3.84 \(\int x \tanh ^{-1}(\tanh (a+b x))^6 \, dx\)

Optimal. Leaf size=34 \[ \frac {x \tanh ^{-1}(\tanh (a+b x))^7}{7 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^8}{56 b^2} \]

[Out]

1/7*x*arctanh(tanh(b*x+a))^7/b-1/56*arctanh(tanh(b*x+a))^8/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} \[ \frac {x \tanh ^{-1}(\tanh (a+b x))^7}{7 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^8}{56 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^6,x]

[Out]

(x*ArcTanh[Tanh[a + b*x]]^7)/(7*b) - ArcTanh[Tanh[a + b*x]]^8/(56*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))^6 \, dx &=\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{7 b}-\frac {\int \tanh ^{-1}(\tanh (a+b x))^7 \, dx}{7 b}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{7 b}-\frac {\operatorname {Subst}\left (\int x^7 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{7 b^2}\\ &=\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{7 b}-\frac {\tanh ^{-1}(\tanh (a+b x))^8}{56 b^2}\\ \end {align*}

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Mathematica [B]  time = 0.14, size = 177, normalized size = 5.21 \[ -\frac {(a+b x) \left (-56 \left (2 a^2+a b x-b^2 x^2\right ) \tanh ^{-1}(\tanh (a+b x))^5+(7 a-b x) (a+b x)^6-8 (6 a-b x) (a+b x)^5 \tanh ^{-1}(\tanh (a+b x))+28 (5 a-b x) (a+b x)^4 \tanh ^{-1}(\tanh (a+b x))^2-56 (4 a-b x) (a+b x)^3 \tanh ^{-1}(\tanh (a+b x))^3+70 (3 a-b x) (a+b x)^2 \tanh ^{-1}(\tanh (a+b x))^4+28 (a-b x) \tanh ^{-1}(\tanh (a+b x))^6\right )}{56 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[Tanh[a + b*x]]^6,x]

[Out]

-1/56*((a + b*x)*((7*a - b*x)*(a + b*x)^6 - 8*(6*a - b*x)*(a + b*x)^5*ArcTanh[Tanh[a + b*x]] + 28*(5*a - b*x)*
(a + b*x)^4*ArcTanh[Tanh[a + b*x]]^2 - 56*(4*a - b*x)*(a + b*x)^3*ArcTanh[Tanh[a + b*x]]^3 + 70*(3*a - b*x)*(a
+ b*x)^2*ArcTanh[Tanh[a + b*x]]^4 - 56*(2*a^2 + a*b*x - b^2*x^2)*ArcTanh[Tanh[a + b*x]]^5 + 28*(a - b*x)*ArcT
anh[Tanh[a + b*x]]^6))/b^2

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fricas [B]  time = 0.47, size = 68, normalized size = 2.00 \[ \frac {1}{8} \, b^{6} x^{8} + \frac {6}{7} \, a b^{5} x^{7} + \frac {5}{2} \, a^{2} b^{4} x^{6} + 4 \, a^{3} b^{3} x^{5} + \frac {15}{4} \, a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + \frac {1}{2} \, a^{6} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^6,x, algorithm="fricas")

[Out]

1/8*b^6*x^8 + 6/7*a*b^5*x^7 + 5/2*a^2*b^4*x^6 + 4*a^3*b^3*x^5 + 15/4*a^4*b^2*x^4 + 2*a^5*b*x^3 + 1/2*a^6*x^2

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giac [B]  time = 0.19, size = 68, normalized size = 2.00 \[ \frac {1}{8} \, b^{6} x^{8} + \frac {6}{7} \, a b^{5} x^{7} + \frac {5}{2} \, a^{2} b^{4} x^{6} + 4 \, a^{3} b^{3} x^{5} + \frac {15}{4} \, a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + \frac {1}{2} \, a^{6} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^6,x, algorithm="giac")

[Out]

1/8*b^6*x^8 + 6/7*a*b^5*x^7 + 5/2*a^2*b^4*x^6 + 4*a^3*b^3*x^5 + 15/4*a^4*b^2*x^4 + 2*a^5*b*x^3 + 1/2*a^6*x^2

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maple [B]  time = 0.17, size = 110, normalized size = 3.24 \[ \frac {x^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{6}}{2}-3 b \left (\frac {x^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{5}}{3}-\frac {5 b \left (\frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{4}-b \left (\frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{5}-\frac {3 b \left (\frac {x^{6} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{6}-\frac {b \left (\frac {x^{7} \arctanh \left (\tanh \left (b x +a \right )\right )}{7}-\frac {x^{8} b}{56}\right )}{3}\right )}{5}\right )\right )}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(tanh(b*x+a))^6,x)

[Out]

1/2*x^2*arctanh(tanh(b*x+a))^6-3*b*(1/3*x^3*arctanh(tanh(b*x+a))^5-5/3*b*(1/4*x^4*arctanh(tanh(b*x+a))^4-b*(1/
5*x^5*arctanh(tanh(b*x+a))^3-3/5*b*(1/6*x^6*arctanh(tanh(b*x+a))^2-1/3*b*(1/7*x^7*arctanh(tanh(b*x+a))-1/56*x^
8*b)))))

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maxima [B]  time = 0.72, size = 110, normalized size = 3.24 \[ -b x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{5} + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{6} + \frac {1}{56} \, {\left (70 \, b x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} - {\left (56 \, b x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - {\left (28 \, b x^{6} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{8} - 8 \, b x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b\right )} b\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^6,x, algorithm="maxima")

[Out]

-b*x^3*arctanh(tanh(b*x + a))^5 + 1/2*x^2*arctanh(tanh(b*x + a))^6 + 1/56*(70*b*x^4*arctanh(tanh(b*x + a))^4 -
(56*b*x^5*arctanh(tanh(b*x + a))^3 - (28*b*x^6*arctanh(tanh(b*x + a))^2 + (b^2*x^8 - 8*b*x^7*arctanh(tanh(b*x
+ a)))*b)*b)*b)*b

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mupad [B]  time = 1.10, size = 104, normalized size = 3.06 \[ \frac {b^6\,x^8}{56}-\frac {b^5\,x^7\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{7}+\frac {b^4\,x^6\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2}-b^3\,x^5\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3+\frac {5\,b^2\,x^4\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{4}-b\,x^3\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^5+\frac {x^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^6}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(tanh(a + b*x))^6,x)

[Out]

(x^2*atanh(tanh(a + b*x))^6)/2 + (b^6*x^8)/56 + (5*b^2*x^4*atanh(tanh(a + b*x))^4)/4 - b^3*x^5*atanh(tanh(a +
b*x))^3 + (b^4*x^6*atanh(tanh(a + b*x))^2)/2 - b*x^3*atanh(tanh(a + b*x))^5 - (b^5*x^7*atanh(tanh(a + b*x)))/7

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sympy [A]  time = 6.54, size = 41, normalized size = 1.21 \[ \begin {cases} \frac {x \operatorname {atanh}^{7}{\left (\tanh {\left (a + b x \right )} \right )}}{7 b} - \frac {\operatorname {atanh}^{8}{\left (\tanh {\left (a + b x \right )} \right )}}{56 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {atanh}^{6}{\left (\tanh {\relax (a )} \right )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a))**6,x)

[Out]

Piecewise((x*atanh(tanh(a + b*x))**7/(7*b) - atanh(tanh(a + b*x))**8/(56*b**2), Ne(b, 0)), (x**2*atanh(tanh(a)
)**6/2, True))

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