3.2.2 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x^4 (d+c d x)} \, dx\) [102]

Optimal. Leaf size=334 \[ -\frac {b^2 c^2}{3 d x}+\frac {b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac {5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac {b^2 c^3 \log (x)}{d}+\frac {b^2 c^3 \log \left (1-c^2 x^2\right )}{2 d}+\frac {8 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{3 d}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {4 b^2 c^3 \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{3 d}+\frac {b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{d}+\frac {b^2 c^3 \text {PolyLog}\left (3,-1+\frac {2}{1+c x}\right )}{2 d} \]

[Out]

-1/3*b^2*c^2/d/x+1/3*b^2*c^3*arctanh(c*x)/d-1/3*b*c*(a+b*arctanh(c*x))/d/x^2+b*c^2*(a+b*arctanh(c*x))/d/x+5/6*
c^3*(a+b*arctanh(c*x))^2/d-1/3*(a+b*arctanh(c*x))^2/d/x^3+1/2*c*(a+b*arctanh(c*x))^2/d/x^2-c^2*(a+b*arctanh(c*
x))^2/d/x-b^2*c^3*ln(x)/d+1/2*b^2*c^3*ln(-c^2*x^2+1)/d+8/3*b*c^3*(a+b*arctanh(c*x))*ln(2-2/(c*x+1))/d-c^3*(a+b
*arctanh(c*x))^2*ln(2-2/(c*x+1))/d-4/3*b^2*c^3*polylog(2,-1+2/(c*x+1))/d+b*c^3*(a+b*arctanh(c*x))*polylog(2,-1
+2/(c*x+1))/d+1/2*b^2*c^3*polylog(3,-1+2/(c*x+1))/d

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Rubi [A]
time = 0.69, antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 15, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.682, Rules used = {6081, 6037, 6129, 331, 212, 6135, 6079, 2497, 272, 36, 29, 31, 6095, 6203, 6745} \begin {gather*} \frac {b c^3 \text {Li}_2\left (\frac {2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac {5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac {c^3 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}+\frac {8 b c^3 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 d}-\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}-\frac {4 b^2 c^3 \text {Li}_2\left (\frac {2}{c x+1}-1\right )}{3 d}+\frac {b^2 c^3 \text {Li}_3\left (\frac {2}{c x+1}-1\right )}{2 d}-\frac {b^2 c^3 \log (x)}{d}+\frac {b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac {b^2 c^2}{3 d x}+\frac {b^2 c^3 \log \left (1-c^2 x^2\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/(x^4*(d + c*d*x)),x]

[Out]

-1/3*(b^2*c^2)/(d*x) + (b^2*c^3*ArcTanh[c*x])/(3*d) - (b*c*(a + b*ArcTanh[c*x]))/(3*d*x^2) + (b*c^2*(a + b*Arc
Tanh[c*x]))/(d*x) + (5*c^3*(a + b*ArcTanh[c*x])^2)/(6*d) - (a + b*ArcTanh[c*x])^2/(3*d*x^3) + (c*(a + b*ArcTan
h[c*x])^2)/(2*d*x^2) - (c^2*(a + b*ArcTanh[c*x])^2)/(d*x) - (b^2*c^3*Log[x])/d + (b^2*c^3*Log[1 - c^2*x^2])/(2
*d) + (8*b*c^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/(3*d) - (c^3*(a + b*ArcTanh[c*x])^2*Log[2 - 2/(1 + c
*x)])/d - (4*b^2*c^3*PolyLog[2, -1 + 2/(1 + c*x)])/(3*d) + (b*c^3*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 +
c*x)])/d + (b^2*c^3*PolyLog[3, -1 + 2/(1 + c*x)])/(2*d)

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 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 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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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 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 6081

