3.181 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^{5/2}} \, dx\)

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

[Out]

-2/3*arctanh(tanh(b*x+a))^2/x^(3/2)-8/3*b*arctanh(tanh(b*x+a))/x^(1/2)+16/3*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} \[ -\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{3 x^{3/2}}-\frac {8 b \tanh ^{-1}(\tanh (a+b x))}{3 \sqrt {x}}+\frac {16 b^2 \sqrt {x}}{3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2/x^(5/2),x]

[Out]

(16*b^2*Sqrt[x])/3 - (8*b*ArcTanh[Tanh[a + b*x]])/(3*Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]]^2)/(3*x^(3/2))

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

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

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^2/x^(5/2),x]

[Out]

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

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fricas [A]  time = 0.59, size = 24, normalized size = 0.50 \[ \frac {2 \, {\left (3 \, b^{2} x^{2} - 6 \, a b x - a^{2}\right )}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(5/2),x, algorithm="fricas")

[Out]

2/3*(3*b^2*x^2 - 6*a*b*x - a^2)/x^(3/2)

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giac [A]  time = 0.17, size = 23, normalized size = 0.48 \[ 2 \, b^{2} \sqrt {x} - \frac {2 \, {\left (6 \, a b x + a^{2}\right )}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(5/2),x, algorithm="giac")

[Out]

2*b^2*sqrt(x) - 2/3*(6*a*b*x + a^2)/x^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^2/x^(5/2),x)

[Out]

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

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maxima [A]  time = 0.34, size = 36, normalized size = 0.75 \[ \frac {16}{3} \, b^{2} \sqrt {x} - \frac {8 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{3 \, \sqrt {x}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.13, size = 122, normalized size = 2.54 \[ 2\,b^2\,\sqrt {x}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{6\,x^{3/2}}+\frac {2\,b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^2/x^(5/2),x)

[Out]

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/(6*x^(3/2)) + (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))/x^(1/2)

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sympy [A]  time = 10.50, size = 48, normalized size = 1.00 \[ \frac {16 b^{2} \sqrt {x}}{3} - \frac {8 b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{3 \sqrt {x}} - \frac {2 \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**2/x**(5/2),x)

[Out]

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

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