3.214 \(\int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=228 \[ \frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{63 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}+\frac{63 b^2}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac{21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

(-63*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(11/2)) + (63*b^2)/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^5) + (21*b)/(4*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*
x]])^4) + 63/(20*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + 9/(4*b*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2)
+ 7/(4*b^2*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2*b*x^(7/2)*ArcTanh[Tanh[a + b*x]]^2) + 7/(4*b^2*x^(9/
2)*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.20206, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ \frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{63 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}+\frac{63 b^2}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac{21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

(-63*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(11/2)) + (63*b^2)/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^5) + (21*b)/(4*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*
x]])^4) + 63/(20*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + 9/(4*b*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2)
+ 7/(4*b^2*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2*b*x^(7/2)*ArcTanh[Tanh[a + b*x]]^2) + 7/(4*b^2*x^(9/
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^{7/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{7 \int \frac{1}{x^{9/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{63 \int \frac{1}{x^{11/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{63 \int \frac{1}{x^{9/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{63 \int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{(63 b) \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac{21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{\left (63 b^2\right ) \int \frac{1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^4}\\ &=\frac{63 b^2}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac{21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{\left (63 b^3\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 )^4}\\ &=-\frac{63 b^{5/2} \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 )^{11/2}}+\frac{63 b^2}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac{21 b}{4 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{63}{20 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{9}{4 b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{7}{4 b^2 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7}{4 b^2 x^{9/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end{align*}

Mathematica [A]  time = 0.327386, size = 174, normalized size = 0.76 \[ \frac{1}{20} \left (\frac{8 \left (-7 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2+36 b^2 x^2\right )}{x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5}+\frac{75 b^3 \sqrt{x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \tanh ^{-1}(\tanh (a+b x))}-\frac{10 b^3 \sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}-\frac{315 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{11/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

((75*b^3*Sqrt[x])/((b*x - ArcTanh[Tanh[a + b*x]])^5*ArcTanh[Tanh[a + b*x]]) - (315*b^(5/2)*ArcTan[(Sqrt[b]*Sqr
t[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(11/2) - (10*b^3*Sqrt[x])/(Arc
Tanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (8*(36*b^2*x^2 - 7*b*x*ArcTanh[Tanh[a + b*x]] + A
rcTanh[Tanh[a + b*x]]^2))/(x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^5))/20

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Maple [A]  time = 0.145, size = 229, normalized size = 1. \begin{align*} -{\frac{2}{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}{x}^{-{\frac{5}{2}}}}-12\,{\frac{{b}^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{5}\sqrt{x}}}+2\,{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}{x}^{3/2}}}-{\frac{15\,{b}^{4}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{5} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{17\,a{b}^{3}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{5} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}-{\frac{17\,{b}^{3} \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 ) ^{5} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}-{\frac{63\,{b}^{3}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{5}}\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/arctanh(tanh(b*x+a))^3,x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)^3/x^(5/2)-12/(arctanh(tanh(b*x+a))-b*x)^5*b^2/x^(1/2)+2/(arctanh(tanh(b*x+a))-
b*x)^4*b/x^(3/2)-15/4/(arctanh(tanh(b*x+a))-b*x)^5*b^4/arctanh(tanh(b*x+a))^2*x^(3/2)-17/4/(arctanh(tanh(b*x+a
))-b*x)^5*b^3/arctanh(tanh(b*x+a))^2*a*x^(1/2)-17/4/(arctanh(tanh(b*x+a))-b*x)^5*b^3/arctanh(tanh(b*x+a))^2*x^
(1/2)*(arctanh(tanh(b*x+a))-b*x-a)-63/4/(arctanh(tanh(b*x+a))-b*x)^5*b^3/((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))

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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^(7/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.16307, size = 598, normalized size = 2.62 \begin{align*} \left [\frac{315 \,{\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\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 (315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{x}}{40 \,{\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, \frac{315 \,{\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{x}}{20 \,{\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

[1/40*(315*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a))
- 2*(315*b^4*x^4 + 525*a*b^3*x^3 + 168*a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4)*sqrt(x))/(a^5*b^2*x^5 + 2*a^6*b*x^4 +
a^7*x^3), 1/20*(315*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (315*b^
4*x^4 + 525*a*b^3*x^3 + 168*a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4)*sqrt(x))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)]

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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**(7/2)/atanh(tanh(b*x+a))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.14484, size = 108, normalized size = 0.47 \begin{align*} -\frac{63 \, b^{3} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{5}} - \frac{15 \, b^{4} x^{\frac{3}{2}} + 17 \, a b^{3} \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{5}} - \frac{2 \,{\left (30 \, b^{2} x^{2} - 5 \, a b x + a^{2}\right )}}{5 \, a^{5} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

-63/4*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 1/4*(15*b^4*x^(3/2) + 17*a*b^3*sqrt(x))/((b*x + a)^2*a
^5) - 2/5*(30*b^2*x^2 - 5*a*b*x + a^2)/(a^5*x^(5/2))