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

Optimal. Leaf size=123 \[ -\frac{3}{2} b^3 c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+3 b^2 c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{3}{2} b c^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.294355, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5916, 5982, 5988, 5932, 2447, 5948} \[ -\frac{3}{2} b^3 c^2 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+3 b^2 c^2 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{3}{2} b c^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3/x^3,x]

[Out]

(3*b*c^2*(a + b*ArcTanh[c*x])^2)/2 - (3*b*c*(a + b*ArcTanh[c*x])^2)/(2*x) + (c^2*(a + b*ArcTanh[c*x])^3)/2 - (
a + b*ArcTanh[c*x])^3/(2*x^2) + 3*b^2*c^2*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - (3*b^3*c^2*PolyLog[2, -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 5982

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTanh[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{x^3} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}+\frac{1}{2} (3 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}+\frac{1}{2} (3 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx+\frac{1}{2} \left (3 b c^3\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x}+\frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}+\left (3 b^2 c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx\\ &=\frac{3}{2} b c^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x}+\frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}+\left (3 b^2 c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx\\ &=\frac{3}{2} b c^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x}+\frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}+3 b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-\left (3 b^3 c^3\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac{3}{2} b c^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x}+\frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{2 x^2}+3 b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-\frac{3}{2} b^3 c^2 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.271884, size = 192, normalized size = 1.56 \[ \frac{-6 b^3 c^2 x^2 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+a \left (-2 a^2-3 a b c^2 x^2 \log (1-c x)+3 a b c^2 x^2 \log (c x+1)-6 a b c x+12 b^2 c^2 x^2 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )\right )-6 b \tanh ^{-1}(c x) \left (a^2+2 a b c x-2 b^2 c^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )+6 b^2 (c x-1) \tanh ^{-1}(c x)^2 (a c x+a+b c x)+2 b^3 \left (c^2 x^2-1\right ) \tanh ^{-1}(c x)^3}{4 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(6*b^2*(-1 + c*x)*(a + a*c*x + b*c*x)*ArcTanh[c*x]^2 + 2*b^3*(-1 + c^2*x^2)*ArcTanh[c*x]^3 - 6*b*ArcTanh[c*x]*
(a^2 + 2*a*b*c*x - 2*b^2*c^2*x^2*Log[1 - E^(-2*ArcTanh[c*x])]) + a*(-2*a^2 - 6*a*b*c*x - 3*a*b*c^2*x^2*Log[1 -
c*x] + 3*a*b*c^2*x^2*Log[1 + c*x] + 12*b^2*c^2*x^2*Log[(c*x)/Sqrt[1 - c^2*x^2]]) - 6*b^3*c^2*x^2*PolyLog[2, E
^(-2*ArcTanh[c*x])])/(4*x^2)

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Maple [C]  time = 0.316, size = 5098, normalized size = 41.5 \begin{align*} \text{output 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^3,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3}{4} \,{\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac{2}{x}\right )} c - \frac{2 \, \operatorname{artanh}\left (c x\right )}{x^{2}}\right )} a^{2} b + \frac{3}{8} \,{\left ({\left (2 \,{\left (\log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) - \log \left (c x + 1\right )^{2} - \log \left (c x - 1\right )^{2} - 4 \, \log \left (c x - 1\right ) + 8 \, \log \left (x\right )\right )} c^{2} + 4 \,{\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac{2}{x}\right )} c \operatorname{artanh}\left (c x\right )\right )} a b^{2} - \frac{1}{16} \, b^{3}{\left (\frac{{\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )^{3} + 3 \,{\left (2 \, c x -{\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{x^{2}} + 2 \, \int -\frac{{\left (c x - 1\right )} \log \left (c x + 1\right )^{3} + 3 \,{\left (2 \, c^{2} x^{2} -{\left (c x - 1\right )} \log \left (c x + 1\right )^{2} -{\left (c^{3} x^{3} - c x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{c x^{4} - x^{3}}\,{d x}\right )} - \frac{3 \, a b^{2} \operatorname{artanh}\left (c x\right )^{2}}{2 \, x^{2}} - \frac{a^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*a^2*b + 3/8*((2*(log(c*x - 1) - 2)*log(c*
x + 1) - log(c*x + 1)^2 - log(c*x - 1)^2 - 4*log(c*x - 1) + 8*log(x))*c^2 + 4*(c*log(c*x + 1) - c*log(c*x - 1)
- 2/x)*c*arctanh(c*x))*a*b^2 - 1/16*b^3*(((c^2*x^2 - 1)*log(-c*x + 1)^3 + 3*(2*c*x - (c^2*x^2 - 1)*log(c*x +
1))*log(-c*x + 1)^2)/x^2 + 2*integrate(-((c*x - 1)*log(c*x + 1)^3 + 3*(2*c^2*x^2 - (c*x - 1)*log(c*x + 1)^2 -
(c^3*x^3 - c*x)*log(c*x + 1))*log(-c*x + 1))/(c*x^4 - x^3), x)) - 3/2*a*b^2*arctanh(c*x)^2/x^2 - 1/2*a^3/x^2

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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^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/x^3,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^3, 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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*x))**3/x**3, 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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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