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

Optimal. Leaf size=44 \[ 8 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{\sqrt {x}}-\frac {16}{3} b^2 x^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 44, 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} \[ 8 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))-\frac {2 \tanh ^{-1}(\tanh (a+b x))^2}{\sqrt {x}}-\frac {16}{3} b^2 x^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 40, normalized size = 0.91 \[ -\frac {2 \left (-12 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right )}{3 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(8*b^2*x^2 - 12*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(3*Sqrt[x])

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fricas [A]  time = 0.48, size = 23, normalized size = 0.52 \[ \frac {2 \, {\left (b^{2} x^{2} + 6 \, a b x - 3 \, a^{2}\right )}}{3 \, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 24, normalized size = 0.55 \[ \frac {2}{3} \, b^{2} x^{\frac {3}{2}} + 4 \, a b \sqrt {x} - \frac {2 \, a^{2}}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 37, normalized size = 0.84 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{\sqrt {x}}+8 b \left (\arctanh \left (\tanh \left (b x +a \right )\right ) \sqrt {x}-\frac {2 b \,x^{\frac {3}{2}}}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 36, normalized size = 0.82 \[ -\frac {16}{3} \, b^{2} x^{\frac {3}{2}} + 8 \, b \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.16, size = 122, normalized size = 2.77 \[ \frac {2\,b^2\,x^{3/2}}{3}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,\sqrt {x}}-2\,b\,\sqrt {x}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^2/x^(3/2),x)

[Out]

(2*b^2*x^(3/2))/3 - (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/(2*x^(1/2)) - 2*b*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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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