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

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Rubi [A]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, 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 31

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

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]

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 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 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]

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*}

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Mathematica [A]  time = 0.09, 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]

-1/2*(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 + Sq
rt[-b^2])*Log[Sqrt[-b^2] + b*Sinh[c + d*x]]))/(Sqrt[-b^2]*(a^2 + b^2)*d)

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fricas [A]  time = 0.50, size = 92, normalized size = 1.33 \[ \frac {2 \, a \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + b \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - b \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} \]

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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giac [A]  time = 0.16, size = 121, normalized size = 1.75 \[ \frac {\frac {2 \, b^{2} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{a^{2} + b^{2}} - \frac {b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}}}{2 \, d} \]

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]

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

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maple [A]  time = 0.00, size = 100, normalized size = 1.45 \[ \frac {b \ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d \left (a^{2}+b^{2}\right )}-\frac {b \ln \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a^{2}+b^{2}\right )}+\frac {2 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}+b^{2}\right )} \]

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 = 0.44, size = 95, normalized size = 1.38 \[ -\frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} \]

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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mupad [B]  time = 1.23, size = 129, normalized size = 1.87 \[ \frac {b\,\ln \left (2\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-4\,b^3-a^2\,b+4\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a^2\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+8\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^2+d\,b^2}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{b\,d+a\,d\,1{}\mathrm {i}}-\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d+b\,d\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)*(a + b*sinh(c + d*x))),x)

[Out]

(b*log(2*a^3*exp(d*x)*exp(c) - 4*b^3 - a^2*b + 4*b^3*exp(2*c)*exp(2*d*x) + a^2*b*exp(2*c)*exp(2*d*x) + 8*a*b^2
*exp(d*x)*exp(c)))/(a^2*d + b^2*d) - (log(exp(c + d*x)*1i + 1)*1i)/(a*d + b*d*1i) - log(exp(c + d*x) + 1i)/(a*
d*1i + b*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

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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