3.1.90 \(\int \frac {(d+c d x)^3 (a+b \tanh ^{-1}(c x))^2}{x^3} \, dx\) [90]

Optimal. Leaf size=385 \[ -\frac {b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {9}{2} c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+b^2 c^2 d^3 \log (x)-2 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c^2 d^3 \log \left (1-c^2 x^2\right )+6 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b^2 c^2 d^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )-3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )+3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1-c x}\right )-3 b^2 c^2 d^3 \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )+\frac {3}{2} b^2 c^2 d^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {PolyLog}\left (3,-1+\frac {2}{1-c x}\right ) \]

[Out]

-b*c*d^3*(a+b*arctanh(c*x))/x+9/2*c^2*d^3*(a+b*arctanh(c*x))^2-1/2*d^3*(a+b*arctanh(c*x))^2/x^2-3*c*d^3*(a+b*a
rctanh(c*x))^2/x+c^3*d^3*x*(a+b*arctanh(c*x))^2-6*c^2*d^3*(a+b*arctanh(c*x))^2*arctanh(-1+2/(-c*x+1))+b^2*c^2*
d^3*ln(x)-2*b*c^2*d^3*(a+b*arctanh(c*x))*ln(2/(-c*x+1))-1/2*b^2*c^2*d^3*ln(-c^2*x^2+1)+6*b*c^2*d^3*(a+b*arctan
h(c*x))*ln(2-2/(c*x+1))-b^2*c^2*d^3*polylog(2,1-2/(-c*x+1))-3*b*c^2*d^3*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x
+1))+3*b*c^2*d^3*(a+b*arctanh(c*x))*polylog(2,-1+2/(-c*x+1))-3*b^2*c^2*d^3*polylog(2,-1+2/(c*x+1))+3/2*b^2*c^2
*d^3*polylog(3,1-2/(-c*x+1))-3/2*b^2*c^2*d^3*polylog(3,-1+2/(-c*x+1))

________________________________________________________________________________________

Rubi [A]
time = 0.58, antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 20, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.909, Rules used = {6087, 6021, 6131, 6055, 2449, 2352, 6037, 6129, 272, 36, 29, 31, 6095, 6135, 6079, 2497, 6033, 6199, 6205, 6745} \begin {gather*} c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2-3 b c^2 d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+3 b c^2 d^3 \text {Li}_2\left (\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac {9}{2} c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-2 b c^2 d^3 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+6 b c^2 d^3 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}-\frac {b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}-b^2 c^2 d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )-3 b^2 c^2 d^3 \text {Li}_2\left (\frac {2}{c x+1}-1\right )+\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (\frac {2}{1-c x}-1\right )-\frac {1}{2} b^2 c^2 d^3 \log \left (1-c^2 x^2\right )+b^2 c^2 d^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d^3*(a + b*ArcTanh[c*x]))/x) + (9*c^2*d^3*(a + b*ArcTanh[c*x])^2)/2 - (d^3*(a + b*ArcTanh[c*x])^2)/(2*x
^2) - (3*c*d^3*(a + b*ArcTanh[c*x])^2)/x + c^3*d^3*x*(a + b*ArcTanh[c*x])^2 + 6*c^2*d^3*(a + b*ArcTanh[c*x])^2
*ArcTanh[1 - 2/(1 - c*x)] + b^2*c^2*d^3*Log[x] - 2*b*c^2*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)] - (b^2*c^2*
d^3*Log[1 - c^2*x^2])/2 + 6*b*c^2*d^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - b^2*c^2*d^3*PolyLog[2, 1 - 2
/(1 - c*x)] - 3*b*c^2*d^3*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)] + 3*b*c^2*d^3*(a + b*ArcTanh[c*x])*
PolyLog[2, -1 + 2/(1 - c*x)] - 3*b^2*c^2*d^3*PolyLog[2, -1 + 2/(1 + c*x)] + (3*b^2*c^2*d^3*PolyLog[3, 1 - 2/(1
- c*x)])/2 - (3*b^2*c^2*d^3*PolyLog[3, -1 + 2/(1 - c*x)])/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 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 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 6033

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1
- c*x)], x] - Dist[2*b*c*p, Int[(a + b*ArcTanh[c*x])^(p - 1)*(ArcTanh[1 - 2/(1 - c*x)]/(1 - c^2*x^2)), x], x]
/; FreeQ[{a, b, c}, x] && IGtQ[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 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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 6129

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), 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] && LtQ[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 6135

