3.3.33 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx\) [233]

Optimal. Leaf size=136 \[ -\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 \sqrt {b}}+\frac {5}{8} \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {5}{12} \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{3} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2} \]

[Out]

-5/8*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^3/b^(1/2)-5/12*(b*x-arctan
h(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(3/2)*x^(1/2)+1/3*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+5/8*(b*x-arctanh(tan
h(b*x+a)))^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2200, 2196} \begin {gather*} -\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 )^3}{8 \sqrt {b}}+\frac {5}{8} \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {5}{12} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {1}{3} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/(8*Sqrt[b]) + (5
*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/8 - (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a +
b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))/12 + (Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(5/2))/3

Rule 2196

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

Rule 2200

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]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \, dx &=\frac {1}{3} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac {1}{6} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{\sqrt {x}} \, dx\\ &=-\frac {5}{12} \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{3} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2}+\frac {1}{8} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {x}} \, dx\\ &=\frac {5}{8} \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {5}{12} \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{3} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac {1}{16} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\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 )^3}{8 \sqrt {b}}+\frac {5}{8} \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {5}{12} \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{3} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 101, normalized size = 0.74 \begin {gather*} \frac {1}{24} \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (15 b^2 x^2-40 b x \tanh ^{-1}(\tanh (a+b x))+33 \tanh ^{-1}(\tanh (a+b x))^2\right )+\frac {5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}\right )}{8 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 40*b*x*ArcTanh[Tanh[a + b*x]] + 33*ArcTanh[Tanh[a + b*x]]^
2))/24 + (5*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*Sqrt
[b])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(285\) vs. \(2(108)=216\).
time = 0.12, size = 286, normalized size = 2.10

method	result	size
derivativedivides	\(\frac {\sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{3}+\frac {5 a \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{12}+\frac {5 a^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{8}+\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{3}}{8 \sqrt {b}}+\frac {15 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8 \sqrt {b}}+\frac {5 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{4}+\frac {15 a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{8 \sqrt {b}}+\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{12}+\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{8}+\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{8 \sqrt {b}}\)	\(286\)
default	\(\frac {\sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{3}+\frac {5 a \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{12}+\frac {5 a^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{8}+\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{3}}{8 \sqrt {b}}+\frac {15 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8 \sqrt {b}}+\frac {5 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{4}+\frac {15 a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{8 \sqrt {b}}+\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{12}+\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{8}+\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{8 \sqrt {b}}\)	\(286\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+5/12*a*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+5/8*a^2*x^(1/2)*arctanh(tanh(
b*x+a))^(1/2)+5/8/b^(1/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a^3+15/8*a^2/b^(1/2)*ln(b^(1/2)*x^(1/
2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+5/4*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh
(tanh(b*x+a))^(1/2)+15/8*a/b^(1/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)
^2+5/12*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+5/8*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(
1/2)*arctanh(tanh(b*x+a))^(1/2)+5/8/b^(1/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a
))-b*x-a)^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 141, normalized size = 1.04 \begin {gather*} \left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*a^3*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*b^3*x^2 + 26*a*b^2*x + 33*a^2*b)
*sqrt(b*x + a)*sqrt(x))/b, -1/24*(15*a^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*b^3*x^2 + 26
*a*b^2*x + 33*a^2*b)*sqrt(b*x + a)*sqrt(x))/b]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3006 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0]%%%}+%%%{-4,[0,1,1]%%%}+%%%{-4,[0,1,0]%%%}+%%%{-4,[0,0,1]%%%},0,%%%{6,[2,0,0]%%%}+%%%{12,[1,1,1]%%%}+%%%{
4,[1,1,0]%%%}+%%%{4,[

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{5/2}}{\sqrt {x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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