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

Optimal. Leaf size=63 \[ -16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+\frac {32}{5} b^3 x^{5/2} \]

[Out]

32/5*b^3*x^(5/2)-16*b^2*x^(3/2)*arctanh(tanh(b*x+a))-2*arctanh(tanh(b*x+a))^3/x^(1/2)+12*b*arctanh(tanh(b*x+a)
)^2*x^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, 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} \[ -16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+\frac {32}{5} b^3 x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^3*x^(5/2))/5 - 16*b^2*x^(3/2)*ArcTanh[Tanh[a + b*x]] + 12*b*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2 - (2*ArcTan
h[Tanh[a + b*x]]^3)/Sqrt[x]

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 \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^{3/2}} \, dx &=-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+(6 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{\sqrt {x}} \, dx\\ &=12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}-\left (24 b^2\right ) \int \sqrt {x} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+\left (16 b^3\right ) \int x^{3/2} \, dx\\ &=\frac {32}{5} b^3 x^{5/2}-16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.90 \[ \frac {2 \left (-40 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+30 b x \tanh ^{-1}(\tanh (a+b x))^2-5 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{5 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(16*b^3*x^3 - 40*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 30*b*x*ArcTanh[Tanh[a + b*x]]^2 - 5*ArcTanh[Tanh[a + b*x]
]^3))/(5*Sqrt[x])

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fricas [A]  time = 0.63, size = 34, normalized size = 0.54 \[ \frac {2 \, {\left (b^{3} x^{3} + 5 \, a b^{2} x^{2} + 15 \, a^{2} b x - 5 \, a^{3}\right )}}{5 \, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^3*x^3 + 5*a*b^2*x^2 + 15*a^2*b*x - 5*a^3)/sqrt(x)

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giac [A]  time = 0.42, size = 35, normalized size = 0.56 \[ \frac {2}{5} \, b^{3} x^{\frac {5}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} + 6 \, a^{2} b \sqrt {x} - \frac {2 \, a^{3}}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 64, normalized size = 1.02 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{\sqrt {x}}+12 b \left (\frac {b^{2} x^{\frac {5}{2}}}{5}+\frac {2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b \,x^{\frac {3}{2}}}{3}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctanh(tanh(b*x+a))^3/x^(1/2)+12*b*(1/5*b^2*x^(5/2)+2/3*(arctanh(tanh(b*x+a))-b*x)*b*x^(3/2)+(arctanh(tanh
(b*x+a))-b*x)^2*x^(1/2))

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maxima [A]  time = 0.35, size = 55, normalized size = 0.87 \[ 12 \, b \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{\sqrt {x}} + \frac {16}{5} \, {\left (2 \, b^{2} x^{\frac {5}{2}} - 5 \, b x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*b*sqrt(x)*arctanh(tanh(b*x + a))^2 - 2*arctanh(tanh(b*x + a))^3/sqrt(x) + 16/5*(2*b^2*x^(5/2) - 5*b*x^(3/2)
*arctanh(tanh(b*x + a)))*b

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mupad [B]  time = 1.21, size = 182, normalized size = 2.89 \[ \frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,\sqrt {x}}+\frac {2\,b^3\,x^{5/2}}{5}+\frac {3\,b\,\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2}-b^2\,x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^3/x^(3/2),x)

[Out]

(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)^3/(4*x^(1/
2)) + (2*b^3*x^(5/2))/5 + (3*b*x^(1/2)*(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)/2 - b^2*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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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