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

Optimal. Leaf size=257 \[ \frac {3 b^2 \text {Li}_3\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e}-\frac {3 b^2 \text {Li}_3\left (\frac {2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e}-\frac {3 b \text {Li}_2\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e}+\frac {3 b \text {Li}_2\left (\frac {2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{-c-d x+1}\right )}{4 d e}+\frac {3 b^3 \text {Li}_4\left (\frac {2}{-c-d x+1}-1\right )}{4 d e} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6107, 12, 5914, 6052, 5948, 6058, 6062, 6610} \[ \frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e}-\frac {3 b^2 \text {PolyLog}\left (3,\frac {2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e}-\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e}+\frac {3 b \text {PolyLog}\left (2,\frac {2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e}-\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{-c-d x+1}\right )}{4 d e}+\frac {3 b^3 \text {PolyLog}\left (4,\frac {2}{-c-d x+1}-1\right )}{4 d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 6062

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

Rule 6107

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && IGtQ[p, 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 {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{c e+d e x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {(6 b) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (1-\frac {2}{1-x}\right ) \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2 \log \left (2-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2 \log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 d e}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c-d x}\right )}{2 d e}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \text {Li}_2\left (1-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \text {Li}_2\left (-1+\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 d e}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c-d x}\right )}{2 d e}+\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1-c-d x}\right )}{2 d e}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_3\left (-1+\frac {2}{1-c-d x}\right )}{2 d e}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 d e}+\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 d e}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c-d x}\right )}{2 d e}+\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1-c-d x}\right )}{2 d e}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_3\left (-1+\frac {2}{1-c-d x}\right )}{2 d e}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1-c-d x}\right )}{4 d e}+\frac {3 b^3 \text {Li}_4\left (-1+\frac {2}{1-c-d x}\right )}{4 d e}\\ \end {align*}

