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

Optimal. Leaf size=108 \[ \frac {2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}} \]

[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/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2163, 2161} \[ \frac {2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcTanh[Tanh[a + b*x]]^(5/2)),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/(3*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2)) + 2/((b*x - ArcTanh[Tanh[a + b*x]])^2*S
qrt[ArcTanh[Tanh[a + b*x]]])

Rule 2161

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]

Rule 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 91, normalized size = 0.84 \[ \frac {2 \left (4 \tanh ^{-1}(\tanh (a+b x))-b x\right )}{3 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

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

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fricas [A]  time = 0.51, size = 177, normalized size = 1.64 \[ \left [\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b x + 4 \, a^{2}\right )} \sqrt {b x + a}}{3 \, {\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}}, \frac {2 \, {\left (3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x + 4 \, a^{2}\right )} \sqrt {b x + a}\right )}}{3 \, {\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3*a*b*x + 4*a^2)*s
qrt(b*x + a))/(a^3*b^2*x^2 + 2*a^4*b*x + a^5), 2/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a)*arctan(sqrt(b*x + a)*
sqrt(-a)/a) + (3*a*b*x + 4*a^2)*sqrt(b*x + a))/(a^3*b^2*x^2 + 2*a^4*b*x + a^5)]

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giac [A]  time = 0.24, size = 45, normalized size = 0.42 \[ \frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {2 \, {\left (3 \, b x + 4 \, a\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + 2/3*(3*b*x + 4*a)/((b*x + a)^(3/2)*a^2)

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maple [A]  time = 0.15, size = 93, normalized size = 0.86 \[ \frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(arctanh(tanh(b*x+a))-b*x)^2/arctanh(tanh(b*x+a))^(1/2)+2/3/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^
(3/2)-2/(arctanh(tanh(b*x+a))-b*x)^(5/2)*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 \[ \int \frac {1}{x \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.26, size = 886, normalized size = 8.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*
x) + 1)) + 2*b*x)^5 + 40*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*e
xp(2*b*x) + 1)) + 2*b*x)^3 - 80*a^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp
(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 - 32*a^5 - 10*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)
) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4 + 80*a^4*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*
x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*1i)/(2*x*(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)))*4i)/(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(5/2) + (16*(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))/(exp(2*a)*exp(2*b*x)
+ 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))*(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)^2) - (16*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(e
xp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(3*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a
)*exp(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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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