3.178 \(\int \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\)

Optimal. Leaf size=48 \[ -\frac {8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {16}{105} b^2 x^{7/2} \]

[Out]

16/105*b^2*x^(7/2)-8/15*b*x^(5/2)*arctanh(tanh(b*x+a))+2/3*x^(3/2)*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} \[ -\frac {8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {16}{105} b^2 x^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

(16*b^2*x^(7/2))/105 - (8*b*x^(5/2)*ArcTanh[Tanh[a + b*x]])/15 + (2*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2)/3

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 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2 \, dx &=\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2-\frac {1}{3} (4 b) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-\frac {8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {1}{15} \left (8 b^2\right ) \int x^{5/2} \, dx\\ &=\frac {16}{105} b^2 x^{7/2}-\frac {8}{15} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 40, normalized size = 0.83 \[ \frac {2}{105} x^{3/2} \left (-28 b x \tanh ^{-1}(\tanh (a+b x))+35 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

(2*x^(3/2)*(8*b^2*x^2 - 28*b*x*ArcTanh[Tanh[a + b*x]] + 35*ArcTanh[Tanh[a + b*x]]^2))/105

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fricas [A]  time = 0.54, size = 27, normalized size = 0.56 \[ \frac {2}{105} \, {\left (15 \, b^{2} x^{3} + 42 \, a b x^{2} + 35 \, a^{2} x\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*b^2*x^3 + 42*a*b*x^2 + 35*a^2*x)*sqrt(x)

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giac [A]  time = 0.74, size = 24, normalized size = 0.50 \[ \frac {2}{7} \, b^{2} x^{\frac {7}{2}} + \frac {4}{5} \, a b x^{\frac {5}{2}} + \frac {2}{3} \, a^{2} x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b^2*x^(7/2) + 4/5*a*b*x^(5/2) + 2/3*a^2*x^(3/2)

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maple [A]  time = 0.25, size = 38, normalized size = 0.79 \[ \frac {2 x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{3}-\frac {8 b \left (\frac {x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{5}-\frac {2 b \,x^{\frac {7}{2}}}{35}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^(3/2)*arctanh(tanh(b*x+a))^2-8/3*b*(1/5*x^(5/2)*arctanh(tanh(b*x+a))-2/35*b*x^(7/2))

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maxima [A]  time = 0.34, size = 36, normalized size = 0.75 \[ \frac {16}{105} \, b^{2} x^{\frac {7}{2}} - \frac {8}{15} \, b x^{\frac {5}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) + \frac {2}{3} \, x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/105*b^2*x^(7/2) - 8/15*b*x^(5/2)*arctanh(tanh(b*x + a)) + 2/3*x^(3/2)*arctanh(tanh(b*x + a))^2

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mupad [B]  time = 1.12, size = 122, normalized size = 2.54 \[ \frac {x^{3/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{6}+\frac {2\,b^2\,x^{7/2}}{7}-\frac {2\,b\,x^{5/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*atanh(tanh(a + b*x))^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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