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

Optimal. Leaf size=76 \[ \frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 76, 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, 2162} \[ \frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2162

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-2*ArcTanh[Sqrt
[v]/Rt[-((b*u - a*v)/a), 2]])/(a*Rt[-((b*u - a*v)/a), 2]), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /;
PiecewiseLinearQ[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^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{b x-\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {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}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 73, normalized size = 0.96 \[ -\frac {2}{\sqrt {x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}-\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 93, normalized size = 1.22 \[ \left [\frac {x \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, \sqrt {x}}{a x}, \frac {2 \, {\left (x \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - \sqrt {x}\right )}}{a x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 31, normalized size = 0.41 \[ -\frac {2 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {2}{a \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 76, normalized size = 1.00 \[ -\frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}-\frac {2 b \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 31, normalized size = 0.41 \[ -\frac {2 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {2}{a \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.97, size = 464, normalized size = 6.11 \[ \frac {4}{\sqrt {x}\,\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 {2\,\sqrt {2}\,\sqrt {b}\,\ln \left (\frac {\sqrt {b}\,\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}\,\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 )-4\,\sqrt {b}\,\sqrt {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}+2\,\sqrt {2}\,b\,x\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 )}^2-4\,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 )+4\,a^2\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 )}\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 )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/(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)
) + (2*2^(1/2)*b^(1/2)*log((b^(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)*(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) - 4*b^(1/2)*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)^(1/2) + 2*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)^2 - 4*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*a^2))/(2*(log((2*exp(2*a)*exp(2*b*x))/(ex
p(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))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(3/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {3}{2}} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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