3.174 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^{7/2}} \, dx\)

Optimal. Leaf size=27 \[ -\frac {2 \tanh ^{-1}(\tanh (a+b x))}{5 x^{5/2}}-\frac {4 b}{15 x^{3/2}} \]

[Out]

-4/15*b/x^(3/2)-2/5*arctanh(tanh(b*x+a))/x^(5/2)

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Rubi [A]  time = 0.01, antiderivative size = 27, 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 = {2168, 30} \[ -\frac {2 \tanh ^{-1}(\tanh (a+b x))}{5 x^{5/2}}-\frac {4 b}{15 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]/x^(7/2),x]

[Out]

(-4*b)/(15*x^(3/2)) - (2*ArcTanh[Tanh[a + b*x]])/(5*x^(5/2))

Rule 30

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

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 {\tanh ^{-1}(\tanh (a+b x))}{x^{7/2}} \, dx &=-\frac {2 \tanh ^{-1}(\tanh (a+b x))}{5 x^{5/2}}+\frac {1}{5} (2 b) \int \frac {1}{x^{5/2}} \, dx\\ &=-\frac {4 b}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))}{5 x^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.85 \[ -\frac {2 \left (3 \tanh ^{-1}(\tanh (a+b x))+2 b x\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]/x^(7/2),x]

[Out]

(-2*(2*b*x + 3*ArcTanh[Tanh[a + b*x]]))/(15*x^(5/2))

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fricas [A]  time = 0.49, size = 13, normalized size = 0.48 \[ -\frac {2 \, {\left (5 \, b x + 3 \, a\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))/x^(7/2),x, algorithm="fricas")

[Out]

-2/15*(5*b*x + 3*a)/x^(5/2)

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giac [A]  time = 0.22, size = 13, normalized size = 0.48 \[ -\frac {2 \, {\left (5 \, b x + 3 \, a\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))/x^(7/2),x, algorithm="giac")

[Out]

-2/15*(5*b*x + 3*a)/x^(5/2)

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maple [A]  time = 0.25, size = 20, normalized size = 0.74 \[ -\frac {4 b}{15 x^{\frac {3}{2}}}-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )}{5 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))/x^(7/2),x)

[Out]

-4/15*b/x^(3/2)-2/5*arctanh(tanh(b*x+a))/x^(5/2)

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maxima [A]  time = 0.33, size = 19, normalized size = 0.70 \[ -\frac {4 \, b}{15 \, x^{\frac {3}{2}}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))/x^(7/2),x, algorithm="maxima")

[Out]

-4/15*b/x^(3/2) - 2/5*arctanh(tanh(b*x + a))/x^(5/2)

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mupad [B]  time = 1.28, size = 57, normalized size = 2.11 \[ \frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{5\,x^{5/2}}-\frac {4\,b}{15\,x^{3/2}}-\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{5\,x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))/x^(7/2),x)

[Out]

log(1/(exp(2*a)*exp(2*b*x) + 1))/(5*x^(5/2)) - (4*b)/(15*x^(3/2)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*
b*x) + 1))/(5*x^(5/2))

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sympy [A]  time = 155.35, size = 27, normalized size = 1.00 \[ - \frac {4 b}{15 x^{\frac {3}{2}}} - \frac {2 \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))/x**(7/2),x)

[Out]

-4*b/(15*x**(3/2)) - 2*atanh(tanh(a + b*x))/(5*x**(5/2))

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