3.159 \(\int \frac {\tanh ^{-1}(c x)^2}{x (d+e x)} \, dx\)

Optimal. Leaf size=275 \[ \frac {\text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 d}+\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{d}-\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac {\text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d}-\frac {\text {Li}_3\left (\frac {2}{1-c x}-1\right )}{2 d}-\frac {\text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 d}-\frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right ) \tanh ^{-1}(c x)}{d}+\frac {\text {Li}_2\left (\frac {2}{1-c x}-1\right ) \tanh ^{-1}(c x)}{d}-\frac {\text {Li}_2\left (1-\frac {2}{c x+1}\right ) \tanh ^{-1}(c x)}{d}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \tanh ^{-1}(c x)^2}{d}+\frac {\log \left (\frac {2}{c x+1}\right ) \tanh ^{-1}(c x)^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {5940, 5914, 6052, 5948, 6058, 6610, 5922} \[ \frac {\text {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}+\frac {\tanh ^{-1}(c x) \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac {\text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d}-\frac {\text {PolyLog}\left (3,\frac {2}{1-c x}-1\right )}{2 d}-\frac {\text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 d}-\frac {\tanh ^{-1}(c x) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{d}+\frac {\tanh ^{-1}(c x) \text {PolyLog}\left (2,\frac {2}{1-c x}-1\right )}{d}-\frac {\tanh ^{-1}(c x) \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{d}-\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \tanh ^{-1}(c x)^2}{d}+\frac {\log \left (\frac {2}{c x+1}\right ) \tanh ^{-1}(c x)^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[c*x]^2/(x*(d + e*x)),x]

[Out]

(2*ArcTanh[c*x]^2*ArcTanh[1 - 2/(1 - c*x)])/d + (ArcTanh[c*x]^2*Log[2/(1 + c*x)])/d - (ArcTanh[c*x]^2*Log[(2*c
*(d + e*x))/((c*d + e)*(1 + c*x))])/d - (ArcTanh[c*x]*PolyLog[2, 1 - 2/(1 - c*x)])/d + (ArcTanh[c*x]*PolyLog[2
, -1 + 2/(1 - c*x)])/d - (ArcTanh[c*x]*PolyLog[2, 1 - 2/(1 + c*x)])/d + (ArcTanh[c*x]*PolyLog[2, 1 - (2*c*(d +
e*x))/((c*d + e)*(1 + c*x))])/d + PolyLog[3, 1 - 2/(1 - c*x)]/(2*d) - PolyLog[3, -1 + 2/(1 - c*x)]/(2*d) - Po
lyLog[3, 1 - 2/(1 + c*x)]/(2*d) + PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))]/(2*d)

Rule 5914

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1
- c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTanh[c*x])^(p - 1)*ArcTanh[1 - 2/(1 - c*x)])/(1 - c^2*x^2), x], x]
/; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 5922

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^2*Log[
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*PolyLog
[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 5940

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 5948

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 6052

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

Rule 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[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 6610

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 {\tanh ^{-1}(c x)^2}{x (d+e x)} \, dx &=\int \left (\frac {\tanh ^{-1}(c x)^2}{d x}-\frac {e \tanh ^{-1}(c x)^2}{d (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {\tanh ^{-1}(c x)^2}{x} \, dx}{d}-\frac {e \int \frac {\tanh ^{-1}(c x)^2}{d+e x} \, dx}{d}\\ &=\frac {2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d}+\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2}{1+c x}\right )}{d}-\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d}+\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d}+\frac {\text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}-\frac {(4 c) \int \frac {\tanh ^{-1}(c x) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac {2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d}+\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2}{1+c x}\right )}{d}-\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d}+\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d}+\frac {\text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}+\frac {(2 c) \int \frac {\tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}-\frac {(2 c) \int \frac {\tanh ^{-1}(c x) \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac {2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d}+\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2}{1+c x}\right )}{d}-\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{d}+\frac {\tanh ^{-1}(c x) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{d}-\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d}+\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d}+\frac {\text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}+\frac {c \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}-\frac {c \int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac {2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d}+\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2}{1+c x}\right )}{d}-\frac {\tanh ^{-1}(c x)^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{d}+\frac {\tanh ^{-1}(c x) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{d}-\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d}+\frac {\tanh ^{-1}(c x) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}+\frac {\text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d}-\frac {\text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{2 d}-\frac {\text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d}+\frac {\text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}\\ \end {align*}

