3.2.16 \(\int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x} \, dx\) [116]

Optimal. Leaf size=63 \[ -2 \text {ArcTan}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}+2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \]

[Out]

-2*arctan(arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^(1/2)+2*arct
anh(tanh(b*x+a))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2190, 2192} \begin {gather*} 2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-2 \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))} \text {ArcTan}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x,x]

[Out]

-2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]
+ 2*Sqrt[ArcTanh[Tanh[a + b*x]]]

Rule 2190

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 2192

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]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 61, normalized size = 0.97 \begin {gather*} 2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-2 \tanh ^{-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))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 54, normalized size = 0.86

method	result	size
default	\(2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\, \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctanh(tanh(b*x+a))^(1/2)-2*(arctanh(tanh(b*x+a))-b*x)^(1/2)*arctanh(arctanh(tanh(b*x+a))^(1/2)/(arctanh(ta
nh(b*x+a))-b*x)^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(arctanh(tanh(b*x + a)))/x, x)

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Fricas [A]
time = 0.34, size = 73, normalized size = 1.16 \begin {gather*} \left [\sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, \sqrt {b x + a}, 2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + 2 \, \sqrt {b x + a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.38, size = 40, normalized size = 0.63 \begin {gather*} \sqrt {2} {\left (\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \sqrt {2} \sqrt {b x + a}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*(sqrt(2)*a*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + sqrt(2)*sqrt(b*x + a))

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Mupad [B]
time = 2.36, size = 308, normalized size = 4.89 \begin {gather*} 2\,\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}+\ln \left (-\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,\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 {\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}-b\,x}+b\,x}{x\,\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}-b\,x}}\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}-b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(1/2)/x,x)

[Out]

2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2) + log(
-(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((exp(2*a)*e
xp(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))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2 - b*x)^(1/2) + b*x)/(x*(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 - b*x)^(1/2)))*(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 - b*x)^(1/2)

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