3.1.53 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^5} \, dx\) [53]

Optimal. Leaf size=42 \[ -\frac {b^2}{12 x^2}-\frac {b \tanh ^{-1}(\tanh (a+b x))}{6 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^2}{4 x^4} \]

[Out]

-1/12*b^2/x^2-1/6*b*arctanh(tanh(b*x+a))/x^3-1/4*arctanh(tanh(b*x+a))^2/x^4

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.52, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2202, 2198} \begin {gather*} \frac {\tanh ^{-1}(\tanh (a+b x))^3}{4 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{12 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2/x^5,x]

[Out]

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

Rule 2198

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-u^(m + 1))*(
v^(n + 1)/((m + 1)*(b*u - a*v))), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] &&
EqQ[m + n + 2, 0] && NeQ[m, -1]

Rule 2202

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-u^(m + 1))*(
v^(n + 1)/((m + 1)*(b*u - a*v))), x] + Dist[b*((m + n + 2)/((m + 1)*(b*u - a*v))), Int[u^(m + 1)*v^n, x], x] /
; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^5} \, dx &=\frac {\tanh ^{-1}(\tanh (a+b x))^3}{4 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^4} \, dx}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b \tanh ^{-1}(\tanh (a+b x))^3}{12 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {\tanh ^{-1}(\tanh (a+b x))^3}{4 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 37, normalized size = 0.88 \begin {gather*} -\frac {b^2 x^2+2 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^2/x^5,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.21, size = 38, normalized size = 0.90

method	result	size
default	\(-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{4 x^{4}}+\frac {b \left (-\frac {b}{6 x^{2}}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{3 x^{3}}\right )}{2}\)	\(38\)
risch	\(\text {Expression too large to display}\)	\(1978\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^2/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*arctanh(tanh(b*x+a))^2/x^4+1/2*b*(-1/6*b/x^2-1/3*arctanh(tanh(b*x+a))/x^3)

________________________________________________________________________________________

Maxima [A]
time = 0.35, size = 36, normalized size = 0.86 \begin {gather*} -\frac {b^{2}}{12 \, x^{2}} - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{6 \, x^{3}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^5,x, algorithm="maxima")

[Out]

-1/12*b^2/x^2 - 1/6*b*arctanh(tanh(b*x + a))/x^3 - 1/4*arctanh(tanh(b*x + a))^2/x^4

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 24, normalized size = 0.57 \begin {gather*} -\frac {6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^5,x, algorithm="fricas")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)/x^4

________________________________________________________________________________________

Sympy [A]
time = 0.39, size = 39, normalized size = 0.93 \begin {gather*} - \frac {b^{2}}{12 x^{2}} - \frac {b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{6 x^{3}} - \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**2/x**5,x)

[Out]

-b**2/(12*x**2) - b*atanh(tanh(a + b*x))/(6*x**3) - atanh(tanh(a + b*x))**2/(4*x**4)

________________________________________________________________________________________

Giac [A]
time = 0.38, size = 24, normalized size = 0.57 \begin {gather*} -\frac {6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^5,x, algorithm="giac")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)/x^4

________________________________________________________________________________________

Mupad [B]
time = 0.95, size = 36, normalized size = 0.86 \begin {gather*} -\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{4\,x^4}-\frac {b^2}{12\,x^2}-\frac {b\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{6\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^2/x^5,x)

[Out]

- atanh(tanh(a + b*x))^2/(4*x^4) - b^2/(12*x^2) - (b*atanh(tanh(a + b*x)))/(6*x^3)

________________________________________________________________________________________