3.1.14 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{(d+e x)^3} \, dx\) [14]

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

[Out]

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

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Rubi [A]
time = 0.36, antiderivative size = 480, normalized size of antiderivative = 1.00, 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 = {6065, 6055, 2449, 2352, 6063, 720, 31, 647, 6057, 2497} \begin {gather*} \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 \left (a+b \tanh ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}+\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^3 d \text {Li}_2\left (1-\frac {2}{c x+1}\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) (c x+1)}\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}{c x+1}\right )}{4 e (c d-e)^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} \end {gather*}

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 31

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

Rule 647

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 720

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 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 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 6057

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 6063

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 6065

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]

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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 6.00, size = 470, normalized size = 0.98 \begin {gather*} -\frac {a^2}{2 e (d+e x)^2}-\frac {a b c^2 \left (\frac {2 \tanh ^{-1}(c x)}{(c d+c e x)^2}+\frac {\log (1-c x)}{(c d+e)^2}+\frac {-\log (1+c x)+\frac {2 e \left (-c^2 d^2+e^2+2 c^2 d (d+e x) \log (c (d+e x))\right )}{c (c d+e)^2 (d+e x)}}{(-c d+e)^2}\right )}{2 e}+\frac {b^2 c^2 \left (-\frac {2 e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )} \tanh ^{-1}(c x)^2}{\sqrt {1-\frac {c^2 d^2}{e^2}} e}-\frac {e \left (-1+c^2 x^2\right ) \tanh ^{-1}(c x)^2}{c^2 (d+e x)^2}+\frac {2 x \tanh ^{-1}(c x) \left (-e+c d \tanh ^{-1}(c x)\right )}{c d (d+e x)}+\frac {2 e \left (-e \tanh ^{-1}(c x)+c d \log \left (\frac {c (d+e x)}{\sqrt {1-c^2 x^2}}\right )\right )}{c^3 d^3-c d e^2}+\frac {2 c d \left (i \pi \log \left (1+e^{2 \tanh ^{-1}(c x)}\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 )-i \pi \left (\tanh ^{-1}(c x)-\frac {1}{2} \log \left (1-c^2 x^2\right )\right )-2 \tanh ^{-1}\left (\frac {c d}{e}\right ) \left (\tanh ^{-1}(c x)+\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 )\right )+\text {PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )\right )}{c^2 d^2-e^2}\right )}{2 (c d-e) (c d+e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a^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)*Arc
Tanh[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*ArcTan
h[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[ArcT
anh[(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 = 1.36, size = 828, normalized size = 1.72 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, 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/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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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/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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