3.1.30 \(\int \frac {1}{\sqrt {-\tanh ^2(c+d x)}} \, dx\) [30]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 = {3739, 3556} \begin {gather*} \frac {\tanh (c+d x) \log (\sinh (c+d x))}{d \sqrt {-\tanh ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-Tanh[c + d*x]^2],x]

[Out]

(Log[Sinh[c + d*x]]*Tanh[c + d*x])/(d*Sqrt[-Tanh[c + d*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*} \int \frac {1}{\sqrt {-\tanh ^2(c+d x)}} \, dx &=\frac {\tanh (c+d x) \int \coth (c+d x) \, dx}{\sqrt {-\tanh ^2(c+d x)}}\\ &=\frac {\log (\sinh (c+d x)) \tanh (c+d x)}{d \sqrt {-\tanh ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 39, normalized size = 1.26 \begin {gather*} \frac {(\log (\cosh (c+d x))+\log (\tanh (c+d x))) \tanh (c+d x)}{d \sqrt {-\tanh ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-Tanh[c + d*x]^2],x]

[Out]

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

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Maple [A]
time = 1.64, size = 56, normalized size = 1.81

method	result	size
derivativedivides	\(\frac {\tanh \left (d x +c \right ) \left (2 \ln \left (\tanh \left (d x +c \right )\right )-\ln \left (\tanh \left (d x +c \right )-1\right )-\ln \left (\tanh \left (d x +c \right )+1\right )\right )}{2 d \sqrt {-\left (\tanh ^{2}\left (d x +c \right )\right )}}\)	\(56\)
default	\(\frac {\tanh \left (d x +c \right ) \left (2 \ln \left (\tanh \left (d x +c \right )\right )-\ln \left (\tanh \left (d x +c \right )-1\right )-\ln \left (\tanh \left (d x +c \right )+1\right )\right )}{2 d \sqrt {-\left (\tanh ^{2}\left (d x +c \right )\right )}}\)	\(56\)
risch	\(\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right ) x}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}}\, \left (1+{\mathrm e}^{2 d x +2 c}\right )}-\frac {2 \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (d x +c \right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}}\, \left (1+{\mathrm e}^{2 d x +2 c}\right ) d}+\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right ) \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}}\, \left (1+{\mathrm e}^{2 d x +2 c}\right ) d}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.48, size = 45, normalized size = 1.45 \begin {gather*} \frac {i \, {\left (d x + c\right )}}{d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.42, size = 23, normalized size = 0.74 \begin {gather*} \frac {i \, d x - i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {- \tanh ^{2}{\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 0.41, size = 63, normalized size = 2.03 \begin {gather*} -\frac {\frac {i \, d x + i \, c}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} - \frac {i \, \log \left (-i \, e^{\left (2 \, d x + 2 \, c\right )} + i\right )}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.22, size = 24, normalized size = 0.77 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\mathrm {tanh}\left (c+d\,x\right )}{\sqrt {-{\mathrm {tanh}\left (c+d\,x\right )}^2}}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-tanh(c + d*x)^2)^(1/2),x)

[Out]

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

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