3.1.95 \(\int \frac {x^3 (a+b \tanh ^{-1}(c x))^2}{d+c d x} \, dx\) [95]

Optimal. Leaf size=329 \[ -\frac {a b x}{c^3 d}+\frac {b^2 x}{3 c^3 d}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^4 d}-\frac {b^2 x \tanh ^{-1}(c x)}{c^3 d}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac {11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac {8 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^4 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d}-\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d}-\frac {4 b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{3 c^4 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^4 d}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c^4 d} \]

[Out]

-a*b*x/c^3/d+1/3*b^2*x/c^3/d-1/3*b^2*arctanh(c*x)/c^4/d-b^2*x*arctanh(c*x)/c^3/d+1/3*b*x^2*(a+b*arctanh(c*x))/
c^2/d+11/6*(a+b*arctanh(c*x))^2/c^4/d+x*(a+b*arctanh(c*x))^2/c^3/d-1/2*x^2*(a+b*arctanh(c*x))^2/c^2/d+1/3*x^3*
(a+b*arctanh(c*x))^2/c/d-8/3*b*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^4/d+(a+b*arctanh(c*x))^2*ln(2/(c*x+1))/c^4/
d-1/2*b^2*ln(-c^2*x^2+1)/c^4/d-4/3*b^2*polylog(2,1-2/(-c*x+1))/c^4/d-b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1
))/c^4/d-1/2*b^2*polylog(3,1-2/(c*x+1))/c^4/d

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Rubi [A]
time = 0.59, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 14, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {6077, 6037, 6127, 327, 212, 6131, 6055, 2449, 2352, 6021, 266, 6095, 6203, 6745} \begin {gather*} -\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d}+\frac {11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}-\frac {8 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c^4 d}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d}-\frac {a b x}{c^3 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac {4 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c^4 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^4 d}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^4 d}+\frac {b^2 x}{3 c^3 d}-\frac {b^2 x \tanh ^{-1}(c x)}{c^3 d}-\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*x)/(c^3*d)) + (b^2*x)/(3*c^3*d) - (b^2*ArcTanh[c*x])/(3*c^4*d) - (b^2*x*ArcTanh[c*x])/(c^3*d) + (b*x^2*
(a + b*ArcTanh[c*x]))/(3*c^2*d) + (11*(a + b*ArcTanh[c*x])^2)/(6*c^4*d) + (x*(a + b*ArcTanh[c*x])^2)/(c^3*d) -
(x^2*(a + b*ArcTanh[c*x])^2)/(2*c^2*d) + (x^3*(a + b*ArcTanh[c*x])^2)/(3*c*d) - (8*b*(a + b*ArcTanh[c*x])*Log
[2/(1 - c*x)])/(3*c^4*d) + ((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^4*d) - (b^2*Log[1 - c^2*x^2])/(2*c^4*d
) - (4*b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(3*c^4*d) - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^4*
d) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^4*d)

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 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 6077

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

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 6127

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

Rule 6131

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x] - Dist[b*(p/2), Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2
/(1 + c*x))^2, 0]

