3.2.13 \(\int \frac {x^2 (a+b \tanh ^{-1}(c x))^2}{(d+c d x)^3} \, dx\) [113]

Optimal. Leaf size=265 \[ -\frac {b^2}{16 c^3 d^3 (1+c x)^2}+\frac {13 b^2}{16 c^3 d^3 (1+c x)}-\frac {13 b^2 \tanh ^{-1}(c x)}{16 c^3 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c^3 d^3} \]

[Out]

-1/16*b^2/c^3/d^3/(c*x+1)^2+13/16*b^2/c^3/d^3/(c*x+1)-13/16*b^2*arctanh(c*x)/c^3/d^3-1/4*b*(a+b*arctanh(c*x))/
c^3/d^3/(c*x+1)^2+7/4*b*(a+b*arctanh(c*x))/c^3/d^3/(c*x+1)-7/8*(a+b*arctanh(c*x))^2/c^3/d^3-1/2*(a+b*arctanh(c
*x))^2/c^3/d^3/(c*x+1)^2+2*(a+b*arctanh(c*x))^2/c^3/d^3/(c*x+1)-(a+b*arctanh(c*x))^2*ln(2/(c*x+1))/c^3/d^3+b*(
a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/c^3/d^3+1/2*b^2*polylog(3,1-2/(c*x+1))/c^3/d^3

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Rubi [A]
time = 0.41, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {6087, 6065, 6063, 641, 46, 213, 6095, 6055, 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^3 d^3}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (c x+1)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (c x+1)^2}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^3 d^3}+\frac {13 b^2}{16 c^3 d^3 (c x+1)}-\frac {b^2}{16 c^3 d^3 (c x+1)^2}-\frac {13 b^2 \tanh ^{-1}(c x)}{16 c^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*b^2/(c^3*d^3*(1 + c*x)^2) + (13*b^2)/(16*c^3*d^3*(1 + c*x)) - (13*b^2*ArcTanh[c*x])/(16*c^3*d^3) - (b*(a
+ b*ArcTanh[c*x]))/(4*c^3*d^3*(1 + c*x)^2) + (7*b*(a + b*ArcTanh[c*x]))/(4*c^3*d^3*(1 + c*x)) - (7*(a + b*Arc
Tanh[c*x])^2)/(8*c^3*d^3) - (a + b*ArcTanh[c*x])^2/(2*c^3*d^3*(1 + c*x)^2) + (2*(a + b*ArcTanh[c*x])^2)/(c^3*d
^3*(1 + c*x)) - ((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^3*d^3) + (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2
/(1 + c*x)])/(c^3*d^3) + (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^3*d^3)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

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 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]

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 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^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^3} \, dx &=\int \left (\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^3 (1+c x)^3}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^3 (1+c x)^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx}{c^2 d^3}+\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1+c x} \, dx}{c^2 d^3}-\frac {2 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{c^2 d^3}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^3}+\frac {a+b \tanh ^{-1}(c x)}{4 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}+\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^2 d^3}-\frac {(4 b) \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 c^2 d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 c^2 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 c^2 d^3}-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^2 d^3}+\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{c^2 d^3}-\frac {b^2 \int \frac {\text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}+\frac {b^2 \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 c^2 d^3}+\frac {b^2 \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 c^2 d^3}-\frac {\left (2 b^2\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^2 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{4 c^2 d^3}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{4 c^2 d^3}-\frac {\left (2 b^2\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{c^2 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}+\frac {b^2 \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}+\frac {b^2 \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}-\frac {\left (2 b^2\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}\\ &=-\frac {b^2}{16 c^3 d^3 (1+c x)^2}+\frac {13 b^2}{16 c^3 d^3 (1+c x)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{16 c^2 d^3}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{8 c^2 d^3}+\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac {b^2}{16 c^3 d^3 (1+c x)^2}+\frac {13 b^2}{16 c^3 d^3 (1+c x)}-\frac {13 b^2 \tanh ^{-1}(c x)}{16 c^3 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.92, size = 310, normalized size = 1.17 \begin {gather*} \frac {-\frac {8 a^2}{(1+c x)^2}+\frac {32 a^2}{1+c x}+16 a^2 \log (1+c x)+16 b^2 \left (\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}{64} \left (-\cosh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (2 \tanh ^{-1}(c x)\right )\right ) \left (-24+\cosh \left (2 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (-12+\cosh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+8 \tanh ^{-1}(c x)^2 \left (-6+\cosh \left (2 \tanh ^{-1}(c x)\right ) \left (1+8 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+\left (-1+8 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right ) \sinh \left (2 \tanh ^{-1}(c x)\right )\right )\right )\right )+a b \left (12 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )+16 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-12 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (6 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )-8 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-6 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )\right )\right )}{16 c^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*a^2)/(1 + c*x)^2 + (32*a^2)/(1 + c*x) + 16*a^2*Log[1 + c*x] + 16*b^2*(ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcT
anh[c*x])] + PolyLog[3, -E^(-2*ArcTanh[c*x])]/2 + ((-Cosh[2*ArcTanh[c*x]] + Sinh[2*ArcTanh[c*x]])*(-24 + Cosh[
2*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(-12 + Cosh[2*ArcTanh[c*x]] - Sinh[2*ArcTanh[c*x]]) - Sinh[2*ArcTanh[c*x]] +
8*ArcTanh[c*x]^2*(-6 + Cosh[2*ArcTanh[c*x]]*(1 + 8*Log[1 + E^(-2*ArcTanh[c*x])]) + (-1 + 8*Log[1 + E^(-2*ArcTa
nh[c*x])])*Sinh[2*ArcTanh[c*x]])))/64) + a*b*(12*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] + 16*PolyLog[2, -
E^(-2*ArcTanh[c*x])] - 12*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(6*Cosh[2*ArcTanh[c*x]]
- Cosh[4*ArcTanh[c*x]] - 8*Log[1 + E^(-2*ArcTanh[c*x])] - 6*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]])))/(1
6*c^3*d^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 9.69, size = 1135, normalized size = 4.28

