3.1.64 \(\int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx\) [64]

Optimal. Leaf size=53 \[ \frac {\log (1-\cosh (x))}{2 (a+b)}-\frac {\log (1+\cosh (x))}{2 (a-b)}+\frac {b \log (b+a \cosh (x))}{a^2-b^2} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3957, 2800, 815} \begin {gather*} \frac {b \log (a \cosh (x)+b)}{a^2-b^2}+\frac {\log (1-\cosh (x))}{2 (a+b)}-\frac {\log (\cosh (x)+1)}{2 (a-b)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csch[x]/(a + b*Sech[x]),x]

[Out]

Log[1 - Cosh[x]]/(2*(a + b)) - Log[1 + Cosh[x]]/(2*(a - b)) + (b*Log[b + a*Cosh[x]])/(a^2 - b^2)

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx &=-\int \frac {\coth (x)}{-b-a \cosh (x)} \, dx\\ &=\text {Subst}\left (\int \frac {x}{(-b+x) \left (a^2-x^2\right )} \, dx,x,-a \cosh (x)\right )\\ &=\text {Subst}\left (\int \left (\frac {1}{2 (a-b) (a-x)}-\frac {b}{(a-b) (a+b) (b-x)}+\frac {1}{2 (a+b) (a+x)}\right ) \, dx,x,-a \cosh (x)\right )\\ &=\frac {\log (1-\cosh (x))}{2 (a+b)}-\frac {\log (1+\cosh (x))}{2 (a-b)}+\frac {b \log (b+a \cosh (x))}{a^2-b^2}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 37, normalized size = 0.70 \begin {gather*} \frac {b \log (b+a \cosh (x))-b \log (\sinh (x))+a \log \left (\tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]/(a + b*Sech[x]),x]

[Out]

(b*Log[b + a*Cosh[x]] - b*Log[Sinh[x]] + a*Log[Tanh[x/2]])/(a^2 - b^2)

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Maple [A]
time = 0.79, size = 48, normalized size = 0.91

method	result	size
default	\(\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b}+\frac {b \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+a +b \right )}{\left (a +b \right ) \left (a -b \right )}\)	\(48\)
risch	\(-\frac {x}{a +b}+\frac {x}{a -b}-\frac {2 b x}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{a +b}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{a -b}+\frac {b \ln \left ({\mathrm e}^{2 x}+\frac {2 b \,{\mathrm e}^{x}}{a}+1\right )}{a^{2}-b^{2}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)/(a+b*sech(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 59, normalized size = 1.11 \begin {gather*} \frac {b \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{2} - b^{2}} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*sech(x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 58, normalized size = 1.09 \begin {gather*} \frac {b \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a + b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + {\left (a - b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} - b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*sech(x)),x, algorithm="fricas")

[Out]

(b*log(2*(a*cosh(x) + b)/(cosh(x) - sinh(x))) - (a + b)*log(cosh(x) + sinh(x) + 1) + (a - b)*log(cosh(x) + sin
h(x) - 1))/(a^2 - b^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {csch}{\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*sech(x)),x)

[Out]

Integral(csch(x)/(a + b*sech(x)), x)

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Giac [A]
time = 0.38, size = 65, normalized size = 1.23 \begin {gather*} \frac {a b \log \left ({\left | a {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{3} - a b^{2}} - \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{2 \, {\left (a - b\right )}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{2 \, {\left (a + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*sech(x)),x, algorithm="giac")

[Out]

a*b*log(abs(a*(e^(-x) + e^x) + 2*b))/(a^3 - a*b^2) - 1/2*log(e^(-x) + e^x + 2)/(a - b) + 1/2*log(e^(-x) + e^x
- 2)/(a + b)

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Mupad [B]
time = 1.74, size = 148, normalized size = 2.79 \begin {gather*} \frac {\ln \left (128\,a\,b-32\,a^2-128\,b^2+32\,a^2\,{\mathrm {e}}^x+128\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a+b}-\frac {\ln \left (-128\,a\,b-32\,a^2-128\,b^2-32\,a^2\,{\mathrm {e}}^x-128\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a-b}+\frac {b\,\ln \left (16\,a\,b^2-4\,a^3\,{\mathrm {e}}^{2\,x}-4\,a^3+32\,b^3\,{\mathrm {e}}^x-8\,a^2\,b\,{\mathrm {e}}^x+16\,a\,b^2\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)*(a + b/cosh(x))),x)

[Out]

log(128*a*b - 32*a^2 - 128*b^2 + 32*a^2*exp(x) + 128*b^2*exp(x) - 128*a*b*exp(x))/(a + b) - log(- 128*a*b - 32
*a^2 - 128*b^2 - 32*a^2*exp(x) - 128*b^2*exp(x) - 128*a*b*exp(x))/(a - b) + (b*log(16*a*b^2 - 4*a^3*exp(2*x) -
4*a^3 + 32*b^3*exp(x) - 8*a^2*b*exp(x) + 16*a*b^2*exp(2*x)))/(a^2 - b^2)

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