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

Optimal. Leaf size=48 \[ -\frac {8}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))+\frac {2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {16}{315} b^2 x^{9/2} \]

[Out]

16/315*b^2*x^(9/2)-8/35*b*x^(7/2)*arctanh(tanh(b*x+a))+2/5*x^(5/2)*arctanh(tanh(b*x+a))^2

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -\frac {8}{35} b x^{7/2} \tanh ^{-1}(\tanh (a+b x))+\frac {2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {16}{315} b^2 x^{9/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*x^(9/2))/315 - (8*b*x^(7/2)*ArcTanh[Tanh[a + b*x]])/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)/5

Rule 30

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

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]))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 40, normalized size = 0.83 \[ \frac {2}{315} x^{5/2} \left (-36 b x \tanh ^{-1}(\tanh (a+b x))+63 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(8*b^2*x^2 - 36*b*x*ArcTanh[Tanh[a + b*x]] + 63*ArcTanh[Tanh[a + b*x]]^2))/315

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fricas [A]  time = 0.47, size = 29, normalized size = 0.60 \[ \frac {2}{315} \, {\left (35 \, b^{2} x^{4} + 90 \, a b x^{3} + 63 \, a^{2} x^{2}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^2*x^4 + 90*a*b*x^3 + 63*a^2*x^2)*sqrt(x)

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giac [A]  time = 0.15, size = 24, normalized size = 0.50 \[ \frac {2}{9} \, b^{2} x^{\frac {9}{2}} + \frac {4}{7} \, a b x^{\frac {7}{2}} + \frac {2}{5} \, a^{2} x^{\frac {5}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*b^2*x^(9/2) + 4/7*a*b*x^(7/2) + 2/5*a^2*x^(5/2)

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maple [A]  time = 0.25, size = 38, normalized size = 0.79 \[ \frac {2 x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {8 b \left (\frac {x^{\frac {7}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{7}-\frac {2 b \,x^{\frac {9}{2}}}{63}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^(5/2)*arctanh(tanh(b*x+a))^2-8/5*b*(1/7*x^(7/2)*arctanh(tanh(b*x+a))-2/63*b*x^(9/2))

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maxima [A]  time = 0.35, size = 36, normalized size = 0.75 \[ \frac {16}{315} \, b^{2} x^{\frac {9}{2}} - \frac {8}{35} \, b x^{\frac {7}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) + \frac {2}{5} \, x^{\frac {5}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/315*b^2*x^(9/2) - 8/35*b*x^(7/2)*arctanh(tanh(b*x + a)) + 2/5*x^(5/2)*arctanh(tanh(b*x + a))^2

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mupad [B]  time = 1.14, size = 122, normalized size = 2.54 \[ \frac {x^{5/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{10}+\frac {2\,b^2\,x^{9/2}}{9}-\frac {2\,b\,x^{7/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*atanh(tanh(a + b*x))^2,x)

[Out]

(x^(5/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2
)/10 + (2*b^2*x^(9/2))/9 - (2*b*x^(7/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2
*a)*exp(2*b*x) + 1)) + 2*b*x))/7

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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