3.446 \(\int \frac{\sec (x)}{a+c \sec (x)+b \tan (x)} \, dx\)

Optimal. Leaf size=51 \[ -\frac{2 \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}} \]

[Out]

(-2*ArcTanh[(b - (a - c)*Tan[x/2])/Sqrt[a^2 + b^2 - c^2]])/Sqrt[a^2 + b^2 - c^2]

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Rubi [A]  time = 0.0760878, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3165, 3124, 618, 206} \[ -\frac{2 \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(a + c*Sec[x] + b*Tan[x]),x]

[Out]

(-2*ArcTanh[(b - (a - c)*Tan[x/2])/Sqrt[a^2 + b^2 - c^2]])/Sqrt[a^2 + b^2 - c^2]

Rule 3165

Int[sec[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(m_)
, x_Symbol] :> Int[1/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + n, 0]
&& IntegerQ[n]

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 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 206

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

Rubi steps

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

Mathematica [A]  time = 0.0450145, size = 50, normalized size = 0.98 \[ -\frac{2 \tanh ^{-1}\left (\frac{(c-a) \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/(a + c*Sec[x] + b*Tan[x]),x]

[Out]

(-2*ArcTanh[(b + (-a + c)*Tan[x/2])/Sqrt[a^2 + b^2 - c^2]])/Sqrt[a^2 + b^2 - c^2]

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Maple [A]  time = 0.059, size = 53, normalized size = 1. \begin{align*} -2\,{\frac{1}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)/(a+c*sec(x)+b*tan(x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+c*sec(x)+b*tan(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14003, size = 799, normalized size = 15.67 \begin{align*} \left [\frac{\log \left (-\frac{2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} -{\left (a^{2} - b^{2}\right )} c^{2} + 2 \,{\left (a^{3} + a b^{2}\right )} c \cos \left (x\right ) -{\left (a^{4} - b^{4} - 2 \,{\left (a^{2} - b^{2}\right )} c^{2}\right )} \cos \left (x\right )^{2} + 2 \,{\left ({\left (a^{2} b + b^{3}\right )} c -{\left (a^{3} b + a b^{3} - 2 \, a b c^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right ) - 2 \,{\left (2 \, a b c \cos \left (x\right )^{2} - a b c +{\left (a^{2} b + b^{3}\right )} \cos \left (x\right ) -{\left (a^{3} + a b^{2} +{\left (a^{2} - b^{2}\right )} c \cos \left (x\right )\right )} \sin \left (x\right )\right )} \sqrt{a^{2} + b^{2} - c^{2}}}{2 \, a c \cos \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2} + c^{2} + 2 \,{\left (a b \cos \left (x\right ) + b c\right )} \sin \left (x\right )}\right )}{2 \, \sqrt{a^{2} + b^{2} - c^{2}}}, \frac{\sqrt{-a^{2} - b^{2} + c^{2}} \arctan \left (\frac{{\left (a c \cos \left (x\right ) + b c \sin \left (x\right ) + a^{2} + b^{2}\right )} \sqrt{-a^{2} - b^{2} + c^{2}}}{{\left (a^{2} b + b^{3} - b c^{2}\right )} \cos \left (x\right ) -{\left (a^{3} + a b^{2} - a c^{2}\right )} \sin \left (x\right )}\right )}{a^{2} + b^{2} - c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+c*sec(x)+b*tan(x)),x, algorithm="fricas")

[Out]

[1/2*log(-(2*a^4 + 3*a^2*b^2 + b^4 - (a^2 - b^2)*c^2 + 2*(a^3 + a*b^2)*c*cos(x) - (a^4 - b^4 - 2*(a^2 - b^2)*c
^2)*cos(x)^2 + 2*((a^2*b + b^3)*c - (a^3*b + a*b^3 - 2*a*b*c^2)*cos(x))*sin(x) - 2*(2*a*b*c*cos(x)^2 - a*b*c +
(a^2*b + b^3)*cos(x) - (a^3 + a*b^2 + (a^2 - b^2)*c*cos(x))*sin(x))*sqrt(a^2 + b^2 - c^2))/(2*a*c*cos(x) + (a
^2 - b^2)*cos(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c)*sin(x)))/sqrt(a^2 + b^2 - c^2), sqrt(-a^2 - b^2 + c^2)*a
rctan((a*c*cos(x) + b*c*sin(x) + a^2 + b^2)*sqrt(-a^2 - b^2 + c^2)/((a^2*b + b^3 - b*c^2)*cos(x) - (a^3 + a*b^
2 - a*c^2)*sin(x)))/(a^2 + b^2 - c^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+c*sec(x)+b*tan(x)),x)

[Out]

Integral(sec(x)/(a + b*tan(x) + c*sec(x)), x)

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Giac [A]  time = 1.16505, size = 99, normalized size = 1.94 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, c\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) - c \tan \left (\frac{1}{2} \, x\right ) - b}{\sqrt{-a^{2} - b^{2} + c^{2}}}\right )\right )}}{\sqrt{-a^{2} - b^{2} + c^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(a+c*sec(x)+b*tan(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(2*a - 2*c) + arctan((a*tan(1/2*x) - c*tan(1/2*x) - b)/sqrt(-a^2 - b^2 + c^2))
)/sqrt(-a^2 - b^2 + c^2)