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

Optimal. Leaf size=174 \[ -\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{3/2}}+\frac{5}{32} x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{5}{24} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac{5 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b} \]

[Out]

(-5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^4)/(64*b^(3/2)) + (
5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/32 - (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a
+ b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/(64*b) - (5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a +
b*x]]^(3/2))/24 + (x^(3/2)*ArcTanh[Tanh[a + b*x]]^(5/2))/4

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Rubi [A]  time = 0.0975145, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ -\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{3/2}}+\frac{5}{32} x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{5}{24} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac{5 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*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]])^4)/(64*b^(3/2)) + (
5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/32 - (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a
+ b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/(64*b) - (5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a +
b*x]]^(3/2))/24 + (x^(3/2)*ArcTanh[Tanh[a + b*x]]^(5/2))/4

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 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac{1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac{1}{8} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx\\ &=-\frac{5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}+\frac{1}{16} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx\\ &=\frac{5}{32} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac{1}{64} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac{\sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=\frac{5}{32} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{64 b}-\frac{5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}+\frac{\left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{128 b}\\ &=-\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 )^4}{64 b^{3/2}}+\frac{5}{32} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{64 b}-\frac{5}{24} x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{4} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}\\ \end{align*}

Mathematica [A]  time = 0.081641, size = 121, normalized size = 0.7 \[ \frac{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-55 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+73 b x \tanh ^{-1}(\tanh (a+b x))^2+15 \tanh ^{-1}(\tanh (a+b x))^3+15 b^3 x^3\right )}{192 b}-\frac{5 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{64 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^3*x^3 - 55*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 73*b*x*ArcTanh[Tanh[a
+ b*x]]^2 + 15*ArcTanh[Tanh[a + b*x]]^3))/(192*b) - (5*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4*Log[b*Sqrt[x] + Sqr
t[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(64*b^(3/2))

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Maple [B]  time = 0.117, size = 471, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*arctanh(tanh(b*x+a))^(5/2),x)

[Out]

1/4*x^(1/2)*arctanh(tanh(b*x+a))^(7/2)/b-1/24/b*a*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)-5/96/b*a^2*x^(1/2)*arctan
h(tanh(b*x+a))^(3/2)-5/64/b*a^3*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-5/64/b^(3/2)*ln(b^(1/2)*x^(1/2)+arctanh(tan
h(b*x+a))^(1/2))*a^4-5/16/b^(3/2)*a^3*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x
-a)-15/64/b*a^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-15/32/b^(3/2)*a^2*ln(b^(1/2)*x
^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^2-5/48/b*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2
)*arctanh(tanh(b*x+a))^(3/2)-15/64/b*a*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-5/16/
b^(3/2)*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^3-1/24/b*(arctanh(tanh(b
*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)-5/96/b*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*
x+a))^(3/2)-5/64/b*(arctanh(tanh(b*x+a))-b*x-a)^3*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-5/64/b^(3/2)*ln(b^(1/2)*x
^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x} \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^(1/2)*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.21766, size = 425, normalized size = 2.44 \begin{align*} \left [\frac{15 \, a^{4} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{384 \, b^{2}}, \frac{15 \, a^{4} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{192 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

[1/384*(15*a^4*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(48*b^4*x^3 + 136*a*b^3*x^2 + 118*
a^2*b^2*x + 15*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^2, 1/192*(15*a^4*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt
(x))) + (48*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*b^2*x + 15*a^3*b)*sqrt(b*x + a)*sqrt(x))/b^2]

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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**(1/2)*atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.31174, size = 275, normalized size = 1.58 \begin{align*} \frac{1}{384} \, \sqrt{2}{\left (48 \, \sqrt{2}{\left (\sqrt{b x + a}{\left (2 \, x + \frac{a}{b}\right )} \sqrt{x} + \frac{a^{2} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{3}{2}}}\right )} a^{2} + 16 \, \sqrt{2}{\left (\sqrt{b x + a}{\left (2 \,{\left (4 \, x + \frac{a}{b}\right )} x - \frac{3 \, a^{2}}{b^{2}}\right )} \sqrt{x} - \frac{3 \, a^{3} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{5}{2}}}\right )} a b + \sqrt{2}{\left ({\left (2 \,{\left (4 \,{\left (6 \, x + \frac{a}{b}\right )} x - \frac{5 \, a^{2}}{b^{2}}\right )} x + \frac{15 \, a^{3}}{b^{3}}\right )} \sqrt{b x + a} \sqrt{x} + \frac{15 \, a^{4} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{7}{2}}}\right )} b^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

1/384*sqrt(2)*(48*sqrt(2)*(sqrt(b*x + a)*(2*x + a/b)*sqrt(x) + a^2*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/
b^(3/2))*a^2 + 16*sqrt(2)*(sqrt(b*x + a)*(2*(4*x + a/b)*x - 3*a^2/b^2)*sqrt(x) - 3*a^3*log(abs(-sqrt(b)*sqrt(x
) + sqrt(b*x + a)))/b^(5/2))*a*b + sqrt(2)*((2*(4*(6*x + a/b)*x - 5*a^2/b^2)*x + 15*a^3/b^3)*sqrt(b*x + a)*sqr
t(x) + 15*a^4*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(7/2))*b^2)