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1135\)
default	\(\text {Expression too large to display}\)	\(1135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(a^2/d^3*ln(c*x+1)-1/2*I*b^2/d^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2*Pi-1/2*I*b^2/d^3*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*arctanh(c*x)^2*Pi-I*b^2
/d^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*arctanh(c*x)^2*Pi+1/2*I*b^2/d^3*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*arctanh(c*x)^2*Pi-1/2*I*b^2
/d^3*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*arctanh(c*x)^2*Pi-1/64*b^2/d^3/(c*x+1)
^2+3/8*b^2/d^3/(c*x+1)+1/2*I*b^2/d^3*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*Pi-1/2*I*b^2/d^3*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*Pi-a*b/d^3*arctanh(c*x)/(c*x+1)^2+4*a*b/d^3*arctanh(c*x)/(c*x+1)+
2*a*b/d^3*arctanh(c*x)*ln(c*x+1)+a*b/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)-a*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)-1
/64*b^2/d^3/(c*x+1)^2*c^2*x^2+1/32*b^2/d^3/(c*x+1)^2*c*x-3/8*b^2/d^3/(c*x+1)*c*x-1/4*a*b/d^3/(c*x+1)^2+7/4*a*b
/d^3/(c*x+1)-1/2*a*b/d^3*ln(c*x+1)^2-a*b/d^3*dilog(1/2*c*x+1/2)-7/8*a*b/d^3*ln(c*x+1)+7/8*a*b/d^3*ln(c*x-1)-b^
2/d^3*arctanh(c*x)^2*ln(2)-2*b^2/d^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-b^2/d^3*arctanh(c*x)*polylo
g(2,-(c*x+1)^2/(-c^2*x^2+1))-1/2*b^2/d^3*arctanh(c*x)^2/(c*x+1)^2+2*b^2/d^3*arctanh(c*x)^2/(c*x+1)+b^2/d^3*arc
tanh(c*x)^2*ln(c*x+1)+3/4*b^2/d^3*arctanh(c*x)/(c*x+1)-1/16*b^2/d^3*arctanh(c*x)/(c*x+1)^2+1/2*b^2/d^3*polylog
(3,-(c*x+1)^2/(-c^2*x^2+1))-7/8*b^2/d^3*arctanh(c*x)^2+2/3*b^2/d^3*arctanh(c*x)^3-1/2*a^2/d^3/(c*x+1)^2+2*a^2/
d^3/(c*x+1)-3/4*b^2/d^3*arctanh(c*x)/(c*x+1)*c*x-1/16*b^2/d^3*arctanh(c*x)/(c*x+1)^2*c^2*x^2+1/8*b^2/d^3*arcta
nh(c*x)/(c*x+1)^2*c*x)

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

[Out]

1/2*a^2*((4*c*x + 3)/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) + 2*log(c*x + 1)/(c^3*d^3)) + 1/8*(4*b^2*c*x + 3*b^
2 + 2*(b^2*c^2*x^2 + 2*b^2*c*x + b^2)*log(c*x + 1))*log(-c*x + 1)^2/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) - in
tegrate(-1/4*((b^2*c^3*x^3 - b^2*c^2*x^2)*log(c*x + 1)^2 + 4*(a*b*c^3*x^3 - a*b*c^2*x^2)*log(c*x + 1) - (4*a*b
*c^3*x^3 + 7*b^2*c*x - 4*(a*b*c^2 - b^2*c^2)*x^2 + 3*b^2 + 2*(2*b^2*c^3*x^3 + 2*b^2*c^2*x^2 + 3*b^2*c*x + b^2)
*log(c*x + 1))*log(-c*x + 1))/(c^6*d^3*x^4 + 2*c^5*d^3*x^3 - 2*c^3*d^3*x - c^2*d^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(x^2*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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