3.1.62 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^4} \, dx\) [62]

Optimal. Leaf size=55 \[ -\frac {b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x) \]

[Out]

-b^2*arctanh(tanh(b*x+a))/x-1/2*b*arctanh(tanh(b*x+a))^2/x^2-1/3*arctanh(tanh(b*x+a))^3/x^3+b^3*ln(x)

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Rubi [A]
time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 29} \begin {gather*} -\frac {b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}-\frac {b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}+b^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^4,x]

[Out]

-((b^2*ArcTanh[Tanh[a + b*x]])/x) - (b*ArcTanh[Tanh[a + b*x]]^2)/(2*x^2) - ArcTanh[Tanh[a + b*x]]^3/(3*x^3) +
b^3*Log[x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2199

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[u^(m + 1)*(v^
n/(a*(m + 1))), x] - Dist[b*(n/(a*(m + 1))), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^4} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^3} \, dx\\ &=-\frac {b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^2 \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^2} \, dx\\ &=-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \int \frac {1}{x} \, dx\\ &=-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 60, normalized size = 1.09 \begin {gather*} \frac {-6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-3 b x \tanh ^{-1}(\tanh (a+b x))^2-2 \tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3 (11+6 \log (x))}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^3/x^4,x]

[Out]

(-6*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 3*b*x*ArcTanh[Tanh[a + b*x]]^2 - 2*ArcTanh[Tanh[a + b*x]]^3 + b^3*x^3*(11
+ 6*Log[x]))/(6*x^3)

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Maple [A]
time = 0.93, size = 52, normalized size = 0.95

method	result	size
default	\(-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3 x^{3}}+b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{2 x^{2}}+b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{x}+b \ln \left (x \right )\right )\right )\)	\(52\)
risch	\(\text {Expression too large to display}\)	\(7816\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*arctanh(tanh(b*x+a))^3/x^3+b*(-1/2*arctanh(tanh(b*x+a))^2/x^2+b*(-arctanh(tanh(b*x+a))/x+b*ln(x)))

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Maxima [A]
time = 0.37, size = 52, normalized size = 0.95 \begin {gather*} {\left (b^{2} \log \left (x\right ) - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x}\right )} b - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x^{2}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^4,x, algorithm="maxima")

[Out]

(b^2*log(x) - b*arctanh(tanh(b*x + a))/x)*b - 1/2*b*arctanh(tanh(b*x + a))^2/x^2 - 1/3*arctanh(tanh(b*x + a))^
3/x^3

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Fricas [A]
time = 0.33, size = 37, normalized size = 0.67 \begin {gather*} \frac {6 \, b^{3} x^{3} \log \left (x\right ) - 18 \, a b^{2} x^{2} - 9 \, a^{2} b x - 2 \, a^{3}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^4,x, algorithm="fricas")

[Out]

1/6*(6*b^3*x^3*log(x) - 18*a*b^2*x^2 - 9*a^2*b*x - 2*a^3)/x^3

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Sympy [A]
time = 0.29, size = 51, normalized size = 0.93 \begin {gather*} b^{3} \log {\left (x \right )} - \frac {b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} - \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**3/x**4,x)

[Out]

b**3*log(x) - b**2*atanh(tanh(a + b*x))/x - b*atanh(tanh(a + b*x))**2/(2*x**2) - atanh(tanh(a + b*x))**3/(3*x*
*3)

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Giac [A]
time = 0.39, size = 35, normalized size = 0.64 \begin {gather*} b^{3} \log \left ({\left | x \right |}\right ) - \frac {18 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^4,x, algorithm="giac")

[Out]

b^3*log(abs(x)) - 1/6*(18*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)/x^3

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Mupad [B]
time = 1.00, size = 51, normalized size = 0.93 \begin {gather*} b^3\,\ln \left (x\right )-\frac {b^2\,x^2\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+\frac {b\,x\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2}+\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^3/x^4,x)

[Out]

b^3*log(x) - (atanh(tanh(a + b*x))^3/3 + (b*x*atanh(tanh(a + b*x))^2)/2 + b^2*x^2*atanh(tanh(a + b*x)))/x^3

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