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx &=-\frac {\int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx}{c}+\frac {\int x^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}+\frac {\int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx}{c^2}-\frac {(2 b) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 d}-\frac {\int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^2 d}\\ &=-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx}{c^3}+\frac {\int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^3 d}+\frac {(2 b) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c^2 d}-\frac {(2 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c^2 d}+\frac {b \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c d}\\ &=\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^4 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d}-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{3 c^3 d}-\frac {b \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^3 d}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{c^3 d}-\frac {(2 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d}-\frac {(2 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c^2 d}-\frac {b^2 \int \frac {x^2}{1-c^2 x^2} \, dx}{3 c d}\\ &=-\frac {a b x}{c^3 d}+\frac {b^2 x}{3 c^3 d}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac {11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^4 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d}-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^3 d}-\frac {b^2 \int \frac {1}{1-c^2 x^2} \, dx}{3 c^3 d}+\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{3 c^3 d}-\frac {b^2 \int \tanh ^{-1}(c x) \, dx}{c^3 d}+\frac {b^2 \int \frac {\text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d}\\ &=-\frac {a b x}{c^3 d}+\frac {b^2 x}{3 c^3 d}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^4 d}-\frac {b^2 x \tanh ^{-1}(c x)}{c^3 d}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac {11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac {8 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^4 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{3 c^4 d}+\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^3 d}+\frac {b^2 \int \frac {x}{1-c^2 x^2} \, dx}{c^2 d}\\ &=-\frac {a b x}{c^3 d}+\frac {b^2 x}{3 c^3 d}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^4 d}-\frac {b^2 x \tanh ^{-1}(c x)}{c^3 d}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac {11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac {8 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^4 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d}-\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c^4 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c^4 d}\\ &=-\frac {a b x}{c^3 d}+\frac {b^2 x}{3 c^3 d}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^4 d}-\frac {b^2 x \tanh ^{-1}(c x)}{c^3 d}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c^2 d}+\frac {11 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^4 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c d}-\frac {8 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^4 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d}-\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d}-\frac {4 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c^4 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 347, normalized size = 1.05 \begin {gather*} \frac {a^2 x}{c^3 d}-\frac {a^2 x^2}{2 c^2 d}+\frac {a^2 x^3}{3 c d}-\frac {a^2 \log (1+c x)}{c^4 d}+\frac {a b \left (-3 c x+8 c x \tanh ^{-1}(c x)+\left (1-c^2 x^2\right ) \left (-1+3 \tanh ^{-1}(c x)-2 c x \tanh ^{-1}(c x)\right )+6 \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-8 \log \left (\frac {1}{\sqrt {1-c^2 x^2}}\right )-3 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{3 c^4 d}+\frac {b^2 \left (2 c x-6 c x \tanh ^{-1}(c x)-2 \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)-8 \tanh ^{-1}(c x)^2+8 c x \tanh ^{-1}(c x)^2+3 \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)^2-2 c x \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)^2-16 \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+6 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+6 \log \left (\frac {1}{\sqrt {1-c^2 x^2}}\right )+\left (8-6 \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-3 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{6 c^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/(c^3*d) - (a^2*x^2)/(2*c^2*d) + (a^2*x^3)/(3*c*d) - (a^2*Log[1 + c*x])/(c^4*d) + (a*b*(-3*c*x + 8*c*x*
ArcTanh[c*x] + (1 - c^2*x^2)*(-1 + 3*ArcTanh[c*x] - 2*c*x*ArcTanh[c*x]) + 6*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh
[c*x])] - 8*Log[1/Sqrt[1 - c^2*x^2]] - 3*PolyLog[2, -E^(-2*ArcTanh[c*x])]))/(3*c^4*d) + (b^2*(2*c*x - 6*c*x*Ar
cTanh[c*x] - 2*(1 - c^2*x^2)*ArcTanh[c*x] - 8*ArcTanh[c*x]^2 + 8*c*x*ArcTanh[c*x]^2 + 3*(1 - c^2*x^2)*ArcTanh[
c*x]^2 - 2*c*x*(1 - c^2*x^2)*ArcTanh[c*x]^2 - 16*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + 6*ArcTanh[c*x]^2*
Log[1 + E^(-2*ArcTanh[c*x])] + 6*Log[1/Sqrt[1 - c^2*x^2]] + (8 - 6*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x
])] - 3*PolyLog[3, -E^(-2*ArcTanh[c*x])]))/(6*c^4*d)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 8.39, size = 1200, normalized size = 3.65

