3.154 \(\int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=78 \[ -\frac {2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}} \]

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

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Rubi [A]  time = 0.04, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2163, 2161} \[ -\frac {2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*ArcTanh[Tanh[a + b*x]]^(3/2)),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]])^(3
/2) - 2/((b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[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))^{3/2}} \, dx &=-\frac {2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {\int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\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 )^{3/2}}-\frac {2}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 75, normalized size = 0.96 \[ \frac {2}{\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*ArcTanh[Tanh[a + b*x]]^(3/2)),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]])^(3/2) + 2/(Sqrt[ArcTanh[Tanh[a + b*x]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]]))

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fricas [A]  time = 0.63, size = 110, normalized size = 1.41 \[ \left [\frac {{\left (b x + a\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, \sqrt {b x + a} a}{a^{2} b x + a^{3}}, \frac {2 \, {\left ({\left (b x + a\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \sqrt {b x + a} a\right )}}{a^{2} b x + a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 37, normalized size = 0.47 \[ \frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {2}{\sqrt {b x + a} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.15, size = 68, normalized size = 0.87 \[ \frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {2 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(x*arctanh(tanh(b*x + a))^(3/2)), x)

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mupad [B]  time = 6.14, size = 614, normalized size = 7.87 \[ -\frac {8\,\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 (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\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 )}+\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 )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )\,1{}\mathrm {i}}{2\,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 )\,2{}\mathrm {i}}{{\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 )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*atanh(tanh(a + b*x))^(3/2)),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)^3 - 8*a^3 - 6*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)^2 + 12*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)*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)))*2i)/(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)^(3/2) - (8*(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/(e
xp(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))/(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 \[ \int \frac {1}{x \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(1/(x*atanh(tanh(a + b*x))**(3/2)), x)

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