3.2.55 \(\int \frac {x (a+b \tanh ^{-1}(c x))^2}{d+e x} \, dx\) [155]

Optimal. Leaf size=279 \[ \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c e}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c e}-\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 d \text {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 e^2}+\frac {b^2 d \text {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6087, 6021, 6131, 6055, 2449, 2352, 6059} \begin {gather*} -\frac {b d \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{e^2}+\frac {d \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e}-\frac {b^2 d \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 e^2}+\frac {b^2 d \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e^2}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 6059

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^2)*(Lo
g[2/(1 + c*x)]/e), x] + (Simp[(a + b*ArcTanh[c*x])^2*(Log[2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + Simp
[b*(a + b*ArcTanh[c*x])*(PolyLog[2, 1 - 2/(1 + c*x)]/e), x] - Simp[b*(a + b*ArcTanh[c*x])*(PolyLog[2, 1 - 2*c*
((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + Simp[b^2*(PolyLog[3, 1 - 2/(1 + c*x)]/(2*e)), x] - Simp[b^2*(PolyL
og[3, 1 - 2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/(2*e)), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 10.32, size = 1036, normalized size = 3.71 \begin {gather*} \frac {6 a^2 e x-6 a^2 d \log (d+e x)+\frac {6 a b \left (-i c d \pi \tanh ^{-1}(c x)+2 c e x \tanh ^{-1}(c x)-2 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \tanh ^{-1}(c x)+c d \tanh ^{-1}(c x)^2-e \tanh ^{-1}(c x)^2+\sqrt {1-\frac {c^2 d^2}{e^2}} e e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )} \tanh ^{-1}(c x)^2+2 c d \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+i c d \pi \log \left (1+e^{2 \tanh ^{-1}(c x)}\right )-2 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+e \log \left (1-c^2 x^2\right )+\frac {1}{2} i c d \pi \log \left (1-c^2 x^2\right )+2 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-c d \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+c d \text {PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )\right )}{c}+\frac {b^2 \left (-6 e \tanh ^{-1}(c x)^2+6 c e x \tanh ^{-1}(c x)^2+8 c d \tanh ^{-1}(c x)^3-4 e \tanh ^{-1}(c x)^3+4 \sqrt {1-\frac {c^2 d^2}{e^2}} e e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )} \tanh ^{-1}(c x)^3-12 e \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+6 c d \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+6 i c d \pi \tanh ^{-1}(c x) \log \left (\frac {1}{2} \left (e^{-\tanh ^{-1}(c x)}+e^{\tanh ^{-1}(c x)}\right )\right )+6 c d \tanh ^{-1}(c x)^2 \log \left (1+\frac {(c d+e) e^{2 \tanh ^{-1}(c x)}}{c d-e}\right )-6 c d \tanh ^{-1}(c x)^2 \log \left (1-e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )-6 c d \tanh ^{-1}(c x)^2 \log \left (1+e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )-6 c d \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-12 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \tanh ^{-1}(c x) \log \left (\frac {1}{2} i e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )-\tanh ^{-1}(c x)} \left (-1+e^{2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )\right )-6 c d \tanh ^{-1}(c x)^2 \log \left (\frac {1}{2} e^{-\tanh ^{-1}(c x)} \left (e \left (-1+e^{2 \tanh ^{-1}(c x)}\right )+c d \left (1+e^{2 \tanh ^{-1}(c x)}\right )\right )\right )+6 c d \tanh ^{-1}(c x)^2 \log \left (\frac {c (d+e x)}{\sqrt {1-c^2 x^2}}\right )+3 i c d \pi \tanh ^{-1}(c x) \log \left (1-c^2 x^2\right )+12 c d \tanh ^{-1}\left (\frac {c d}{e}\right ) \tanh ^{-1}(c x) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )+6 \left (e-c d \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+6 c d \tanh ^{-1}(c x) \text {PolyLog}\left (2,-\frac {(c d+e) e^{2 \tanh ^{-1}(c x)}}{c d-e}\right )-12 c d \tanh ^{-1}(c x) \text {PolyLog}\left (2,-e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )-12 c d \tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )-6 c d \tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-3 c d \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )-3 c d \text {PolyLog}\left (3,-\frac {(c d+e) e^{2 \tanh ^{-1}(c x)}}{c d-e}\right )+12 c d \text {PolyLog}\left (3,-e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )+12 c d \text {PolyLog}\left (3,e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )+3 c d \text {PolyLog}\left (3,e^{2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )\right )}{c}}{6 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*e*x - 6*a^2*d*Log[d + e*x] + (6*a*b*((-I)*c*d*Pi*ArcTanh[c*x] + 2*c*e*x*ArcTanh[c*x] - 2*c*d*ArcTanh[(c
*d)/e]*ArcTanh[c*x] + c*d*ArcTanh[c*x]^2 - e*ArcTanh[c*x]^2 + (Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^Arc
Tanh[(c*d)/e] + 2*c*d*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + I*c*d*Pi*Log[1 + E^(2*ArcTanh[c*x])] - 2*c*d
*ArcTanh[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 2*c*d*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTan
h[(c*d)/e] + ArcTanh[c*x]))] + e*Log[1 - c^2*x^2] + (I/2)*c*d*Pi*Log[1 - c^2*x^2] + 2*c*d*ArcTanh[(c*d)/e]*Log
[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - c*d*PolyLog[2, -E^(-2*ArcTanh[c*x])] + c*d*PolyLog[2, E^(-2*(ArcTa
nh[(c*d)/e] + ArcTanh[c*x]))]))/c + (b^2*(-6*e*ArcTanh[c*x]^2 + 6*c*e*x*ArcTanh[c*x]^2 + 8*c*d*ArcTanh[c*x]^3
- 4*e*ArcTanh[c*x]^3 + (4*Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^3)/E^ArcTanh[(c*d)/e] - 12*e*ArcTanh[c*x]*Log
[1 + E^(-2*ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + (6*I)*c*d*Pi*ArcTanh[c*x]*Log[
(E^(-ArcTanh[c*x]) + E^ArcTanh[c*x])/2] + 6*c*d*ArcTanh[c*x]^2*Log[1 + ((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d - e
)] - 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]^2*Log[1 + E^(ArcTa
nh[(c*d)/e] + ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 12*c*d*
ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[(I/2)*E^(-ArcTanh[(c*d)/e] - ArcTanh[c*x])*(-1 + E^(2*(ArcTanh[(c*d)/e] + Ar
cTanh[c*x])))] - 6*c*d*ArcTanh[c*x]^2*Log[(e*(-1 + E^(2*ArcTanh[c*x])) + c*d*(1 + E^(2*ArcTanh[c*x])))/(2*E^Ar
cTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[(c*(d + e*x))/Sqrt[1 - c^2*x^2]] + (3*I)*c*d*Pi*ArcTanh[c*x]*Log[1 - c
^2*x^2] + 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] + 6*(e - c*d*ArcTa
nh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]*PolyLog[2, -(((c*d + e)*E^(2*ArcTanh[c*x]))/(c*
d - e))] - 12*c*d*ArcTanh[c*x]*PolyLog[2, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 12*c*d*ArcTanh[c*x]*PolyLog[
2, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x])
)] - 3*c*d*PolyLog[3, -E^(-2*ArcTanh[c*x])] - 3*c*d*PolyLog[3, -(((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d - e))] +
12*c*d*PolyLog[3, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 12*c*d*PolyLog[3, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x]
)] + 3*c*d*PolyLog[3, E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/c)/(6*e^2)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 14.52, size = 14121, normalized size = 50.61

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(14121\)
default	\(\text {Expression too large to display}\)	\(14121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________