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

Rule 6199

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

Rule 6205

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-(a + b*ArcT
anh[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 {(d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx &=\int \left (c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^3}+\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2}+\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d^3 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx+\left (3 c d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx+\left (3 c^2 d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx+\left (c^3 d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+\left (b c d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (6 b c^2 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx-\left (12 b c^3 d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (2 b c^4 d^3\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=4 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+\left (b c d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (6 b c^2 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx+\left (b c^3 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx-\left (2 b c^3 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx+\left (6 b c^3 d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (6 b c^3 d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {9}{2} c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )-2 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )+6 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )+\left (b^2 c^2 d^3\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx+\left (2 b^2 c^3 d^3\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx+\left (3 b^2 c^3 d^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (3 b^2 c^3 d^3\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (6 b^2 c^3 d^3\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {9}{2} c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )-2 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )+6 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-3 b^2 c^2 d^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )+\frac {1}{2} \left (b^2 c^2 d^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )-\left (2 b^2 c^2 d^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )\\ &=-\frac {b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {9}{2} c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )-2 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )+6 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b^2 c^2 d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )-3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-3 b^2 c^2 d^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )+\frac {1}{2} \left (b^2 c^2 d^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 c^4 d^3\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {9}{2} c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+6 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+b^2 c^2 d^3 \log (x)-2 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c^2 d^3 \log \left (1-c^2 x^2\right )+6 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b^2 c^2 d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )-3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+3 b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-3 b^2 c^2 d^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.76, size = 461, normalized size = 1.20 \begin {gather*} \frac {1}{2} d^3 \left (-\frac {a^2}{x^2}-\frac {6 a^2 c}{x}+2 a^2 c^3 x+6 a^2 c^2 \log (x)-\frac {a b \left (2 \tanh ^{-1}(c x)+c x (2+c x \log (1-c x)-c x \log (1+c x))\right )}{x^2}+\frac {b^2 \left (-2 c x \tanh ^{-1}(c x)+\left (-1+c^2 x^2\right ) \tanh ^{-1}(c x)^2+2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )\right )}{x^2}+2 a b c^2 \left (2 c x \tanh ^{-1}(c x)+\log \left (1-c^2 x^2\right )\right )-\frac {6 a b c \left (2 \tanh ^{-1}(c x)+c x \left (-2 \log (c x)+\log \left (1-c^2 x^2\right )\right )\right )}{x}+2 b^2 c^2 \left (\tanh ^{-1}(c x) \left ((-1+c x) \tanh ^{-1}(c x)-2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )+\frac {6 b^2 c \left (\tanh ^{-1}(c x) \left ((-1+c x) \tanh ^{-1}(c x)+2 c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )-c x \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )\right )}{x}-6 a b c^2 (\text {PolyLog}(2,-c x)-\text {PolyLog}(2,c x))+6 b^2 c^2 \left (\frac {i \pi ^3}{24}-\frac {2}{3} \tanh ^{-1}(c x)^3-\tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(-(a^2/x^2) - (6*a^2*c)/x + 2*a^2*c^3*x + 6*a^2*c^2*Log[x] - (a*b*(2*ArcTanh[c*x] + c*x*(2 + c*x*Log[1 -
c*x] - c*x*Log[1 + c*x])))/x^2 + (b^2*(-2*c*x*ArcTanh[c*x] + (-1 + c^2*x^2)*ArcTanh[c*x]^2 + 2*c^2*x^2*Log[(c*
x)/Sqrt[1 - c^2*x^2]]))/x^2 + 2*a*b*c^2*(2*c*x*ArcTanh[c*x] + Log[1 - c^2*x^2]) - (6*a*b*c*(2*ArcTanh[c*x] + c
*x*(-2*Log[c*x] + Log[1 - c^2*x^2])))/x + 2*b^2*c^2*(ArcTanh[c*x]*((-1 + c*x)*ArcTanh[c*x] - 2*Log[1 + E^(-2*A
rcTanh[c*x])]) + PolyLog[2, -E^(-2*ArcTanh[c*x])]) + (6*b^2*c*(ArcTanh[c*x]*((-1 + c*x)*ArcTanh[c*x] + 2*c*x*L
og[1 - E^(-2*ArcTanh[c*x])]) - c*x*PolyLog[2, E^(-2*ArcTanh[c*x])]))/x - 6*a*b*c^2*(PolyLog[2, -(c*x)] - PolyL
og[2, c*x]) + 6*b^2*c^2*((I/24)*Pi^3 - (2*ArcTanh[c*x]^3)/3 - ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + Ar
cTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + ArcTanh[c*x]*PolyLo
g[2, E^(2*ArcTanh[c*x])] + PolyLog[3, -E^(-2*ArcTanh[c*x])]/2 - PolyLog[3, E^(2*ArcTanh[c*x])]/2)))/2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 6.84, size = 1276, normalized size = 3.31

