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

Optimal. Leaf size=48 \[ -\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}-\frac {16 b^2}{15 \sqrt {x}} \]

[Out]

-8/15*b*arctanh(tanh(b*x+a))/x^(3/2)-2/5*arctanh(tanh(b*x+a))^2/x^(5/2)-16/15*b^2/x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}-\frac {16 b^2}{15 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^2)/(15*Sqrt[x]) - (8*b*ArcTanh[Tanh[a + b*x]])/(15*x^(3/2)) - (2*ArcTanh[Tanh[a + b*x]]^2)/(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))^2}{x^{7/2}} \, dx &=-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}+\frac {1}{5} (4 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^{5/2}} \, dx\\ &=-\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}+\frac {1}{15} \left (8 b^2\right ) \int \frac {1}{x^{3/2}} \, dx\\ &=-\frac {16 b^2}{15 \sqrt {x}}-\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 40, normalized size = 0.83 \[ -\frac {2 \left (4 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(8*b^2*x^2 + 4*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2))

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fricas [A]  time = 0.53, size = 24, normalized size = 0.50 \[ -\frac {2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*b^2*x^2 + 10*a*b*x + 3*a^2)/x^(5/2)

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giac [A]  time = 0.25, size = 24, normalized size = 0.50 \[ -\frac {2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*b^2*x^2 + 10*a*b*x + 3*a^2)/x^(5/2)

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maple [A]  time = 0.25, size = 38, normalized size = 0.79 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5 x^{\frac {5}{2}}}+\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{3 x^{\frac {3}{2}}}-\frac {2 b}{3 \sqrt {x}}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*arctanh(tanh(b*x+a))^2/x^(5/2)+8/5*b*(-1/3*arctanh(tanh(b*x+a))/x^(3/2)-2/3*b/x^(1/2))

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maxima [A]  time = 0.35, size = 36, normalized size = 0.75 \[ -\frac {16 \, b^{2}}{15 \, \sqrt {x}} - \frac {8 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{15 \, x^{\frac {3}{2}}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/15*b^2/sqrt(x) - 8/15*b*arctanh(tanh(b*x + a))/x^(3/2) - 2/5*arctanh(tanh(b*x + a))^2/x^(5/2)

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mupad [B]  time = 1.13, size = 122, normalized size = 2.54 \[ \frac {2\,b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{3\,x^{3/2}}-\frac {2\,b^2}{\sqrt {x}}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{10\,x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^2/x^(7/2),x)

[Out]

(2*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(3*x
^(3/2)) - (2*b^2)/x^(1/2) - (log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*
x) + 1)) + 2*b*x)^2/(10*x^(5/2))

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sympy [A]  time = 154.70, size = 49, normalized size = 1.02 \[ - \frac {16 b^{2}}{15 \sqrt {x}} - \frac {8 b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{15 x^{\frac {3}{2}}} - \frac {2 \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*b**2/(15*sqrt(x)) - 8*b*atanh(tanh(a + b*x))/(15*x**(3/2)) - 2*atanh(tanh(a + b*x))**2/(5*x**(5/2))

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