^2} \, dx-\frac {\left (2 b d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{5 c^2}-\frac {\left (b^2 d^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{3 c^2}+\frac {\left (2 b^2 d^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{3 c^2}+\frac {\left (b^2 d^2\right ) \int \tanh ^{-1}(c x) \, dx}{c^2}-\frac {1}{6} \left (b^2 c d^2\right ) \operatorname {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )-\frac {1}{10} \left (b^2 c^2 d^2\right ) \int \left (-\frac {1}{c^4}-\frac {x^2}{c^2}+\frac {1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac {a b d^2 x}{c^2}+\frac {19 b^2 d^2 x}{30 c^2}+\frac {1}{30} b^2 d^2 x^3-\frac {b^2 d^2 \tanh ^{-1}(c x)}{3 c^3}+\frac {b^2 d^2 x \tanh ^{-1}(c x)}{c^2}+\frac {8 b d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {1}{3} b d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{10} b c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^3}+\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {16 b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{15 c^3}-\frac {\left (2 b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{3 c^3}-\frac {\left (b^2 d^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{10 c^2}-\frac {\left (b^2 d^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{5 c^2}+\frac {\left (2 b^2 d^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^2}-\frac {\left (b^2 d^2\right ) \int \frac {x}{1-c^2 x^2} \, dx}{c}-\frac {1}{6} \left (b^2 c d^2\right ) \operatorname {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 {a b d^2 x}{c^2}+\frac {19 b^2 d^2 x}{30 c^2}+\frac {b^2 d^2 x^2}{6 c}+\frac {1}{30} b^2 d^2 x^3-\frac {19 b^2 d^2 \tanh ^{-1}(c x)}{30 c^3}+\frac {b^2 d^2 x \tanh ^{-1}(c x)}{c^2}+\frac {8 b d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {1}{3} b d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{10} b c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^3}+\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {16 b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{15 c^3}+\frac {2 b^2 d^2 \log \left (1-c^2 x^2\right )}{3 c^3}-\frac {b^2 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c^3}-\frac {\left (2 b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{5 c^3}\\ &=\frac {a b d^2 x}{c^2}+\frac {19 b^2 d^2 x}{30 c^2}+\frac {b^2 d^2 x^2}{6 c}+\frac {1}{30} b^2 d^2 x^3-\frac {19 b^2 d^2 \tanh ^{-1}(c x)}{30 c^3}+\frac {b^2 d^2 x \tanh ^{-1}(c x)}{c^2}+\frac {8 b d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {1}{3} b d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{10} b c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^3}+\frac {1}{3} d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c^2 d^2 x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {16 b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{15 c^3}+\frac {2 b^2 d^2 \log \left (1-c^2 x^2\right )}{3 c^3}-\frac {8 b^2 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{15 c^3}\\ \end {align*}

