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

Optimal. Leaf size=73 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{3/2} \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {x}}{b \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 73, 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 = {2168, 2162} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{3/2} \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {x}}{b \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 70, normalized size = 0.96 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac {\sqrt {x}}{b \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 115, normalized size = 1.58 \[ \left [-\frac {2 \, a b \sqrt {x} + \sqrt {-a b} {\left (b x + a\right )} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{2 \, {\left (a b^{3} x + a^{2} b^{2}\right )}}, -\frac {a b \sqrt {x} + \sqrt {a b} {\left (b x + a\right )} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )}{a b^{3} x + a^{2} b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.18, size = 36, normalized size = 0.49 \[ \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} - \frac {\sqrt {x}}{{\left (b x + a\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 61, normalized size = 0.84 \[ -\frac {\sqrt {x}}{b \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{b \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^(1/2)/b/arctanh(tanh(b*x+a))+1/b/((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 = 37, normalized size = 0.51 \[ -\frac {\sqrt {x}}{b^{2} x + a b} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.77, size = 344, normalized size = 4.71 \[ \frac {\sqrt {2}\,\ln \left (\frac {b^{7/2}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}+2\,\sqrt {2}\,b\,x\right )}{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}\right )}{2\,b^{3/2}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}-\frac {2\,\sqrt {x}}{b\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/atanh(tanh(a + b*x))^2,x)

[Out]

(2^(1/2)*log((b^(7/2)*(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))/(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - l
og(2/(exp(2*a)*exp(2*b*x) + 1)))))/(2*b^(3/2)*(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*x^(1/2))/(b*(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))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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