3.780 \(\int \frac {1}{a+c \text {sech}(x)+b \tanh (x)} \, dx\)

Optimal. Leaf size=107 \[ -\frac {2 a c \tan ^{-1}\left (\frac {(a-c) \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2\right ) \sqrt {a^2-b^2-c^2}}-\frac {b \log (a \cosh (x)+b \sinh (x)+c)}{a^2-b^2}+\frac {a x}{a^2-b^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3159, 3138, 3124, 618, 204} \[ -\frac {2 a c \tan ^{-1}\left (\frac {(a-c) \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2\right ) \sqrt {a^2-b^2-c^2}}-\frac {b \log (a \cosh (x)+b \sinh (x)+c)}{a^2-b^2}+\frac {a x}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*Sech[x] + b*Tanh[x])^(-1),x]

[Out]

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

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3124

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

Rule 3138

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

Rule 3159

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Int[Cos[d + e*x
]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

Rubi steps

\begin {align*} \int \frac {1}{a+c \text {sech}(x)+b \tanh (x)} \, dx &=\int \frac {\cosh (x)}{c+a \cosh (x)+b \sinh (x)} \, dx\\ &=\frac {a x}{a^2-b^2}-\frac {b \log (c+a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {(a c) \int \frac {1}{c+a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=\frac {a x}{a^2-b^2}-\frac {b \log (c+a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {(2 a c) \operatorname {Subst}\left (\int \frac {1}{a+c+2 b x-(-a+c) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2}\\ &=\frac {a x}{a^2-b^2}-\frac {b \log (c+a \cosh (x)+b \sinh (x))}{a^2-b^2}+\frac {(4 a c) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2-c^2\right )-x^2} \, dx,x,2 b+2 (a-c) \tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2}\\ &=\frac {a x}{a^2-b^2}-\frac {2 a c \tan ^{-1}\left (\frac {b+(a-c) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2\right ) \sqrt {a^2-b^2-c^2}}-\frac {b \log (c+a \cosh (x)+b \sinh (x))}{a^2-b^2}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 86, normalized size = 0.80 \[ \frac {-\frac {2 a c \tan ^{-1}\left (\frac {(a-c) \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2-c^2}}\right )}{\sqrt {a^2-b^2-c^2}}-b \log (a \cosh (x)+b \sinh (x)+c)+a x}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*Sech[x] + b*Tanh[x])^(-1),x]

[Out]

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

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fricas [A]  time = 0.46, size = 429, normalized size = 4.01 \[ \left [\frac {\sqrt {-a^{2} + b^{2} + c^{2}} a c \log \left (\frac {2 \, {\left (a + b\right )} c \cosh \relax (x) + {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \relax (x)^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \relax (x)^{2} - a^{2} + b^{2} + 2 \, c^{2} + 2 \, {\left ({\left (a + b\right )} c + {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x) - 2 \, \sqrt {-a^{2} + b^{2} + c^{2}} {\left ({\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x) + c\right )}}{{\left (a + b\right )} \cosh \relax (x)^{2} + {\left (a + b\right )} \sinh \relax (x)^{2} + 2 \, c \cosh \relax (x) + 2 \, {\left ({\left (a + b\right )} \cosh \relax (x) + c\right )} \sinh \relax (x) + a - b}\right ) + {\left (a^{3} + a^{2} b - a b^{2} - b^{3} - {\left (a + b\right )} c^{2}\right )} x - {\left (a^{2} b - b^{3} - b c^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x) + c\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} - b^{2}\right )} c^{2}}, \frac {2 \, \sqrt {a^{2} - b^{2} - c^{2}} a c \arctan \left (-\frac {{\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right ) + {\left (a^{3} + a^{2} b - a b^{2} - b^{3} - {\left (a + b\right )} c^{2}\right )} x - {\left (a^{2} b - b^{3} - b c^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x) + c\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} - b^{2}\right )} c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c*sech(x)+b*tanh(x)),x, algorithm="fricas")

[Out]

[(sqrt(-a^2 + b^2 + c^2)*a*c*log((2*(a + b)*c*cosh(x) + (a^2 + 2*a*b + b^2)*cosh(x)^2 + (a^2 + 2*a*b + b^2)*si
nh(x)^2 - a^2 + b^2 + 2*c^2 + 2*((a + b)*c + (a^2 + 2*a*b + b^2)*cosh(x))*sinh(x) - 2*sqrt(-a^2 + b^2 + c^2)*(
(a + b)*cosh(x) + (a + b)*sinh(x) + c))/((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + 2*c*cosh(x) + 2*((a + b)*cosh
(x) + c)*sinh(x) + a - b)) + (a^3 + a^2*b - a*b^2 - b^3 - (a + b)*c^2)*x - (a^2*b - b^3 - b*c^2)*log(2*(a*cosh
(x) + b*sinh(x) + c)/(cosh(x) - sinh(x))))/(a^4 - 2*a^2*b^2 + b^4 - (a^2 - b^2)*c^2), (2*sqrt(a^2 - b^2 - c^2)
*a*c*arctan(-((a + b)*cosh(x) + (a + b)*sinh(x) + c)/sqrt(a^2 - b^2 - c^2)) + (a^3 + a^2*b - a*b^2 - b^3 - (a
+ b)*c^2)*x - (a^2*b - b^3 - b*c^2)*log(2*(a*cosh(x) + b*sinh(x) + c)/(cosh(x) - sinh(x))))/(a^4 - 2*a^2*b^2 +
b^4 - (a^2 - b^2)*c^2)]

