\(\int \sqrt {a \tanh ^2(x)} \, dx\) [25]

Optimal result

Integrand size = 10, antiderivative size = 16 \[ \int \sqrt {a \tanh ^2(x)} \, dx=\coth (x) \log (\cosh (x)) \sqrt {a \tanh ^2(x)} \]

[Out]

coth(x)*ln(cosh(x))*(a*tanh(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3739, 3556} \[ \int \sqrt {a \tanh ^2(x)} \, dx=\coth (x) \sqrt {a \tanh ^2(x)} \log (\cosh (x)) \]

[In]

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

[Out]

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

Rule 3556

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

Rule 3739

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
]])

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \sqrt {a \tanh ^2(x)} \, dx=\coth (x) \log (\cosh (x)) \sqrt {a \tanh ^2(x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.62

method	result	size
derivativedivides	\(-\frac {\sqrt {a \tanh \left (x \right )^{2}}\, \left (\ln \left (\tanh \left (x \right )-1\right )+\ln \left (1+\tanh \left (x \right )\right )\right )}{2 \tanh \left (x \right )}\)	\(26\)
default	\(-\frac {\sqrt {a \tanh \left (x \right )^{2}}\, \left (\ln \left (\tanh \left (x \right )-1\right )+\ln \left (1+\tanh \left (x \right )\right )\right )}{2 \tanh \left (x \right )}\)	\(26\)
risch	\(-\frac {\sqrt {\frac {a \left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right ) x}{{\mathrm e}^{2 x}-1}+\frac {\sqrt {\frac {a \left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right ) \ln \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{2 x}-1}\)	\(81\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 4.50 \[ \int \sqrt {a \tanh ^2(x)} \, dx=-\frac {{\left (x e^{\left (2 \, x\right )} - {\left (e^{\left (2 \, x\right )} + 1\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + x\right )} \sqrt {\frac {a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}}}{e^{\left (2 \, x\right )} - 1} \]

[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)

Sympy [F]

\[ \int \sqrt {a \tanh ^2(x)} \, dx=\int \sqrt {a \tanh ^{2}{\left (x \right )}}\, dx \]

[In]

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

[Out]

Integral(sqrt(a*tanh(x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \sqrt {a \tanh ^2(x)} \, dx=-\sqrt {a} x - \sqrt {a} \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.94 \[ \int \sqrt {a \tanh ^2(x)} \, dx=-{\left (x \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \log \left (e^{\left (2 \, x\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right )\right )} \sqrt {a} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a \tanh ^2(x)} \, dx=\int \sqrt {a\,{\mathrm {tanh}\left (x\right )}^2} \,d x \]

[In]

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

[Out]

int((a*tanh(x)^2)^(1/2), x)