3.2.6 \(\int \frac {x}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx\) [106]

Optimal. Leaf size=34 \[ -\frac {x}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {1}{2 b^2 \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2199, 2188, 30} \begin {gather*} -\frac {1}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, 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 {x}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {x}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {\int \frac {1}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{2 b}\\ &=-\frac {x}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{2 b^2}\\ &=-\frac {x}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {1}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 27, normalized size = 0.79 \begin {gather*} -\frac {b x+\tanh ^{-1}(\tanh (a+b x))}{2 b^2 \tanh ^{-1}(\tanh (a+b x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 43, normalized size = 1.26

method	result	size
default	\(-\frac {1}{b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {b x -\arctanh \left (\tanh \left (b x +a \right )\right )}{2 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}\)	\(43\)
risch	\(-\frac {2 i \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+4 i b x +4 i \ln \left ({\mathrm e}^{b x +a}\right )\right )}{b^{2} \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+4 i \ln \left ({\mathrm e}^{b x +a}\right )\right )^{2}}\)	\(540\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.84, size = 32, normalized size = 0.94 \begin {gather*} -\frac {2 \, b x + a}{2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.32, size = 32, normalized size = 0.94 \begin {gather*} -\frac {2 \, b x + a}{2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 37.24, size = 42, normalized size = 1.24 \begin {gather*} \begin {cases} - \frac {x}{2 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {1}{2 b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 \operatorname {atanh}^{3}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x/(2*b*atanh(tanh(a + b*x))**2) - 1/(2*b**2*atanh(tanh(a + b*x))), Ne(b, 0)), (x**2/(2*atanh(tanh(
a))**3), True))

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Giac [A]
time = 0.39, size = 18, normalized size = 0.53 \begin {gather*} -\frac {2 \, b x + a}{2 \, {\left (b x + a\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x + a)/((b*x + a)^2*b^2)

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Mupad [B]
time = 0.08, size = 25, normalized size = 0.74 \begin {gather*} -\frac {\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+b\,x}{2\,b^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/atanh(tanh(a + b*x))^3,x)

[Out]

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

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