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

Optimal. Leaf size=81 \[ -\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x}+3 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}-3 b \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))} \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2159, 2161} \[ -\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x}+3 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}-3 b \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))} \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

Rule 2161

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTan[Sqrt[v
]/Rt[(b*u - a*v)/a, 2]])/(a*Rt[(b*u - a*v)/a, 2]), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; Piecewis
eLinearQ[u, v, x]

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

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Mathematica [A]  time = 0.05, size = 79, normalized size = 0.98 \[ -\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x}+3 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}-3 b \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))-b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 102, normalized size = 1.26 \[ \left [\frac {3 \, \sqrt {a} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, b x - a\right )} \sqrt {b x + a}}{2 \, x}, \frac {3 \, \sqrt {-a} b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, b x - a\right )} \sqrt {b x + a}}{x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*sqrt(a)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(2*b*x - a)*sqrt(b*x + a))/x, (3*sqrt(-a)
*b*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (2*b*x - a)*sqrt(b*x + a))/x]

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giac [A]  time = 0.40, size = 69, normalized size = 0.85 \[ \frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} a b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {2} \sqrt {b x + a} b^{2} - \frac {\sqrt {2} \sqrt {b x + a} a b}{x}\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*(3*sqrt(2)*a*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2*sqrt(2)*sqrt(b*x + a)*b^2 - sqrt(2)*s
qrt(b*x + a)*a*b/x)/b

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maple [A]  time = 0.15, size = 85, normalized size = 1.05 \[ 2 b \left (\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{2}+\frac {b x}{2}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b x}-\frac {3 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\, \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.22, size = 459, normalized size = 5.67 \[ 3\,b\,\sqrt {\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}+\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\sqrt {\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{2\,x}-\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\sqrt {\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{2\,x}+b\,\ln \left (-\frac {4\,\sqrt {2}\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-4\,\sqrt {2}\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+8\,\sqrt {\frac {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\sqrt {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,b\,x}+4\,\sqrt {2}\,b\,x}{x\,\sqrt {\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,b\,x}}\right )\,\sqrt {\frac {9\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{8}-\frac {9\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{8}-\frac {9\,b\,x}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(3/2)/x^2,x)

[Out]

3*b*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2) + (log
(1/(exp(2*a)*exp(2*b*x) + 1))*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*
b*x) + 1))/2)^(1/2))/(2*x) - (log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*(log((exp(2*a)*exp(2*b*x))/
(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(2*x) + b*log(-(4*2^(1/2)*log(1/(exp
(2*a)*exp(2*b*x) + 1)) - 4*2^(1/2)*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 8*(log((exp(2*a)*exp
(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*(log((exp(2*a)*exp(2*b*x))/(
exp(2*a)*exp(2*b*x) + 1)) - log(1/(exp(2*a)*exp(2*b*x) + 1)) - 2*b*x)^(1/2) + 4*2^(1/2)*b*x)/(x*(log((exp(2*a)
*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(1/(exp(2*a)*exp(2*b*x) + 1)) - 2*b*x)^(1/2)))*((9*log((exp(2*a)*
exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/8 - (9*log(1/(exp(2*a)*exp(2*b*x) + 1)))/8 - (9*b*x)/4)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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