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

Optimal. Leaf size=62 \[ \frac {2 (a c-b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tanh \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{c \sqrt {c-d} \sqrt {c+d}}+\frac {b \tan ^{-1}(\sinh (x))}{c} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2828, 3001, 3770, 2659, 208} \[ \frac {2 (a c-b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tanh \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{c \sqrt {c-d} \sqrt {c+d}}+\frac {b \tan ^{-1}(\sinh (x))}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTan[Sinh[x]])/c + (2*(a*c - b*d)*ArcTanh[(Sqrt[c - d]*Tanh[x/2])/Sqrt[c + d]])/(c*Sqrt[c - d]*Sqrt[c + d
])

Rule 208

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[((a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^n, x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 63, normalized size = 1.02 \[ \frac {2 \left (\frac {(b d-a c) \tan ^{-1}\left (\frac {(c-d) \tanh \left (\frac {x}{2}\right )}{\sqrt {d^2-c^2}}\right )}{\sqrt {d^2-c^2}}+b \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 249, normalized size = 4.02 \[ \left [-\frac {{\left (a c - b d\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {d^{2} \cosh \relax (x)^{2} + d^{2} \sinh \relax (x)^{2} + 2 \, c d \cosh \relax (x) + 2 \, c^{2} - d^{2} + 2 \, {\left (d^{2} \cosh \relax (x) + c d\right )} \sinh \relax (x) + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cosh \relax (x) + d \sinh \relax (x) + c\right )}}{d \cosh \relax (x)^{2} + d \sinh \relax (x)^{2} + 2 \, c \cosh \relax (x) + 2 \, {\left (d \cosh \relax (x) + c\right )} \sinh \relax (x) + d}\right ) - 2 \, {\left (b c^{2} - b d^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )}{c^{3} - c d^{2}}, -\frac {2 \, {\left ({\left (a c - b d\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cosh \relax (x) + d \sinh \relax (x) + c\right )}}{c^{2} - d^{2}}\right ) - {\left (b c^{2} - b d^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )\right )}}{c^{3} - c d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((a*c - b*d)*sqrt(c^2 - d^2)*log((d^2*cosh(x)^2 + d^2*sinh(x)^2 + 2*c*d*cosh(x) + 2*c^2 - d^2 + 2*(d^2*cosh(
x) + c*d)*sinh(x) + 2*sqrt(c^2 - d^2)*(d*cosh(x) + d*sinh(x) + c))/(d*cosh(x)^2 + d*sinh(x)^2 + 2*c*cosh(x) +
2*(d*cosh(x) + c)*sinh(x) + d)) - 2*(b*c^2 - b*d^2)*arctan(cosh(x) + sinh(x)))/(c^3 - c*d^2), -2*((a*c - b*d)*
sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cosh(x) + d*sinh(x) + c)/(c^2 - d^2)) - (b*c^2 - b*d^2)*arctan(co
sh(x) + sinh(x)))/(c^3 - c*d^2)]

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giac [A]  time = 0.14, size = 53, normalized size = 0.85 \[ \frac {2 \, b \arctan \left (e^{x}\right )}{c} + \frac {2 \, {\left (a c - b d\right )} \arctan \left (\frac {d e^{x} + c}{\sqrt {-c^{2} + d^{2}}}\right )}{\sqrt {-c^{2} + d^{2}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 89, normalized size = 1.44 \[ \frac {2 \arctanh \left (\frac {\left (c -d \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) a}{\sqrt {\left (c +d \right ) \left (c -d \right )}}-\frac {2 \arctanh \left (\frac {\left (c -d \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) b d}{c \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*d^2>0)', see `assume?`
for more details)Is 4*c^2-4*d^2 positive or negative?

