3.3.34 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{3/2}} \, dx\) [234]

Optimal. Leaf size=121 \[ \frac {15}{4} \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {15}{4} b \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {5}{2} b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt {x}} \]

[Out]

15/4*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^2*b^(1/2)-2*arctanh(tanh(b
*x+a))^(5/2)/x^(1/2)+5/2*b*arctanh(tanh(b*x+a))^(3/2)*x^(1/2)-15/4*b*(b*x-arctanh(tanh(b*x+a)))*x^(1/2)*arctan
h(tanh(b*x+a))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 121, normalized size of antiderivative = 1.00, 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 = {2199, 2200, 2196} \begin {gather*} -\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt {x}}+\frac {5}{2} b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {15}{4} b \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {15}{4} \sqrt {b} \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 )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2199

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

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Mathematica [A]
time = 0.05, size = 101, normalized size = 0.83 \begin {gather*} -\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (15 b^2 x^2-25 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2\right )}{4 \sqrt {x}}+\frac {15}{4} \sqrt {b} \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 25*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/
Sqrt[x] + (15*Sqrt[b]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]
])/4

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(459\) vs. \(2(95)=190\).
time = 0.12, size = 460, normalized size = 3.80

method	result	size
derivativedivides	\(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {2 b \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}+\frac {5 b a \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 b \,a^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 \sqrt {b}\, \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{3}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {45 \sqrt {b}\, 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 )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 b 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 )}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {45 \sqrt {b}\, 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}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {5 b \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}}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 b \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 )}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 \sqrt {b}\, \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}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(460\)
default	\(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {2 b \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}+\frac {5 b a \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 b \,a^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 \sqrt {b}\, \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{3}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {45 \sqrt {b}\, 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 )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 b 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 )}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {45 \sqrt {b}\, 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}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {5 b \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}}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 b \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 )}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}+\frac {15 \sqrt {b}\, \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}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(460\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/(arctanh(tanh(b*x+a))-b*x)/x^(1/2)*arctanh(tanh(b*x+a))^(7/2)+2*b/(arctanh(tanh(b*x+a))-b*x)*x^(1/2)*arctan
h(tanh(b*x+a))^(5/2)+5/2*b/(arctanh(tanh(b*x+a))-b*x)*a*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+15/4*b/(arctanh(tan
h(b*x+a))-b*x)*a^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+15/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*ln(b^(1/2)*x^(1/
2)+arctanh(tanh(b*x+a))^(1/2))*a^3+45/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*a^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh
(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+15/2*b/(arctanh(tanh(b*x+a))-b*x)*a*(arctanh(tanh(b*x+a))-b*x-a)*
x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+45/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b
*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^2+5/2*b/(arctanh(tanh(b*x+a))-b*x)*(arctanh(tanh(b*x+a))-b*x-a)*x^(
1/2)*arctanh(tanh(b*x+a))^(3/2)+15/4*b/(arctanh(tanh(b*x+a))-b*x)*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arcta
nh(tanh(b*x+a))^(1/2)+15/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*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^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 137, normalized size = 1.13 \begin {gather*} \left [\frac {15 \, a^{2} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{8 \, x}, -\frac {15 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{4 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*a^2*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^2*x^2 + 9*a*b*x - 8*a^2)*sqrt
(b*x + a)*sqrt(x))/x, -1/4*(15*a^2*sqrt(-b)*x*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (2*b^2*x^2 + 9*a*b*
x - 8*a^2)*sqrt(b*x + a)*sqrt(x))/x]

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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**(3/2),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^(3/2),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}}{x^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^(3/2),x)

[Out]

int(atanh(tanh(a + b*x))^(5/2)/x^(3/2), x)

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