3.13 \(\int \frac{1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx\)

Optimal. Leaf size=121 \[ \frac{2 a \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f \sqrt{a-b} \sqrt{a+b} (a c-b d)}-\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (a c-b d)} \]

[Out]

(2*a*ArcTan[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a*c - b*d)*f) - (2*d*ArcTan
h[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(Sqrt[c - d]*Sqrt[c + d]*(a*c - b*d)*f)

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Rubi [A]  time = 0.265811, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2828, 3001, 2659, 205, 208} \[ \frac{2 a \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f \sqrt{a-b} \sqrt{a+b} (a c-b d)}-\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (a c-b d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[e + f*x])*(c + d*Sec[e + f*x])),x]

[Out]

(2*a*ArcTan[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a*c - b*d)*f) - (2*d*ArcTan
h[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(Sqrt[c - d]*Sqrt[c + d]*(a*c - b*d)*f)

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 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 205

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

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]

Rubi steps

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

Mathematica [A]  time = 0.212984, size = 106, normalized size = 0.88 \[ \frac{\frac{2 d \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}-\frac{2 a \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}}{a c f-b d f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Cos[e + f*x])*(c + d*Sec[e + f*x])),x]

[Out]

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

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Maple [A]  time = 0.071, size = 110, normalized size = 0.9 \begin{align*} -2\,{\frac{d}{f \left ( ac-bd \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{ \left ( c-d \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) }+2\,{\frac{a}{f \left ( ac-bd \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x)

[Out]

-2/f*d/(a*c-b*d)/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tan(1/2*f*x+1/2*e)/((c+d)*(c-d))^(1/2))+2/f*a/(a*c-b*d)/((a
+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(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(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 11.5093, size = 2182, normalized size = 18.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x, algorithm="fricas")

[Out]

[-1/2*((a^2 - b^2)*sqrt(c^2 - d^2)*d*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2
)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2)) - (a*c^2 -
a*d^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(f*x + e) + (2*a^2 - b^2)*cos(f*x + e)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(f*
x + e) + b)*sin(f*x + e) - a^2 + 2*b^2)/(b^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + a^2)))/(((a^3 - a*b^2)*c^3
- (a^2*b - b^3)*c^2*d - (a^3 - a*b^2)*c*d^2 + (a^2*b - b^3)*d^3)*f), -1/2*(2*(a^2 - b^2)*sqrt(-c^2 + d^2)*d*ar
ctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) - (a*c^2 - a*d^2)*sqrt(-a^2 + b^2)*log
((2*a*b*cos(f*x + e) + (2*a^2 - b^2)*cos(f*x + e)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(f*x + e) + b)*sin(f*x + e) - a
^2 + 2*b^2)/(b^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + a^2)))/(((a^3 - a*b^2)*c^3 - (a^2*b - b^3)*c^2*d - (a^3
- a*b^2)*c*d^2 + (a^2*b - b^3)*d^3)*f), -1/2*((a^2 - b^2)*sqrt(c^2 - d^2)*d*log((2*c*d*cos(f*x + e) - (c^2 -
2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2
+ 2*c*d*cos(f*x + e) + d^2)) - 2*(a*c^2 - a*d^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(f*x + e) + b)/(sqrt(a^2 - b^2
)*sin(f*x + e))))/(((a^3 - a*b^2)*c^3 - (a^2*b - b^3)*c^2*d - (a^3 - a*b^2)*c*d^2 + (a^2*b - b^3)*d^3)*f), -((
a^2 - b^2)*sqrt(-c^2 + d^2)*d*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) - (a*c
^2 - a*d^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(f*x + e) + b)/(sqrt(a^2 - b^2)*sin(f*x + e))))/(((a^3 - a*b^2)*c^3
- (a^2*b - b^3)*c^2*d - (a^3 - a*b^2)*c*d^2 + (a^2*b - b^3)*d^3)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \cos{\left (e + f x \right )}\right ) \left (c + d \sec{\left (e + f x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x)

[Out]

Integral(1/((a + b*cos(e + f*x))*(c + d*sec(e + f*x))), x)

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Giac [B]  time = 1.33298, size = 690, normalized size = 5.7 \begin{align*} \frac{\frac{{\left (\sqrt{a^{2} - b^{2}} a c{\left | a - b \right |} - \sqrt{a^{2} - b^{2}}{\left (2 \, a - b\right )} d{\left | a - b \right |} + \sqrt{a^{2} - b^{2}}{\left | a c - b d \right |}{\left | a - b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{\frac{b c - a d + \sqrt{{\left (a c + b c + a d + b d\right )}{\left (a c - b c - a d + b d\right )} +{\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )}{\left (a c - b d\right )}^{2} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} c{\left | a c - b d \right |} -{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d{\left | a c - b d \right |}} + \frac{{\left (\sqrt{-c^{2} + d^{2}} a{\left (c - 2 \, d\right )}{\left | -c + d \right |} + \sqrt{-c^{2} + d^{2}} b d{\left | -c + d \right |} - \sqrt{-c^{2} + d^{2}}{\left | a c - b d \right |}{\left | -c + d \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{\frac{b c - a d - \sqrt{{\left (a c + b c + a d + b d\right )}{\left (a c - b c - a d + b d\right )} +{\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a c - b d\right )}^{2}{\left (c^{2} - 2 \, c d + d^{2}\right )} +{\left (c^{2} d - 2 \, c d^{2} + d^{3}\right )} a{\left | a c - b d \right |} -{\left (c^{3} - 2 \, c^{2} d + c d^{2}\right )} b{\left | a c - b d \right |}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e)),x, algorithm="giac")

[Out]

((sqrt(a^2 - b^2)*a*c*abs(a - b) - sqrt(a^2 - b^2)*(2*a - b)*d*abs(a - b) + sqrt(a^2 - b^2)*abs(a*c - b*d)*abs
(a - b))*(pi*floor(1/2*(f*x + e)/pi + 1/2) + arctan(tan(1/2*f*x + 1/2*e)/sqrt((b*c - a*d + sqrt((a*c + b*c + a
*d + b*d)*(a*c - b*c - a*d + b*d) + (b*c - a*d)^2))/(a*c - b*c - a*d + b*d))))/((a^2 - 2*a*b + b^2)*(a*c - b*d
)^2 + (a^2*b - 2*a*b^2 + b^3)*c*abs(a*c - b*d) - (a^3 - 2*a^2*b + a*b^2)*d*abs(a*c - b*d)) + (sqrt(-c^2 + d^2)
*a*(c - 2*d)*abs(-c + d) + sqrt(-c^2 + d^2)*b*d*abs(-c + d) - sqrt(-c^2 + d^2)*abs(a*c - b*d)*abs(-c + d))*(pi
*floor(1/2*(f*x + e)/pi + 1/2) + arctan(tan(1/2*f*x + 1/2*e)/sqrt((b*c - a*d - sqrt((a*c + b*c + a*d + b*d)*(a
*c - b*c - a*d + b*d) + (b*c - a*d)^2))/(a*c - b*c - a*d + b*d))))/((a*c - b*d)^2*(c^2 - 2*c*d + d^2) + (c^2*d
- 2*c*d^2 + d^3)*a*abs(a*c - b*d) - (c^3 - 2*c^2*d + c*d^2)*b*abs(a*c - b*d)))/f