3.3.52 \(\int \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\) [252]

Optimal. Leaf size=52 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2199, 2196} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2196

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

Rule 2199

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

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Mathematica [A]
time = 0.09, size = 55, normalized size = 1.06 \begin {gather*} -\frac {2 \sqrt {x}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 42, normalized size = 0.81

method	result	size
derivativedivides	\(-\frac {2 \sqrt {x}}{b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {2 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\)	\(42\)
default	\(-\frac {2 \sqrt {x}}{b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {2 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 119, normalized size = 2.29 \begin {gather*} \left [\frac {{\left (b x + a\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, \sqrt {b x + a} b \sqrt {x}}{b^{3} x + a b^{2}}, -\frac {2 \, {\left ({\left (b x + a\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + \sqrt {b x + a} b \sqrt {x}\right )}}{b^{3} x + a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{\operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.40, size = 39, normalized size = 0.75 \begin {gather*} -\frac {2 \, \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {3}{2}}} - \frac {2 \, \sqrt {x}}{\sqrt {b x + a} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {x}}{{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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