\(\int \sqrt {-\tanh ^2(c+d x)} \, dx\) [29]

Optimal result

Integrand size = 14, antiderivative size = 31 \[ \int \sqrt {-\tanh ^2(c+d x)} \, dx=\frac {\coth (c+d x) \log (\cosh (c+d x)) \sqrt {-\tanh ^2(c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3739, 3556} \[ \int \sqrt {-\tanh ^2(c+d x)} \, dx=\frac {\sqrt {-\tanh ^2(c+d x)} \coth (c+d x) \log (\cosh (c+d x))}{d} \]

[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 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 (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] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \sqrt {-\tanh ^2(c+d x)} \, dx=\frac {\coth (c+d x) \log (\cosh (c+d x)) \sqrt {-\tanh ^2(c+d x)}}{d} \]

[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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(-\frac {\sqrt {-\tanh \left (d x +c \right )^{2}}\, \left (\ln \left (\tanh \left (d x +c \right )-1\right )+\ln \left (\tanh \left (d x +c \right )+1\right )\right )}{2 d \tanh \left (d x +c \right )}\)	\(45\)
default	\(-\frac {\sqrt {-\tanh \left (d x +c \right )^{2}}\, \left (\ln \left (\tanh \left (d x +c \right )-1\right )+\ln \left (\tanh \left (d x +c \right )+1\right )\right )}{2 d \tanh \left (d x +c \right )}\)	\(45\)
risch	\(\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, x}{{\mathrm e}^{2 d x +2 c}-1}-\frac {2 \left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, \left (d x +c \right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right ) d}+\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right ) d}\)	\(192\)

[In]

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

[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)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \sqrt {-\tanh ^2(c+d x)} \, dx=\frac {-i \, d x + i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{d} \]

[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

Sympy [F]

\[ \int \sqrt {-\tanh ^2(c+d x)} \, dx=\int \sqrt {- \tanh ^{2}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral(sqrt(-tanh(c + d*x)**2), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \sqrt {-\tanh ^2(c+d x)} \, dx=-\frac {i \, {\left (d x + c\right )}}{d} - \frac {i \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} \]

[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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \sqrt {-\tanh ^2(c+d x)} \, dx=\frac {i \, {\left (d x + c\right )} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right ) - i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}{d} \]

[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

Mupad [F(-1)]

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

[In]

int((-tanh(c + d*x)^2)^(1/2),x)

[Out]

int((-tanh(c + d*x)^2)^(1/2), x)