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

Optimal. Leaf size=69 \[ \frac{16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32}{315} b^3 x^{9/2} \]

[Out]

(-32*b^3*x^(9/2))/315 + (16*b^2*x^(7/2)*ArcTanh[Tanh[a + b*x]])/35 - (4*b*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)/5
+ (2*x^(3/2)*ArcTanh[Tanh[a + b*x]]^3)/3

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Rubi [A]  time = 0.0369685, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 30} \[ \frac{16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32}{315} b^3 x^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(-32*b^3*x^(9/2))/315 + (16*b^2*x^(7/2)*ArcTanh[Tanh[a + b*x]])/35 - (4*b*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)/5
+ (2*x^(3/2)*ArcTanh[Tanh[a + b*x]]^3)/3

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 30

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

Rubi steps

\begin{align*} \int \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-(2 b) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac{4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3+\frac{1}{5} \left (8 b^2\right ) \int x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac{16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{35} \left (16 b^3\right ) \int x^{7/2} \, dx\\ &=-\frac{32}{315} b^3 x^{9/2}+\frac{16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3\\ \end{align*}

Mathematica [A]  time = 0.0276033, size = 57, normalized size = 0.83 \[ -\frac{2}{315} x^{3/2} \left (-72 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+126 b x \tanh ^{-1}(\tanh (a+b x))^2-105 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^(3/2)*(16*b^3*x^3 - 72*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 126*b*x*ArcTanh[Tanh[a + b*x]]^2 - 105*ArcTanh[T
anh[a + b*x]]^3))/315

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Maple [A]  time = 0.046, size = 56, normalized size = 0.8 \begin{align*}{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{3}{x}^{{\frac{3}{2}}}}-4\,b \left ( 1/5\,{x}^{5/2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}-4/5\,b \left ( 1/7\,{x}^{7/2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -{\frac{2\,b{x}^{9/2}}{63}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3*x^(1/2),x)

[Out]

2/3*x^(3/2)*arctanh(tanh(b*x+a))^3-4*b*(1/5*x^(5/2)*arctanh(tanh(b*x+a))^2-4/5*b*(1/7*x^(7/2)*arctanh(tanh(b*x
+a))-2/63*b*x^(9/2)))

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Maxima [A]  time = 1.05312, size = 74, normalized size = 1.07 \begin{align*} -\frac{4}{5} \, b x^{\frac{5}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac{2}{3} \, x^{\frac{3}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{16}{315} \,{\left (2 \, b^{2} x^{\frac{9}{2}} - 9 \, b x^{\frac{7}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3*x^(1/2),x, algorithm="maxima")

[Out]

-4/5*b*x^(5/2)*arctanh(tanh(b*x + a))^2 + 2/3*x^(3/2)*arctanh(tanh(b*x + a))^3 - 16/315*(2*b^2*x^(9/2) - 9*b*x
^(7/2)*arctanh(tanh(b*x + a)))*b

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Fricas [A]  time = 2.03469, size = 97, normalized size = 1.41 \begin{align*} \frac{2}{315} \,{\left (35 \, b^{3} x^{4} + 135 \, a b^{2} x^{3} + 189 \, a^{2} b x^{2} + 105 \, a^{3} x\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3*x^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*b^3*x^4 + 135*a*b^2*x^3 + 189*a^2*b*x^2 + 105*a^3*x)*sqrt(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**3*x**(1/2),x)

[Out]

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

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Giac [A]  time = 1.16223, size = 47, normalized size = 0.68 \begin{align*} \frac{2}{9} \, b^{3} x^{\frac{9}{2}} + \frac{6}{7} \, a b^{2} x^{\frac{7}{2}} + \frac{6}{5} \, a^{2} b x^{\frac{5}{2}} + \frac{2}{3} \, a^{3} x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3*x^(1/2),x, algorithm="giac")

[Out]

2/9*b^3*x^(9/2) + 6/7*a*b^2*x^(7/2) + 6/5*a^2*b*x^(5/2) + 2/3*a^3*x^(3/2)