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giac [A]  time = 0.13, size = 106, normalized size = 0.99 \[ -\frac {2 \, a c \arctan \left (\frac {a e^{x} + b e^{x} + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )}{\sqrt {a^{2} - b^{2} - c^{2}} {\left (a^{2} - b^{2}\right )}} - \frac {b \log \left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + 2 \, c e^{x} + a - b\right )}{a^{2} - b^{2}} + \frac {x}{a - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c*sech(x)+b*tanh(x)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.24, size = 422, normalized size = 3.94 \[ -\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 a}+\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a -2 b}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) c +2 \tanh \left (\frac {x}{2}\right ) b +a +c \right ) a b}{\left (a +b \right ) \left (a -b \right ) \left (a -c \right )}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) c +2 \tanh \left (\frac {x}{2}\right ) b +a +c \right ) c b}{\left (a +b \right ) \left (a -b \right ) \left (a -c \right )}-\frac {2 \arctan \left (\frac {2 \left (a -c \right ) \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) a c}{\left (a +b \right ) \left (a -b \right ) \sqrt {a^{2}-b^{2}-c^{2}}}-\frac {2 \arctan \left (\frac {2 \left (a -c \right ) \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) b^{2}}{\left (a +b \right ) \left (a -b \right ) \sqrt {a^{2}-b^{2}-c^{2}}}+\frac {2 \arctan \left (\frac {2 \left (a -c \right ) \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) b^{2} a}{\left (a +b \right ) \left (a -b \right ) \sqrt {a^{2}-b^{2}-c^{2}}\, \left (a -c \right )}-\frac {2 \arctan \left (\frac {2 \left (a -c \right ) \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) b^{2} c}{\left (a +b \right ) \left (a -b \right ) \sqrt {a^{2}-b^{2}-c^{2}}\, \left (a -c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c*sech(x)+b*tanh(x)),x)

[Out]

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

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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(1/(a+c*sech(x)+b*tanh(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(c^2+b^2-a^2>0)', see `assume?`
for more details)Is c^2+b^2-a^2 positive or negative?

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mupad [B]  time = 6.28, size = 472, normalized size = 4.41 \[ \frac {x}{a-b}+\frac {\ln \left (a-b+2\,c\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )\,\left (-2\,a^2\,b+2\,b^3+2\,b\,c^2\right )}{2\,\left (a^4-2\,a^2\,b^2-a^2\,c^2+b^4+b^2\,c^2\right )}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,a\,c}{{\left (a+b\right )}^2\,\left (a^2-b^2\right )\,{\left (a-b\right )}^2\,\sqrt {a^2\,c^2}}-\frac {2\,\left (a^2\,c\,\sqrt {a^2\,c^2}-b^2\,c\,\sqrt {a^2\,c^2}\right )}{a\,{\left (a+b\right )}^2\,{\left (a^2-b^2\right )}^2\,{\left (a-b\right )}^2\,\left (-a^2+b^2+c^2\right )}\right )-\frac {2\,\left (a^3\,\sqrt {a^2\,c^2}+b^3\,\sqrt {a^2\,c^2}-a\,b^2\,\sqrt {a^2\,c^2}-a^2\,b\,\sqrt {a^2\,c^2}\right )}{a\,{\left (a+b\right )}^2\,{\left (a^2-b^2\right )}^2\,{\left (a-b\right )}^2\,\left (-a^2+b^2+c^2\right )}\right )\,\left (\frac {a^3\,\sqrt {-{\left (a^2-b^2\right )}^2\,\left (-a^2+b^2+c^2\right )}}{2}-\frac {b^3\,\sqrt {-{\left (a^2-b^2\right )}^2\,\left (-a^2+b^2+c^2\right )}}{2}-\frac {a\,b^2\,\sqrt {-{\left (a^2-b^2\right )}^2\,\left (-a^2+b^2+c^2\right )}}{2}+\frac {a^2\,b\,\sqrt {-{\left (a^2-b^2\right )}^2\,\left (-a^2+b^2+c^2\right )}}{2}\right )\right )\,\sqrt {a^2\,c^2}}{\sqrt {-{\left (a^2-b^2\right )}^2\,\left (-a^2+b^2+c^2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tanh(x) + c/cosh(x)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \tanh {\relax (x )} + c \operatorname {sech}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c*sech(x)+b*tanh(x)),x)

[Out]

Integral(1/(a + b*tanh(x) + c*sech(x)), x)

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