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

Optimal. Leaf size=172 \[ \frac{7 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac{7 b^2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}-\frac{7 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]  time = 0.136852, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ \frac{7 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac{7 b^2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}-\frac{7 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.266968, size = 139, normalized size = 0.81 \[ -\frac{2 \left (-16 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+58 b^2 x^2\right )}{15 x^{5/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}-\frac{b^3 \sqrt{x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}-\frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])
^(9/2) - (b^3*Sqrt[x])/(ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) - (2*(58*b^2*x^2 - 16*b*x*
ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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Maple [A]  time = 0.178, size = 151, normalized size = 0.9 \begin{align*} -{\frac{{b}^{3}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}-7\,{\frac{{b}^{3}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}} \right ) }-{\frac{2}{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}{x}^{-{\frac{5}{2}}}}-6\,{\frac{{b}^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}\sqrt{x}}}+{\frac{4\,b}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(arctanh(tanh(b*x+a))-b*x)^4*b^3*x^(1/2)/arctanh(tanh(b*x+a))-7/(arctanh(tanh(b*x+a))-b*x)^4*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))-2/5/(arctanh(tanh(b*x+a))-b*
x)^2/x^(5/2)-6/(arctanh(tanh(b*x+a))-b*x)^4*b^2/x^(1/2)+4/3/(arctanh(tanh(b*x+a))-b*x)^3*b/x^(3/2)

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.19201, size = 460, normalized size = 2.67 \begin{align*} \left [\frac{105 \,{\left (b^{3} x^{4} + a 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 (105 \, b^{3} x^{3} + 70 \, a b^{2} x^{2} - 14 \, a^{2} b x + 6 \, a^{3}\right )} \sqrt{x}}{30 \,{\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, \frac{105 \,{\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (105 \, b^{3} x^{3} + 70 \, a b^{2} x^{2} - 14 \, a^{2} b x + 6 \, a^{3}\right )} \sqrt{x}}{15 \,{\left (a^{4} b x^{4} + a^{5} 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))^2,x, algorithm="fricas")

[Out]

[1/30*(105*(b^3*x^4 + a*b^2*x^3)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(105*b^3*x^3
+ 70*a*b^2*x^2 - 14*a^2*b*x + 6*a^3)*sqrt(x))/(a^4*b*x^4 + a^5*x^3), 1/15*(105*(b^3*x^4 + a*b^2*x^3)*sqrt(b/a
)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (105*b^3*x^3 + 70*a*b^2*x^2 - 14*a^2*b*x + 6*a^3)*sqrt(x))/(a^4*b*x^4 + a^
5*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.12369, size = 95, normalized size = 0.55 \begin{align*} -\frac{7 \, b^{3} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{4}} - \frac{b^{3} \sqrt{x}}{{\left (b x + a\right )} a^{4}} - \frac{2 \,{\left (45 \, b^{2} x^{2} - 10 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{4} 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))^2,x, algorithm="giac")

[Out]

-7*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - b^3*sqrt(x)/((b*x + a)*a^4) - 2/15*(45*b^2*x^2 - 10*a*b*x
+ 3*a^2)/(a^4*x^(5/2))