3.493 \(\int \frac{\cos (c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=76 \[ \frac{2 b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{b x}{a^2}+\frac{\sin (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.101524, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3853, 12, 3783, 2659, 208} \[ \frac{2 b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{b x}{a^2}+\frac{\sin (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^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 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{\cos (c+d x)}{a+b \sec (c+d x)} \, dx &=\frac{\sin (c+d x)}{a d}-\frac{\int \frac{b}{a+b \sec (c+d x)} \, dx}{a}\\ &=\frac{\sin (c+d x)}{a d}-\frac{b \int \frac{1}{a+b \sec (c+d x)} \, dx}{a}\\ &=-\frac{b x}{a^2}+\frac{\sin (c+d x)}{a d}+\frac{b \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^2}\\ &=-\frac{b x}{a^2}+\frac{\sin (c+d x)}{a d}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 d}\\ &=-\frac{b x}{a^2}+\frac{2 b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 \sqrt{a-b} \sqrt{a+b} d}+\frac{\sin (c+d x)}{a d}\\ \end{align*}

Mathematica [A]  time = 0.138573, size = 72, normalized size = 0.95 \[ \frac{-\frac{2 b^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a \sin (c+d x)-b (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 102, normalized size = 1.3 \begin{align*} 2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{b\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+2\,{\frac{{b}^{2}}{d{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/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(cos(d*x+c)/(a+b*sec(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79363, size = 599, normalized size = 7.88 \begin{align*} \left [\frac{\sqrt{a^{2} - b^{2}} b^{2} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \,{\left (a^{2} b - b^{3}\right )} d x + 2 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d}, \frac{\sqrt{-a^{2} + b^{2}} b^{2} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) -{\left (a^{2} b - b^{3}\right )} d x +{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - a^{2} b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26341, size = 170, normalized size = 2.24 \begin{align*} \frac{\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )} b^{2}}{\sqrt{-a^{2} + b^{2}} a^{2}} - \frac{{\left (d x + c\right )} b}{a^{2}} + \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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