3.1.20 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x^2} \, dx\) [20]

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

[Out]

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

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Rubi [A]
time = 0.10, 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 = {6037, 6135, 6079, 2497} \begin {gather*} 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 ) \end {gather*}

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 2497

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 6037

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

Rule 6079

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 6135

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*}

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Mathematica [A]
time = 0.10, size = 94, normalized size = 1.32 \begin {gather*} \frac {b^2 (-1+c x) \tanh ^{-1}(c x)^2+2 b \tanh ^{-1}(c x) \left (-a+b c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )-a \left (a-2 b c x \log (c x)+b c x \log \left (1-c^2 x^2\right )\right )-b^2 c x \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[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] Leaf count of result is larger than twice the leaf count of optimal. \(243\) vs. \(2(71)=142\).
time = 0.10, size = 244, normalized size = 3.44

method	result	size
derivativedivides	\(c \left (-\frac {a^{2}}{c x}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{c x}-b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )+2 b^{2} \ln \left (c x \right ) \arctanh \left (c x \right )-b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )+b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{4}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{4}-b^{2} \dilog \left (c x \right )-b^{2} \dilog \left (c x +1\right )-b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {2 a b \arctanh \left (c x \right )}{c x}-a b \ln \left (c x -1\right )+2 a b \ln \left (c x \right )-a b \ln \left (c x +1\right )\right )\)	\(244\)
default	\(c \left (-\frac {a^{2}}{c x}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{c x}-b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )+2 b^{2} \ln \left (c x \right ) \arctanh \left (c x \right )-b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )+b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{4}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{4}-b^{2} \dilog \left (c x \right )-b^{2} \dilog \left (c x +1\right )-b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {2 a b \arctanh \left (c x \right )}{c x}-a b \ln \left (c x -1\right )+2 a b \ln \left (c x \right )-a b \ln \left (c x +1\right )\right )\)	\(244\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}}{x^{2}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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)

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