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1200\)
default	\(\text {Expression too large to display}\)	\(1200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(-a^2/d*ln(c*x+1)+2/3*a*b/d*c^3*x^3*arctanh(c*x)+b^2/d*arctanh(c*x)^2*c*x+a*b/d*dilog(1/2*c*x+1/2)+1/2*a
*b/d*ln(c*x+1)^2+5/6*a*b/d*ln(c*x-1)+11/6*a*b/d*ln(c*x+1)-8/3*b^2/d*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(
1/2))-8/3*b^2/d*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-b^2/d*arctanh(c*x)^2*ln(c*x+1)+b^2/d*arctanh(c
*x)^2*ln(2)+b^2/d*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+2*b^2/d*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+
1)^(1/2))-a*b/d*c*x+1/3*a*b/d*c^2*x^2+1/3*b^2/d*c^3*x^3*arctanh(c*x)^2-b^2/d*arctanh(c*x)*c*x+1/3*b^2/d*arctan
h(c*x)*c^2*x^2-1/2*b^2/d*arctanh(c*x)^2*c^2*x^2-2*a*b/d*arctanh(c*x)*ln(c*x+1)+a*b/d*ln(-1/2*c*x+1/2)*ln(1/2*c
*x+1/2)-a*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/3*b^2/d+b^2/d*ln(1+(c*x+1)^2/(-c^2*x^2+1))-8/3*b^2/d*dilog(1+I*(c*x
+1)/(-c^2*x^2+1)^(1/2))-1/2*b^2/d*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-8/3*b^2/d*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^
(1/2))-4/3*b^2/d*arctanh(c*x)+11/6*b^2/d*arctanh(c*x)^2-2/3*b^2/d*arctanh(c*x)^3-1/2*I*b^2/d*arctanh(c*x)^2*Pi
*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*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)))+a^2/d*c*x-1/2*a^2/d*c^2*x^2+1/3*a^2/d*c^3*x^3+1/3*b^2/d*c*x-1/2*I*b^2/d*arctanh(c*x)^2*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+I*b^2/d*arctanh(c*x)^2*Pi*cs
gn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2+1/2*I*b^2/d*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)/
(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))+1/2*I*b^2/d*arctanh(c*x)^2*Pi*csgn(I/(1+(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)))^2-4/3*a*b/d-a*b/d*arctanh(c*x)*c^2*x^2+2*a*b/d
*arctanh(c*x)*c*x+1/2*I*b^2/d*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3+1/2
*I*b^2/d*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2/(c*d*x+d),x, algorithm="maxima")

[Out]

1/6*a^2*((2*c^2*x^3 - 3*c*x^2 + 6*x)/(c^3*d) - 6*log(c*x + 1)/(c^4*d)) + 1/24*(2*b^2*c^3*x^3 - 3*b^2*c^2*x^2 +
6*b^2*c*x - 6*b^2*log(c*x + 1))*log(-c*x + 1)^2/(c^4*d) - integrate(-1/12*(3*(b^2*c^4*x^4 - b^2*c^3*x^3)*log(
c*x + 1)^2 + 12*(a*b*c^4*x^4 - a*b*c^3*x^3)*log(c*x + 1) - (3*b^2*c^2*x^2 + 2*(6*a*b*c^4 + b^2*c^4)*x^4 + 6*b^
2*c*x - (12*a*b*c^3 + b^2*c^3)*x^3 + 6*(b^2*c^4*x^4 - b^2*c^3*x^3 - b^2*c*x - b^2)*log(c*x + 1))*log(-c*x + 1)
)/(c^5*d*x^2 - c^3*d), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2} x^{3}}{c x + 1}\, dx + \int \frac {b^{2} x^{3} \operatorname {atanh}^{2}{\left (c x \right )}}{c x + 1}\, dx + \int \frac {2 a b x^{3} \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x**3/(c*x + 1), x) + Integral(b**2*x**3*atanh(c*x)**2/(c*x + 1), x) + Integral(2*a*b*x**3*atanh
(c*x)/(c*x + 1), x))/d

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2/(c*d*x+d),x, algorithm="giac")

[Out]

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

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

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