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

Optimal. Leaf size=80 \[ -\frac {32 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2157, 30} \[ -\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}+\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^3*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (4*x^2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b^2) + (16*x*ArcTanh[Tanh
[a + b*x]]^(7/2))/(35*b^3) - (32*ArcTanh[Tanh[a + b*x]]^(9/2))/(315*b^4)

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

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.82 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (-126 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+72 b x \tanh ^{-1}(\tanh (a+b x))^2-16 \tanh ^{-1}(\tanh (a+b x))^3+105 b^3 x^3\right )}{315 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(105*b^3*x^3 - 126*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 72*b*x*ArcTanh[Tanh[a + b*
x]]^2 - 16*ArcTanh[Tanh[a + b*x]]^3))/(315*b^4)

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fricas [A]  time = 0.46, size = 53, normalized size = 0.66 \[ \frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt(b*x + a)/b^4

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giac [A]  time = 0.19, size = 125, normalized size = 1.56 \[ \frac {\sqrt {2} {\left (\frac {9 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a}{b^{3}} + \frac {\sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )}}{b^{3}}\right )}}{315 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*sqrt(2)*(9*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^3 + 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^3)/b

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maple [A]  time = 0.15, size = 124, normalized size = 1.55 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (-3 \arctanh \left (\tanh \left (b x +a \right )\right )+3 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )+\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 53, normalized size = 0.66 \[ \frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt(b*x + a)/b^4

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mupad [B]  time = 0.99, size = 648, normalized size = 8.10 \[ \frac {2\,x^4\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{9}-\frac {x^3\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}+\frac {2\,b\,x}{9}\right )}{7\,b}-\frac {16\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\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\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}+\frac {2\,b\,x}{9}\right )}{35\,b^4}-\frac {6\,x^2\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\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 )\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}+\frac {2\,b\,x}{9}\right )}{35\,b^2}-\frac {8\,x\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\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\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}+\frac {2\,b\,x}{9}\right )}{35\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atanh(tanh(a + b*x))^(1/2),x)

[Out]

(2*x^4*(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))/
9 - (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)
*(log(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/9 + (2*b*x)/9))/
(7*b) - (16*(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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^3*(l
og(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/9 + (2*b*x)/9))/(35
*b^4) - (6*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)*(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)*(
log(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/9 + (2*b*x)/9))/(3
5*b^2) - (8*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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^2*
(log(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/9 + (2*b*x)/9))/(
35*b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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