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

Optimal. Leaf size=74 \[ \frac{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{16 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{32 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac{2 x^3}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

(-2*x^3)/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (12*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/b^2 - (16*x*ArcTanh[Tanh[a +
b*x]]^(3/2))/b^3 + (32*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b^4)

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Rubi [A]  time = 0.0486649, antiderivative size = 74, 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} \[ \frac{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac{16 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac{32 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac{2 x^3}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0396789, size = 66, normalized size = 0.89 \[ \frac{2 \left (30 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-40 b x \tanh ^{-1}(\tanh (a+b x))^2+16 \tanh ^{-1}(\tanh (a+b x))^3-5 b^3 x^3\right )}{5 b^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-5*b^3*x^3 + 30*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 40*b*x*ArcTanh[Tanh[a + b*x]]^2 + 16*ArcTanh[Tanh[a + b*x
]]^3))/(5*b^4*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Maple [B]  time = 0.038, size = 201, normalized size = 2.7 \begin{align*} 2\,{\frac{1}{{b}^{4}} \left ( 1/5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}- \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}a- \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +3\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{a}^{2}+6\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-{\frac{-{a}^{3}-3\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) -3\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}- \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}} \right ) } \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*(1/5*arctanh(tanh(b*x+a))^(5/2)-arctanh(tanh(b*x+a))^(3/2)*a-arctanh(tanh(b*x+a))^(3/2)*(arctanh(tanh(b*
x+a))-b*x-a)+3*arctanh(tanh(b*x+a))^(1/2)*a^2+6*a*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/2)+3*(a
rctanh(tanh(b*x+a))-b*x-a)^2*arctanh(tanh(b*x+a))^(1/2)-(-a^3-3*a^2*(arctanh(tanh(b*x+a))-b*x-a)-3*a*(arctanh(
tanh(b*x+a))-b*x-a)^2-(arctanh(tanh(b*x+a))-b*x-a)^3)/arctanh(tanh(b*x+a))^(1/2))

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Maxima [A]  time = 1.79652, size = 70, normalized size = 0.95 \begin{align*} \frac{2 \,{\left (b^{4} x^{4} - a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 24 \, a^{3} b x + 16 \, a^{4}\right )}}{5 \,{\left (b x + a\right )}^{\frac{3}{2}} 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]

2/5*(b^4*x^4 - a*b^3*x^3 + 6*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 = 2.38357, size = 108, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} \sqrt{b x + a}}{5 \,{\left (b^{5} x + a b^{4}\right )}} \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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{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 = 1.17376, size = 82, normalized size = 1.11 \begin{align*} \frac{2 \, a^{3}}{\sqrt{b x + a} b^{4}} + \frac{2 \,{\left ({\left (b x + a\right )}^{\frac{5}{2}} b^{16} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{16} + 15 \, \sqrt{b x + a} a^{2} b^{16}\right )}}{5 \, b^{20}} \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]

2*a^3/(sqrt(b*x + a)*b^4) + 2/5*((b*x + a)^(5/2)*b^16 - 5*(b*x + a)^(3/2)*a*b^16 + 15*sqrt(b*x + a)*a^2*b^16)/
b^20