3.14 \(\int \frac{(a+b \tanh ^{-1}(c x))^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=480 \[ -\frac{b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{(c d-e)^2 (c d+e)^2}+\frac{b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{(c d-e)^2 (c d+e)^2}+\frac{b^2 c^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{4 e (c d+e)^2}+\frac{b^2 c^2 \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{4 e (c d-e)^2}+\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}+\frac{2 b c^3 d \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{(c d-e)^2 (c d+e)^2}-\frac{2 b c^3 d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{(c d-e)^2 (c d+e)^2}+\frac{b c^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 e (c d+e)^2}-\frac{b c^2 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 e (c d-e)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac{b^2 c^2 \log (1-c x)}{2 (c d-e) (c d+e)^2}-\frac{b^2 c^2 \log (c x+1)}{2 (c d-e)^2 (c d+e)}+\frac{b^2 c^2 e \log (d+e x)}{(c d-e)^2 (c d+e)^2} \]

[Out]

(b*c*(a + b*ArcTanh[c*x]))/((c^2*d^2 - e^2)*(d + e*x)) - (a + b*ArcTanh[c*x])^2/(2*e*(d + e*x)^2) + (b*c^2*(a
+ b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(2*e*(c*d + e)^2) + (b^2*c^2*Log[1 - c*x])/(2*(c*d - e)*(c*d + e)^2) - (b*
c^2*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(2*(c*d - e)^2*e) + (2*b*c^3*d*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)
])/((c*d - e)^2*(c*d + e)^2) - (b^2*c^2*Log[1 + c*x])/(2*(c*d - e)^2*(c*d + e)) + (b^2*c^2*e*Log[d + e*x])/((c
*d - e)^2*(c*d + e)^2) - (2*b*c^3*d*(a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/((c*d - e
)^2*(c*d + e)^2) + (b^2*c^2*PolyLog[2, 1 - 2/(1 - c*x)])/(4*e*(c*d + e)^2) + (b^2*c^2*PolyLog[2, 1 - 2/(1 + c*
x)])/(4*(c*d - e)^2*e) - (b^2*c^3*d*PolyLog[2, 1 - 2/(1 + c*x)])/((c*d - e)^2*(c*d + e)^2) + (b^2*c^3*d*PolyLo
g[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/((c*d - e)^2*(c*d + e)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.498787, antiderivative size = 480, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {5928, 5918, 2402, 2315, 5926, 706, 31, 633, 5920, 2447} \[ -\frac{b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{(c d-e)^2 (c d+e)^2}+\frac{b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{(c d-e)^2 (c d+e)^2}+\frac{b^2 c^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{4 e (c d+e)^2}+\frac{b^2 c^2 \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{4 e (c d-e)^2}+\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}+\frac{2 b c^3 d \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{(c d-e)^2 (c d+e)^2}-\frac{2 b c^3 d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{(c d-e)^2 (c d+e)^2}+\frac{b c^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 e (c d+e)^2}-\frac{b c^2 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 e (c d-e)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac{b^2 c^2 \log (1-c x)}{2 (c d-e) (c d+e)^2}-\frac{b^2 c^2 \log (c x+1)}{2 (c d-e)^2 (c d+e)}+\frac{b^2 c^2 e \log (d+e x)}{(c d-e)^2 (c d+e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*(a + b*ArcTanh[c*x]))/((c^2*d^2 - e^2)*(d + e*x)) - (a + b*ArcTanh[c*x])^2/(2*e*(d + e*x)^2) + (b*c^2*(a
+ b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(2*e*(c*d + e)^2) + (b^2*c^2*Log[1 - c*x])/(2*(c*d - e)*(c*d + e)^2) - (b*
c^2*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(2*(c*d - e)^2*e) + (2*b*c^3*d*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)
])/((c*d - e)^2*(c*d + e)^2) - (b^2*c^2*Log[1 + c*x])/(2*(c*d - e)^2*(c*d + e)) + (b^2*c^2*e*Log[d + e*x])/((c
*d - e)^2*(c*d + e)^2) - (2*b*c^3*d*(a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/((c*d - e
)^2*(c*d + e)^2) + (b^2*c^2*PolyLog[2, 1 - 2/(1 - c*x)])/(4*e*(c*d + e)^2) + (b^2*c^2*PolyLog[2, 1 - 2/(1 + c*
x)])/(4*(c*d - e)^2*e) - (b^2*c^3*d*PolyLog[2, 1 - 2/(1 + c*x)])/((c*d - e)^2*(c*d + e)^2) + (b^2*c^3*d*PolyLo
g[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/((c*d - e)^2*(c*d + e)^2)

