3.179 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{\sqrt {x}} \, dx\)

Optimal. Leaf size=46 \[ -\frac {8}{3} b x^{3/2} \tanh ^{-1}(\tanh (a+b x))+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2+\frac {16}{15} b^2 x^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 46, 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} \[ -\frac {8}{3} b x^{3/2} \tanh ^{-1}(\tanh (a+b x))+2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2+\frac {16}{15} b^2 x^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2/Sqrt[x],x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 40, normalized size = 0.87 \[ \frac {2}{15} \sqrt {x} \left (-20 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^2/Sqrt[x],x]

[Out]

(2*Sqrt[x]*(8*b^2*x^2 - 20*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh[a + b*x]]^2))/15

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fricas [A]  time = 0.49, size = 24, normalized size = 0.52 \[ \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 10 \, a b x + 15 \, a^{2}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 24, normalized size = 0.52 \[ \frac {2}{5} \, b^{2} x^{\frac {5}{2}} + \frac {4}{3} \, a b x^{\frac {3}{2}} + 2 \, a^{2} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(1/2),x, algorithm="giac")

[Out]

2/5*b^2*x^(5/2) + 4/3*a*b*x^(3/2) + 2*a^2*sqrt(x)

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maple [A]  time = 0.26, size = 47, normalized size = 1.02 \[ \frac {2 b^{2} x^{\frac {5}{2}}}{5}+\frac {4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b \,x^{\frac {3}{2}}}{3}+2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^2/x^(1/2),x)

[Out]

2/5*b^2*x^(5/2)+4/3*(arctanh(tanh(b*x+a))-b*x)*b*x^(3/2)+2*(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)

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maxima [A]  time = 0.34, size = 36, normalized size = 0.78 \[ \frac {16}{15} \, b^{2} x^{\frac {5}{2}} - \frac {8}{3} \, b x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) + 2 \, \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.15, size = 122, normalized size = 2.65 \[ \frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2}+\frac {2\,b^2\,x^{5/2}}{5}-\frac {2\,b\,x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^2/x^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**2/x**(1/2),x)

[Out]

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

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