3.231 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{13/2}} \, dx\)

Optimal. Leaf size=148 \[ \frac{32 b^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{1155 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

(32*b^3*ArcTanh[Tanh[a + b*x]]^(5/2))/(1155*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*ArcTanh[Tanh[a
+ b*x]]^(5/2))/(231*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (4*b*ArcTanh[Tanh[a + b*x]]^(5/2))/(33*x^(9/2
)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(5/2))/(11*x^(11/2)*(b*x - ArcTanh[Tanh[a + b*
x]]))

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Rubi [A]  time = 0.0755237, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac{32 b^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{1155 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(13/2),x]

[Out]

(32*b^3*ArcTanh[Tanh[a + b*x]]^(5/2))/(1155*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*ArcTanh[Tanh[a
+ b*x]]^(5/2))/(231*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (4*b*ArcTanh[Tanh[a + b*x]]^(5/2))/(33*x^(9/2
)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(5/2))/(11*x^(11/2)*(b*x - ArcTanh[Tanh[a + b*
x]]))

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{13/2}} \, dx &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(6 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx}{11 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (8 b^2\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{9/2}} \, dx}{33 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (16 b^3\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx}{231 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac{32 b^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{1155 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0661309, size = 82, normalized size = 0.55 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \left (-495 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+385 b x \tanh ^{-1}(\tanh (a+b x))^2-105 \tanh ^{-1}(\tanh (a+b x))^3+231 b^3 x^3\right )}{1155 x^{11/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(13/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(231*b^3*x^3 - 495*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 385*b*x*ArcTanh[Tanh[a + b
*x]]^2 - 105*ArcTanh[Tanh[a + b*x]]^3))/(1155*x^(11/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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Maple [A]  time = 0.153, size = 151, normalized size = 1. \begin{align*} -{\frac{2}{11\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -11\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{11}{2}}}}-{\frac{12\,b}{11\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -11\,bx} \left ( -{\frac{1}{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -9\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{9}{2}}}}-{\frac{4\,b}{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -9\,bx} \left ( -{\frac{1}{7\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -7\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{7}{2}}}}+{\frac{2\,b}{35\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{5}{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x)

[Out]

-2/11/(arctanh(tanh(b*x+a))-b*x)/x^(11/2)*arctanh(tanh(b*x+a))^(5/2)-12/11*b/(arctanh(tanh(b*x+a))-b*x)*(-1/9/
(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(5/2)-4/9*b/(arctanh(tanh(b*x+a))-b*x)*(-1/7/(arctanh(
tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(5/2)+2/35*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(5/2)*arctanh(tanh(
b*x+a))^(5/2)))

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Maxima [A]  time = 1.50009, size = 76, normalized size = 0.51 \begin{align*} \frac{2 \,{\left (16 \, b^{4} x^{4} - 24 \, a b^{3} x^{3} + 30 \, a^{2} b^{2} x^{2} - 35 \, a^{3} b x - 105 \, a^{4}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{1155 \, a^{4} x^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x, algorithm="maxima")

[Out]

2/1155*(16*b^4*x^4 - 24*a*b^3*x^3 + 30*a^2*b^2*x^2 - 35*a^3*b*x - 105*a^4)*(b*x + a)^(3/2)/(a^4*x^(11/2))

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Fricas [A]  time = 1.95807, size = 162, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (16 \, b^{5} x^{5} - 8 \, a b^{4} x^{4} + 6 \, a^{2} b^{3} x^{3} - 5 \, a^{3} b^{2} x^{2} - 140 \, a^{4} b x - 105 \, a^{5}\right )} \sqrt{b x + a}}{1155 \, a^{4} x^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x, algorithm="fricas")

[Out]

2/1155*(16*b^5*x^5 - 8*a*b^4*x^4 + 6*a^2*b^3*x^3 - 5*a^3*b^2*x^2 - 140*a^4*b*x - 105*a^5)*sqrt(b*x + a)/(a^4*x
^(11/2))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(3/2)/x**(13/2),x)

[Out]

Timed out

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Giac [A]  time = 1.3363, size = 131, normalized size = 0.89 \begin{align*} -\frac{\sqrt{2}{\left (\frac{231 \, \sqrt{2} b^{11}}{a} - 2 \,{\left (\frac{99 \, \sqrt{2} b^{11}}{a^{2}} + 4 \,{\left (\frac{2 \, \sqrt{2}{\left (b x + a\right )} b^{11}}{a^{4}} - \frac{11 \, \sqrt{2} b^{11}}{a^{3}}\right )}{\left (b x + a\right )}\right )}{\left (b x + a\right )}\right )}{\left (b x + a\right )}^{\frac{5}{2}} b}{1155 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{11}{2}}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x, algorithm="giac")

[Out]

-1/1155*sqrt(2)*(231*sqrt(2)*b^11/a - 2*(99*sqrt(2)*b^11/a^2 + 4*(2*sqrt(2)*(b*x + a)*b^11/a^4 - 11*sqrt(2)*b^
11/a^3)*(b*x + a))*(b*x + a))*(b*x + a)^(5/2)*b/(((b*x + a)*b - a*b)^(11/2)*abs(b))