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Mathematica [C]  time = 0.51, size = 599, normalized size = 2.33 \[ \frac {64 a^3 \log (c+d x)-96 i a^2 b \left (-i \text {Li}_2\left (e^{-2 \tanh ^{-1}(c+d x)}\right )-i \text {Li}_2\left (-e^{2 \tanh ^{-1}(c+d x)}\right )-\frac {1}{4} i \left (\pi -2 i \tanh ^{-1}(c+d x)\right )^2+i \tanh ^{-1}(c+d x)^2+\left (\pi -2 i \tanh ^{-1}(c+d x)\right ) \log \left (e^{2 \tanh ^{-1}(c+d x)}+1\right )-\log \left (\frac {2}{\sqrt {1-(c+d x)^2}}\right ) \left (\pi -2 i \tanh ^{-1}(c+d x)\right )+2 i \tanh ^{-1}(c+d x) \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )-2 i \log \left (\frac {2 i (c+d x)}{\sqrt {1-(c+d x)^2}}\right ) \tanh ^{-1}(c+d x)\right )+192 a^2 b \left (-\log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+\log \left (\frac {i (c+d x)}{\sqrt {1-(c+d x)^2}}\right )\right ) \tanh ^{-1}(c+d x)+8 a b^2 \left (24 \tanh ^{-1}(c+d x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+24 \tanh ^{-1}(c+d x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c+d x)}\right )+12 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )-12 \text {Li}_3\left (e^{2 \tanh ^{-1}(c+d x)}\right )-16 \tanh ^{-1}(c+d x)^3-24 \tanh ^{-1}(c+d x)^2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )+24 \tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+i \pi ^3\right )+b^3 \left (96 \tanh ^{-1}(c+d x)^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+96 \tanh ^{-1}(c+d x)^2 \text {Li}_2\left (e^{2 \tanh ^{-1}(c+d x)}\right )+96 \tanh ^{-1}(c+d x) \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )-96 \tanh ^{-1}(c+d x) \text {Li}_3\left (e^{2 \tanh ^{-1}(c+d x)}\right )+48 \text {Li}_4\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+48 \text {Li}_4\left (e^{2 \tanh ^{-1}(c+d x)}\right )-32 \tanh ^{-1}(c+d x)^4-64 \tanh ^{-1}(c+d x)^3 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )+64 \tanh ^{-1}(c+d x)^3 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+\pi ^4\right )}{64 d e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(64*a^3*Log[c + d*x] + 192*a^2*b*ArcTanh[c + d*x]*(-Log[1/Sqrt[1 - (c + d*x)^2]] + Log[(I*(c + d*x))/Sqrt[1 -
(c + d*x)^2]]) - (96*I)*a^2*b*((-1/4*I)*(Pi - (2*I)*ArcTanh[c + d*x])^2 + I*ArcTanh[c + d*x]^2 + (2*I)*ArcTanh
[c + d*x]*Log[1 - E^(-2*ArcTanh[c + d*x])] + (Pi - (2*I)*ArcTanh[c + d*x])*Log[1 + E^(2*ArcTanh[c + d*x])] - (
Pi - (2*I)*ArcTanh[c + d*x])*Log[2/Sqrt[1 - (c + d*x)^2]] - (2*I)*ArcTanh[c + d*x]*Log[((2*I)*(c + d*x))/Sqrt[
1 - (c + d*x)^2]] - I*PolyLog[2, E^(-2*ArcTanh[c + d*x])] - I*PolyLog[2, -E^(2*ArcTanh[c + d*x])]) + 8*a*b^2*(
I*Pi^3 - 16*ArcTanh[c + d*x]^3 - 24*ArcTanh[c + d*x]^2*Log[1 + E^(-2*ArcTanh[c + d*x])] + 24*ArcTanh[c + d*x]^
2*Log[1 - E^(2*ArcTanh[c + d*x])] + 24*ArcTanh[c + d*x]*PolyLog[2, -E^(-2*ArcTanh[c + d*x])] + 24*ArcTanh[c +
d*x]*PolyLog[2, E^(2*ArcTanh[c + d*x])] + 12*PolyLog[3, -E^(-2*ArcTanh[c + d*x])] - 12*PolyLog[3, E^(2*ArcTanh
[c + d*x])]) + b^3*(Pi^4 - 32*ArcTanh[c + d*x]^4 - 64*ArcTanh[c + d*x]^3*Log[1 + E^(-2*ArcTanh[c + d*x])] + 64
*ArcTanh[c + d*x]^3*Log[1 - E^(2*ArcTanh[c + d*x])] + 96*ArcTanh[c + d*x]^2*PolyLog[2, -E^(-2*ArcTanh[c + d*x]
)] + 96*ArcTanh[c + d*x]^2*PolyLog[2, E^(2*ArcTanh[c + d*x])] + 96*ArcTanh[c + d*x]*PolyLog[3, -E^(-2*ArcTanh[
c + d*x])] - 96*ArcTanh[c + d*x]*PolyLog[3, E^(2*ArcTanh[c + d*x])] + 48*PolyLog[4, -E^(-2*ArcTanh[c + d*x])]
+ 48*PolyLog[4, E^(2*ArcTanh[c + d*x])]))/(64*d*e)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (d x + c\right ) + a^{3}}{d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{d e x + c e}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.18, size = 1842, normalized size = 7.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*b^3/e*arctanh(d*x+c)^3*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/d*b^3/e*arctanh(d*x+c)^3*ln((d*x+c+1)^2/(1-(d
*x+c)^2)-1)+3/2/d*b^3/e*arctanh(d*x+c)*polylog(3,-(d*x+c+1)^2/(1-(d*x+c)^2))-3/2/d*b^3/e*arctanh(d*x+c)^2*poly
log(2,-(d*x+c+1)^2/(1-(d*x+c)^2))+3/d*b^3/e*arctanh(d*x+c)^2*polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-6/d*b^3/
e*arctanh(d*x+c)*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+1/d*b^3/e*arctanh(d*x+c)^3*ln(1+(d*x+c+1)/(1-(d*x+c)
^2)^(1/2))+3/d*b^3/e*arctanh(d*x+c)^2*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-6/d*b^3/e*arctanh(d*x+c)*polyl
og(3,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/2/d*a*b^2/e*polylog(3,-(d*x+c+1)^2/(1-(d*x+c)^2))-6/d*a*b^2/e*polylog(3
,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-6/d*a*b^2/e*polylog(3,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-3/2/d*a^2*b/e*dilog(d*x+
c)-3/2/d*a^2*b/e*dilog(d*x+c+1)+1/d*b^3/e*ln(d*x+c)*arctanh(d*x+c)^3-1/2*I/d*b^3/e*Pi*csgn(I*((d*x+c+1)^2/(1-(
d*x+c)^2)-1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*x+c)^3-1/2*I/d*b
^3/e*Pi*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2
)))^2*arctanh(d*x+c)^3+6/d*b^3/e*polylog(4,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+1/d*a^3/e*ln(d*x+c)-3/4/d*b^3/e*pol
ylog(4,-(d*x+c+1)^2/(1-(d*x+c)^2))+6/d*b^3/e*polylog(4,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/d*a*b^2/e*ln(d*x+c)*ar
ctanh(d*x+c)^2+3/d*a*b^2/e*arctanh(d*x+c)^2*ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+6/d*a*b^2/e*arctanh(d*x+c)*pol
ylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/d*a^2*b/e*ln(d*x+c)*arctanh(d*x+c)-3/2/d*a^2*b/e*ln(d*x+c)*ln(d*x+c+1
)-3/d*a*b^2/e*arctanh(d*x+c)*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))-3/d*a*b^2/e*arctanh(d*x+c)^2*ln((d*x+c+1)^2
/(1-(d*x+c)^2)-1)+3/d*a*b^2/e*arctanh(d*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-3/2*I/d*a*b^2/e*Pi*csgn(I/(
1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*
x+c)^2-3/2*I/d*a*b^2/e*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c
+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*x+c)^2+3/2*I/d*a*b^2/e*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I/(1+(d
*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*arctanh(d*x+c)^2
+1/2*I/d*b^3/e*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*x+c+1
)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*arctanh(d*x+c)^3+1/2*I/d*b^3/e*Pi*csgn(I*((d*x+c+1)^2/(1-(
d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3*arctanh(d*x+c)^3+3/2*I/d*a*b^2/e*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+
c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3*arctanh(d*x+c)^2+6/d*a*b^2/e*arctanh(d*x+c)*polylog(2,(d*x+c+1)/(1-(
d*x+c)^2)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{3} \log \left (d e x + c e\right )}{d e} + \int \frac {b^{3} {\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}^{3}}{8 \, {\left (d e x + c e\right )}} + \frac {3 \, a b^{2} {\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}^{2}}{4 \, {\left (d e x + c e\right )}} + \frac {3 \, a^{2} b {\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}}{2 \, {\left (d e x + c e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{c\,e+d\,e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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