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

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

[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)

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Rubi [A]  time = 0.0466606, antiderivative size = 72, normalized size of antiderivative = 1., 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 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]))

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 30

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

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*}

Mathematica [A]  time = 0.0238973, 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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Maple [A]  time = 0.042, size = 74, normalized size = 1. \begin{align*}{\frac{{x}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{4}}-b \left ({\frac{{x}^{5} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{5}}-{\frac{3\,b}{5} \left ({\frac{{x}^{6} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{6}}-{\frac{b}{3} \left ({\frac{{x}^{7}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{7}}-{\frac{{x}^{8}b}{56}} \right ) } \right ) } \right ) \end{align*}

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 = 1.73531, size = 97, normalized size = 1.35 \begin{align*} -\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 \end{align*}

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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Fricas [A]  time = 1.38852, size = 99, normalized size = 1.38 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 6.74145, size = 75, normalized size = 1.04 \begin{align*} \frac{b^{4} x^{8}}{280} - \frac{b^{3} x^{7} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{35} + \frac{b^{2} x^{6} \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{10} - \frac{b x^{5} \operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{5} + \frac{x^{4} \operatorname{atanh}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13812, size = 61, normalized size = 0.85 \begin{align*} \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} \end{align*}

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