3.273 \(\int \frac {(B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=102 \[ \frac {(2 B+3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(2 B+3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

1/5*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(2*B+3*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+1/15*(2*B+3*C)*sin(
d*x+c)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]  time = 0.15, antiderivative size = 102, normalized size of antiderivative = 1.00, 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, 2750, 2650, 2648} \[ \frac {(2 B+3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(2 B+3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

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])^3,x]

[Out]

((B - C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((2*B + 3*C)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^
2) + ((2*B + 3*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*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 2650

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

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))^3} \, dx &=\int \frac {B+C \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx\\ &=\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 B+3 C) \int \frac {1}{(a+a \cos (c+d x))^2} \, dx}{5 a}\\ &=\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 B+3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 B+3 C) \int \frac {1}{a+a \cos (c+d x)} \, dx}{15 a^2}\\ &=\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 B+3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 B+3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 96, normalized size = 0.94 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left ((2 B+3 C) \left (5 \sin \left (c+\frac {3 d x}{2}\right )+\sin \left (2 c+\frac {5 d x}{2}\right )\right )+5 (4 B+3 C) \sin \left (\frac {d x}{2}\right )-15 C \sin \left (c+\frac {d x}{2}\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]

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])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(5*(4*B + 3*C)*Sin[(d*x)/2] - 15*C*Sin[c + (d*x)/2] + (2*B + 3*C)*(5*Sin[c + (3*d*x
)/2] + Sin[2*c + (5*d*x)/2])))/(30*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 0.47, size = 93, normalized size = 0.91 \[ \frac {{\left ({\left (2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 7 \, B + 3 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/15*((2*B + 3*C)*cos(d*x + c)^2 + 3*(2*B + 3*C)*cos(d*x + c) + 7*B + 3*C)*sin(d*x + c)/(a^3*d*cos(d*x + c)^3
+ 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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giac [A]  time = 0.38, size = 75, normalized size = 0.74 \[ \frac {3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]

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))^3,x, algorithm="giac")

[Out]

1/60*(3*B*tan(1/2*d*x + 1/2*c)^5 - 3*C*tan(1/2*d*x + 1/2*c)^5 + 10*B*tan(1/2*d*x + 1/2*c)^3 + 15*B*tan(1/2*d*x
+ 1/2*c) + 15*C*tan(1/2*d*x + 1/2*c))/(a^3*d)

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maple [A]  time = 0.18, size = 64, normalized size = 0.63 \[ \frac {\frac {\left (B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}} \]

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))^3,x)

[Out]

1/4/d/a^3*(1/5*(B-C)*tan(1/2*d*x+1/2*c)^5+2/3*B*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 = 0.71, size = 115, normalized size = 1.13 \[ \frac {\frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

1/60*(B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a^3 + 3*C*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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mupad [B]  time = 1.09, size = 66, normalized size = 0.65 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,B+15\,C+10\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]

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))^3),x)

[Out]

(tan(c/2 + (d*x)/2)*(15*B + 15*C + 10*B*tan(c/2 + (d*x)/2)^2 + 3*B*tan(c/2 + (d*x)/2)^4 - 3*C*tan(c/2 + (d*x)/
2)^4))/(60*a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

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))**3,x)

[Out]

(Integral(B*cos(c + d*x)*sec(c + d*x)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integra
l(C*cos(c + d*x)**2*sec(c + d*x)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3

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