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Mathematica [C]  time = 9.37, size = 521, normalized size = 1.89 \[ \frac {16 e \sqrt {1-\frac {c^2 d^2}{e^2}} \tanh ^{-1}(c x)^3 e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )}+24 c d \tanh ^{-1}(c x)^2 \log \left (\frac {c (d+e x)}{\sqrt {1-c^2 x^2}}\right )+12 i \pi c d \log \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)-48 c d \tanh ^{-1}(c x) \text {Li}_2\left (-e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )-48 c d \tanh ^{-1}(c x) \text {Li}_2\left (e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )+48 c d \text {Li}_3\left (-e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )+48 c d \text {Li}_3\left (e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )-24 c d \tanh ^{-1}(c x)^2 \log \left (1-e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}\right )-24 c d \tanh ^{-1}(c x)^2 \log \left (e^{\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)}+1\right )-24 c d \tanh ^{-1}(c x)^2 \log \left (e^{-\tanh ^{-1}(c x)} \left (c d \left (e^{2 \tanh ^{-1}(c x)}+1\right )+e \left (e^{2 \tanh ^{-1}(c x)}-1\right )\right )\right )-48 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (\frac {1}{2} i e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )-\tanh ^{-1}(c x)} \left (e^{2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}-1\right )\right )+48 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )+24 c d \tanh ^{-1}(c x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c x)}\right )-12 c d \text {Li}_3\left (e^{2 \tanh ^{-1}(c x)}\right )+16 c d \tanh ^{-1}(c x)^3+24 c d \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+c d \log (16777216) \tanh ^{-1}(c x)^2+24 i \pi c d \tanh ^{-1}(c x) \log \left (e^{-\tanh ^{-1}(c x)}+e^{\tanh ^{-1}(c x)}\right )-24 i \pi c d \log (2) \tanh ^{-1}(c x)+i \pi ^3 c d-16 e \tanh ^{-1}(c x)^3}{24 c d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*c*d*Pi^3 + 16*c*d*ArcTanh[c*x]^3 - 16*e*ArcTanh[c*x]^3 + (16*Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^3)/E^Ar
cTanh[(c*d)/e] - (24*I)*c*d*Pi*ArcTanh[c*x]*Log[2] + c*d*ArcTanh[c*x]^2*Log[16777216] + (24*I)*c*d*Pi*ArcTanh[
c*x]*Log[E^(-ArcTanh[c*x]) + E^ArcTanh[c*x]] + 24*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - 24*c*d*ArcT
anh[c*x]^2*Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 24*c*d*ArcTanh[c*x]^2*Log[1 + E^(ArcTanh[(c*d)/e] +
ArcTanh[c*x])] - 48*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] + ArcTanh[c*x])))] - 24*c*d*ArcTanh[c*x]^2*Log[(e*(-1 + E^(2*ArcTanh[c*x])) + c*d*(1 + E^(
2*ArcTanh[c*x])))/E^ArcTanh[c*x]] + 24*c*d*ArcTanh[c*x]^2*Log[(c*(d + e*x))/Sqrt[1 - c^2*x^2]] + (12*I)*c*d*Pi
*ArcTanh[c*x]*Log[1 - c^2*x^2] + 48*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*
x]]] + 24*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - 48*c*d*ArcTanh[c*x]*PolyLog[2, -E^(ArcTanh[(c*d)/e
] + ArcTanh[c*x])] - 48*c*d*ArcTanh[c*x]*PolyLog[2, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 12*c*d*PolyLog[3, E
^(2*ArcTanh[c*x])] + 48*c*d*PolyLog[3, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 48*c*d*PolyLog[3, E^(ArcTanh[(c
*d)/e] + ArcTanh[c*x])])/(24*c*d^2)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (c x\right )^{2}}{e x^{2} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (c x\right )^{2}}{{\left (e x + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.18, size = 1507, normalized size = 5.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctanh(c*x)^2/d*ln(c*x)-arctanh(c*x)^2/d*ln(c*e*x+c*d)+arctanh(c*x)^2/d*ln(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(
1+(c*x+1)^2/(-c^2*x^2+1)))-arctanh(c*x)^2/d*ln((c*x+1)^2/(-c^2*x^2+1)-1)+1/d*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2
*x^2+1)^(1/2))+2/d*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))-2/d*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2)
)+1/d*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+2/d*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-
2/d*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*I/d*Pi*arctanh(c*x)^2*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*
(1+(c*x+1)^2/(-c^2*x^2+1))))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)))/(1+(c*x+1)^2
/(-c^2*x^2+1)))^2-1/2*I/d*Pi*arctanh(c*x)^2*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)
-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1))))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)))/(1
+(c*x+1)^2/(-c^2*x^2+1)))-1/2*I/d*Pi*arctanh(c*x)^2*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^
2*x^2+1)))/(1+(c*x+1)^2/(-c^2*x^2+1)))^3+1/2*I/d*Pi*arctanh(c*x)^2*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(
((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)))/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-1/2*I/d*Pi*arctanh(
c*x)^2*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)))^2-1/2*
I/d*Pi*arctanh(c*x)^2*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+1/2*I/d*Pi*arctanh(c*x)^2*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3+1/2*I/d*P
i*arctanh(c*x)^2*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)))-1/d*e/(c*d+e)*arctanh(c*x)^2*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+
e))-1/d*e/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+1/2/d*e/(c*d+e)*polylog(3,(c
*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-c/(c*d+e)*arctanh(c*x)^2*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-
c/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+1/2*c/(c*d+e)*polylog(3,(c*d+e)*(c*x
+1)^2/(-c^2*x^2+1)/(-c*d+e))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (c x\right )^{2}}{{\left (e x + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atanh}\left (c\,x\right )}^2}{x\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(c*x)^2/(x*(d + e*x)),x)

[Out]

int(atanh(c*x)^2/(x*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{2}{\left (c x \right )}}{x \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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