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

Optimal. Leaf size=128 \[ -\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{2 b^2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

(-2*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(7/2
) + (2*b^2)/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (2*b)/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) +
2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0840996, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2163, 2162} \[ -\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{2 b^2}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(7/2
) + (2*b^2)/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (2*b)/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) +
2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

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

Mathematica [A]  time = 0.146365, size = 107, normalized size = 0.84 \[ \frac{2 \left (-11 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+23 b^2 x^2\right )}{15 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac{2 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])
^(7/2) + (2*(23*b^2*x^2 - 11*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2)*(b*x - ArcT
anh[Tanh[a + b*x]])^3)

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Maple [A]  time = 0.134, size = 120, normalized size = 0.9 \begin{align*} -{\frac{2}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx}{x}^{-{\frac{5}{2}}}}-2\,{\frac{{b}^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\sqrt{x}}}+{\frac{2\,b}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}{x}^{-{\frac{3}{2}}}}-2\,{\frac{{b}^{3}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.2064, size = 336, normalized size = 2.62 \begin{align*} \left [\frac{15 \, b^{2} x^{3} \sqrt{-\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) - 2 \,{\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt{x}}{15 \, a^{3} x^{3}}, \frac{2 \,{\left (15 \, b^{2} x^{3} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt{x}\right )}}{15 \, a^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*b^2*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(15*b^2*x^2 - 5*a*b*x + 3*a
^2)*sqrt(x))/(a^3*x^3), 2/15*(15*b^2*x^3*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*b^2*x^2 - 5*a*b*x + 3
*a^2)*sqrt(x))/(a^3*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)),x)

[Out]

Timed out

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Giac [A]  time = 1.15843, size = 70, normalized size = 0.55 \begin{align*} -\frac{2 \, b^{3} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{3}} - \frac{2 \,{\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} 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)),x, algorithm="giac")

[Out]

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