3.159 \(\int \frac{x^3}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac{4 x^2}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}-\frac{2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0485976, antiderivative size = 76, 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} \[ -\frac{4 x^2}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}-\frac{2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(-2*x^3)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (4*x^2)/(b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (16*x*Sqrt[ArcTanh[
Tanh[a + b*x]]])/b^3 - (32*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*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))^{5/2}} \, dx &=-\frac{2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{2 \int \frac{x^2}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{b}\\ &=-\frac{2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{4 x^2}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{8 \int \frac{x}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=-\frac{2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{4 x^2}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{16 \int \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=-\frac{2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{4 x^2}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{16 \operatorname{Subst}\left (\int \sqrt{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}\\ &=-\frac{2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{4 x^2}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0388968, size = 65, normalized size = 0.86 \[ -\frac{2 \left (6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-24 b x \tanh ^{-1}(\tanh (a+b x))^2+16 \tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3\right )}{3 b^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(-2*(b^3*x^3 + 6*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 24*b*x*ArcTanh[Tanh[a + b*x]]^2 + 16*ArcTanh[Tanh[a + b*x]]^
3))/(3*b^4*ArcTanh[Tanh[a + b*x]]^(3/2))

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Maple [B]  time = 0.042, size = 186, normalized size = 2.5 \begin{align*} 2\,{\frac{1}{{b}^{4}} \left ( 1/3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}-3\,a\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-{\frac{3\,{a}^{2}+6\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \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 ) }}}-1/3\,{\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}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

2/b^4*(1/3*arctanh(tanh(b*x+a))^(3/2)-3*a*arctanh(tanh(b*x+a))^(1/2)-3*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(ta
nh(b*x+a))^(1/2)-(3*a^2+6*a*(arctanh(tanh(b*x+a))-b*x-a)+3*(arctanh(tanh(b*x+a))-b*x-a)^2)/arctanh(tanh(b*x+a)
)^(1/2)-1/3*(-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))^(3/2))

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Maxima [A]  time = 1.7914, size = 70, normalized size = 0.92 \begin{align*} \frac{2 \,{\left (b^{4} x^{4} - 5 \, a b^{3} x^{3} - 30 \, a^{2} b^{2} x^{2} - 40 \, a^{3} b x - 16 \, a^{4}\right )}}{3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/3*(b^4*x^4 - 5*a*b^3*x^3 - 30*a^2*b^2*x^2 - 40*a^3*b*x - 16*a^4)/((b*x + a)^(5/2)*b^4)

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Fricas [A]  time = 1.93832, size = 131, normalized size = 1.72 \begin{align*} \frac{2 \,{\left (b^{3} x^{3} - 6 \, a b^{2} x^{2} - 24 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt{b x + a}}{3 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/3*(b^3*x^3 - 6*a*b^2*x^2 - 24*a^2*b*x - 16*a^3)*sqrt(b*x + a)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)

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Sympy [A]  time = 92.4217, size = 90, normalized size = 1.18 \begin{align*} \begin{cases} - \frac{2 x^{3}}{3 b \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}} - \frac{4 x^{2}}{b^{2} \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}} + \frac{16 x \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}}{b^{3}} - \frac{32 \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}}{3 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 \operatorname{atanh}^{\frac{5}{2}}{\left (\tanh{\left (a \right )} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Piecewise((-2*x**3/(3*b*atanh(tanh(a + b*x))**(3/2)) - 4*x**2/(b**2*sqrt(atanh(tanh(a + b*x)))) + 16*x*sqrt(at
anh(tanh(a + b*x)))/b**3 - 32*atanh(tanh(a + b*x))**(3/2)/(3*b**4), Ne(b, 0)), (x**4/(4*atanh(tanh(a))**(5/2))
, True))

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Giac [A]  time = 1.16221, size = 80, normalized size = 1.05 \begin{align*} -\frac{2 \,{\left (9 \,{\left (b x + a\right )} a^{2} - a^{3}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{4}} + \frac{2 \,{\left ({\left (b x + a\right )}^{\frac{3}{2}} b^{8} - 9 \, \sqrt{b x + a} a b^{8}\right )}}{3 \, b^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/3*(9*(b*x + a)*a^2 - a^3)/((b*x + a)^(3/2)*b^4) + 2/3*((b*x + a)^(3/2)*b^8 - 9*sqrt(b*x + a)*a*b^8)/b^12