3.77 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^5} \, dx\)

Optimal. Leaf size=74 \[ -\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 x^4}-\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^4 \log (x) \]

[Out]

-b^3*arctanh(tanh(b*x+a))/x-1/2*b^2*arctanh(tanh(b*x+a))^2/x^2-1/3*b*arctanh(tanh(b*x+a))^3/x^3-1/4*arctanh(ta
nh(b*x+a))^4/x^4+b^4*ln(x)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 29} \[ -\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 x^4}+b^4 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^5,x]

[Out]

-((b^3*ArcTanh[Tanh[a + b*x]])/x) - (b^2*ArcTanh[Tanh[a + b*x]]^2)/(2*x^2) - (b*ArcTanh[Tanh[a + b*x]]^3)/(3*x
^3) - ArcTanh[Tanh[a + b*x]]^4/(4*x^4) + b^4*Log[x]

Rule 29

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

Rule 2168

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))^4}{x^5} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 x^4}+b \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^4} \, dx\\ &=-\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 x^4}+b^2 \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^3} \, dx\\ &=-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 x^4}+b^3 \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^2} \, dx\\ &=-\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 x^4}+b^4 \int \frac {1}{x} \, dx\\ &=-\frac {b^3 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac {b^2 \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{4 x^4}+b^4 \log (x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 78, normalized size = 1.05 \[ -\frac {12 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+4 b x \tanh ^{-1}(\tanh (a+b x))^3+3 \tanh ^{-1}(\tanh (a+b x))^4-b^4 x^4 (12 \log (x)+25)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^5,x]

[Out]

-1/12*(12*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 6*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 4*b*x*ArcTanh[Tanh[a + b*x]]^3
+ 3*ArcTanh[Tanh[a + b*x]]^4 - b^4*x^4*(25 + 12*Log[x]))/x^4

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 48, normalized size = 0.65 \[ \frac {12 \, b^{4} x^{4} \log \relax (x) - 48 \, a b^{3} x^{3} - 36 \, a^{2} b^{2} x^{2} - 16 \, a^{3} b x - 3 \, a^{4}}{12 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^5,x, algorithm="fricas")

[Out]

1/12*(12*b^4*x^4*log(x) - 48*a*b^3*x^3 - 36*a^2*b^2*x^2 - 16*a^3*b*x - 3*a^4)/x^4

________________________________________________________________________________________

giac [A]  time = 0.18, size = 46, normalized size = 0.62 \[ b^{4} \log \left ({\left | x \right |}\right ) - \frac {48 \, a b^{3} x^{3} + 36 \, a^{2} b^{2} x^{2} + 16 \, a^{3} b x + 3 \, a^{4}}{12 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^5,x, algorithm="giac")

[Out]

b^4*log(abs(x)) - 1/12*(48*a*b^3*x^3 + 36*a^2*b^2*x^2 + 16*a^3*b*x + 3*a^4)/x^4

________________________________________________________________________________________

maple [A]  time = 0.15, size = 69, normalized size = 0.93 \[ -\frac {b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}{x}-\frac {b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{2 x^{2}}-\frac {b \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3 x^{3}}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{4 x^{4}}+b^{4} \ln \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^5,x)

[Out]

-b^3*arctanh(tanh(b*x+a))/x-1/2*b^2*arctanh(tanh(b*x+a))^2/x^2-1/3*b*arctanh(tanh(b*x+a))^3/x^3-1/4*arctanh(ta
nh(b*x+a))^4/x^4+b^4*ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 72, normalized size = 0.97 \[ \frac {1}{2} \, {\left (2 \, {\left (b^{2} \log \relax (x) - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x}\right )} b - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{2}}\right )} b - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{3}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^5,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.14, size = 68, normalized size = 0.92 \[ b^4\,\ln \relax (x)-\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{4\,x^4}-\frac {b^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2\,x^2}-\frac {b^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{x}-\frac {b\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^5,x)

[Out]

b^4*log(x) - atanh(tanh(a + b*x))^4/(4*x^4) - (b^2*atanh(tanh(a + b*x))^2)/(2*x^2) - (b^3*atanh(tanh(a + b*x))
)/x - (b*atanh(tanh(a + b*x))^3)/(3*x^3)

________________________________________________________________________________________

sympy [A]  time = 1.33, size = 70, normalized size = 0.95 \[ b^{4} \log {\relax (x )} - \frac {b^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} - \frac {b \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} - \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**5,x)

[Out]

b**4*log(x) - b**3*atanh(tanh(a + b*x))/x - b**2*atanh(tanh(a + b*x))**2/(2*x**2) - b*atanh(tanh(a + b*x))**3/
(3*x**3) - atanh(tanh(a + b*x))**4/(4*x**4)

________________________________________________________________________________________