3.69 \(\int x^3 \tanh ^{-1}(\tanh (a+b x))^4 \, dx\)

Optimal. Leaf size=72 \[ -\frac {\tanh ^{-1}(\tanh (a+b x))^8}{280 b^4}+\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{35 b^3}-\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^6}{10 b^2}+\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^5}{5 b} \]

[Out]

1/5*x^3*arctanh(tanh(b*x+a))^5/b-1/10*x^2*arctanh(tanh(b*x+a))^6/b^2+1/35*x*arctanh(tanh(b*x+a))^7/b^3-1/280*a
rctanh(tanh(b*x+a))^8/b^4

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Rubi [A]  time = 0.05, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ -\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^6}{10 b^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^8}{280 b^4}+\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{35 b^3}+\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^5}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x^3*ArcTanh[Tanh[a + b*x]]^5)/(5*b) - (x^2*ArcTanh[Tanh[a + b*x]]^6)/(10*b^2) + (x*ArcTanh[Tanh[a + b*x]]^7)/
(35*b^3) - ArcTanh[Tanh[a + b*x]]^8/(280*b^4)

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^3 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {3 \int x^2 \tanh ^{-1}(\tanh (a+b x))^5 \, dx}{5 b}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^6}{10 b^2}+\frac {\int x \tanh ^{-1}(\tanh (a+b x))^6 \, dx}{5 b^2}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^6}{10 b^2}+\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{35 b^3}-\frac {\int \tanh ^{-1}(\tanh (a+b x))^7 \, dx}{35 b^3}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^6}{10 b^2}+\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{35 b^3}-\frac {\operatorname {Subst}\left (\int x^7 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{35 b^4}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac {x^2 \tanh ^{-1}(\tanh (a+b x))^6}{10 b^2}+\frac {x \tanh ^{-1}(\tanh (a+b x))^7}{35 b^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^8}{280 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 71, normalized size = 0.99 \[ \frac {1}{280} x^4 \left (-8 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+28 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-56 b x \tanh ^{-1}(\tanh (a+b x))^3+70 \tanh ^{-1}(\tanh (a+b x))^4+b^4 x^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x^4*(b^4*x^4 - 8*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 28*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 56*b*x*ArcTanh[Tanh[a
+ b*x]]^3 + 70*ArcTanh[Tanh[a + b*x]]^4))/280

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fricas [A]  time = 0.62, size = 45, normalized size = 0.62 \[ \frac {1}{8} \, b^{4} x^{8} + \frac {4}{7} \, a b^{3} x^{7} + a^{2} b^{2} x^{6} + \frac {4}{5} \, a^{3} b x^{5} + \frac {1}{4} \, a^{4} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 45, normalized size = 0.62 \[ \frac {1}{8} \, b^{4} x^{8} + \frac {4}{7} \, a b^{3} x^{7} + a^{2} b^{2} x^{6} + \frac {4}{5} \, a^{3} b x^{5} + \frac {1}{4} \, a^{4} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.15, size = 74, normalized size = 1.03 \[ \frac {x^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{4}-b \left (\frac {x^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{5}-\frac {3 b \left (\frac {x^{6} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{6}-\frac {b \left (\frac {x^{7} \arctanh \left (\tanh \left (b x +a \right )\right )}{7}-\frac {x^{8} b}{56}\right )}{3}\right )}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctanh(tanh(b*x+a))^4,x)

[Out]

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 [A]  time = 0.58, size = 72, normalized size = 1.00 \[ -\frac {1}{5} \, b x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{4} \, x^{4} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{280} \, {\left (28 \, b x^{6} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{8} - 8 \, b x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^4,x, algorithm="maxima")

[Out]

-1/5*b*x^5*arctanh(tanh(b*x + a))^3 + 1/4*x^4*arctanh(tanh(b*x + a))^4 + 1/280*(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

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mupad [B]  time = 1.02, size = 70, normalized size = 0.97 \[ \frac {b^4\,x^8}{280}-\frac {b^3\,x^7\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{35}+\frac {b^2\,x^6\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{10}-\frac {b\,x^5\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{5}+\frac {x^4\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atanh(tanh(a + b*x))^4,x)

[Out]

(x^4*atanh(tanh(a + b*x))^4)/4 + (b^4*x^8)/280 + (b^2*x^6*atanh(tanh(a + b*x))^2)/10 - (b*x^5*atanh(tanh(a + b
*x))^3)/5 - (b^3*x^7*atanh(tanh(a + b*x)))/35

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sympy [A]  time = 6.72, size = 78, normalized size = 1.08 \[ \begin {cases} \frac {x^{3} \operatorname {atanh}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} - \frac {x^{2} \operatorname {atanh}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{10 b^{2}} + \frac {x \operatorname {atanh}^{7}{\left (\tanh {\left (a + b x \right )} \right )}}{35 b^{3}} - \frac {\operatorname {atanh}^{8}{\left (\tanh {\left (a + b x \right )} \right )}}{280 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {atanh}^{4}{\left (\tanh {\relax (a )} \right )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atanh(tanh(b*x+a))**4,x)

[Out]

Piecewise((x**3*atanh(tanh(a + b*x))**5/(5*b) - x**2*atanh(tanh(a + b*x))**6/(10*b**2) + x*atanh(tanh(a + b*x)
)**7/(35*b**3) - atanh(tanh(a + b*x))**8/(280*b**4), Ne(b, 0)), (x**4*atanh(tanh(a))**4/4, True))

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