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

Optimal. Leaf size=162 \[ \frac{b c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d}+\frac{b^2 c \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )}{2 d}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.412051, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {5934, 5916, 5988, 5932, 2447, 5948, 6056, 6610} \[ \frac{b c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d}+\frac{b^2 c \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )}{2 d}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5934

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

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]]

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 )^2}{x^2 (d+c d x)} \, dx &=-\left (c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x (d+c d x)} \, dx\right )+\frac{\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx}{d}\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{(2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac{\left (2 b 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}{d}\\ &=\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}+\frac{(2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{d}-\frac{\left (b^2 c^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}+\frac{b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}+\frac{b^2 c \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}-\frac{\left (2 b^2 c^2\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+c x}\right )}{d}-\frac{b^2 c \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}+\frac{b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d}+\frac{b^2 c \text{Li}_3\left (-1+\frac{2}{1+c x}\right )}{2 d}\\ \end{align*}

Mathematica [C]  time = 0.639727, size = 225, normalized size = 1.39 \[ \frac{\frac{a b \left (c x \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+2 c x \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )-2 \tanh ^{-1}(c x) \left (c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+1\right )\right )}{x}+b^2 c \left (-\tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )+\frac{2}{3} \tanh ^{-1}(c x)^3-\frac{\tanh ^{-1}(c x)^2}{c x}+\tanh ^{-1}(c x)^2-\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )+a^2 (-c) \log (x)+a^2 c \log (c x+1)-\frac{a^2}{x}}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a^2/x) - a^2*c*Log[x] + a^2*c*Log[1 + c*x] + (a*b*(-2*ArcTanh[c*x]*(1 + c*x*Log[1 - E^(-2*ArcTanh[c*x])]) +
2*c*x*Log[(c*x)/Sqrt[1 - c^2*x^2]] + c*x*PolyLog[2, E^(-2*ArcTanh[c*x])]))/x + b^2*c*((-I/24)*Pi^3 + ArcTanh[
c*x]^2 - ArcTanh[c*x]^2/(c*x) + (2*ArcTanh[c*x]^3)/3 + 2*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - ArcTanh[c
*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - PolyLog[2, E^(-2*ArcTanh[c*x])] - ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x
])] + PolyLog[3, E^(2*ArcTanh[c*x])]/2))/d

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Maple [C]  time = 0.685, size = 7232, normalized size = 44.6 \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))^2/x^2/(c*d*x+d),x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(c*log(c*x + 1)/d - c*log(x)/d - 1/(d*x)) + 1/4*(b^2*c*x*log(c*x + 1) - b^2)*log(-c*x + 1)^2/(d*x) - integ
rate(-1/4*((b^2*c*x - b^2)*log(c*x + 1)^2 + 4*(a*b*c*x - a*b)*log(c*x + 1) + 2*(b^2*c^2*x^2 + 2*a*b - (2*a*b*c
- b^2*c)*x - (b^2*c^3*x^3 + b^2*c^2*x^2 + b^2*c*x - b^2)*log(c*x + 1))*log(-c*x + 1))/(c^2*d*x^4 - d*x^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}}{c d x^{3} + d 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/(c*d*x+d),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{{\left (c d x + d\right )} 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/(c*d*x+d),x, algorithm="giac")

[Out]

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