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

Optimal. Leaf size=108 \[ \frac{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{b^{7/2}}-\frac{x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{5 x^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0764184, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2159, 2162} \[ \frac{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{b^{7/2}}-\frac{x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{5 x^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x^(3/2))/(3*b^2) + (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]]))/b^3 - (5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x -
ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2))/b^(7/2) - x^(5/2)/(b*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 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[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{x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac{x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{5 \int \frac{x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b}\\ &=\frac{5 x^{3/2}}{3 b^2}-\frac{x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b^2}\\ &=\frac{5 x^{3/2}}{3 b^2}+\frac{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b^3}\\ &=\frac{5 x^{3/2}}{3 b^2}+\frac{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{5 \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 )^{3/2}}{b^{7/2}}-\frac{x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}\\ \end{align*}

Mathematica [A]  time = 0.134398, size = 119, normalized size = 1.1 \[ -\frac{\sqrt{x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^3 \tanh ^{-1}(\tanh (a+b x))}-\frac{4 \sqrt{x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^3}+\frac{5 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{7/2}}+\frac{2 x^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2))/(3*b^2) - (4*Sqrt[x]*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/b^3 + (5*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b
*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(3/2))/b^(7/2) - (Sqrt[x]*(-(b*x) + ArcTanh[T
anh[a + b*x]])^2)/(b^3*ArcTanh[Tanh[a + b*x]])

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Maple [B]  time = 0.139, size = 294, normalized size = 2.7 \begin{align*}{\frac{2}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}-4\,{\frac{a\sqrt{x}}{{b}^{3}}}-4\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{x}}{{b}^{3}}}-{\frac{{a}^{2}}{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}-2\,{\frac{a\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}+5\,{\frac{{a}^{2}}{{b}^{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 ) }+10\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{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 ) }+5\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{{b}^{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(x^(5/2)/arctanh(tanh(b*x+a))^2,x)

[Out]

2/3*x^(3/2)/b^2-4/b^3*a*x^(1/2)-4/b^3*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)-1/b^3*x^(1/2)/arctanh(tanh(b*x+a))*
a^2-2/b^3*x^(1/2)/arctanh(tanh(b*x+a))*a*(arctanh(tanh(b*x+a))-b*x-a)-1/b^3*x^(1/2)/arctanh(tanh(b*x+a))*(arct
anh(tanh(b*x+a))-b*x-a)^2+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))*a^2+10/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))*a*(arctanh(tanh(b*x+a))-b*x-a)+5/b^3/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(ta
nh(b*x+a))-b*x)*b)^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^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(x^(5/2)/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.06784, size = 366, normalized size = 3.39 \begin{align*} \left [\frac{15 \,{\left (a b x + a^{2}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x + 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt{x}}{6 \,{\left (b^{4} x + a b^{3}\right )}}, \frac{15 \,{\left (a b x + a^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt{x}}{3 \,{\left (b^{4} x + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b*x + a^2)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(2*b^2*x^2 - 10*a*b*x
- 15*a^2)*sqrt(x))/(b^4*x + a*b^3), 1/3*(15*(a*b*x + a^2)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (2*b^2*x^2
- 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x + a*b^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(x**(5/2)/atanh(tanh(b*x+a))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.1411, size = 88, normalized size = 0.81 \begin{align*} \frac{5 \, a^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} - \frac{a^{2} \sqrt{x}}{{\left (b x + a\right )} b^{3}} + \frac{2 \,{\left (b^{4} x^{\frac{3}{2}} - 6 \, a b^{3} \sqrt{x}\right )}}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - a^2*sqrt(x)/((b*x + a)*b^3) + 2/3*(b^4*x^(3/2) - 6*a*b^3*s
qrt(x))/b^6