3.29 \(\int \sqrt{-\tanh ^2(c+d x)} \, dx\)

Optimal. Leaf size=31 \[ \frac{\sqrt{-\tanh ^2(c+d x)} \coth (c+d x) \log (\cosh (c+d x))}{d} \]

[Out]

(Coth[c + d*x]*Log[Cosh[c + d*x]]*Sqrt[-Tanh[c + d*x]^2])/d

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Rubi [A]  time = 0.0176316, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3658, 3475} \[ \frac{\sqrt{-\tanh ^2(c+d x)} \coth (c+d x) \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-Tanh[c + d*x]^2],x]

[Out]

(Coth[c + d*x]*Log[Cosh[c + d*x]]*Sqrt[-Tanh[c + d*x]^2])/d

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{-\tanh ^2(c+d x)} \, dx &=\left (\coth (c+d x) \sqrt{-\tanh ^2(c+d x)}\right ) \int \tanh (c+d x) \, dx\\ &=\frac{\coth (c+d x) \log (\cosh (c+d x)) \sqrt{-\tanh ^2(c+d x)}}{d}\\ \end{align*}

Mathematica [A]  time = 0.0330789, size = 31, normalized size = 1. \[ \frac{\sqrt{-\tanh ^2(c+d x)} \coth (c+d x) \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-Tanh[c + d*x]^2],x]

[Out]

(Coth[c + d*x]*Log[Cosh[c + d*x]]*Sqrt[-Tanh[c + d*x]^2])/d

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Maple [A]  time = 0.029, size = 45, normalized size = 1.5 \begin{align*} -{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) +\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{2\,d\tanh \left ( dx+c \right ) }\sqrt{- \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-tanh(d*x+c)^2)^(1/2),x)

[Out]

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

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Maxima [C]  time = 1.62238, size = 38, normalized size = 1.23 \begin{align*} -\frac{i \,{\left (d x + c\right )}}{d} - \frac{i \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tanh(d*x+c)^2)^(1/2),x, algorithm="maxima")

[Out]

-I*(d*x + c)/d - I*log(e^(-2*d*x - 2*c) + 1)/d

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Fricas [C]  time = 2.34377, size = 55, normalized size = 1.77 \begin{align*} \frac{-i \, d x + i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tanh(d*x+c)^2)^(1/2),x, algorithm="fricas")

[Out]

(-I*d*x + I*log(e^(2*d*x + 2*c) + 1))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tanh(d*x+c)**2)**(1/2),x)

[Out]

Integral(sqrt(-tanh(c + d*x)**2), x)

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Giac [C]  time = 1.18398, size = 73, normalized size = 2.35 \begin{align*} \frac{i \,{\left (d x + c\right )} \mathrm{sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right ) - i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) \mathrm{sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-tanh(d*x+c)^2)^(1/2),x, algorithm="giac")

[Out]

(I*(d*x + c)*sgn(-e^(4*d*x + 4*c) + 1) - I*log(e^(2*d*x + 2*c) + 1)*sgn(-e^(4*d*x + 4*c) + 1))/d