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Mathematica [A]  time = 1.03, size = 297, normalized size = 0.95 \[ \frac {d^2 \left (6 a^2 c^5 x^5+15 a^2 c^4 x^4+10 a^2 c^3 x^3+3 a b c^4 x^4+10 a b c^3 x^3+16 a b c^2 x^2+16 a b \log \left (c^2 x^2-1\right )+b \tanh ^{-1}(c x) \left (2 a c^3 x^3 \left (6 c^2 x^2+15 c x+10\right )+b \left (3 c^4 x^4+10 c^3 x^3+16 c^2 x^2+30 c x-19\right )-32 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+30 a b c x+15 a b \log (1-c x)-15 a b \log (c x+1)-9 a b+b^2 c^3 x^3+5 b^2 c^2 x^2+20 b^2 \log \left (1-c^2 x^2\right )+b^2 \left (6 c^5 x^5+15 c^4 x^4+10 c^3 x^3-31\right ) \tanh ^{-1}(c x)^2+16 b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+19 b^2 c x-5 b^2\right )}{30 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*(-9*a*b - 5*b^2 + 30*a*b*c*x + 19*b^2*c*x + 16*a*b*c^2*x^2 + 5*b^2*c^2*x^2 + 10*a^2*c^3*x^3 + 10*a*b*c^3*
x^3 + b^2*c^3*x^3 + 15*a^2*c^4*x^4 + 3*a*b*c^4*x^4 + 6*a^2*c^5*x^5 + b^2*(-31 + 10*c^3*x^3 + 15*c^4*x^4 + 6*c^
5*x^5)*ArcTanh[c*x]^2 + b*ArcTanh[c*x]*(2*a*c^3*x^3*(10 + 15*c*x + 6*c^2*x^2) + b*(-19 + 30*c*x + 16*c^2*x^2 +
10*c^3*x^3 + 3*c^4*x^4) - 32*b*Log[1 + E^(-2*ArcTanh[c*x])]) + 15*a*b*Log[1 - c*x] - 15*a*b*Log[1 + c*x] + 20
*b^2*Log[1 - c^2*x^2] + 16*a*b*Log[-1 + c^2*x^2] + 16*b^2*PolyLog[2, -E^(-2*ArcTanh[c*x])]))/(30*c^3)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} c^{2} d^{2} x^{4} + 2 \, a^{2} c d^{2} x^{3} + a^{2} d^{2} x^{2} + {\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c d^{2} x^{3} + b^{2} d^{2} x^{2}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c^{2} d^{2} x^{4} + 2 \, a b c d^{2} x^{3} + a b d^{2} x^{2}\right )} \operatorname {artanh}\left (c x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(a^2*c^2*d^2*x^4 + 2*a^2*c*d^2*x^3 + a^2*d^2*x^2 + (b^2*c^2*d^2*x^4 + 2*b^2*c*d^2*x^3 + b^2*d^2*x^2)*a
rctanh(c*x)^2 + 2*(a*b*c^2*d^2*x^4 + 2*a*b*c*d^2*x^3 + a*b*d^2*x^2)*arctanh(c*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c d x + d\right )}^{2} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 521, normalized size = 1.67 \[ \frac {d^{2} b^{2} \arctanh \left (c x \right ) x^{3}}{3}+\frac {c \,d^{2} a^{2} x^{4}}{2}+\frac {d^{2} a b \,x^{3}}{3}+\frac {d^{2} b^{2} \arctanh \left (c x \right )^{2} x^{3}}{3}-\frac {8 d^{2} b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{15 c^{3}}+\frac {31 d^{2} b^{2} \ln \left (c x -1\right )^{2}}{120 c^{3}}+\frac {7 d^{2} b^{2} \ln \left (c x +1\right )}{20 c^{3}}+\frac {59 d^{2} b^{2} \ln \left (c x -1\right )}{60 c^{3}}+\frac {c^{2} d^{2} a^{2} x^{5}}{5}-\frac {d^{2} b^{2} \ln \left (c x +1\right )^{2}}{120 c^{3}}+c \,d^{2} a b \arctanh \left (c x \right ) x^{4}+\frac {2 c^{2} d^{2} a b \arctanh \left (c x \right ) x^{5}}{5}+\frac {b^{2} d^{2} x^{3}}{30}+\frac {19 b^{2} d^{2} x}{30 c^{2}}+\frac {b^{2} d^{2} x^{2}}{6 c}+\frac {c \,d^{2} b^{2} \arctanh \left (c x \right )^{2} x^{4}}{2}+\frac {d^{2} b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{30 c^{3}}+\frac {8 d^{2} b^{2} \arctanh \left (c x \right ) x^{2}}{15 c}-\frac {31 d^{2} b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{60 c^{3}}+\frac {d^{2} a b \ln \left (c x +1\right )}{30 c^{3}}+\frac {31 d^{2} a b \ln \left (c x -1\right )}{30 c^{3}}+\frac {c \,d^{2} a b \,x^{4}}{10}+\frac {d^{2} b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{60 c^{3}}-\frac {d^{2} b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{60 c^{3}}+\frac {a b \,d^{2} x}{c^{2}}+\frac {d^{2} a^{2} x^{3}}{3}+\frac {8 d^{2} a b \,x^{2}}{15 c}+\frac {c \,d^{2} b^{2} \arctanh \left (c x \right ) x^{4}}{10}+\frac {c^{2} d^{2} b^{2} \arctanh \left (c x \right )^{2} x^{5}}{5}+\frac {2 d^{2} a b \arctanh \left (c x \right ) x^{3}}{3}+\frac {31 d^{2} b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{30 c^{3}}+\frac {b^{2} d^{2} x \arctanh \left (c x \right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*d*x+d)^2*(a+b*arctanh(c*x))^2,x)

