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

Optimal. Leaf size=71 \[ c \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 b c \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b^2 (-c) \text {Li}_2\left (\frac {2}{c x+1}-1\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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 = 0.15, size = 94, normalized size = 1.32 \[ \frac {-a \left (a+b c x \log \left (1-c^2 x^2\right )-2 b c x \log (c x)\right )+2 b \tanh ^{-1}(c x) \left (b c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-a\right )-b^2 c x \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+b^2 (c x-1) \tanh ^{-1}(c x)^2}{x} \]

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

________________________________________________________________________________________

fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b \operatorname {artanh}\left (c x\right ) + a^{2}}{x^{2}}, x\right ) \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 248, normalized size = 3.49 \[ -\frac {a^{2}}{x}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{x}+2 c \,b^{2} \ln \left (c x \right ) \arctanh \left (c x \right )-c \,b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )-c \,b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )-\frac {c \,b^{2} \ln \left (c x -1\right )^{2}}{4}+c \,b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )+\frac {c \,b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2}+\frac {c \,b^{2} \ln \left (c x +1\right )^{2}}{4}-\frac {c \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2}+\frac {c \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2}-c \,b^{2} \dilog \left (c x \right )-c \,b^{2} \dilog \left (c x +1\right )-c \,b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {2 a b \arctanh \left (c x \right )}{x}+2 c a b \ln \left (c x \right )-c a b \ln \left (c x -1\right )-c a b \ln \left (c x +1\right ) \]

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^2/x^2,x)

[Out]

int((a + b*atanh(c*x))^2/x^2, x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________