3.113 \(\int x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx\)

Optimal. Leaf size=59 \[ \frac {16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

[Out]

2/3*x^2*arctanh(tanh(b*x+a))^(3/2)/b-8/15*x*arctanh(tanh(b*x+a))^(5/2)/b^2+16/105*arctanh(tanh(b*x+a))^(7/2)/b
^3

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2157, 30} \[ \frac {16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (8*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(15*b^2) + (16*ArcTanh[Tanh[a
+ b*x]]^(7/2))/(105*b^3)

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^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \int x \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b}\\ &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {8 \int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{15 b^2}\\ &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {8 \operatorname {Subst}\left (\int x^{5/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{15 b^3}\\ &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.83 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (-28 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+35 b^2 x^2\right )}{105 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(35*b^2*x^2 - 28*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/(10
5*b^3)

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fricas [A]  time = 0.62, size = 42, normalized size = 0.71 \[ \frac {2 \, {\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{105 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x + a)/b^3

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giac [B]  time = 0.15, size = 102, normalized size = 1.73 \[ \frac {\sqrt {2} {\left (\frac {7 \, \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a}{b^{2}} + \frac {3 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}}{b^{2}}\right )}}{105 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 69, normalized size = 1.17 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 42, normalized size = 0.71 \[ \frac {2 \, {\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{105 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x + a)/b^3

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mupad [B]  time = 0.99, size = 485, normalized size = 8.22 \[ \frac {2\,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}}}{7}-\frac {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 )}{7}-\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 )}{7}+\frac {2\,b\,x}{7}\right )}{5\,b}-\frac {8\,\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 )}{7}-\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 )}{7}+\frac {2\,b\,x}{7}\right )}{15\,b^3}-\frac {4\,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 )\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{7}-\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 )}{7}+\frac {2\,b\,x}{7}\right )}{15\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(tanh(a + b*x))^(1/2),x)

[Out]

(2*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)
*(log(2/(exp(2*a)*exp(2*b*x) + 1))/7 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/7 + (2*b*x)/7))/
(5*b) - (8*(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*(lo
g(2/(exp(2*a)*exp(2*b*x) + 1))/7 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/7 + (2*b*x)/7))/(15*
b^3) - (4*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)*(log
(2/(exp(2*a)*exp(2*b*x) + 1))/7 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/7 + (2*b*x)/7))/(15*b
^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(atanh(tanh(a + b*x))), x)

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