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

Optimal. Leaf size=65 \[ \frac {(B+2 C) \sin (c+d x)}{3 d \left (a^2 \cos (c+d x)+a^2\right )}+\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

1/3*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2+1/3*(B+2*C)*sin(d*x+c)/d/(a^2+a^2*cos(d*x+c))

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Rubi [A]  time = 0.12, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {3029, 2750, 2648} \[ \frac {(B+2 C) \sin (c+d x)}{3 d \left (a^2 \cos (c+d x)+a^2\right )}+\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 76, normalized size = 1.17 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left ((B+2 C) \sin \left (c+\frac {3 d x}{2}\right )+3 (B+C) \sin \left (\frac {d x}{2}\right )-3 C \sin \left (c+\frac {d x}{2}\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(3*(B + C)*Sin[(d*x)/2] - 3*C*Sin[c + (d*x)/2] + (B + 2*C)*Sin[c + (3*d*x)/2]))/(3*
a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.45, size = 58, normalized size = 0.89 \[ \frac {{\left ({\left (B + 2 \, C\right )} \cos \left (d x + c\right ) + 2 \, B + C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

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+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*((B + 2*C)*cos(d*x + c) + 2*B + C)*sin(d*x + c)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 0.35, size = 60, normalized size = 0.92 \[ \frac {B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{6 \, a^{2} d} \]

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+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(B*tan(1/2*d*x + 1/2*c)^3 - C*tan(1/2*d*x + 1/2*c)^3 + 3*B*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x + 1/2*c)
)/(a^2*d)

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maple [A]  time = 0.18, size = 60, normalized size = 0.92 \[ \frac {\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}} \]

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+a*cos(d*x+c))^2,x)

[Out]

1/2/d/a^2*(1/3*B*tan(1/2*d*x+1/2*c)^3-1/3*C*tan(1/2*d*x+1/2*c)^3+B*tan(1/2*d*x+1/2*c)+C*tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.67, size = 93, normalized size = 1.43 \[ \frac {\frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} + \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

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+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(B*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 + C*(3*sin(d*x + c)/(cos(
d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

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mupad [B]  time = 1.05, size = 45, normalized size = 0.69 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B+C\right )}{2\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)*(a + a*cos(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d) + (tan(c/2 + (d*x)/2)*(B + C))/(2*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

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+a*cos(d*x+c))**2,x)

[Out]

(Integral(B*cos(c + d*x)*sec(c + d*x)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**2*
sec(c + d*x)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x))/a**2

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