3.445 \(\int \frac{1}{a+c \sec (x)+b \tan (x)} \, dx\)

Optimal. Leaf size=97 \[ \frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt{a^2+b^2-c^2}}+\frac{b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac{a x}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.127024, antiderivative size = 97, normalized size of antiderivative = 1., 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, 206} \[ \frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt{a^2+b^2-c^2}}+\frac{b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac{a x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*Sec[x] + b*Tan[x])^(-1),x]

[Out]

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

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]

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

Mathematica [A]  time = 0.192824, size = 79, normalized size = 0.81 \[ \frac{\frac{2 a c \tanh ^{-1}\left (\frac{(c-a) \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}}+b \log (a \cos (x)+b \sin (x)+c)+a x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*Sec[x] + b*Tan[x])^(-1),x]

[Out]

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

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Maple [B]  time = 0.064, size = 414, normalized size = 4.3 \begin{align*}{\frac{ab}{ \left ({a}^{2}+{b}^{2} \right ) \left ( a-c \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}c-2\,b\tan \left ( x/2 \right ) -a-c \right ) }-{\frac{cb}{ \left ({a}^{2}+{b}^{2} \right ) \left ( a-c \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}c-2\,b\tan \left ( x/2 \right ) -a-c \right ) }+2\,{\frac{ac}{ \left ({a}^{2}+{b}^{2} \right ) \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 ) }-2\,{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) \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 ) }+2\,{\frac{a{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( a-c \right ) }\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 ) }-2\,{\frac{c{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( a-c \right ) }\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 ) }-{\frac{b}{{a}^{2}+{b}^{2}}\ln \left ( 1+ \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2} \right ) }+2\,{\frac{a\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{a}^{2}+{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c*sec(x)+b*tan(x)),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.56773, size = 1277, normalized size = 13.16 \begin{align*} \left [\frac{\sqrt{a^{2} + b^{2} - c^{2}} a c \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 \,{\left (a^{3} + a b^{2} - a c^{2}\right )} x +{\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (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 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} -{\left (a^{2} + b^{2}\right )} c^{2}\right )}}, -\frac{2 \, \sqrt{-a^{2} - b^{2} + c^{2}} a c \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 ) - 2 \,{\left (a^{3} + a b^{2} - a c^{2}\right )} x -{\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (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 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} -{\left (a^{2} + b^{2}\right )} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{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(1/(a+c*sec(x)+b*tan(x)),x)

[Out]

Integral(1/(a + b*tan(x) + c*sec(x)), x)

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Giac [A]  time = 1.17449, size = 213, normalized size = 2.2 \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 )} a c}{{\left (a^{2} + b^{2}\right )} \sqrt{-a^{2} - b^{2} + c^{2}}} + \frac{a x}{a^{2} + b^{2}} + \frac{b \log \left (-a \tan \left (\frac{1}{2} \, x\right )^{2} + c \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a + c\right )}{a^{2} + b^{2}} - \frac{b \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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
)))*a*c/((a^2 + b^2)*sqrt(-a^2 - b^2 + c^2)) + a*x/(a^2 + b^2) + b*log(-a*tan(1/2*x)^2 + c*tan(1/2*x)^2 + 2*b*
tan(1/2*x) + a + c)/(a^2 + b^2) - b*log(tan(1/2*x)^2 + 1)/(a^2 + b^2)