3.1.68 \(\int x^3 (d+c d x) (a+b \tanh ^{-1}(c x))^2 \, dx\) [68]

Optimal. Leaf size=270 \[ \frac {a b d x}{2 c^3}+\frac {3 b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {b^2 d x^3}{30 c}-\frac {3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac {b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{5 c^4}+\frac {b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}-\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{5 c^4} \]

[Out]

1/2*a*b*d*x/c^3+3/10*b^2*d*x/c^3+1/12*b^2*d*x^2/c^2+1/30*b^2*d*x^3/c-3/10*b^2*d*arctanh(c*x)/c^4+1/2*b^2*d*x*a
rctanh(c*x)/c^3+1/5*b*d*x^2*(a+b*arctanh(c*x))/c^2+1/6*b*d*x^3*(a+b*arctanh(c*x))/c+1/10*b*d*x^4*(a+b*arctanh(
c*x))-1/20*d*(a+b*arctanh(c*x))^2/c^4+1/4*d*x^4*(a+b*arctanh(c*x))^2+1/5*c*d*x^5*(a+b*arctanh(c*x))^2-2/5*b*d*
(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^4+1/3*b^2*d*ln(-c^2*x^2+1)/c^4-1/5*b^2*d*polylog(2,1-2/(-c*x+1))/c^4

________________________________________________________________________________________

Rubi [A]
time = 0.46, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 15, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6087, 6037, 6127, 272, 45, 6021, 266, 6095, 308, 212, 327, 6131, 6055, 2449, 2352} \begin {gather*} -\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}-\frac {2 b d \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^4}+\frac {a b d x}{2 c^3}+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^4}-\frac {3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac {3 b^2 d x}{10 c^3}+\frac {b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}+\frac {b^2 d x^3}{30 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*d*x)/(2*c^3) + (3*b^2*d*x)/(10*c^3) + (b^2*d*x^2)/(12*c^2) + (b^2*d*x^3)/(30*c) - (3*b^2*d*ArcTanh[c*x])/
(10*c^4) + (b^2*d*x*ArcTanh[c*x])/(2*c^3) + (b*d*x^2*(a + b*ArcTanh[c*x]))/(5*c^2) + (b*d*x^3*(a + b*ArcTanh[c
*x]))/(6*c) + (b*d*x^4*(a + b*ArcTanh[c*x]))/10 - (d*(a + b*ArcTanh[c*x])^2)/(20*c^4) + (d*x^4*(a + b*ArcTanh[
c*x])^2)/4 + (c*d*x^5*(a + b*ArcTanh[c*x])^2)/5 - (2*b*d*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(5*c^4) + (b^2
*d*Log[1 - c^2*x^2])/(3*c^4) - (b^2*d*PolyLog[2, 1 - 2/(1 - c*x)])/(5*c^4)

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 308

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 327

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 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 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 6087

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 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 (d+c d x) \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\int \left (d x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+c d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int x^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx+(c d) \int x^4 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{2} (b c d) \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^2 d\right ) \int \frac {x^5 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} (2 b d) \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx-\frac {1}{5} (2 b d) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac {(b d) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c}-\frac {(b d) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{2 c}\\ &=\frac {b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{6} \left (b^2 d\right ) \int \frac {x^3}{1-c^2 x^2} \, dx+\frac {(b d) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac {(b d) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{2 c^3}+\frac {(2 b d) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^2}-\frac {(2 b d) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^2}-\frac {1}{10} \left (b^2 c d\right ) \int \frac {x^4}{1-c^2 x^2} \, dx\\ &=\frac {a b d x}{2 c^3}+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac {b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{12} \left (b^2 d\right ) \text {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )-\frac {(2 b d) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{5 c^3}+\frac {\left (b^2 d\right ) \int \tanh ^{-1}(c x) \, dx}{2 c^3}-\frac {\left (b^2 d\right ) \int \frac {x^2}{1-c^2 x^2} \, dx}{5 c}-\frac {1}{10} \left (b^2 c d\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 x}{2 c^3}+\frac {3 b^2 d x}{10 c^3}+\frac {b^2 d x^3}{30 c}+\frac {b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac {b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{5 c^4}-\frac {1}{12} \left (b^2 d\right ) \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 {\left (b^2 d\right ) \int \frac {1}{1-c^2 x^2} \, dx}{10 c^3}-\frac {\left (b^2 d\right ) \int \frac {1}{1-c^2 x^2} \, dx}{5 c^3}+\frac {\left (2 b^2 d\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^3}-\frac {\left (b^2 d\right ) \int \frac {x}{1-c^2 x^2} \, dx}{2 c^2}\\ &=\frac {a b d x}{2 c^3}+\frac {3 b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {b^2 d x^3}{30 c}-\frac {3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac {b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{5 c^4}+\frac {b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}-\frac {\left (2 b^2 d\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{5 c^4}\\ &=\frac {a b d x}{2 c^3}+\frac {3 b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {b^2 d x^3}{30 c}-\frac {3 b^2 d \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 d x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac {b d x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {1}{10} b d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{5 c^4}+\frac {b^2 d \log \left (1-c^2 x^2\right )}{3 c^4}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.50, size = 271, normalized size = 1.00 \begin {gather*} \frac {d \left (-18 a b-5 b^2+30 a b c x+18 b^2 c x+12 a b c^2 x^2+5 b^2 c^2 x^2+10 a b c^3 x^3+2 b^2 c^3 x^3+15 a^2 c^4 x^4+6 a b c^4 x^4+12 a^2 c^5 x^5+3 b^2 \left (-9+5 c^4 x^4+4 c^5 x^5\right ) \tanh ^{-1}(c x)^2+2 b \tanh ^{-1}(c x) \left (3 a c^4 x^4 (5+4 c x)+b \left (-9+15 c x+6 c^2 x^2+5 c^3 x^3+3 c^4 x^4\right )-12 b \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+15 a b \log (1-c x)-15 a b \log (1+c x)+20 b^2 \log \left (1-c^2 x^2\right )+12 a b \log \left (-1+c^2 x^2\right )+12 b^2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{60 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(-18*a*b - 5*b^2 + 30*a*b*c*x + 18*b^2*c*x + 12*a*b*c^2*x^2 + 5*b^2*c^2*x^2 + 10*a*b*c^3*x^3 + 2*b^2*c^3*x^
3 + 15*a^2*c^4*x^4 + 6*a*b*c^4*x^4 + 12*a^2*c^5*x^5 + 3*b^2*(-9 + 5*c^4*x^4 + 4*c^5*x^5)*ArcTanh[c*x]^2 + 2*b*
ArcTanh[c*x]*(3*a*c^4*x^4*(5 + 4*c*x) + b*(-9 + 15*c*x + 6*c^2*x^2 + 5*c^3*x^3 + 3*c^4*x^4) - 12*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] + 12*a*b*Log[-1 + c^2
*x^2] + 12*b^2*PolyLog[2, -E^(-2*ArcTanh[c*x])]))/(60*c^4)