Rule 5928

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcTanh[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^(p
- 1), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] &
& NeQ[q, -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 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 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 5926

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

Rule 706

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

Rule 31

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

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 5920

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

Rule 2447

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]]

Rubi steps

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

Mathematica [C]  time = 7.50752, size = 470, normalized size = 0.98 \[ \frac{b^2 c^2 \left (\frac{2 c d \left (\text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-i \pi \left (\tanh ^{-1}(c x)-\frac{1}{2} \log \left (1-c^2 x^2\right )\right )-2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 \tanh ^{-1}\left (\frac{c d}{e}\right ) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )+\tanh ^{-1}(c x)\right )+i \pi \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )\right )}{c^2 d^2-e^2}+\frac{2 e \left (c d \log \left (\frac{c (d+e x)}{\sqrt{1-c^2 x^2}}\right )-e \tanh ^{-1}(c x)\right )}{c^3 d^3-c d e^2}-\frac{2 \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac{c d}{e}\right )}}{e \sqrt{1-\frac{c^2 d^2}{e^2}}}-\frac{e \left (c^2 x^2-1\right ) \tanh ^{-1}(c x)^2}{c^2 (d+e x)^2}+\frac{2 x \tanh ^{-1}(c x) \left (c d \tanh ^{-1}(c x)-e\right )}{c d (d+e x)}\right )}{2 (c d-e) (c d+e)}-\frac{a^2}{2 e (d+e x)^2}-\frac{a b c^2 \left (\frac{\frac{2 e \left (c^2 \left (-d^2\right )+2 c^2 d (d+e x) \log (c (d+e x))+e^2\right )}{c (c d+e)^2 (d+e x)}-\log (c x+1)}{(e-c d)^2}+\frac{\log (1-c x)}{(c d+e)^2}+\frac{2 \tanh ^{-1}(c x)}{(c d+c e x)^2}\right )}{2 e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-a^2/(2*e*(d + e*x)^2) - (a*b*c^2*((2*ArcTanh[c*x])/(c*d + c*e*x)^2 + Log[1 - c*x]/(c*d + e)^2 + (-Log[1 + c*x
] + (2*e*(-(c^2*d^2) + e^2 + 2*c^2*d*(d + e*x)*Log[c*(d + e*x)]))/(c*(c*d + e)^2*(d + e*x)))/(-(c*d) + e)^2))/
(2*e) + (b^2*c^2*((-2*ArcTanh[c*x]^2)/(Sqrt[1 - (c^2*d^2)/e^2]*e*E^ArcTanh[(c*d)/e]) - (e*(-1 + c^2*x^2)*ArcTa
nh[c*x]^2)/(c^2*(d + e*x)^2) + (2*x*ArcTanh[c*x]*(-e + c*d*ArcTanh[c*x]))/(c*d*(d + e*x)) + (2*e*(-(e*ArcTanh[
c*x]) + c*d*Log[(c*(d + e*x))/Sqrt[1 - c^2*x^2]]))/(c^3*d^3 - c*d*e^2) + (2*c*d*(I*Pi*Log[1 + E^(2*ArcTanh[c*x
])] - 2*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - I*Pi*(ArcTanh[c*x] - Log[1 - c^2*x^2]
/2) - 2*ArcTanh[(c*d)/e]*(ArcTanh[c*x] + Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - Log[I*Sinh[ArcTan
h[(c*d)/e] + ArcTanh[c*x]]]) + PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(c^2*d^2 - e^2)))/(2*(c*
d - e)*(c*d + e))