[Out]

1/2*c*d^2*a^2*x^4+1/3*d^2*a*b*x^3+31/120/c^3*d^2*b^2*ln(c*x-1)^2+1/3*d^2*b^2*arctanh(c*x)^2*x^3+7/20/c^3*d^2*b
^2*ln(c*x+1)+59/60/c^3*d^2*b^2*ln(c*x-1)+1/5*c^2*d^2*a^2*x^5-8/15/c^3*d^2*b^2*dilog(1/2+1/2*c*x)-1/120/c^3*d^2
*b^2*ln(c*x+1)^2+1/3*d^2*b^2*arctanh(c*x)*x^3+c*d^2*a*b*arctanh(c*x)*x^4+2/5*c^2*d^2*a*b*arctanh(c*x)*x^5+1/30
*b^2*d^2*x^3+19/30*b^2*d^2*x/c^2+1/6*b^2*d^2*x^2/c+1/2*c*d^2*b^2*arctanh(c*x)^2*x^4+1/30/c^3*d^2*b^2*arctanh(c
*x)*ln(c*x+1)-31/60/c^3*d^2*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+1/30/c^3*d^2*a*b*ln(c*x+1)+31/30/c^3*d^2*a*b*ln(c*x-
1)+8/15/c*d^2*b^2*arctanh(c*x)*x^2+1/10*c*d^2*b^2*arctanh(c*x)*x^4+1/5*c^2*d^2*b^2*arctanh(c*x)^2*x^5+1/10*c*d
^2*a*b*x^4+2/3*d^2*a*b*arctanh(c*x)*x^3+1/60/c^3*d^2*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/60/c^3*d^2*b^2*ln(-1/2*c
*x+1/2)*ln(1/2+1/2*c*x)+31/30/c^3*d^2*b^2*arctanh(c*x)*ln(c*x-1)+a*b*d^2*x/c^2+b^2*d^2*x*arctanh(c*x)/c^2+1/3*
d^2*a^2*x^3+8/15/c*d^2*a*b*x^2

