3.2.34 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x} \, dx\) [134]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2190, 2192} \begin {gather*} -2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2} \text {ArcTan}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )+2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {2}{3} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x,x]

[Out]

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

Rule 2190

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

Rule 2192

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

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 99, normalized size = 0.82 \begin {gather*} \frac {2}{15} \left (15 b^2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-35 b x \tanh ^{-1}(\tanh (a+b x))^{3/2}+23 \tanh ^{-1}(\tanh (a+b x))^{5/2}-15 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {-b x+\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15*b^2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]] - 35*b*x*ArcTanh[Tanh[a + b*x]]^(3/2) + 23*ArcTanh[Tanh[a + b*x]]^
(5/2) - 15*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(5/2)))/15

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(221\) vs. \(2(105)=210\).
time = 0.06, size = 222, normalized size = 1.83

method	result	size
default	\(\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} a}{3}+\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3}+2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\, a^{2}+4 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (a^{3}+3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\)	\(222\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.33, size = 114, normalized size = 0.94 \begin {gather*} \left [a^{\frac {5}{2}} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x + a}, 2 \, \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x + a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.39, size = 73, normalized size = 0.60 \begin {gather*} \frac {1}{15} \, \sqrt {2} {\left (\frac {15 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 3 \, \sqrt {2} {\left (b x + a\right )}^{\frac {5}{2}} + 5 \, \sqrt {2} {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {2} \sqrt {b x + a} a^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sqrt(2)*(15*sqrt(2)*a^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 3*sqrt(2)*(b*x + a)^(5/2) + 5*sqrt(2)*(
b*x + a)^(3/2)*a + 15*sqrt(2)*sqrt(b*x + a)*a^2)

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Mupad [B]
time = 4.82, size = 789, normalized size = 6.52 \begin {gather*} \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {3\,b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2}-\frac {2\,\left (3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-\frac {8\,b^2\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{5}\right )\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{3\,b}\right )}{b}+\frac {2\,b^2\,x^2\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{5}-\frac {x\,\left (3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-\frac {8\,b^2\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{5}\right )\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{3\,b}+\frac {\sqrt {2}\,\ln \left (\frac {\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-\sqrt {2}\,b\,x+\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\,2{}\mathrm {i}\right )\,16{}\mathrm {i}}{x\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^{5/2}\,1{}\mathrm {i}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x,x)

[Out]

((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((3*b*(
log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/2 - (2*(
3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (8
*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/5)
*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(3*b))
)/b + (2*b^2*x^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/
2)^(1/2))/5 + (2^(1/2)*log((((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2
*b*x) + 1))/2)^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)
) + 2*b*x)^(1/2)*2i + 2^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*
b*x) + 1)) + 2*b*x) - 2^(1/2)*b*x)*16i)/(x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2)))*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(
2*a)*exp(2*b*x) + 1)) + 2*b*x)^(5/2)*1i)/8 - (x*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp
(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (8*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*ex
p(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/5)*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 -
log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(3*b)

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