________________________________________________________________________________________

Maple [A]
time = 0.31, size = 403, normalized size = 1.49 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(d*a^2*(1/5*c^5*x^5+1/4*c^4*x^4)+2/5*d*a*b*arctanh(c*x)*c^5*x^5+9/20*d*a*b*ln(c*x-1)+1/2*d*a*b*arctanh(c
*x)*c^4*x^4-1/20*d*a*b*ln(c*x+1)-9/40*d*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)-1/40*d*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/
40*d*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)+9/20*d*b^2*arctanh(c*x)*ln(c*x-1)-1/20*d*b^2*arctanh(c*x)*ln(c*x+1)-
1/5*d*b^2*dilog(1/2*c*x+1/2)+9/80*d*b^2*ln(c*x-1)^2+1/80*d*b^2*ln(c*x+1)^2+29/60*d*b^2*ln(c*x-1)+11/60*d*b^2*l
n(c*x+1)+1/5*d*b^2*arctanh(c*x)*c^2*x^2+1/2*d*b^2*arctanh(c*x)*c*x+1/12*d*b^2*c^2*x^2+3/10*d*b^2*c*x+1/30*d*b^
2*c^3*x^3+1/10*d*a*b*c^4*x^4+1/6*d*a*b*c^3*x^3+1/5*d*a*b*c^2*x^2+1/2*d*a*b*c*x+1/5*d*b^2*arctanh(c*x)^2*c^5*x^
5+1/4*d*b^2*arctanh(c*x)^2*c^4*x^4+1/10*d*b^2*arctanh(c*x)*c^4*x^4+1/6*d*b^2*arctanh(c*x)*c^3*x^3)

________________________________________________________________________________________

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

[Out]

1/5*a^2*c*d*x^5 + 1/4*b^2*d*x^4*arctanh(c*x)^2 + 1/4*a^2*d*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*d - 1/36000*(24*c^6*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^10 - 15*log(c*
x + 1)/c^11 + 15*log(c*x - 1)/c^11) - 45*c^5*((c^2*x^4 + 2*x^2)/c^8 + 2*log(c^2*x^2 - 1)/c^10) - 1080000*c^5*i
ntegrate(1/150*x^5*log(c*x + 1)/(c^6*x^2 - c^4), x) + 50*c^4*(2*(c^2*x^3 + 3*x)/c^8 - 3*log(c*x + 1)/c^9 + 3*l
og(c*x - 1)/c^9) - 300*c^3*(x^2/c^6 + log(c^2*x^2 - 1)/c^8) + 900*c^2*(2*x/c^6 - log(c*x + 1)/c^7 + log(c*x -
1)/c^7) - 540000*c*integrate(1/150*x*log(c*x + 1)/(c^6*x^2 - c^4), x) - 60*(30*c^5*x^5*log(c*x + 1)^2 + (12*c^
5*x^5 - 15*c^4*x^4 + 20*c^3*x^3 - 30*c^2*x^2 + 60*c*x - 60*(c^5*x^5 + 1)*log(c*x + 1))*log(-c*x + 1))/c^5 - (7
2*(c*x - 1)^5*(25*log(-c*x + 1)^2 - 10*log(-c*x + 1) + 2) + 1125*(c*x - 1)^4*(8*log(-c*x + 1)^2 - 4*log(-c*x +
1) + 1) + 2000*(c*x - 1)^3*(9*log(-c*x + 1)^2 - 6*log(-c*x + 1) + 2) + 9000*(c*x - 1)^2*(2*log(-c*x + 1)^2 -
2*log(-c*x + 1) + 1) + 9000*(c*x - 1)*(log(-c*x + 1)^2 - 2*log(-c*x + 1) + 2))/c^5 + 1800*log(150*c^6*x^2 - 15
0*c^4)/c^5 - 540000*integrate(1/150*log(c*x + 1)/(c^6*x^2 - c^4), x))*b^2*c*d + 1/12*(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*d + 1/48*(4*c*(2*(c^2*x^3 + 3*x)/c^4 - 3
*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5)*arctanh(c*x) + (4*c^2*x^2 - 2*(3*log(c*x - 1) - 8)*log(c*x + 1) + 3*lo
g(c*x + 1)^2 + 3*log(c*x - 1)^2 + 16*log(c*x - 1))/c^4)*b^2*d

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x))^2*(d + c*d*x),x)

[Out]

int(x^3*(a + b*atanh(c*x))^2*(d + c*d*x), x)

________________________________________________________________________________________