3.79 \(\int \frac {(a+b \tanh ^{-1}(c x^2))^3}{x} \, dx\)

Optimal. Leaf size=207 \[ \frac {3}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b^2 \text {Li}_3\left (\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b \text {Li}_2\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {3}{4} b \text {Li}_2\left (\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {3}{8} b^3 \text {Li}_4\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{8} b^3 \text {Li}_4\left (\frac {2}{1-c x^2}-1\right ) \]

[Out]

-(a+b*arctanh(c*x^2))^3*arctanh(-1+2/(-c*x^2+1))-3/4*b*(a+b*arctanh(c*x^2))^2*polylog(2,1-2/(-c*x^2+1))+3/4*b*
(a+b*arctanh(c*x^2))^2*polylog(2,-1+2/(-c*x^2+1))+3/4*b^2*(a+b*arctanh(c*x^2))*polylog(3,1-2/(-c*x^2+1))-3/4*b
^2*(a+b*arctanh(c*x^2))*polylog(3,-1+2/(-c*x^2+1))-3/8*b^3*polylog(4,1-2/(-c*x^2+1))+3/8*b^3*polylog(4,-1+2/(-
c*x^2+1))

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Rubi [A]  time = 0.56, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6095, 5914, 6052, 5948, 6058, 6062, 6610} \[ \frac {3}{4} b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b^2 \text {PolyLog}\left (3,\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b \text {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {3}{4} b \text {PolyLog}\left (2,\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {3}{8} b^3 \text {PolyLog}\left (4,1-\frac {2}{1-c x^2}\right )+\frac {3}{8} b^3 \text {PolyLog}\left (4,\frac {2}{1-c x^2}-1\right )+\tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])^3/x,x]

[Out]

(a + b*ArcTanh[c*x^2])^3*ArcTanh[1 - 2/(1 - c*x^2)] - (3*b*(a + b*ArcTanh[c*x^2])^2*PolyLog[2, 1 - 2/(1 - c*x^
2)])/4 + (3*b*(a + b*ArcTanh[c*x^2])^2*PolyLog[2, -1 + 2/(1 - c*x^2)])/4 + (3*b^2*(a + b*ArcTanh[c*x^2])*PolyL
og[3, 1 - 2/(1 - c*x^2)])/4 - (3*b^2*(a + b*ArcTanh[c*x^2])*PolyLog[3, -1 + 2/(1 - c*x^2)])/4 - (3*b^3*PolyLog
[4, 1 - 2/(1 - c*x^2)])/8 + (3*b^3*PolyLog[4, -1 + 2/(1 - c*x^2)])/8

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 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && 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}\left (c x^2\right )\right )^3}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{x} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-(3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )+\frac {1}{2} (3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} (3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {1}{2} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c x^2}\right )-\frac {1}{4} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )+\frac {1}{4} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c x^2}\right )-\frac {3}{8} b^3 \text {Li}_4\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{8} b^3 \text {Li}_4\left (-1+\frac {2}{1-c x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 211, normalized size = 1.02 \[ \frac {3}{8} b \left (2 \text {Li}_2\left (\frac {c x^2+1}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-2 \text {Li}_2\left (\frac {c x^2+1}{c x^2-1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+b \left (-2 \text {Li}_3\left (\frac {c x^2+1}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+2 \text {Li}_3\left (\frac {c x^2+1}{c x^2-1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+b \left (\text {Li}_4\left (\frac {c x^2+1}{1-c x^2}\right )-\text {Li}_4\left (\frac {c x^2+1}{c x^2-1}\right )\right )\right )\right )+\tanh ^{-1}\left (\frac {2}{c x^2-1}+1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*ArcTanh[c*x^2])^3*ArcTanh[1 + 2/(-1 + c*x^2)] + (3*b*(2*(a + b*ArcTanh[c*x^2])^2*PolyLog[2, (1 + c*x^2)
/(1 - c*x^2)] - 2*(a + b*ArcTanh[c*x^2])^2*PolyLog[2, (1 + c*x^2)/(-1 + c*x^2)] + b*(-2*(a + b*ArcTanh[c*x^2])
*PolyLog[3, (1 + c*x^2)/(1 - c*x^2)] + 2*(a + b*ArcTanh[c*x^2])*PolyLog[3, (1 + c*x^2)/(-1 + c*x^2)] + b*(Poly
Log[4, (1 + c*x^2)/(1 - c*x^2)] - PolyLog[4, (1 + c*x^2)/(-1 + c*x^2)]))))/8

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (c x^{2}\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (c x^{2}\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (c x^{2}\right ) + a^{3}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.21, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{3}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))^3/x,x)

[Out]

int((a + b*atanh(c*x^2))^3/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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