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

Optimal. Leaf size=104 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b} \]

[Out]

-1/4*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^2/b^(3/2)+1/2*x^(3/2)*arct
anh(tanh(b*x+a))^(1/2)-1/4*(b*x-arctanh(tanh(b*x+a)))*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)/b

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Rubi [A]  time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2165

Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTanh[(
Rt[a*b, 2]*Sqrt[u])/(a*Sqrt[v])])/Rt[a*b, 2], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[u, v,
x]

Rule 2169

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 + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

Rubi steps

\begin {align*} \int \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {1}{2} x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {1}{4} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=\frac {1}{2} x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b}-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2 \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 b}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 84, normalized size = 0.81 \[ \frac {\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))+b x\right )}{4 b}-\frac {\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 114, normalized size = 1.10 \[ \left [\frac {a^{2} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x + a b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, b^{2}}, \frac {a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (2 \, b^{2} x + a b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(a^2*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^2*x + a*b)*sqrt(b*x + a)*sqrt(x))/
b^2, 1/4*(a^2*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (2*b^2*x + a*b)*sqrt(b*x + a)*sqrt(x))/b^2
]

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giac [A]  time = 0.18, size = 48, normalized size = 0.46 \[ \frac {1}{4} \, \sqrt {b x + a} {\left (2 \, x + \frac {a}{b}\right )} \sqrt {x} + \frac {a^{2} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{4 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(b*x + a)*(2*x + a/b)*sqrt(x) + 1/4*a^2*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(3/2)

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maple [B]  time = 0.27, size = 174, normalized size = 1.67 \[ \frac {\sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{2 b}-\frac {a \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{4 b}-\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{2}}{4 b^{\frac {3}{2}}}-\frac {a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{2 b^{\frac {3}{2}}}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{4 b}-\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{4 b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)/b-1/4/b*a*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-1/4/b^(3/2)*ln(b^(1/2)*x^(
1/2)+arctanh(tanh(b*x+a))^(1/2))*a^2-1/2/b^(3/2)*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tan
h(b*x+a))-b*x-a)-1/4/b*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-1/4/b^(3/2)*ln(b^(1/2)*
x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x} \sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x)*sqrt(arctanh(tanh(b*x + a))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {x}\,\sqrt {\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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