3.203 \(\int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=97 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt {b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {1}{b \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2162} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt {b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {1}{b \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 80, normalized size = 0.82 \[ \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt {b} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 116, normalized size = 1.20 \[ \left [\frac {2 \, a b \sqrt {x} - \sqrt {-a b} {\left (b x + a\right )} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{2 \, {\left (a^{2} b^{2} x + a^{3} b\right )}}, \frac {a b \sqrt {x} - \sqrt {a b} {\left (b x + a\right )} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )}{a^{2} b^{2} x + a^{3} b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 35, normalized size = 0.36 \[ \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} + \frac {\sqrt {x}}{{\left (b x + a\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.27, size = 82, normalized size = 0.85 \[ \frac {\sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 35, normalized size = 0.36 \[ \frac {\sqrt {x}}{a b x + a^{2}} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.18, size = 516, normalized size = 5.32 \[ \frac {\sqrt {2}\,\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 (4\,a^2\,b+b\,{\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\,b\,\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 )}{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 )}{\sqrt {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 )}^{3/2}}-\frac {4\,\sqrt {x}}{\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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(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)*(4*a^2*b + b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e
xp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 - 4*a*b*(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*(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))))))/(b^(1/2)*(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)^(3/2)) - (4*x^(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))/(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 \[ \int \frac {1}{\sqrt {x} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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