3.797 \(\int \frac{(B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=67 \[ \frac{2 (b B-a C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b d \sqrt{a-b} \sqrt{a+b}}+\frac{C x}{b} \]

[Out]

(C*x)/b + (2*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)

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Rubi [A]  time = 0.144358, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3029, 2735, 2659, 205} \[ \frac{2 (b B-a C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b d \sqrt{a-b} \sqrt{a+b}}+\frac{C x}{b} \]

Antiderivative was successfully verified.

[In]

Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x]),x]

[Out]

(C*x)/b + (2*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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]

Rubi steps

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

Mathematica [A]  time = 0.120246, size = 68, normalized size = 1.01 \[ \frac{\frac{2 (a C-b B) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+C (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.049, size = 113, normalized size = 1.7 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{db}}+2\,{\frac{B}{d\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{aC}{db\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

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

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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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45848, size = 524, normalized size = 7.82 \begin{align*} \left [\frac{2 \,{\left (C a^{2} - C b^{2}\right )} d x +{\left (C a - B b\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \,{\left (a^{2} b - b^{3}\right )} d}, \frac{{\left (C a^{2} - C b^{2}\right )} d x -{\left (C a - B b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{{\left (a^{2} b - b^{3}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.67828, size = 135, normalized size = 2.01 \begin{align*} \frac{\frac{{\left (d x + c\right )} C}{b} - \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 )}{\left (C a - B b\right )}}{\sqrt{a^{2} - b^{2}} b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

((d*x + c)*C/b - 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)))*(C*a - B*b)/(sqrt(a^2 - b^2)*b))/d