3.307 \(\int \frac{\text{sech}(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=69 \[ \frac{b \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )}+\frac{a \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}-\frac{b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0698349, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2668, 706, 31, 635, 204, 260} \[ \frac{b \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )}+\frac{a \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}-\frac{b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]/(a + b*Sinh[c + d*x]),x]

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.0963634, size = 114, normalized size = 1.65 \[ -\frac{b \left (\left (\sqrt{-b^2}-a\right ) \log \left (\sqrt{-b^2}-b \sinh (c+d x)\right )-2 \sqrt{-b^2} \log (a+b \sinh (c+d x))+\left (a+\sqrt{-b^2}\right ) \log \left (\sqrt{-b^2}+b \sinh (c+d x)\right )\right )}{2 \sqrt{-b^2} d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]/(a + b*Sinh[c + d*x]),x]

[Out]

-(b*((-a + Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Sinh[c + d*x]] - 2*Sqrt[-b^2]*Log[a + b*Sinh[c + d*x]] + (a + Sqrt[-
b^2])*Log[Sqrt[-b^2] + b*Sinh[c + d*x]]))/(2*Sqrt[-b^2]*(a^2 + b^2)*d)

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Maple [A]  time = 0.004, size = 100, normalized size = 1.5 \begin{align*}{\frac{b}{d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }-{\frac{b}{d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{a\arctan \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.6696, size = 128, normalized size = 1.86 \begin{align*} -\frac{2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac{b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15555, size = 247, normalized size = 3.58 \begin{align*} \frac{2 \, a \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + b \log \left (\frac{2 \,{\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - b \log \left (\frac{2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(2*a*arctan(cosh(d*x + c) + sinh(d*x + c)) + b*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) -
b*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/((a^2 + b^2)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral(sech(c + d*x)/(a + b*sinh(c + d*x)), x)

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Giac [A]  time = 1.14886, size = 171, normalized size = 2.48 \begin{align*} \frac{b^{2} \log \left ({\left | b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{2} b d + b^{3} d} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{2 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{2 \,{\left (a^{2} d + b^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

b^2*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^2*b*d + b^3*d) + 1/2*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c)
- 1)*e^(-d*x - c)))*a/(a^2*d + b^2*d) - 1/2*b*log((e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^2*d + b^2*d)