3.183 \(\int x^{7/2} \tanh ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=69 \[ \frac{16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32 b^3 x^{15/2}}{6435} \]

[Out]

(-32*b^3*x^(15/2))/6435 + (16*b^2*x^(13/2)*ArcTanh[Tanh[a + b*x]])/429 - (4*b*x^(11/2)*ArcTanh[Tanh[a + b*x]]^
2)/33 + (2*x^(9/2)*ArcTanh[Tanh[a + b*x]]^3)/9

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Rubi [A]  time = 0.0385126, 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} \[ \frac{16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{32 b^3 x^{15/2}}{6435} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(-32*b^3*x^(15/2))/6435 + (16*b^2*x^(13/2)*ArcTanh[Tanh[a + b*x]])/429 - (4*b*x^(11/2)*ArcTanh[Tanh[a + b*x]]^
2)/33 + (2*x^(9/2)*ArcTanh[Tanh[a + b*x]]^3)/9

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 x^{7/2} \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac{2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{3} (2 b) \int x^{9/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac{4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3+\frac{1}{33} \left (8 b^2\right ) \int x^{11/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac{16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac{1}{429} \left (16 b^3\right ) \int x^{13/2} \, dx\\ &=-\frac{32 b^3 x^{15/2}}{6435}+\frac{16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac{2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3\\ \end{align*}

Mathematica [A]  time = 0.0289775, size = 57, normalized size = 0.83 \[ -\frac{2 x^{9/2} \left (-120 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+390 b x \tanh ^{-1}(\tanh (a+b x))^2-715 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{6435} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(-2*x^(9/2)*(16*b^3*x^3 - 120*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 390*b*x*ArcTanh[Tanh[a + b*x]]^2 - 715*ArcTanh[
Tanh[a + b*x]]^3))/6435

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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}}{9}{x}^{{\frac{9}{2}}}}-{\frac{4\,b}{3} \left ({\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{11}{x}^{{\frac{11}{2}}}}-{\frac{4\,b}{11} \left ({\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{13}{x}^{{\frac{13}{2}}}}-{\frac{2\,b}{195}{x}^{{\frac{15}{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*arctanh(tanh(b*x+a))^3,x)

[Out]

2/9*x^(9/2)*arctanh(tanh(b*x+a))^3-4/3*b*(1/11*x^(11/2)*arctanh(tanh(b*x+a))^2-4/11*b*(1/13*x^(13/2)*arctanh(t
anh(b*x+a))-2/195*x^(15/2)*b))

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Maxima [A]  time = 1.06234, size = 74, normalized size = 1.07 \begin{align*} -\frac{4}{33} \, b x^{\frac{11}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac{2}{9} \, x^{\frac{9}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{16}{6435} \,{\left (2 \, b^{2} x^{\frac{15}{2}} - 15 \, b x^{\frac{13}{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(x^(7/2)*arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-4/33*b*x^(11/2)*arctanh(tanh(b*x + a))^2 + 2/9*x^(9/2)*arctanh(tanh(b*x + a))^3 - 16/6435*(2*b^2*x^(15/2) - 1
5*b*x^(13/2)*arctanh(tanh(b*x + a)))*b

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Fricas [A]  time = 1.92607, size = 105, normalized size = 1.52 \begin{align*} \frac{2}{6435} \,{\left (429 \, b^{3} x^{7} + 1485 \, a b^{2} x^{6} + 1755 \, a^{2} b x^{5} + 715 \, a^{3} x^{4}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

2/6435*(429*b^3*x^7 + 1485*a*b^2*x^6 + 1755*a^2*b*x^5 + 715*a^3*x^4)*sqrt(x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*atanh(tanh(b*x+a))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.14243, size = 47, normalized size = 0.68 \begin{align*} \frac{2}{15} \, b^{3} x^{\frac{15}{2}} + \frac{6}{13} \, a b^{2} x^{\frac{13}{2}} + \frac{6}{11} \, a^{2} b x^{\frac{11}{2}} + \frac{2}{9} \, a^{3} x^{\frac{9}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

2/15*b^3*x^(15/2) + 6/13*a*b^2*x^(13/2) + 6/11*a^2*b*x^(11/2) + 2/9*a^3*x^(9/2)