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

Optimal. Leaf size=34 \[ \frac {x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^6}{30 b^2} \]

[Out]

1/5*x*arctanh(tanh(b*x+a))^5/b-1/30*arctanh(tanh(b*x+a))^6/b^2

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Rubi [A]  time = 0.01, 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 = {2168, 2157, 30} \[ \frac {x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^6}{30 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x*ArcTanh[Tanh[a + b*x]]^5)/(5*b) - ArcTanh[Tanh[a + b*x]]^6/(30*b^2)

Rule 30

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

Rule 2157

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 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 x \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {\int \tanh ^{-1}(\tanh (a+b x))^5 \, dx}{5 b}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {\operatorname {Subst}\left (\int x^5 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^6}{30 b^2}\\ \end {align*}

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Mathematica [B]  time = 0.10, size = 125, normalized size = 3.68 \[ -\frac {(a+b x) \left (-20 \left (2 a^2+a b x-b^2 x^2\right ) \tanh ^{-1}(\tanh (a+b x))^3+(5 a-b x) (a+b x)^4-6 (4 a-b x) (a+b x)^3 \tanh ^{-1}(\tanh (a+b x))+15 (3 a-b x) (a+b x)^2 \tanh ^{-1}(\tanh (a+b x))^2+15 (a-b x) \tanh ^{-1}(\tanh (a+b x))^4\right )}{30 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*((a + b*x)*((5*a - b*x)*(a + b*x)^4 - 6*(4*a - b*x)*(a + b*x)^3*ArcTanh[Tanh[a + b*x]] + 15*(3*a - b*x)*
(a + b*x)^2*ArcTanh[Tanh[a + b*x]]^2 - 20*(2*a^2 + a*b*x - b^2*x^2)*ArcTanh[Tanh[a + b*x]]^3 + 15*(a - b*x)*Ar
cTanh[Tanh[a + b*x]]^4))/b^2

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fricas [A]  time = 0.70, size = 46, normalized size = 1.35 \[ \frac {1}{6} \, b^{4} x^{6} + \frac {4}{5} \, a b^{3} x^{5} + \frac {3}{2} \, a^{2} b^{2} x^{4} + \frac {4}{3} \, a^{3} b x^{3} + \frac {1}{2} \, a^{4} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^4*x^6 + 4/5*a*b^3*x^5 + 3/2*a^2*b^2*x^4 + 4/3*a^3*b*x^3 + 1/2*a^4*x^2

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giac [A]  time = 0.17, size = 46, normalized size = 1.35 \[ \frac {1}{6} \, b^{4} x^{6} + \frac {4}{5} \, a b^{3} x^{5} + \frac {3}{2} \, a^{2} b^{2} x^{4} + \frac {4}{3} \, a^{3} b x^{3} + \frac {1}{2} \, a^{4} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^4*x^6 + 4/5*a*b^3*x^5 + 3/2*a^2*b^2*x^4 + 4/3*a^3*b*x^3 + 1/2*a^4*x^2

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maple [B]  time = 0.16, size = 74, normalized size = 2.18 \[ \frac {x^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{2}-2 b \left (\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3}-b \left (\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{4}-\frac {b \left (\frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}{5}-\frac {x^{6} b}{30}\right )}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctanh(tanh(b*x+a))^4-2*b*(1/3*x^3*arctanh(tanh(b*x+a))^3-b*(1/4*x^4*arctanh(tanh(b*x+a))^2-1/2*b*(1/
5*x^5*arctanh(tanh(b*x+a))-1/30*x^6*b)))

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maxima [B]  time = 0.58, size = 72, normalized size = 2.12 \[ -\frac {2}{3} \, b x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{30} \, {\left (15 \, b x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{6} - 6 \, b x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*b*x^3*arctanh(tanh(b*x + a))^3 + 1/2*x^2*arctanh(tanh(b*x + a))^4 + 1/30*(15*b*x^4*arctanh(tanh(b*x + a))
^2 + (b^2*x^6 - 6*b*x^5*arctanh(tanh(b*x + a)))*b)*b

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mupad [B]  time = 1.00, size = 70, normalized size = 2.06 \[ \frac {b^4\,x^6}{30}-\frac {b^3\,x^5\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{5}+\frac {b^2\,x^4\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2}-\frac {2\,b\,x^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3}+\frac {x^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(tanh(a + b*x))^4,x)

[Out]

(x^2*atanh(tanh(a + b*x))^4)/2 + (b^4*x^6)/30 + (b^2*x^4*atanh(tanh(a + b*x))^2)/2 - (2*b*x^3*atanh(tanh(a + b
*x))^3)/3 - (b^3*x^5*atanh(tanh(a + b*x)))/5

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sympy [A]  time = 2.33, size = 41, normalized size = 1.21 \[ \begin {cases} \frac {x \operatorname {atanh}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} - \frac {\operatorname {atanh}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{30 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {atanh}^{4}{\left (\tanh {\relax (a )} \right )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*atanh(tanh(a + b*x))**5/(5*b) - atanh(tanh(a + b*x))**6/(30*b**2), Ne(b, 0)), (x**2*atanh(tanh(a)
)**4/2, True))

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