3.213 \(\int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=201 \[ \frac{5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{35 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac{35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{35 b}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4} \]

[Out]

(-35*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(9/2)) + (35*b)/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^4) + 35/(12*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3
) + 7/(4*b*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + 5/(4*b^2*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2
*b*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2) + 5/(4*b^2*x^(7/2)*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.158671, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ \frac{5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{35 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac{35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{35 b}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

(-35*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(9/2)) + (35*b)/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^4) + 35/(12*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3
) + 7/(4*b*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + 5/(4*b^2*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2
*b*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2) + 5/(4*b^2*x^(7/2)*ArcTanh[Tanh[a + b*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]))

Rule 2163

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

Rule 2162

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-2*ArcTanh[Sqrt
[v]/Rt[-((b*u - a*v)/a), 2]])/(a*Rt[-((b*u - a*v)/a), 2]), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /;
PiecewiseLinearQ[u, v, x]

Rubi steps

\begin{align*} \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{5 \int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{35 \int \frac{1}{x^{9/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac{5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{35 \int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{35 \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{(35 b) \int \frac{1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac{35 b}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (35 b^2\right ) \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=-\frac{35 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac{35 b}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{35}{12 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{7}{4 b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{5}{4 b^2 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{5}{4 b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end{align*}

Mathematica [A]  time = 0.255728, size = 156, normalized size = 0.78 \[ \frac{1}{12} \left (\frac{6 b^2 \sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}+\frac{33 b^2 \sqrt{x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}+\frac{105 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}}+\frac{80 b x-8 \tanh ^{-1}(\tanh (a+b x))}{x^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

((105*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]
])^(9/2) + (80*b*x - 8*ArcTanh[Tanh[a + b*x]])/(x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (33*b^2*Sqrt[x]
)/(ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (6*b^2*Sqrt[x])/(ArcTanh[Tanh[a + b*x]]^2*(-(
b*x) + ArcTanh[Tanh[a + b*x]])^3))/12

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Maple [A]  time = 0.142, size = 207, normalized size = 1. \begin{align*}{\frac{11\,{b}^{3}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{13\,a{b}^{2}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}+{\frac{13\,{b}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}+{\frac{35\,{b}^{2}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}}-{\frac{2}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}{x}^{-{\frac{3}{2}}}}+6\,{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x)

[Out]

11/4/(arctanh(tanh(b*x+a))-b*x)^4*b^3/arctanh(tanh(b*x+a))^2*x^(3/2)+13/4/(arctanh(tanh(b*x+a))-b*x)^4*b^2/arc
tanh(tanh(b*x+a))^2*a*x^(1/2)+13/4/(arctanh(tanh(b*x+a))-b*x)^4*b^2/arctanh(tanh(b*x+a))^2*x^(1/2)*(arctanh(ta
nh(b*x+a))-b*x-a)+35/4/(arctanh(tanh(b*x+a))-b*x)^4*b^2/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/
((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))-2/3/(arctanh(tanh(b*x+a))-b*x)^3/x^(3/2)+6/(arctanh(tanh(b*x+a))-b*x)^4*
b/x^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.12883, size = 545, normalized size = 2.71 \begin{align*} \left [\frac{105 \,{\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x + 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) + 2 \,{\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{x}}{24 \,{\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, -\frac{105 \,{\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{x}}{12 \,{\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

[1/24*(105*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) +
2*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), -1/12*(10
5*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (105*b^3*x^3 + 175*a*b^2*x^2
+ 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^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(1/x**(5/2)/atanh(tanh(b*x+a))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15018, size = 96, normalized size = 0.48 \begin{align*} \frac{35 \, b^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{4}} + \frac{2 \,{\left (9 \, b x - a\right )}}{3 \, a^{4} x^{\frac{3}{2}}} + \frac{11 \, b^{3} x^{\frac{3}{2}} + 13 \, a b^{2} \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

35/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + 2/3*(9*b*x - a)/(a^4*x^(3/2)) + 1/4*(11*b^3*x^(3/2) + 1
3*a*b^2*sqrt(x))/((b*x + a)^2*a^4)