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

Optimal. Leaf size=69 \[ \frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32}{315} b^3 x^{9/2} \]

[Out]

-32/315*b^3*x^(9/2)+16/35*b^2*x^(7/2)*arctanh(tanh(b*x+a))-4/5*b*x^(5/2)*arctanh(tanh(b*x+a))^2+2/3*x^(3/2)*ar
ctanh(tanh(b*x+a))^3

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Rubi [A]  time = 0.04, antiderivative size = 69, 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} \[ \frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32}{315} b^3 x^{9/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*b^3*x^(9/2))/315 + (16*b^2*x^(7/2)*ArcTanh[Tanh[a + b*x]])/35 - (4*b*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)/5
+ (2*x^(3/2)*ArcTanh[Tanh[a + b*x]]^3)/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))^3 \, dx &=\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-(2 b) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} \left (8 b^2\right ) \int x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {1}{35} \left (16 b^3\right ) \int x^{7/2} \, dx\\ &=-\frac {32}{315} b^3 x^{9/2}+\frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.83 \[ -\frac {2}{315} x^{3/2} \left (-72 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+126 b x \tanh ^{-1}(\tanh (a+b x))^2-105 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^(3/2)*(16*b^3*x^3 - 72*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 126*b*x*ArcTanh[Tanh[a + b*x]]^2 - 105*ArcTanh[T
anh[a + b*x]]^3))/315

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fricas [A]  time = 0.68, size = 38, normalized size = 0.55 \[ \frac {2}{315} \, {\left (35 \, b^{3} x^{4} + 135 \, a b^{2} x^{3} + 189 \, a^{2} b x^{2} + 105 \, a^{3} x\right )} \sqrt {x} \]

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]

2/315*(35*b^3*x^4 + 135*a*b^2*x^3 + 189*a^2*b*x^2 + 105*a^3*x)*sqrt(x)

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giac [A]  time = 0.16, size = 35, normalized size = 0.51 \[ \frac {2}{9} \, b^{3} x^{\frac {9}{2}} + \frac {6}{7} \, a b^{2} x^{\frac {7}{2}} + \frac {6}{5} \, a^{2} b x^{\frac {5}{2}} + \frac {2}{3} \, a^{3} x^{\frac {3}{2}} \]

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]

2/9*b^3*x^(9/2) + 6/7*a*b^2*x^(7/2) + 6/5*a^2*b*x^(5/2) + 2/3*a^3*x^(3/2)

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maple [A]  time = 0.26, size = 56, normalized size = 0.81 \[ \frac {2 x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3}-4 b \left (\frac {x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {4 b \left (\frac {x^{\frac {7}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{7}-\frac {2 b \,x^{\frac {9}{2}}}{63}\right )}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^(3/2)*arctanh(tanh(b*x+a))^3-4*b*(1/5*x^(5/2)*arctanh(tanh(b*x+a))^2-4/5*b*(1/7*x^(7/2)*arctanh(tanh(b*x
+a))-2/63*b*x^(9/2)))

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maxima [A]  time = 0.36, size = 55, normalized size = 0.80 \[ -\frac {4}{5} \, b x^{\frac {5}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {2}{3} \, x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {16}{315} \, {\left (2 \, b^{2} x^{\frac {9}{2}} - 9 \, b x^{\frac {7}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \]

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]

-4/5*b*x^(5/2)*arctanh(tanh(b*x + a))^2 + 2/3*x^(3/2)*arctanh(tanh(b*x + a))^3 - 16/315*(2*b^2*x^(9/2) - 9*b*x
^(7/2)*arctanh(tanh(b*x + a)))*b

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mupad [B]  time = 1.16, size = 182, normalized size = 2.64 \[ \frac {2\,b^3\,x^{9/2}}{9}-\frac {x^{3/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{12}+\frac {3\,b\,x^{5/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{10}-\frac {3\,b^2\,x^{7/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*atanh(tanh(a + b*x))^3,x)

[Out]

(2*b^3*x^(9/2))/9 - (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)^3)/12 + (3*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)^2)/10 - (3*b^2*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 \[ \int \sqrt {x} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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