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1276\)
default	\(\text {Expression too large to display}\)	\(1276\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*(3/2*I*d^3*b^2*arctanh(c*x)^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csg
n(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))+2*d^3*a*b*arctanh(c*x)*c*x-d^3*a*b*arctanh(c*x)/c^2
/x^2+d^3*b^2*arctanh(c*x)^2*c*x-3/2*d^3*b^2*arctanh(c*x)^2-5/2*d^3*a*b*ln(c*x-1)-3/2*d^3*a*b*ln(c*x+1)-b^2*d^3
*arctanh(c*x)-1/2*d^3*a^2/c^2/x^2-3/2*I*d^3*b^2*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)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-3/2*I*d^3*b^2*arctanh(c*x)^2*Pi*csgn(I*((c*x+1)^2/(-c^2
*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2+6*d^3*a*b*arctanh(c*x)*ln(c*x)-3*d
^3*a*b*ln(c*x)*ln(c*x+1)-3*d^3*b^2*arctanh(c*x)^2/c/x-6*d^3*b^2*dilog((c*x+1)/(-c^2*x^2+1)^(1/2))+6*d^3*b^2*di
log(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+d^3*a^2*c*x+d^3*b^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2)-1)+d^3*b^2*ln(1+(c*x+1)/(-
c^2*x^2+1)^(1/2))-6*d^3*a*b*arctanh(c*x)/c/x+3/2*I*d^3*b^2*arctanh(c*x)^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)
/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-2*d^3*b^2*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-2*d^3*b^2*dilog(1+I*(c*x+1)/(-c
^2*x^2+1)^(1/2))+3/2*d^3*b^2*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-6*d^3*b^2*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2)
)-6*d^3*b^2*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+3*d^3*a^2*ln(c*x)-1/2*d^3*b^2*arctanh(c*x)^2/c^2/x^2-d^3*b^
2*arctanh(c*x)/c/x-d^3*a*b/c/x+6*d^3*a*b*ln(c*x)+6*d^3*b^2*arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-3*d^3
*a^2/c/x-3*d^3*a*b*dilog(c*x)-3*d^3*a*b*dilog(c*x+1)+3*d^3*b^2*arctanh(c*x)^2*ln(c*x)-2*d^3*b^2*arctanh(c*x)*l
n(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-2*d^3*b^2*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3*d^3*b^2*arctanh(
c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+3*d^3*b^2*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+6*d^3*b^2*arctan
h(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))+3*d^3*b^2*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+6*d^3*
b^2*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))-3*d^3*b^2*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1
)))

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

[Out]

a^2*c^3*d^3*x + (2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b*c^2*d^3 + 3*a^2*c^2*d^3*log(x) - 3*(c*(log(c^2*x^
2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a*b*c*d^3 + 1/2*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(
c*x)/x^2)*a*b*d^3 - 3*a^2*c*d^3/x - 1/2*a^2*d^3/x^2 + 1/8*(2*b^2*c^3*d^3*x^3 - 6*b^2*c*d^3*x - b^2*d^3)*log(-c
*x + 1)^2/x^2 - integrate(-1/4*((b^2*c^4*d^3*x^4 + 2*b^2*c^3*d^3*x^3 - 2*b^2*c*d^3*x - b^2*d^3)*log(c*x + 1)^2
+ 12*(a*b*c^3*d^3*x^3 - a*b*c^2*d^3*x^2)*log(c*x + 1) - (2*b^2*c^4*d^3*x^4 + 12*a*b*c^3*d^3*x^3 - b^2*c*d^3*x
- 6*(2*a*b*c^2*d^3 + b^2*c^2*d^3)*x^2 + 2*(b^2*c^4*d^3*x^4 + 2*b^2*c^3*d^3*x^3 - 2*b^2*c*d^3*x - b^2*d^3)*log
(c*x + 1))*log(-c*x + 1))/(c*x^4 - x^3), 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((c*d*x+d)^3*(a+b*arctanh(c*x))^2/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{3} \left (\int a^{2} c^{3}\, dx + \int \frac {a^{2}}{x^{3}}\, dx + \int \frac {3 a^{2} c}{x^{2}}\, dx + \int \frac {3 a^{2} c^{2}}{x}\, dx + \int b^{2} c^{3} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x^{3}}\, dx + \int 2 a b c^{3} \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {3 b^{2} c \operatorname {atanh}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {3 b^{2} c^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x}\, dx + \int \frac {6 a b c \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {6 a b c^{2} \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*(Integral(a**2*c**3, x) + Integral(a**2/x**3, x) + Integral(3*a**2*c/x**2, x) + Integral(3*a**2*c**2/x, x
) + Integral(b**2*c**3*atanh(c*x)**2, x) + Integral(b**2*atanh(c*x)**2/x**3, x) + Integral(2*a*b*c**3*atanh(c*
x), x) + Integral(2*a*b*atanh(c*x)/x**3, x) + Integral(3*b**2*c*atanh(c*x)**2/x**2, x) + Integral(3*b**2*c**2*
atanh(c*x)**2/x, x) + Integral(6*a*b*c*atanh(c*x)/x**2, x) + Integral(6*a*b*c**2*atanh(c*x)/x, x))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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