3.1.26 \(\int \frac {1}{\sqrt {a \tanh ^2(x)}} \, dx\) [26]

Optimal. Leaf size=16 \[ \frac {\log (\sinh (x)) \tanh (x)}{\sqrt {a \tanh ^2(x)}} \]

[Out]

ln(sinh(x))*tanh(x)/(a*tanh(x)^2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3739, 3556} \begin {gather*} \frac {\tanh (x) \log (\sinh (x))}{\sqrt {a \tanh ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Tanh[x]^2],x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} \frac {\log (\sinh (x)) \tanh (x)}{\sqrt {a \tanh ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Tanh[x]^2],x]

[Out]

(Log[Sinh[x]]*Tanh[x])/Sqrt[a*Tanh[x]^2]

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Maple [A]
time = 0.79, size = 29, normalized size = 1.81

method	result	size
derivativedivides	\(-\frac {\tanh \left (x \right ) \left (\ln \left (1+\tanh \left (x \right )\right )+\ln \left (\tanh \left (x \right )-1\right )-2 \ln \left (\tanh \left (x \right )\right )\right )}{2 \sqrt {a \left (\tanh ^{2}\left (x \right )\right )}}\)	\(29\)
default	\(-\frac {\tanh \left (x \right ) \left (\ln \left (1+\tanh \left (x \right )\right )+\ln \left (\tanh \left (x \right )-1\right )-2 \ln \left (\tanh \left (x \right )\right )\right )}{2 \sqrt {a \left (\tanh ^{2}\left (x \right )\right )}}\)	\(29\)
risch	\(-\frac {\left ({\mathrm e}^{2 x}-1\right ) x}{\sqrt {\frac {a \left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}+\frac {\left ({\mathrm e}^{2 x}-1\right ) \ln \left ({\mathrm e}^{2 x}-1\right )}{\sqrt {\frac {a \left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).
time = 0.49, size = 31, normalized size = 1.94 \begin {gather*} -\frac {x}{\sqrt {a}} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{\sqrt {a}} - \frac {\log \left (e^{\left (-x\right )} - 1\right )}{\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/sqrt(a) - log(e^(-x) + 1)/sqrt(a) - log(e^(-x) - 1)/sqrt(a)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (14) = 28\).
time = 0.40, size = 76, normalized size = 4.75 \begin {gather*} -\frac {{\left (x e^{\left (2 \, x\right )} - {\left (e^{\left (2 \, x\right )} + 1\right )} \log \left (\frac {2 \, \sinh \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}}}{a e^{\left (2 \, x\right )} - a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a \tanh ^{2}{\left (x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a*tanh(x)**2), x)

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Giac [A]
time = 0.42, size = 1, normalized size = 0.06 \begin {gather*} 0 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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Mupad [B]
time = 1.20, size = 14, normalized size = 0.88 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {\mathrm {tanh}\left (x\right )}{\sqrt {{\mathrm {tanh}\left (x\right )}^2}}\right )}{\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(tanh(x)/(tanh(x)^2)^(1/2))/a^(1/2)

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