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

Optimal. Leaf size=111 \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{7/2}}-\frac{2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

(5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]))/b^(7/2) - (2*x^(5/2
))/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*Sqrt[x]*Sqrt[Ar
cTanh[Tanh[a + b*x]]])/b^3

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Rubi [A]  time = 0.0659394, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{7/2}}-\frac{2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]))/b^(7/2) - (2*x^(5/2
))/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*Sqrt[x]*Sqrt[Ar
cTanh[Tanh[a + b*x]]])/b^3

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 2169

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 + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

Rule 2165

Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTanh[(
Rt[a*b, 2]*Sqrt[u])/(a*Sqrt[v])])/Rt[a*b, 2], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[u, v,
x]

Rubi steps

\begin{align*} \int \frac{x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac{2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 \int \frac{x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 b}\\ &=-\frac{2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{10 x^{3/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 \int \frac{\sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=-\frac{2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{10 x^{3/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac{\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{2 b^3}\\ &=\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{7/2}}-\frac{2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{10 x^{3/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0903393, size = 101, normalized size = 0.91 \[ \frac{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{b^{7/2}}-\frac{\sqrt{x} \left (10 b x \tanh ^{-1}(\tanh (a+b x))-15 \tanh ^{-1}(\tanh (a+b x))^2+2 b^2 x^2\right )}{3 b^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[x]*(2*b^2*x^2 + 10*b*x*ArcTanh[Tanh[a + b*x]] - 15*ArcTanh[Tanh[a + b*x]]^2))/(3*b^3*ArcTanh[Tanh[a + b
*x]]^(3/2)) + (5*(b*x - ArcTanh[Tanh[a + b*x]])*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/b^(7/2)

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Maple [B]  time = 0.127, size = 180, normalized size = 1.6 \begin{align*}{\frac{1}{b}{x}^{{\frac{5}{2}}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,a}{3\,{b}^{2}}{x}^{{\frac{3}{2}}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}+5\,{\frac{a\sqrt{x}}{{b}^{3}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}-5\,{\frac{a\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }{{b}^{7/2}}}+{\frac{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx-5\,a}{3\,{b}^{2}}{x}^{{\frac{3}{2}}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}+5\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{x}}{{b}^{3}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}-5\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }{{b}^{7/2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{5}{2}}}{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(5/2)/arctanh(tanh(b*x + a))^(5/2), x)

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Fricas [A]  time = 2.09868, size = 512, normalized size = 4.61 \begin{align*} \left [\frac{15 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{6 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac{15 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b^2*x^2 + 2*a^2*b*x + a^3)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*b^3*x^2
+ 20*a*b^2*x + 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/3*(15*(a*b^2*x^2 + 2*a^2*b
*x + a^3)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (3*b^3*x^2 + 20*a*b^2*x + 15*a^2*b)*sqrt(b*x +
a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.20249, size = 82, normalized size = 0.74 \begin{align*} \frac{{\left (x{\left (\frac{3 \, x}{b} + \frac{20 \, a}{b^{2}}\right )} + \frac{15 \, a^{2}}{b^{3}}\right )} \sqrt{x}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}}} + \frac{5 \, a \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x*(3*x/b + 20*a/b^2) + 15*a^2/b^3)*sqrt(x)/(b*x + a)^(3/2) + 5*a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)
))/b^(7/2)