3.25 \(\int \sqrt{a \tanh ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ \coth (x) \sqrt{a \tanh ^2(x)} \log (\cosh (x)) \]

[Out]

Coth[x]*Log[Cosh[x]]*Sqrt[a*Tanh[x]^2]

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Rubi [A]  time = 0.0153141, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3658, 3475} \[ \coth (x) \sqrt{a \tanh ^2(x)} \log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*Tanh[x]^2],x]

[Out]

Coth[x]*Log[Cosh[x]]*Sqrt[a*Tanh[x]^2]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{a \tanh ^2(x)} \, dx &=\left (\coth (x) \sqrt{a \tanh ^2(x)}\right ) \int \tanh (x) \, dx\\ &=\coth (x) \log (\cosh (x)) \sqrt{a \tanh ^2(x)}\\ \end{align*}

Mathematica [A]  time = 0.0063562, size = 16, normalized size = 1. \[ \coth (x) \sqrt{a \tanh ^2(x)} \log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*Tanh[x]^2],x]

[Out]

Coth[x]*Log[Cosh[x]]*Sqrt[a*Tanh[x]^2]

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Maple [A]  time = 0.033, size = 26, normalized size = 1.6 \begin{align*} -{\frac{\ln \left ( \tanh \left ( x \right ) -1 \right ) +\ln \left ( 1+\tanh \left ( x \right ) \right ) }{2\,\tanh \left ( x \right ) }\sqrt{a \left ( \tanh \left ( x \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*tanh(x)^2)^(1/2)*(ln(tanh(x)-1)+ln(1+tanh(x)))/tanh(x)

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Maxima [A]  time = 1.57406, size = 26, normalized size = 1.62 \begin{align*} -\sqrt{a} x - \sqrt{a} \log \left (e^{\left (-2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a)*x - sqrt(a)*log(e^(-2*x) + 1)

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Fricas [B]  time = 2.24951, size = 196, normalized size = 12.25 \begin{align*} -\frac{{\left (x e^{\left (2 \, x\right )} -{\left (e^{\left (2 \, x\right )} + 1\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + x\right )} \sqrt{\frac{a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}}}{e^{\left (2 \, x\right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*e^(2*x) - (e^(2*x) + 1)*log(2*cosh(x)/(cosh(x) - sinh(x))) + x)*sqrt((a*e^(4*x) - 2*a*e^(2*x) + a)/(e^(4*x
) + 2*e^(2*x) + 1))/(e^(2*x) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \tanh ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tanh(x)**2)**(1/2),x)

[Out]

Integral(sqrt(a*tanh(x)**2), x)

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Giac [B]  time = 1.19594, size = 42, normalized size = 2.62 \begin{align*} -{\left (x \mathrm{sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \log \left (e^{\left (2 \, x\right )} + 1\right ) \mathrm{sgn}\left (e^{\left (4 \, x\right )} - 1\right )\right )} \sqrt{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*sgn(e^(4*x) - 1) - log(e^(2*x) + 1)*sgn(e^(4*x) - 1))*sqrt(a)