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

Optimal. Leaf size=38 \[ \frac {2 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac {2 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (4*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2)

Rule 30

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

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

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

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Mathematica [A]  time = 0.08, size = 32, normalized size = 0.84 \[ \frac {2 \left (7 b x-2 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 41, normalized size = 1.08 \[ \frac {2 \, {\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{35 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*x^3 + 8*a*b^2*x^2 + a^2*b*x - 2*a^3)*sqrt(b*x + a)/b^2

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giac [B]  time = 0.55, size = 131, normalized size = 3.45 \[ \frac {\sqrt {2} {\left (\frac {35 \, \sqrt {2} {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2}}{b} + \frac {14 \, \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a}{b} + \frac {3 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}}{b}\right )}}{105 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*sqrt(2)*(35*sqrt(2)*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2/b + 14*sqrt(2)*(3*(b*x + a)^(5/2) - 10*(b*
x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a/b + 3*sqrt(2)*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a
)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)/b)/b

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maple [A]  time = 0.14, size = 42, normalized size = 1.11 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 31, normalized size = 0.82 \[ \frac {2 \, {\left (5 \, b^{2} x^{2} + 3 \, a b x - 2 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{35 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.08, size = 823, normalized size = 21.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(tanh(a + b*x))^(3/2),x)

[Out]

(2*b*x^3*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)
)/7 + (x^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/
2)*((12*b*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x
))/7 - 2*b*(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*b) + (x*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^
(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
+ (4*((12*b*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 +
b*x))/7 - 2*b*(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))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(5
*b)))/(3*b) + (2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/
2)^(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 + (4*((12*b*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2
+ b*x))/7 - 2*b*(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))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))
/(5*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))
/(3*b^2)

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sympy [A]  time = 26.90, size = 49, normalized size = 1.29 \[ \begin {cases} \frac {2 x \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} - \frac {4 \operatorname {atanh}^{\frac {7}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{35 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\relax (a )} \right )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*x*atanh(tanh(a + b*x))**(5/2)/(5*b) - 4*atanh(tanh(a + b*x))**(7/2)/(35*b**2), Ne(b, 0)), (x**2*a
tanh(tanh(a))**(3/2)/2, True))

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