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

Optimal. Leaf size=102 \[ \frac {(3 B+7 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(3 B-8 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

-1/5*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(3*B-8*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+1/15*(3*B+7*C)*sin
(d*x+c)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]  time = 0.13, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3019, 2750, 2648} \[ \frac {(3 B+7 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(3 B-8 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B - C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((3*B - 8*C)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])
^2) + ((3*B + 7*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 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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {B \cos (c+d x)+C \cos ^2(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 {\int \frac {-3 a (B-C)-5 a C \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(3 B-8 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 B+7 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 {(3 B-8 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 B+7 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 = 0.34, size = 135, normalized size = 1.32 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-15 (B+2 C) \sin \left (c+\frac {d x}{2}\right )+15 B \sin \left (c+\frac {3 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+5 (3 B+8 C) \sin \left (\frac {d x}{2}\right )+20 C \sin \left (c+\frac {3 d x}{2}\right )-15 C \sin \left (2 c+\frac {3 d x}{2}\right )+7 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(5*(3*B + 8*C)*Sin[(d*x)/2] - 15*(B + 2*C)*Sin[c + (d*x)/2] + 15*B*Sin[c + (3*d*x)/
2] + 20*C*Sin[c + (3*d*x)/2] - 15*C*Sin[2*c + (3*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] + 7*C*Sin[2*c + (5*d*x)/2]
))/(30*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 0.57, size = 93, normalized size = 0.91 \[ \frac {{\left ({\left (3 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + 3 \, B + 2 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((3*B + 7*C)*cos(d*x + c)^2 + 3*(3*B + 2*C)*cos(d*x + c) + 3*B + 2*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 = 0.34, size = 75, normalized size = 0.74 \[ -\frac {3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(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*C*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 = 0.12, size = 64, normalized size = 0.63 \[ \frac {\frac {\left (-B +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(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*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 = 0.52, size = 115, normalized size = 1.13 \[ \frac {\frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, B {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(C*(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*B*(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 = 1.07, size = 66, normalized size = 0.65 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,B+15\,C-3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + a*cos(c + d*x))^3,x)

[Out]

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

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sympy [A]  time = 5.44, size = 119, normalized size = 1.17 \[ \begin {cases} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} + \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} - \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right )}{\left (a \cos {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((-B*tan(c/2 + d*x/2)**5/(20*a**3*d) + B*tan(c/2 + d*x/2)/(4*a**3*d) + C*tan(c/2 + d*x/2)**5/(20*a**3
*d) - C*tan(c/2 + d*x/2)**3/(6*a**3*d) + C*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)
/(a*cos(c) + a)**3, True))

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