3.3.35 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{5/2}} \, dx\) [235]

Optimal. Leaf size=106 \[ -5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+5 b^2 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 \sqrt {x}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}} \]

[Out]

-5*b^(3/2)*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))-2/3*arctanh(tanh(b*x
+a))^(5/2)/x^(3/2)-10/3*b*arctanh(tanh(b*x+a))^(3/2)/x^(1/2)+5*b^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 106, 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*} -5 b^{3/2} \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 )+5 b^2 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}}-\frac {10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 97, normalized size = 0.92 \begin {gather*} \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (15 b^2 x^2-10 b x \tanh ^{-1}(\tanh (a+b x))-2 \tanh ^{-1}(\tanh (a+b x))^2\right )}{3 x^{3/2}}+5 b^{3/2} \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \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^(5/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(500\) vs. \(2(82)=164\).
time = 0.12, size = 501, normalized size = 4.73 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(7/2)-8/3*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)*
arctanh(tanh(b*x+a))^(7/2)+8/3*b^2/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+10/3*b^2/(a
rctanh(tanh(b*x+a))-b*x)^2*a*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+5*b^2/(arctanh(tanh(b*x+a))-b*x)^2*a^2*x^(1/2)
*arctanh(tanh(b*x+a))^(1/2)+5*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/
2))*a^3+15*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^2*a^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(ta
nh(b*x+a))-b*x-a)+10*b^2/(arctanh(tanh(b*x+a))-b*x)^2*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+
a))^(1/2)+15*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^2*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(ta
nh(b*x+a))-b*x-a)^2+10/3*b^2/(arctanh(tanh(b*x+a))-b*x)^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*
x+a))^(3/2)+5*b^2/(arctanh(tanh(b*x+a))-b*x)^2*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/
2)+5*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^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^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.33, size = 138, normalized size = 1.30 \begin {gather*} \left [\frac {15 \, a b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {15 \, a \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a*b^(3/2)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*b^2*x^2 - 14*a*b*x - 2*a^2)*sqr
t(b*x + a)*sqrt(x))/x^2, -1/3*(15*a*sqrt(-b)*b*x^2*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (3*b^2*x^2 - 1
4*a*b*x - 2*a^2)*sqrt(b*x + a)*sqrt(x))/x^2]

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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**(5/2),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^(5/2),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^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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