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

Optimal. Leaf size=102 \[ -3 b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{3}{2} b^3 c \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )+c \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x}+3 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.268323, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5916, 5988, 5932, 5948, 6056, 6610} \[ -3 b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{3}{2} b^3 c \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )+c \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x}+3 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3/x^2,x]

[Out]

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5988

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

Rule 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 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 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[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{\left (a+b \tanh ^{-1}(c x)\right )^3}{x^2} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x}+(3 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x \left (1-c^2 x^2\right )} \, dx\\ &=c \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x}+(3 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x (1+c x)} \, dx\\ &=c \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x}+3 b c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )-\left (6 b^2 c^2\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=c \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x}+3 b c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )-3 b^2 c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )+\left (3 b^3 c^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=c \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x}+3 b c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )-3 b^2 c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )-\frac{3}{2} b^3 c \text{Li}_3\left (-1+\frac{2}{1+c x}\right )\\ \end{align*}

Mathematica [C]  time = 0.320924, size = 196, normalized size = 1.92 \[ 3 a b^2 c \left (\tanh ^{-1}(c x) \left (-\frac{\tanh ^{-1}(c x)}{c x}+\tanh ^{-1}(c x)+2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )\right )+b^3 c \left (3 \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-\frac{3}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )-\frac{\tanh ^{-1}(c x)^3}{c x}-\tanh ^{-1}(c x)^3+3 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+\frac{i \pi ^3}{8}\right )-\frac{3}{2} a^2 b c \log \left (1-c^2 x^2\right )+3 a^2 b c \log (x)-\frac{3 a^2 b \tanh ^{-1}(c x)}{x}-\frac{a^3}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a^3/x) - (3*a^2*b*ArcTanh[c*x])/x + 3*a^2*b*c*Log[x] - (3*a^2*b*c*Log[1 - c^2*x^2])/2 + 3*a*b^2*c*(ArcTanh[c
*x]*(ArcTanh[c*x] - ArcTanh[c*x]/(c*x) + 2*Log[1 - E^(-2*ArcTanh[c*x])]) - PolyLog[2, E^(-2*ArcTanh[c*x])]) +
b^3*c*((I/8)*Pi^3 - ArcTanh[c*x]^3 - ArcTanh[c*x]^3/(c*x) + 3*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + 3*A
rcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - (3*PolyLog[3, E^(2*ArcTanh[c*x])])/2)

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Maple [C]  time = 0.183, size = 1583, normalized size = 15.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*c*a*b^2*arctanh(c*x)*ln(c*x)-3*c*a*b^2*arctanh(c*x)*ln(c*x-1)-3*c*a*b^2*ln(c*x)*ln(c*x+1)-3*c*b^3*arctanh(c*
x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+6*c*b^3*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))+3*c*b^3*arctanh(
c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+6*c*b^3*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))-3/2*c*b^3*a
rctanh(c*x)^2*ln(c*x-1)-3/2*c*b^3*arctanh(c*x)^2*ln(c*x+1)+3*c*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2
))-3*c*a*b^2*dilog(c*x)-3*c*a*b^2*dilog(c*x+1)-3*a^2*b/x*arctanh(c*x)-3*a*b^2/x*arctanh(c*x)^2+3*c*a^2*b*ln(c*
x)-3/2*c*a^2*b*ln(c*x-1)-3/2*c*a^2*b*ln(c*x+1)-3/4*c*a*b^2*ln(c*x-1)^2+3/4*c*a*b^2*ln(c*x+1)^2+3*c*b^3*arctanh
(c*x)^2*ln(2)+3*c*a*b^2*dilog(1/2+1/2*c*x)+3*c*b^3*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+3*c*b^3*ln(
c*x)*arctanh(c*x)^2-b^3/x*arctanh(c*x)^3-6*c*b^3*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))-6*c*b^3*polylog(3,(c*x
+1)/(-c^2*x^2+1)^(1/2))-c*b^3*arctanh(c*x)^3+3/2*I*c*b^3*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/((c*x+1)
^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2-3/4*I*c*b^3*P
i*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c
^2*x^2+1)+1))*arctanh(c*x)^2-a^3/x+3/4*I*c*b^3*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x
^2-1))*arctanh(c*x)^2-3/4*I*c*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c
^2*x^2+1)+1))^2*arctanh(c*x)^2+3/2*I*c*b^3*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*
arctanh(c*x)^2+3/4*I*c*b^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2+3/4*I*c*b^3*Pi*csgn(I*(c*x+1)^2/(
c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2+3/2*I*c*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*ar
ctanh(c*x)^2-3/2*I*c*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2-3*c*a*b^2*arctanh(c*x)*ln(c*x+
1)+3/2*c*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-3/2*c*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+3/2*c*a*b^2*ln(-1/2*c*x+1/2)*l
n(1/2+1/2*c*x)+3/2*I*c*b^3*Pi*arctanh(c*x)^2+3/2*I*c*b^3*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^
2/(c^2*x^2-1))^2*arctanh(c*x)^2-3/2*I*c*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+
1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2-3/2*I*c*b^3*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((
c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+3/4*I*c*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x
^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{3}{2} \,{\left (c{\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac{2 \, \operatorname{artanh}\left (c x\right )}{x}\right )} a^{2} b - \frac{a^{3}}{x} - \frac{{\left (b^{3} c x - b^{3}\right )} \log \left (-c x + 1\right )^{3} + 3 \,{\left (2 \, a b^{2} +{\left (b^{3} c x + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \, x} - \int -\frac{{\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{3} + 6 \,{\left (a b^{2} c x - a b^{2}\right )} \log \left (c x + 1\right )^{2} + 3 \,{\left (4 \, a b^{2} c x -{\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{2} + 2 \,{\left (b^{3} c^{2} x^{2} + 2 \, a b^{2} -{\left (2 \, a b^{2} c - b^{3} c\right )} x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \,{\left (c x^{3} - x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a^2*b - a^3/x - 1/8*((b^3*c*x - b^3)*log(-c*x + 1)^3
+ 3*(2*a*b^2 + (b^3*c*x + b^3)*log(c*x + 1))*log(-c*x + 1)^2)/x - integrate(-1/8*((b^3*c*x - b^3)*log(c*x + 1
)^3 + 6*(a*b^2*c*x - a*b^2)*log(c*x + 1)^2 + 3*(4*a*b^2*c*x - (b^3*c*x - b^3)*log(c*x + 1)^2 + 2*(b^3*c^2*x^2
+ 2*a*b^2 - (2*a*b^2*c - b^3*c)*x)*log(c*x + 1))*log(-c*x + 1))/(c*x^3 - x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \operatorname{artanh}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (c x\right ) + a^{3}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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