3.132 \(\int x \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=38 \[ \frac {2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2} \]

[Out]

2/7*x*arctanh(tanh(b*x+a))^(7/2)/b-4/63*arctanh(tanh(b*x+a))^(9/2)/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} \[ \frac {2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(2*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (4*ArcTanh[Tanh[a + b*x]]^(9/2))/(63*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))^{5/2} \, dx &=\frac {2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {2 \int \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{7 b}\\ &=\frac {2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {2 \operatorname {Subst}\left (\int x^{7/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{7 b^2}\\ &=\frac {2 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 32, normalized size = 0.84 \[ \frac {2 \left (9 b x-2 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(2*(9*b*x - 2*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*b^2)

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fricas [A]  time = 0.60, size = 52, normalized size = 1.37 \[ \frac {2 \, {\left (7 \, b^{4} x^{4} + 19 \, a b^{3} x^{3} + 15 \, a^{2} b^{2} x^{2} + a^{3} b x - 2 \, a^{4}\right )} \sqrt {b x + a}}{63 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/63*(7*b^4*x^4 + 19*a*b^3*x^3 + 15*a^2*b^2*x^2 + a^3*b*x - 2*a^4)*sqrt(b*x + a)/b^2

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giac [B]  time = 0.18, size = 197, normalized size = 5.18 \[ \frac {\sqrt {2} {\left (\frac {105 \, \sqrt {2} {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{3}}{b} + \frac {63 \, \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{2}}{b} + \frac {27 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a}{b} + \frac {\sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )}}{b}\right )}}{315 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

1/315*sqrt(2)*(105*sqrt(2)*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^3/b + 63*sqrt(2)*(3*(b*x + a)^(5/2) - 10*(b
*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a^2/b + 27*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)*a/b + sqrt(2)*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x +
a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)/b)/b

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maple [A]  time = 0.14, size = 42, normalized size = 1.11 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(tanh(b*x+a))^(5/2),x)

[Out]

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

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maxima [A]  time = 0.52, size = 31, normalized size = 0.82 \[ \frac {2 \, {\left (7 \, b^{2} x^{2} + 5 \, a b x - 2 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{63 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/63*(7*b^2*x^2 + 5*a*b*x - 2*a^2)*(b*x + a)^(5/2)/b^2

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mupad [B]  time = 1.11, size = 773, normalized size = 20.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(tanh(a + b*x))^(5/2),x)

[Out]

(log(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3*(log((exp(2*a)*exp(2*
b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(63*b^2) - (log((exp(2*a)*exp(
2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^4*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)
*exp(2*b*x) + 1))/2)^(1/2))/(252*b^2) - (log(1/(exp(2*a)*exp(2*b*x) + 1))^4*(log((exp(2*a)*exp(2*b*x))/(exp(2*
a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(252*b^2) + (log(1/(exp(2*a)*exp(2*b*x) + 1
))^3*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)
)/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(63*b^2) - (x*log(1/(exp(2*a)*exp(2*b*x) + 1))^3*(log((exp(2*
a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(28*b) + (x*log((exp(
2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/
(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(28*b) - (log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log((exp(2*a)*exp(2*b*x))/(e
xp(2*a)*exp(2*b*x) + 1))^2*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x
) + 1))/2)^(1/2))/(42*b^2) - (3*x*log(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b
*x) + 1))^2*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2
))/(28*b) + (3*x*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*(log(
(exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(28*b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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