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

Optimal. Leaf size=65 \[ \frac {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32}{35} b^3 x^{7/2} \]

[Out]

-32/35*b^3*x^(7/2)+16/5*b^2*x^(5/2)*arctanh(tanh(b*x+a))-4*b*x^(3/2)*arctanh(tanh(b*x+a))^2+2*arctanh(tanh(b*x
+a))^3*x^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 65, 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} \[ \frac {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32}{35} b^3 x^{7/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*b^3*x^(7/2))/35 + (16*b^2*x^(5/2)*ArcTanh[Tanh[a + b*x]])/5 - 4*b*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2 + 2*Sq
rt[x]*ArcTanh[Tanh[a + b*x]]^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 \frac {\tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}} \, dx &=2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3-(6 b) \int \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3+\left (8 b^2\right ) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3-\frac {1}{5} \left (16 b^3\right ) \int x^{5/2} \, dx\\ &=-\frac {32}{35} b^3 x^{7/2}+\frac {16}{5} b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))-4 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.88 \[ \frac {2}{35} \sqrt {x} \left (56 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-70 b x \tanh ^{-1}(\tanh (a+b x))^2+35 \tanh ^{-1}(\tanh (a+b x))^3-16 b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(-16*b^3*x^3 + 56*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 70*b*x*ArcTanh[Tanh[a + b*x]]^2 + 35*ArcTanh[Tan
h[a + b*x]]^3))/35

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fricas [A]  time = 0.53, size = 35, normalized size = 0.54 \[ \frac {2}{35} \, {\left (5 \, b^{3} x^{3} + 21 \, a b^{2} x^{2} + 35 \, a^{2} b x + 35 \, a^{3}\right )} \sqrt {x} \]

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]

2/35*(5*b^3*x^3 + 21*a*b^2*x^2 + 35*a^2*b*x + 35*a^3)*sqrt(x)

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giac [A]  time = 0.29, size = 35, normalized size = 0.54 \[ \frac {2}{7} \, b^{3} x^{\frac {7}{2}} + \frac {6}{5} \, a b^{2} x^{\frac {5}{2}} + 2 \, a^{2} b x^{\frac {3}{2}} + 2 \, a^{3} \sqrt {x} \]

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]

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

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maple [A]  time = 0.25, size = 69, normalized size = 1.06 \[ \frac {2 b^{3} x^{\frac {7}{2}}}{7}+\frac {6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b^{2} x^{\frac {5}{2}}}{5}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} b \,x^{\frac {3}{2}}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b^3*x^(7/2)+6/5*(arctanh(tanh(b*x+a))-b*x)*b^2*x^(5/2)+2*(arctanh(tanh(b*x+a))-b*x)^2*b*x^(3/2)+2*(arctanh
(tanh(b*x+a))-b*x)^3*x^(1/2)

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maxima [A]  time = 0.36, size = 55, normalized size = 0.85 \[ -4 \, b x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + 2 \, \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {16}{35} \, {\left (2 \, b^{2} x^{\frac {7}{2}} - 7 \, b x^{\frac {5}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*b*x^(3/2)*arctanh(tanh(b*x + a))^2 + 2*sqrt(x)*arctanh(tanh(b*x + a))^3 - 16/35*(2*b^2*x^(7/2) - 7*b*x^(5/2
)*arctanh(tanh(b*x + a)))*b

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mupad [B]  time = 1.19, size = 182, normalized size = 2.80 \[ \frac {2\,b^3\,x^{7/2}}{7}-\frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4}+\frac {b\,x^{3/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2}-\frac {3\,b^2\,x^{5/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^3/x^(1/2),x)

[Out]

(2*b^3*x^(7/2))/7 - (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)^3)/4 + (b*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)/2 - (3*b^2*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 \[ \int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{\sqrt {x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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