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Maple [A]  time = 0.069, size = 824, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*a*b/(c*d+e)/(c*d-e)/(c*e*x+c*d)-c^3*b^2*d/(c*d+e)^2/(c*d-e)^2*dilog((c*e*x-e)/(-c*d-e))+1/4*c^2*b^2/e/(c*d
-e)^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+c^2*b^2*arctanh(c*x)/(c*d+e)/(c*d-e)/(c*e*x+c*d)-1/2*c^2*b^2/e*arctanh(c*x)/(
c*d+e)^2*ln(c*x-1)+1/2*c^2*b^2/e*arctanh(c*x)/(c*d-e)^2*ln(c*x+1)+1/4*c^2*b^2/e/(c*d+e)^2*ln(c*x-1)*ln(1/2+1/2
*c*x)-1/4*c^2*b^2/e/(c*d-e)^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-1/2*c^2*a*b/e/(c*d+e)^2*ln(c*x-1)+c^2*b^2*e/(c*
d+e)^2/(c*d-e)^2*ln(c*e*x+c*d)-1/2*c^2*a^2/(c*e*x+c*d)^2/e-1/8*c^2*b^2/e/(c*d+e)^2*ln(c*x-1)^2-1/4*c^2*b^2/e/(
c*d-e)^2*dilog(1/2+1/2*c*x)-1/8*c^2*b^2/e/(c*d-e)^2*ln(c*x+1)^2-1/2*c^2*b^2/(c*e*x+c*d)^2/e*arctanh(c*x)^2+1/4
*c^2*b^2/e/(c*d+e)^2*dilog(1/2+1/2*c*x)-2*c^3*b^2*arctanh(c*x)*d/(c*d+e)^2/(c*d-e)^2*ln(c*e*x+c*d)+c^3*b^2*d/(
c*d+e)^2/(c*d-e)^2*ln((c*e*x+e)/(-c*d+e))*ln(c*e*x+c*d)-c^3*b^2*d/(c*d+e)^2/(c*d-e)^2*ln((c*e*x-e)/(-c*d-e))*l
n(c*e*x+c*d)-2*c^3*a*b*d/(c*d+e)^2/(c*d-e)^2*ln(c*e*x+c*d)+1/2*c^2*a*b/e/(c*d-e)^2*ln(c*x+1)-c^2*a*b/(c*e*x+c*
d)^2/e*arctanh(c*x)+c^2*b^2/(c*d+e)/(c*d-e)/(2*c*d+2*e)*ln(c*x-1)-c^2*b^2/(c*d+e)/(c*d-e)/(2*c*d-2*e)*ln(c*x+1
)+c^3*b^2*d/(c*d+e)^2/(c*d-e)^2*dilog((c*e*x+e)/(-c*d+e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \,{\left ({\left (\frac{4 \, c^{2} d \log \left (e x + d\right )}{c^{4} d^{4} - 2 \, c^{2} d^{2} e^{2} + e^{4}} - \frac{c \log \left (c x + 1\right )}{c^{2} d^{2} e - 2 \, c d e^{2} + e^{3}} + \frac{c \log \left (c x - 1\right )}{c^{2} d^{2} e + 2 \, c d e^{2} + e^{3}} - \frac{2}{c^{2} d^{3} - d e^{2} +{\left (c^{2} d^{2} e - e^{3}\right )} x}\right )} c + \frac{2 \, \operatorname{artanh}\left (c x\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right )} a b - \frac{1}{8} \, b^{2}{\left (\frac{\log \left (-c x + 1\right )^{2}}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e} + 2 \, \int -\frac{{\left (c e x - e\right )} \log \left (c x + 1\right )^{2} +{\left (c e x + c d - 2 \,{\left (c e x - e\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{c e^{4} x^{4} - d^{3} e +{\left (3 \, c d e^{3} - e^{4}\right )} x^{3} + 3 \,{\left (c d^{2} e^{2} - d e^{3}\right )} x^{2} +{\left (c d^{3} e - 3 \, d^{2} e^{2}\right )} x}\,{d x}\right )} - \frac{a^{2}}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((4*c^2*d*log(e*x + d)/(c^4*d^4 - 2*c^2*d^2*e^2 + e^4) - c*log(c*x + 1)/(c^2*d^2*e - 2*c*d*e^2 + e^3) + c
*log(c*x - 1)/(c^2*d^2*e + 2*c*d*e^2 + e^3) - 2/(c^2*d^3 - d*e^2 + (c^2*d^2*e - e^3)*x))*c + 2*arctanh(c*x)/(e
^3*x^2 + 2*d*e^2*x + d^2*e))*a*b - 1/8*b^2*(log(-c*x + 1)^2/(e^3*x^2 + 2*d*e^2*x + d^2*e) + 2*integrate(-((c*e
*x - e)*log(c*x + 1)^2 + (c*e*x + c*d - 2*(c*e*x - e)*log(c*x + 1))*log(-c*x + 1))/(c*e^4*x^4 - d^3*e + (3*c*d
*e^3 - e^4)*x^3 + 3*(c*d^2*e^2 - d*e^3)*x^2 + (c*d^3*e - 3*d^2*e^2)*x), x)) - 1/2*a^2/(e^3*x^2 + 2*d*e^2*x + d
^2*e)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{artanh}\left (c x\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x\right ) + a^{2}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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