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mupad [B]  time = 6.35, size = 636, normalized size = 10.26 \[ \frac {\ln \left (\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c-b\,d\right )\,\left (\frac {32\,\left (a^2\,c^2\,d-2\,a\,b\,c\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3-2\,b^2\,c^2\,d+3\,{\mathrm {e}}^x\,b^2\,c\,d^2+2\,b^2\,d^3\right )}{d^5}+\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (\frac {32\,c^2\,\left (2\,b\,d^2-4\,a\,c^2\,{\mathrm {e}}^x+a\,d^2\,{\mathrm {e}}^x-2\,a\,c\,d+3\,b\,c\,d\,{\mathrm {e}}^x\right )}{d^5}-\frac {32\,c^2\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c-b\,d\right )\,\left (4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2-2\,d^3\right )}{d^5\,\left (c\,d^2-c^3\right )}\right )\,\left (a\,c-b\,d\right )}{c\,d^2-c^3}\right )}{c\,d^2-c^3}-\frac {32\,b\,\left (a\,c-b\,d\right )\,\left (2\,b\,d-a\,d\,{\mathrm {e}}^x+4\,b\,c\,{\mathrm {e}}^x\right )}{d^5}\right )\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c-b\,d\right )}{c\,d^2-c^3}-\frac {\ln \left (-\frac {32\,b\,\left (a\,c-b\,d\right )\,\left (2\,b\,d-a\,d\,{\mathrm {e}}^x+4\,b\,c\,{\mathrm {e}}^x\right )}{d^5}-\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c-b\,d\right )\,\left (\frac {32\,\left (a^2\,c^2\,d-2\,a\,b\,c\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3-2\,b^2\,c^2\,d+3\,{\mathrm {e}}^x\,b^2\,c\,d^2+2\,b^2\,d^3\right )}{d^5}-\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (\frac {32\,c^2\,\left (2\,b\,d^2-4\,a\,c^2\,{\mathrm {e}}^x+a\,d^2\,{\mathrm {e}}^x-2\,a\,c\,d+3\,b\,c\,d\,{\mathrm {e}}^x\right )}{d^5}+\frac {32\,c^2\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c-b\,d\right )\,\left (4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2-2\,d^3\right )}{d^5\,\left (c\,d^2-c^3\right )}\right )\,\left (a\,c-b\,d\right )}{c\,d^2-c^3}\right )}{c\,d^2-c^3}\right )\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c-b\,d\right )}{c\,d^2-c^3}-\frac {b\,\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}}{c}+\frac {b\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*log(exp(x) + 1i)*1i)/c - (b*log(exp(x) - 1i)*1i)/c + (log((((c + d)*(c - d))^(1/2)*(a*c - b*d)*((32*(2*b^2*
d^3 + a^2*c^2*d - 2*b^2*c^2*d - 4*b^2*c^3*exp(x) + 3*b^2*c*d^2*exp(x) - 2*a*b*c*d^2))/d^5 + (((c + d)*(c - d))
^(1/2)*((32*c^2*(2*b*d^2 - 4*a*c^2*exp(x) + a*d^2*exp(x) - 2*a*c*d + 3*b*c*d*exp(x)))/d^5 - (32*c^2*((c + d)*(
c - d))^(1/2)*(a*c - b*d)*(3*c^2*d - 2*d^3 + 4*c^3*exp(x) - 3*c*d^2*exp(x)))/(d^5*(c*d^2 - c^3)))*(a*c - b*d))
/(c*d^2 - c^3)))/(c*d^2 - c^3) - (32*b*(a*c - b*d)*(2*b*d - a*d*exp(x) + 4*b*c*exp(x)))/d^5)*((c + d)*(c - d))
^(1/2)*(a*c - b*d))/(c*d^2 - c^3) - (log(- (32*b*(a*c - b*d)*(2*b*d - a*d*exp(x) + 4*b*c*exp(x)))/d^5 - (((c +
d)*(c - d))^(1/2)*(a*c - b*d)*((32*(2*b^2*d^3 + a^2*c^2*d - 2*b^2*c^2*d - 4*b^2*c^3*exp(x) + 3*b^2*c*d^2*exp(
x) - 2*a*b*c*d^2))/d^5 - (((c + d)*(c - d))^(1/2)*((32*c^2*(2*b*d^2 - 4*a*c^2*exp(x) + a*d^2*exp(x) - 2*a*c*d
+ 3*b*c*d*exp(x)))/d^5 + (32*c^2*((c + d)*(c - d))^(1/2)*(a*c - b*d)*(3*c^2*d - 2*d^3 + 4*c^3*exp(x) - 3*c*d^2
*exp(x)))/(d^5*(c*d^2 - c^3)))*(a*c - b*d))/(c*d^2 - c^3)))/(c*d^2 - c^3))*((c + d)*(c - d))^(1/2)*(a*c - b*d)
)/(c*d^2 - c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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