3.121 \(\int x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=80 \[ -\frac{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}-\frac{32 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac{16 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}+\frac{2 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

[Out]

(2*x^3*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (12*x^2*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2) + (16*x*ArcTanh[Ta
nh[a + b*x]]^(9/2))/(105*b^3) - (32*ArcTanh[Tanh[a + b*x]]^(11/2))/(1155*b^4)

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Rubi [A]  time = 0.0460745, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2157, 30} \[ -\frac{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}-\frac{32 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac{16 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}+\frac{2 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*x^3*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (12*x^2*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2) + (16*x*ArcTanh[Ta
nh[a + b*x]]^(9/2))/(105*b^3) - (32*ArcTanh[Tanh[a + b*x]]^(11/2))/(1155*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))^{3/2} \, dx &=\frac{2 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{6 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b}\\ &=\frac{2 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{24 \int x \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^2}\\ &=\frac{2 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{16 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{16 \int \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{105 b^3}\\ &=\frac{2 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{16 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{16 \operatorname{Subst}\left (\int x^{9/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{105 b^4}\\ &=\frac{2 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{16 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{32 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0334686, size = 66, normalized size = 0.82 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \left (-198 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+88 b x \tanh ^{-1}(\tanh (a+b x))^2-16 \tanh ^{-1}(\tanh (a+b x))^3+231 b^3 x^3\right )}{1155 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(231*b^3*x^3 - 198*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 88*b*x*ArcTanh[Tanh[a + b*
x]]^2 - 16*ArcTanh[Tanh[a + b*x]]^3))/(1155*b^4)

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Maple [A]  time = 0.033, size = 124, normalized size = 1.6 \begin{align*} 2\,{\frac{1/11\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{11/2}+1/9\, \left ( -3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +3\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{9/2}+1/7\, \left ( \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) \left ( -2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx \right ) + \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}+1/5\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctanh(tanh(b*x+a))^(3/2),x)

[Out]

2/b^4*(1/11*arctanh(tanh(b*x+a))^(11/2)+1/9*(-3*arctanh(tanh(b*x+a))+3*b*x)*arctanh(tanh(b*x+a))^(9/2)+1/7*((b
*x-arctanh(tanh(b*x+a)))*(-2*arctanh(tanh(b*x+a))+2*b*x)+(b*x-arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^(7
/2)+1/5*(b*x-arctanh(tanh(b*x+a)))^3*arctanh(tanh(b*x+a))^(5/2))

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Maxima [A]  time = 1.76476, size = 72, normalized size = 0.9 \begin{align*} \frac{2 \,{\left (105 \, b^{4} x^{4} + 35 \, a b^{3} x^{3} - 30 \, a^{2} b^{2} x^{2} + 24 \, a^{3} b x - 16 \, a^{4}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{1155 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/1155*(105*b^4*x^4 + 35*a*b^3*x^3 - 30*a^2*b^2*x^2 + 24*a^3*b*x - 16*a^4)*(b*x + a)^(3/2)/b^4

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Fricas [A]  time = 1.79939, size = 147, normalized size = 1.84 \begin{align*} \frac{2 \,{\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt{b x + a}}{1155 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^4*b*x - 16*a^5)*sqrt(b*x + a)/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(x**3*atanh(tanh(a + b*x))**(3/2), x)

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Giac [A]  time = 1.15894, size = 169, normalized size = 2.11 \begin{align*} \frac{\sqrt{2}{\left (\frac{11 \, \sqrt{2}{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} a}{b^{3}} + \frac{\sqrt{2}{\left (315 \,{\left (b x + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}\right )}}{b^{3}}\right )}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

1/3465*sqrt(2)*(11*sqrt(2)*(35*(b*x + a)^(9/2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*(b*x +
a)^(3/2)*a^3)*a/b^3 + sqrt(2)*(315*(b*x + a)^(11/2) - 1540*(b*x + a)^(9/2)*a + 2970*(b*x + a)^(7/2)*a^2 - 2772
*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4)/b^3)/b