3.6.72 \(\int \frac {a+b \text {sech}(x)}{c+d \cosh (x)} \, dx\) [572]

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

[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 = 0.10, 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 = {2907, 3080, 3855, 2738, 214} \begin {gather*} \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 \text {ArcTan}(\sinh (x))}{c} \end {gather*}

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 214

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

Rule 2738

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 2907

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 3080

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 3855

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)) \text {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 = 0.09, size = 63, normalized size = 1.02 \begin {gather*} \frac {2 \left (b \text {ArcTan}\left (\tanh \left (\frac {x}{2}\right )\right )+\frac {(-a c+b d) \text {ArcTan}\left (\frac {(c-d) \tanh \left (\frac {x}{2}\right )}{\sqrt {-c^2+d^2}}\right )}{\sqrt {-c^2+d^2}}\right )}{c} \end {gather*}

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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Maple [A]
time = 1.01, size = 59, normalized size = 0.95

method	result	size
default	\(-\frac {2 \left (-a c +b d \right ) \arctanh \left (\frac {\left (c -d \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{c \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{c}\)	\(59\)
risch	\(\frac {i b \ln \left ({\mathrm e}^{x}+i\right )}{c}-\frac {i b \ln \left ({\mathrm e}^{x}-i\right )}{c}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}-d^{2}}\, c -c^{2}+d^{2}}{\sqrt {c^{2}-d^{2}}\, d}\right ) a}{\sqrt {c^{2}-d^{2}}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}-d^{2}}\, c -c^{2}+d^{2}}{\sqrt {c^{2}-d^{2}}\, d}\right ) b d}{\sqrt {c^{2}-d^{2}}\, c}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}-d^{2}}\, c +c^{2}-d^{2}}{\sqrt {c^{2}-d^{2}}\, d}\right ) a}{\sqrt {c^{2}-d^{2}}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}-d^{2}}\, c +c^{2}-d^{2}}{\sqrt {c^{2}-d^{2}}\, d}\right ) b d}{\sqrt {c^{2}-d^{2}}\, c}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 de

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Fricas [A]
time = 0.58, size = 249, normalized size = 4.02 \begin {gather*} \left [-\frac {{\left (a c - b d\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {d^{2} \cosh \left (x\right )^{2} + d^{2} \sinh \left (x\right )^{2} + 2 \, c d \cosh \left (x\right ) + 2 \, c^{2} - d^{2} + 2 \, {\left (d^{2} \cosh \left (x\right ) + c d\right )} \sinh \left (x\right ) + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cosh \left (x\right ) + d \sinh \left (x\right ) + c\right )}}{d \cosh \left (x\right )^{2} + d \sinh \left (x\right )^{2} + 2 \, c \cosh \left (x\right ) + 2 \, {\left (d \cosh \left (x\right ) + c\right )} \sinh \left (x\right ) + d}\right ) - 2 \, {\left (b c^{2} - b d^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\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 \left (x\right ) + d \sinh \left (x\right ) + c\right )}}{c^{2} - d^{2}}\right ) - {\left (b c^{2} - b d^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )\right )}}{c^{3} - c d^{2}}\right ] \end {gather*}

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

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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Giac [A]
time = 0.41, size = 53, normalized size = 0.85 \begin {gather*} \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} \end {gather*}

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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Mupad [B]
time = 6.35, size = 636, normalized size = 10.26 \begin {gather*} \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} \end {gather*}

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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