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

Optimal. Leaf size=110 \[ -\frac{5 x^{3/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{15 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}{4 b^{7/2}}-\frac{x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{15 \sqrt{x}}{4 b^3} \]

[Out]

(15*Sqrt[x])/(4*b^3) - (15*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Ta
nh[a + b*x]]])/(4*b^(7/2)) - x^(5/2)/(2*b*ArcTanh[Tanh[a + b*x]]^2) - (5*x^(3/2))/(4*b^2*ArcTanh[Tanh[a + b*x]
])

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Rubi [A]  time = 0.0711217, antiderivative size = 110, 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} \[ -\frac{5 x^{3/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{15 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}{4 b^{7/2}}-\frac{x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{15 \sqrt{x}}{4 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[x])/(4*b^3) - (15*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Ta
nh[a + b*x]]])/(4*b^(7/2)) - x^(5/2)/(2*b*ArcTanh[Tanh[a + b*x]]^2) - (5*x^(3/2))/(4*b^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 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))^3} \, dx &=-\frac{x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{5 \int \frac{x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac{x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{5 x^{3/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{15 \int \frac{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac{15 \sqrt{x}}{4 b^3}-\frac{x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{5 x^{3/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (15 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^3}\\ &=\frac{15 \sqrt{x}}{4 b^3}-\frac{15 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}{4 b^{7/2}}-\frac{x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{5 x^{3/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}\\ \end{align*}

Mathematica [A]  time = 0.0876208, size = 104, normalized size = 0.95 \[ \frac{1}{4} \left (-\frac{5 x^{3/2}}{b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{15 \sqrt{\tanh ^{-1}(\tanh (a+b x))-b x} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{7/2}}-\frac{2 x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{15 \sqrt{x}}{b^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.138, size = 249, normalized size = 2.3 \begin{align*} 2\,{\frac{\sqrt{x}}{{b}^{3}}}+{\frac{9\,a}{4\,{b}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -9\,bx-9\,a}{4\,{b}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{7\,{a}^{2}}{4\,{b}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}+{\frac{7\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{2\,{b}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}+{\frac{7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{4\,{b}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}-{\frac{15\,a}{4\,{b}^{3}}\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{15\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -15\,bx-15\,a}{4\,{b}^{3}}\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(x^(5/2)/arctanh(tanh(b*x+a))^3,x)

[Out]

2*x^(1/2)/b^3+9/4/b^2/arctanh(tanh(b*x+a))^2*x^(3/2)*a+9/4/b^2/arctanh(tanh(b*x+a))^2*x^(3/2)*(arctanh(tanh(b*
x+a))-b*x-a)+7/4/b^3/arctanh(tanh(b*x+a))^2*x^(1/2)*a^2+7/2/b^3/arctanh(tanh(b*x+a))^2*x^(1/2)*a*(arctanh(tanh
(b*x+a))-b*x-a)+7/4/b^3/arctanh(tanh(b*x+a))^2*x^(1/2)*(arctanh(tanh(b*x+a))-b*x-a)^2-15/4/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))*a-15/4/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))*(arctanh(tanh(b*x+a))-b*x-a)

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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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03374, size = 443, normalized size = 4.03 \begin{align*} \left [\frac{15 \,{\left (b^{2} x^{2} + 2 \, 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 (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt{x}}{8 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac{15 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) -{\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt{x}}{4 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} 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))^3,x, algorithm="fricas")

[Out]

[1/8*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(8*b^2*x^2
+ 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a/b)*a
rctan(b*sqrt(x)*sqrt(a/b)/a) - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.135, size = 80, normalized size = 0.73 \begin{align*} -\frac{15 \, a \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{3}} + \frac{2 \, \sqrt{x}}{b^{3}} + \frac{9 \, a b x^{\frac{3}{2}} + 7 \, a^{2} \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3 + 1/4*(9*a*b*x^(3/2) + 7*a^2*sqrt(x))/((b*
x + a)^2*b^3)