3.30 \(\int \frac {1}{\sqrt {-\tanh ^2(c+d x)}} \, dx\)

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

[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 = 0.02, 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 = {3658, 3475} \[ \frac {\tanh (c+d x) \log (\sinh (c+d x))}{d \sqrt {-\tanh ^2(c+d x)}} \]

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 3475

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

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
]])

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 = 0.08, size = 39, normalized size = 1.26 \[ \frac {\tanh (c+d x) (\log (\tanh (c+d x))+\log (\cosh (c+d x)))}{d \sqrt {-\tanh ^2(c+d x)}} \]

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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fricas [C]  time = 0.50, size = 23, normalized size = 0.74 \[ \frac {i \, d x - i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{d} \]

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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giac [C]  time = 0.14, size = 63, normalized size = 2.03 \[ \frac {\frac {-2 i \, d x - 2 i \, c}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} + \frac {2 i \, \log \left (-i \, e^{\left (2 \, d x + 2 \, c\right )} + i\right )}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 52, normalized size = 1.68 \[ -\frac {\tanh \left (d x +c \right ) \left (\ln \left (\tanh \left (d x +c \right )-1\right )+\ln \left (1+\tanh \left (d x +c \right )\right )-2 \ln \left (\tanh \left (d x +c \right )\right )\right )}{2 d \sqrt {-\left (\tanh ^{2}\left (d x +c \right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.45, size = 45, normalized size = 1.45 \[ \frac {i \, {\left (d x + c\right )}}{d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} \]

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \tanh ^{2}{\left (c + d x \right )}}}\, dx \]

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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