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maxima [B]  time = 0.66, size = 604, normalized size = 1.94 \[ \frac {1}{5} \, a^{2} c^{2} d^{2} x^{5} + \frac {1}{2} \, a^{2} c d^{2} x^{4} + \frac {1}{10} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} a b c^{2} d^{2} + \frac {1}{3} \, a^{2} d^{2} x^{3} + \frac {1}{6} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\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 b c d^{2} + \frac {1}{3} \, {\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 )} a b d^{2} + \frac {8 \, {\left (\log \left (c x + 1\right ) \log \left (-\frac {1}{2} \, c x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c x + \frac {1}{2}\right )\right )} b^{2} d^{2}}{15 \, c^{3}} + \frac {7 \, b^{2} d^{2} \log \left (c x + 1\right )}{20 \, c^{3}} + \frac {59 \, b^{2} d^{2} \log \left (c x - 1\right )}{60 \, c^{3}} + \frac {4 \, b^{2} c^{3} d^{2} x^{3} + 20 \, b^{2} c^{2} d^{2} x^{2} + 76 \, b^{2} c d^{2} x + {\left (6 \, b^{2} c^{5} d^{2} x^{5} + 15 \, b^{2} c^{4} d^{2} x^{4} + 10 \, b^{2} c^{3} d^{2} x^{3} + b^{2} d^{2}\right )} \log \left (c x + 1\right )^{2} + {\left (6 \, b^{2} c^{5} d^{2} x^{5} + 15 \, b^{2} c^{4} d^{2} x^{4} + 10 \, b^{2} c^{3} d^{2} x^{3} - 31 \, b^{2} d^{2}\right )} \log \left (-c x + 1\right )^{2} + 2 \, {\left (3 \, b^{2} c^{4} d^{2} x^{4} + 10 \, b^{2} c^{3} d^{2} x^{3} + 16 \, b^{2} c^{2} d^{2} x^{2} + 30 \, b^{2} c d^{2} x\right )} \log \left (c x + 1\right ) - 2 \, {\left (3 \, b^{2} c^{4} d^{2} x^{4} + 10 \, b^{2} c^{3} d^{2} x^{3} + 16 \, b^{2} c^{2} d^{2} x^{2} + 30 \, b^{2} c d^{2} x + {\left (6 \, b^{2} c^{5} d^{2} x^{5} + 15 \, b^{2} c^{4} d^{2} x^{4} + 10 \, b^{2} c^{3} d^{2} x^{3} + b^{2} d^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{120 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*c^2*d^2*x^5 + 1/2*a^2*c*d^2*x^4 + 1/10*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2
- 1)/c^6))*a*b*c^2*d^2 + 1/3*a^2*d^2*x^3 + 1/6*(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*b*c*d^2 + 1/3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a*b*d^2
+ 8/15*(log(c*x + 1)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*b^2*d^2/c^3 + 7/20*b^2*d^2*log(c*x + 1)/c^3 +
59/60*b^2*d^2*log(c*x - 1)/c^3 + 1/120*(4*b^2*c^3*d^2*x^3 + 20*b^2*c^2*d^2*x^2 + 76*b^2*c*d^2*x + (6*b^2*c^5*
d^2*x^5 + 15*b^2*c^4*d^2*x^4 + 10*b^2*c^3*d^2*x^3 + b^2*d^2)*log(c*x + 1)^2 + (6*b^2*c^5*d^2*x^5 + 15*b^2*c^4*
d^2*x^4 + 10*b^2*c^3*d^2*x^3 - 31*b^2*d^2)*log(-c*x + 1)^2 + 2*(3*b^2*c^4*d^2*x^4 + 10*b^2*c^3*d^2*x^3 + 16*b^
2*c^2*d^2*x^2 + 30*b^2*c*d^2*x)*log(c*x + 1) - 2*(3*b^2*c^4*d^2*x^4 + 10*b^2*c^3*d^2*x^3 + 16*b^2*c^2*d^2*x^2
+ 30*b^2*c*d^2*x + (6*b^2*c^5*d^2*x^5 + 15*b^2*c^4*d^2*x^4 + 10*b^2*c^3*d^2*x^3 + b^2*d^2)*log(c*x + 1))*log(-
c*x + 1))/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c*x))^2*(d + c*d*x)^2,x)

[Out]

int(x^2*(a + b*atanh(c*x))^2*(d + c*d*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{2} \left (\int a^{2} x^{2}\, dx + \int 2 a^{2} c x^{3}\, dx + \int a^{2} c^{2} x^{4}\, dx + \int b^{2} x^{2} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int 2 a b x^{2} \operatorname {atanh}{\left (c x \right )}\, dx + \int 2 b^{2} c x^{3} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int b^{2} c^{2} x^{4} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int 4 a b c x^{3} \operatorname {atanh}{\left (c x \right )}\, dx + \int 2 a b c^{2} x^{4} \operatorname {atanh}{\left (c x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(a**2*x**2, x) + Integral(2*a**2*c*x**3, x) + Integral(a**2*c**2*x**4, x) + Integral(b**2*x**2*a
tanh(c*x)**2, x) + Integral(2*a*b*x**2*atanh(c*x), x) + Integral(2*b**2*c*x**3*atanh(c*x)**2, x) + Integral(b*
*2*c**2*x**4*atanh(c*x)**2, x) + Integral(4*a*b*c*x**3*atanh(c*x), x) + Integral(2*a*b*c**2*x**4*atanh(c*x), x
))

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