3.236 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=93 \[ 2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac {2 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {x}}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 93, 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 = {2168, 2165} \[ 2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac {2 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {x}}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(7/2),x]

[Out]

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

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]

Rule 2168

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]))

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 95, normalized size = 1.02 \[ -\frac {2 \left (-15 b^{5/2} x^{5/2} \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )+15 b^2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}+5 b x \tanh ^{-1}(\tanh (a+b x))^{3/2}+3 \tanh ^{-1}(\tanh (a+b x))^{5/2}\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(7/2),x]

[Out]

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

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fricas [A]  time = 0.90, size = 137, normalized size = 1.47 \[ \left [\frac {15 \, b^{\frac {5}{2}} x^{3} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{15 \, x^{3}}, -\frac {2 \, {\left (15 \, \sqrt {-b} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}\right )}}{15 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x, algorithm="fricas")

[Out]

[1/15*(15*b^(5/2)*x^3*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(23*b^2*x^2 + 11*a*b*x + 3*a^2)*sqr
t(b*x + a)*sqrt(x))/x^3, -2/15*(15*sqrt(-b)*b^2*x^3*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (23*b^2*x^2 +
11*a*b*x + 3*a^2)*sqrt(b*x + a)*sqrt(x))/x^3]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),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,[1,0,1]%%%}+%%%{6,[0,2,2]%%%}+%%%{4,[0,2,1]%%%}+%%%{6,[0,2,0]%%%}+%%%{4,[0,1,2]%%%}+%%%{4,
[0,1,1]%%%}+%%%{6,[0,0,2]%%%},0,%%%{-4,[3,0,0]%%%}+%%%{-12,[2,1,1]%%%}+%%%{4,[2,1,0]%%%}+%%%{4,[2,0,1]%%%}+%%%
{-12,[1,2,2]%%%}+%%%{8,[1,2,1]%%%}+%%%{4,[1,2,0]%%%}+%%%{8,[1,1,2]%%%}+%%%{-40,[1,1,1]%%%}+%%%{4,[1,0,2]%%%}+%
%%{-4,[0,3,3]%%%}+%%%{4,[0,3,2]%%%}+%%%{4,[0,3,1]%%%}+%%%{-4,[0,3,0]%%%}+%%%{4,[0,2,3]%%%}+%%%{-40,[0,2,2]%%%}
+%%%{4,[0,2,1]%%%}+%%%{4,[0,1,3]%%%}+%%%{4,[0,1,2]%%%}+%%%{-4,[0,0,3]%%%},0,%%%{1,[4,0,0]%%%}+%%%{4,[3,1,1]%%%
}+%%%{-4,[3,1,0]%%%}+%%%{-4,[3,0,1]%%%}+%%%{6,[2,2,2]%%%}+%%%{-12,[2,2,1]%%%}+%%%{6,[2,2,0]%%%}+%%%{-12,[2,1,2
]%%%}+%%%{4,[2,1,1]%%%}+%%%{6,[2,0,2]%%%}+%%%{4,[1,3,3]%%%}+%%%{-12,[1,3,2]%%%}+%%%{12,[1,3,1]%%%}+%%%{-4,[1,3
,0]%%%}+%%%{-12,[1,2,3]%%%}+%%%{8,[1,2,2]%%%}+%%%{4,[1,2,1]%%%}+%%%{12,[1,1,3]%%%}+%%%{4,[1,1,2]%%%}+%%%{-4,[1
,0,3]%%%}+%%%{1,[0,4,4]%%%}+%%%{-4,[0,4,3]%%%}+%%%{6,[0,4,2]%%%}+%%%{-4,[0,4,1]%%%}+%%%{1,[0,4,0]%%%}+%%%{-4,[
0,3,4]%%%}+%%%{4,[0,3,3]%%%}+%%%{4,[0,3,2]%%%}+%%%{-4,[0,3,1]%%%}+%%%{6,[0,2,4]%%%}+%%%{4,[0,2,3]%%%}+%%%{6,[0
,2,2]%%%}+%%%{-4,[0,1,4]%%%}+%%%{-4,[0,1,3]%%%}+%%%{1,[0,0,4]%%%}] at parameters values [0,85.3561567818,61.79
37478349]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,[1,0,1]%%%}+%%%{6,[0,2,2]%%%}+%%%{4,[0,2,
1]%%%}+%%%{6,[0,2,0]%%%}+%%%{4,[0,1,2]%%%}+%%%{4,[0,1,1]%%%}+%%%{6,[0,0,2]%%%},0,%%%{-4,[3,0,0]%%%}+%%%{-12,[2
,1,1]%%%}+%%%{4,[2,1,0]%%%}+%%%{4,[2,0,1]%%%}+%%%{-12,[1,2,2]%%%}+%%%{8,[1,2,1]%%%}+%%%{4,[1,2,0]%%%}+%%%{8,[1
,1,2]%%%}+%%%{-40,[1,1,1]%%%}+%%%{4,[1,0,2]%%%}+%%%{-4,[0,3,3]%%%}+%%%{4,[0,3,2]%%%}+%%%{4,[0,3,1]%%%}+%%%{-4,
[0,3,0]%%%}+%%%{4,[0,2,3]%%%}+%%%{-40,[0,2,2]%%%}+%%%{4,[0,2,1]%%%}+%%%{4,[0,1,3]%%%}+%%%{4,[0,1,2]%%%}+%%%{-4
,[0,0,3]%%%},0,%%%{1,[4,0,0]%%%}+%%%{4,[3,1,1]%%%}+%%%{-4,[3,1,0]%%%}+%%%{-4,[3,0,1]%%%}+%%%{6,[2,2,2]%%%}+%%%
{-12,[2,2,1]%%%}+%%%{6,[2,2,0]%%%}+%%%{-12,[2,1,2]%%%}+%%%{4,[2,1,1]%%%}+%%%{6,[2,0,2]%%%}+%%%{4,[1,3,3]%%%}+%
%%{-12,[1,3,2]%%%}+%%%{12,[1,3,1]%%%}+%%%{-4,[1,3,0]%%%}+%%%{-12,[1,2,3]%%%}+%%%{8,[1,2,2]%%%}+%%%{4,[1,2,1]%%
%}+%%%{12,[1,1,3]%%%}+%%%{4,[1,1,2]%%%}+%%%{-4,[1,0,3]%%%}+%%%{1,[0,4,4]%%%}+%%%{-4,[0,4,3]%%%}+%%%{6,[0,4,2]%
%%}+%%%{-4,[0,4,1]%%%}+%%%{1,[0,4,0]%%%}+%%%{-4,[0,3,4]%%%}+%%%{4,[0,3,3]%%%}+%%%{4,[0,3,2]%%%}+%%%{-4,[0,3,1]
%%%}+%%%{6,[0,2,4]%%%}+%%%{4,[0,2,3]%%%}+%%%{6,[0,2,2]%%%}+%%%{-4,[0,1,4]%%%}+%%%{-4,[0,1,3]%%%}+%%%{1,[0,0,4]
%%%}] at parameters values [0,71.707969239,78.6493344628]sqrt(2)/2/abs(b)*b^2/b*(2*((-345*sqrt(2)*b^5*a^2/225/
a^2*sqrt(a+b*x)*sqrt(a+b*x)+525*sqrt(2)*b^5*a^3/225/a^2)*sqrt(a+b*x)*sqrt(a+b*x)-225*sqrt(2)*b^5*a^4/225/a^2)*
sqrt(a+b*x)*sqrt(-a*b+b*(a+b*x))/(-a*b+b*(a+b*x))^3-2*b^3*sqrt(2)/sqrt(b)*ln(abs(sqrt(-a*b+b*(a+b*x))-sqrt(b)*
sqrt(a+b*x))))

