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

Optimal. Leaf size=116 \[ \frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{7/2}}+\frac{2 x^{5/2}}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.0791174, antiderivative size = 116, 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 = {2159, 2162} \[ \frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{7/2}}+\frac{2 x^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/ArcTanh[Tanh[a + b*x]],x]

[Out]

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

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

Mathematica [A]  time = 0.105778, size = 108, normalized size = 0.93 \[ \frac{2 \left (-35 b^{3/2} x^{3/2} \tanh ^{-1}(\tanh (a+b x))+15 \sqrt{b} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-15 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )+23 b^{5/2} x^{5/2}\right )}{15 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(23*b^(5/2)*x^(5/2) - 35*b^(3/2)*x^(3/2)*ArcTanh[Tanh[a + b*x]] + 15*Sqrt[b]*Sqrt[x]*ArcTanh[Tanh[a + b*x]]
^2 - 15*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(5/2
)))/(15*b^(7/2))

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Maple [B]  time = 0.126, size = 330, normalized size = 2.8 \begin{align*}{\frac{2}{5\,b}{x}^{{\frac{5}{2}}}}-{\frac{2\,a}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}-{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -2\,bx-2\,a}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}+2\,{\frac{\sqrt{x}{a}^{2}}{{b}^{3}}}+4\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{x}}{{b}^{3}}}+2\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}\sqrt{x}}{{b}^{3}}}-2\,{\frac{{a}^{3}}{{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 ) }-6\,{\frac{{a}^{2} \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 ) }-6\,{\frac{a \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 ) }-2\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{{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)),x)

[Out]

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

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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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.20097, size = 308, normalized size = 2.66 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt{x}}{15 \, b^{3}}, -\frac{2 \,{\left (15 \, a^{2} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) -{\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt{x}\right )}}{15 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**(5/2)/atanh(tanh(a + b*x)), x)

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Giac [A]  time = 1.13205, size = 80, normalized size = 0.69 \begin{align*} -\frac{2 \, a^{3} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} + \frac{2 \,{\left (3 \, b^{4} x^{\frac{5}{2}} - 5 \, a b^{3} x^{\frac{3}{2}} + 15 \, a^{2} b^{2} \sqrt{x}\right )}}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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