3.2.71 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))}{\sqrt {x}} \, dx\) [171]

Optimal. Leaf size=25 \[ -\frac {4}{3} b x^{3/2}+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x)) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 30} \begin {gather*} 2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{3} b x^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]/Sqrt[x],x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

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Mathematica [A]
time = 0.01, size = 23, normalized size = 0.92 \begin {gather*} \frac {2}{3} \sqrt {x} \left (-2 b x+3 \tanh ^{-1}(\tanh (a+b x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 20, normalized size = 0.80

method	result	size
derivativedivides	\(-\frac {4 b \,x^{\frac {3}{2}}}{3}+2 \arctanh \left (\tanh \left (b x +a \right )\right ) \sqrt {x}\)	\(20\)
default	\(-\frac {4 b \,x^{\frac {3}{2}}}{3}+2 \arctanh \left (\tanh \left (b x +a \right )\right ) \sqrt {x}\)	\(20\)
risch	\(2 \sqrt {x}\, \ln \left ({\mathrm e}^{b x +a}\right )-\frac {\left (-3 i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-6 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+3 i \pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-3 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+3 i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+3 i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+3 i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+8 b x \right ) \sqrt {x}}{6}\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 19, normalized size = 0.76 \begin {gather*} -\frac {4}{3} \, b x^{\frac {3}{2}} + 2 \, \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*b*x^(3/2) + 2*sqrt(x)*arctanh(tanh(b*x + a))

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Fricas [A]
time = 0.37, size = 12, normalized size = 0.48 \begin {gather*} \frac {2}{3} \, {\left (b x + 3 \, a\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x + 3*a)*sqrt(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.39, size = 13, normalized size = 0.52 \begin {gather*} \frac {2}{3} \, b x^{\frac {3}{2}} + 2 \, a \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b*x^(3/2) + 2*a*sqrt(x)

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Mupad [B]
time = 1.13, size = 56, normalized size = 2.24 \begin {gather*} \sqrt {x}\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\frac {4\,b\,x^{3/2}}{3}-\sqrt {x}\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))/x^(1/2),x)

[Out]

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

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