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maple [B]  time = 0.25, size = 532, normalized size = 5.72 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {4 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}-\frac {16 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {16 b^{3} \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {4 b^{3} a \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {2 b^{3} a^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {2 b^{\frac {5}{2}} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{3}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {6 b^{\frac {5}{2}} 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 )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {4 b^{3} 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 )}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {6 b^{\frac {5}{2}} 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}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {4 b^{3} \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}}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {2 b^{3} \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 )}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {2 b^{\frac {5}{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 )^{3}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(7/2)-4/15*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)
*arctanh(tanh(b*x+a))^(7/2)-16/15*b^2/(arctanh(tanh(b*x+a))-b*x)^3/x^(1/2)*arctanh(tanh(b*x+a))^(7/2)+16/15*b^
3/(arctanh(tanh(b*x+a))-b*x)^3*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+4/3*b^3/(arctanh(tanh(b*x+a))-b*x)^3*a*x^(1/
2)*arctanh(tanh(b*x+a))^(3/2)+2*b^3/(arctanh(tanh(b*x+a))-b*x)^3*a^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+2*b^(5
/2)/(arctanh(tanh(b*x+a))-b*x)^3*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a^3+6*b^(5/2)/(arctanh(tanh(b*
x+a))-b*x)^3*a^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+4*b^3/(arctanh(ta
nh(b*x+a))-b*x)^3*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+6*b^(5/2)/(arctanh(tanh(b*
x+a))-b*x)^3*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^2+4/3*b^3/(arctanh(
tanh(b*x+a))-b*x)^3*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+2*b^3/(arctanh(tanh(b*x+a)
)-b*x)^3*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+2*b^(5/2)/(arctanh(tanh(b*x+a))-b*x
)^3*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 \[ \int \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{5/2}}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^(7/2),x)

[Out]

int(atanh(tanh(a + b*x))^(5/2)/x^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2)/x**(7/2),x)

[Out]

Timed out

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