3.2.72 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^{3/2}} \, dx\) [172]

Optimal. Leaf size=23 \[ 4 b \sqrt {x}-\frac {2 \tanh ^{-1}(\tanh (a+b x))}{\sqrt {x}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 23, 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*} 4 b \sqrt {x}-\frac {2 \tanh ^{-1}(\tanh (a+b x))}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]/x^(3/2),x]

[Out]

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

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Mathematica [A]
time = 0.02, size = 20, normalized size = 0.87 \begin {gather*} \frac {4 b x-2 \tanh ^{-1}(\tanh (a+b x))}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 20, normalized size = 0.87

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 19, normalized size = 0.83 \begin {gather*} 4 \, b \sqrt {x} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 12, normalized size = 0.52 \begin {gather*} \frac {2 \, {\left (b x - a\right )}}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*x - a)/sqrt(x)

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Sympy [A]
time = 0.39, size = 22, normalized size = 0.96 \begin {gather*} 4 b \sqrt {x} - \frac {2 \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*b*sqrt(x) - 2*atanh(tanh(a + b*x))/sqrt(x)

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Giac [A]
time = 0.38, size = 13, normalized size = 0.57 \begin {gather*} 2 \, b \sqrt {x} - \frac {2 \, a}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*sqrt(x) - 2*a/sqrt(x)

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Mupad [B]
time = 1.12, size = 56, normalized size = 2.43 \begin {gather*} \frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{\sqrt {x}}+4\,b\,\sqrt {x}-\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))/x^(3/2),x)

[Out]

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

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