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

Optimal. Leaf size=100 \[ \frac{2 (a B-b C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b B-a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

[Out]

(2*(a*B - b*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*d) - ((b*B - a
*C)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.153697, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3029, 2754, 12, 2659, 205} \[ \frac{2 (a B-b C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b B-a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

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])^2,x]

[Out]

(2*(a*B - b*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*d) - ((b*B - a
*C)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

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

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))^2} \, dx &=\int \frac{B+C \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx\\ &=-\frac{(b B-a C) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{-a B+b C}{a+b \cos (c+d x)} \, dx}{-a^2+b^2}\\ &=-\frac{(b B-a C) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{(a B-b C) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^2-b^2}\\ &=-\frac{(b B-a C) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{(2 (a B-b 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 )}{\left (a^2-b^2\right ) d}\\ &=\frac{2 (a B-b C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac{(b B-a C) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.321673, size = 97, normalized size = 0.97 \[ \frac{\frac{2 (a B-b C) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{(a C-b B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{d} \]

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])^2,x]

[Out]

((2*(a*B - b*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + ((-(b*B) + a*C)*Sin
[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])))/d

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Maple [B]  time = 0.055, size = 234, normalized size = 2.3 \begin{align*} -2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) bB}{d \left ({a}^{2}-{b}^{2} \right ) \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) aC}{d \left ({a}^{2}-{b}^{2} \right ) \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+2\,{\frac{Ba}{d \left ( a+b \right ) \left ( a-b \right ) \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{Cb}{d \left ( a+b \right ) \left ( a-b \right ) \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))^2,x)

[Out]

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

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.67439, size = 846, normalized size = 8.46 \begin{align*} \left [-\frac{{\left (B a^{2} - C a b +{\left (B a b - C b^{2}\right )} \cos \left (d x + c\right )\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 (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, \frac{{\left (B a^{2} - C a b +{\left (B a b - C b^{2}\right )} \cos \left (d x + c\right )\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 (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\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))^2,x, algorithm="fricas")

[Out]

[-1/2*((B*a^2 - C*a*b + (B*a*b - C*b^2)*cos(d*x + c))*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)) - 2*(C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*sin(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*co
s(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d), ((B*a^2 - C*a*b + (B*a*b - C*b^2)*cos(d*x + c))*sqrt(a^2 - b^2)*arc
tan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*sin(d*x + c))/
((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*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 )}}{\left (a + b \cos{\left (c + d x \right )}\right )^{2}}\, 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))**2,x)

[Out]

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

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Giac [A]  time = 1.38674, size = 212, normalized size = 2.12 \begin{align*} \frac{2 \,{\left (\frac{{\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 (B a - C b\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} + \frac{C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}{\left (a^{2} - b^{2}\right )}}\right )}}{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))^2,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*a - C*b)/(a^2 - b^2)^(3/2) + (C*a*tan(1/2*d*x + 1/2*c) - B*b*tan(1/2*d*x + 1/2*c))/((a*t
an(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)*(a^2 - b^2)))/d