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

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 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 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 {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^4 (d+c d x)} \, dx &=-\left (c \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3 (d+c d x)} \, dx\right )+\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^4} \, dx}{d}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+c^2 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2 (d+c d x)} \, dx-\frac {c \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx}{d}+\frac {(2 b c) \int \frac {a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-c^3 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x (d+c d x)} \, dx+\frac {(2 b c) \int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx}{3 d}+\frac {c^2 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx}{d}-\frac {\left (b c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx}{d}+\frac {\left (2 b c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {\left (b c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d}+\frac {\left (b^2 c^2\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx}{3 d}+\frac {\left (2 b c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{3 d}+\frac {\left (2 b c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{d}-\frac {\left (b c^4\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{d}+\frac {\left (2 b c^4\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}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac {5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac {2 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{3 d}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}+\frac {b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d}+\frac {\left (2 b c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{d}-\frac {\left (b^2 c^3\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac {\left (b^2 c^4\right ) \int \frac {1}{1-c^2 x^2} \, dx}{3 d}-\frac {\left (2 b^2 c^4\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{3 d}-\frac {\left (b^2 c^4\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac {b^2 c^2}{3 d x}+\frac {b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac {5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac {8 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{3 d}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {b^2 c^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{3 d}+\frac {b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d}+\frac {b^2 c^3 \text {Li}_3\left (-1+\frac {2}{1+c x}\right )}{2 d}-\frac {\left (b^2 c^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d}-\frac {\left (2 b^2 c^4\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=-\frac {b^2 c^2}{3 d x}+\frac {b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac {5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac {8 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{3 d}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {4 b^2 c^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{3 d}+\frac {b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d}+\frac {b^2 c^3 \text {Li}_3\left (-1+\frac {2}{1+c x}\right )}{2 d}-\frac {\left (b^2 c^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}-\frac {\left (b^2 c^5\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b^2 c^2}{3 d x}+\frac {b^2 c^3 \tanh ^{-1}(c x)}{3 d}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 d x^2}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d x}+\frac {5 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d x^2}-\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac {b^2 c^3 \log (x)}{d}+\frac {b^2 c^3 \log \left (1-c^2 x^2\right )}{2 d}+\frac {8 b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{3 d}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {4 b^2 c^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{3 d}+\frac {b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d}+\frac {b^2 c^3 \text {Li}_3\left (-1+\frac {2}{1+c x}\right )}{2 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.86, size = 388, normalized size = 1.16 \begin {gather*} \frac {-\frac {8 a^2}{x^3}+\frac {12 a^2 c}{x^2}-\frac {24 a^2 c^2}{x}-24 a^2 c^3 \log (x)+24 a^2 c^3 \log (1+c x)-\frac {8 a b \left (\tanh ^{-1}(c x) \left (2-3 c x+6 c^2 x^2+3 c^3 x^3+6 c^3 x^3 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )-c x \left (-1+3 c x+c^2 x^2+8 c^2 x^2 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )\right )-3 c^3 x^3 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )\right )}{x^3}+b^2 c^3 \left (-i \pi ^3-\frac {8}{c x}+8 \tanh ^{-1}(c x)-\frac {8 \tanh ^{-1}(c x)}{c^2 x^2}+\frac {24 \tanh ^{-1}(c x)}{c x}+20 \tanh ^{-1}(c x)^2-\frac {8 \tanh ^{-1}(c x)^2}{c^3 x^3}+\frac {12 \tanh ^{-1}(c x)^2}{c^2 x^2}-\frac {24 \tanh ^{-1}(c x)^2}{c x}+16 \tanh ^{-1}(c x)^3+64 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-24 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )-24 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )-32 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-24 \tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+12 \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/(x^4*(d + c*d*x)),x]

[Out]

((-8*a^2)/x^3 + (12*a^2*c)/x^2 - (24*a^2*c^2)/x - 24*a^2*c^3*Log[x] + 24*a^2*c^3*Log[1 + c*x] - (8*a*b*(ArcTan
h[c*x]*(2 - 3*c*x + 6*c^2*x^2 + 3*c^3*x^3 + 6*c^3*x^3*Log[1 - E^(-2*ArcTanh[c*x])]) - c*x*(-1 + 3*c*x + c^2*x^
2 + 8*c^2*x^2*Log[(c*x)/Sqrt[1 - c^2*x^2]]) - 3*c^3*x^3*PolyLog[2, E^(-2*ArcTanh[c*x])]))/x^3 + b^2*c^3*((-I)*
Pi^3 - 8/(c*x) + 8*ArcTanh[c*x] - (8*ArcTanh[c*x])/(c^2*x^2) + (24*ArcTanh[c*x])/(c*x) + 20*ArcTanh[c*x]^2 - (
8*ArcTanh[c*x]^2)/(c^3*x^3) + (12*ArcTanh[c*x]^2)/(c^2*x^2) - (24*ArcTanh[c*x]^2)/(c*x) + 16*ArcTanh[c*x]^3 +
64*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - 24*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - 24*Log[(c*x)/Sq
rt[1 - c^2*x^2]] - 32*PolyLog[2, E^(-2*ArcTanh[c*x])] - 24*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] + 12*Po
lyLog[3, E^(2*ArcTanh[c*x])]))/(24*d)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 14.58, size = 1895, normalized size = 5.67

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1895\)
default	\(\text {Expression too large to display}\)	\(1895\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^2/x^4/(c*d*x+d),x,method=_RETURNVERBOSE)

[Out]

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

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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((a+b*arctanh(c*x))^2/x^4/(c*d*x+d),x, algorithm="maxima")

[Out]

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

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/(c*d*x^5 + d*x^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2}}{c x^{5} + x^{4}}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c x^{5} + x^{4}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{c x^{5} + x^{4}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/x**4/(c*d*x+d),x)

[Out]

(Integral(a**2/(c*x**5 + x**4), x) + Integral(b**2*atanh(c*x)**2/(c*x**5 + x**4), x) + Integral(2*a*b*atanh(c*
x)/(c*x**5 + x**4), 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((a+b*arctanh(c*x))^2/x^4/(c*d*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2/((c*d*x + d)*x^4), 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}{x^4\,\left (d+c\,d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^2/(x^4*(d + c*d*x)),x)

[Out]

int((a + b*atanh(c*x))^2/(x^4*(d + c*d*x)), x)

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