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

Optimal. Leaf size=71 \[ b^2 (-c) \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+c \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.145297, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5916, 5988, 5932, 2447} \[ b^2 (-c) \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+c \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/x^2,x]

[Out]

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

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 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]]

Rubi steps

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

Mathematica [A]  time = 0.143522, size = 94, normalized size = 1.32 \[ \frac{-b^2 c x \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-a \left (a+b c x \log \left (1-c^2 x^2\right )-2 b c x \log (c x)\right )+2 b \tanh ^{-1}(c x) \left (b c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-a\right )+b^2 (c x-1) \tanh ^{-1}(c x)^2}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.021, size = 248, normalized size = 3.5 \begin{align*} -{\frac{{a}^{2}}{x}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{x}}-c{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) +2\,c{b}^{2}\ln \left ( cx \right ){\it Artanh} \left ( cx \right ) -c{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) -{\frac{c{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{4}}+c{b}^{2}{\it dilog} \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) +{\frac{c{b}^{2}\ln \left ( cx-1 \right ) }{2}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{c{b}^{2}}{2}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{c{b}^{2}\ln \left ( cx+1 \right ) }{2}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }+{\frac{c{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{4}}-c{b}^{2}{\it dilog} \left ( cx \right ) -c{b}^{2}{\it dilog} \left ( cx+1 \right ) -c{b}^{2}\ln \left ( cx \right ) \ln \left ( cx+1 \right ) -2\,{\frac{ab{\it Artanh} \left ( cx \right ) }{x}}-cab\ln \left ( cx-1 \right ) +2\,cab\ln \left ( cx \right ) -cab\ln \left ( cx+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2/x-b^2/x*arctanh(c*x)^2-c*b^2*arctanh(c*x)*ln(c*x-1)+2*c*b^2*ln(c*x)*arctanh(c*x)-c*b^2*arctanh(c*x)*ln(c*
x+1)-1/4*c*b^2*ln(c*x-1)^2+c*b^2*dilog(1/2+1/2*c*x)+1/2*c*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+1/2*c*b^2*ln(-1/2*c*x+
1/2)*ln(1/2+1/2*c*x)-1/2*c*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/4*c*b^2*ln(c*x+1)^2-c*b^2*dilog(c*x)-c*b^2*dilog(c
*x+1)-c*b^2*ln(c*x)*ln(c*x+1)-2*a*b/x*arctanh(c*x)-c*a*b*ln(c*x-1)+2*c*a*b*ln(c*x)-c*a*b*ln(c*x+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/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 )^{2}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*x))**2/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 )}^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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