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

Optimal. Leaf size=114 \[ \frac{(6 A-29 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{C x}{a^3}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(3 A-7 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

(C*x)/a^3 - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((3*A - 7*C)*Sin[c + d*x])/(1
5*a*d*(a + a*Cos[c + d*x])^2) + ((6*A - 29*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.255591, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3042, 2968, 3019, 2735, 2648} \[ \frac{(6 A-29 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{C x}{a^3}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(3 A-7 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/a^3 - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((3*A - 7*C)*Sin[c + d*x])/(1
5*a*d*(a + a*Cos[c + d*x])^2) + ((6*A - 29*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

Rule 3042

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

Rule 2968

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

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]

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 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]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos (c+d x) (a (3 A-2 C)+5 a C \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{a (3 A-2 C) \cos (c+d x)+5 a C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(3 A-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{\int \frac{-2 a^2 (3 A-7 C)-15 a^2 C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{C x}{a^3}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(3 A-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(6 A-29 C) \int \frac{1}{a+a \cos (c+d x)} \, dx}{15 a^2}\\ &=\frac{C x}{a^3}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(3 A-7 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(6 A-29 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.523532, size = 227, normalized size = 1.99 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (-30 A \sin \left (c+\frac{d x}{2}\right )+30 A \sin \left (c+\frac{3 d x}{2}\right )+6 A \sin \left (2 c+\frac{5 d x}{2}\right )+30 A \sin \left (\frac{d x}{2}\right )+270 C \sin \left (c+\frac{d x}{2}\right )-230 C \sin \left (c+\frac{3 d x}{2}\right )+90 C \sin \left (2 c+\frac{3 d x}{2}\right )-64 C \sin \left (2 c+\frac{5 d x}{2}\right )+150 C d x \cos \left (c+\frac{d x}{2}\right )+75 C d x \cos \left (c+\frac{3 d x}{2}\right )+75 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+15 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+15 C d x \cos \left (3 c+\frac{5 d x}{2}\right )-370 C \sin \left (\frac{d x}{2}\right )+150 C d x \cos \left (\frac{d x}{2}\right )\right )}{480 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*C*d*x*Cos[(d*x)/2] + 150*C*d*x*Cos[c + (d*x)/2] + 75*C*d*x*Cos[c + (3*d*x)/2
] + 75*C*d*x*Cos[2*c + (3*d*x)/2] + 15*C*d*x*Cos[2*c + (5*d*x)/2] + 15*C*d*x*Cos[3*c + (5*d*x)/2] + 30*A*Sin[(
d*x)/2] - 370*C*Sin[(d*x)/2] - 30*A*Sin[c + (d*x)/2] + 270*C*Sin[c + (d*x)/2] + 30*A*Sin[c + (3*d*x)/2] - 230*
C*Sin[c + (3*d*x)/2] + 90*C*Sin[2*c + (3*d*x)/2] + 6*A*Sin[2*c + (5*d*x)/2] - 64*C*Sin[2*c + (5*d*x)/2]))/(480
*a^3*d)

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Maple [A]  time = 0.027, size = 117, normalized size = 1. \begin{align*} -{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+1/3/d/a^3*C*tan(1/2*d*x+1/2*c)^3+1/4/d/a^
3*A*tan(1/2*d*x+1/2*c)-7/4/d/a^3*C*tan(1/2*d*x+1/2*c)+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.51142, size = 189, normalized size = 1.66 \begin{align*} -\frac{C{\left (\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \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}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac{3 \, A{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - 3*A*(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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Fricas [A]  time = 1.61388, size = 351, normalized size = 3.08 \begin{align*} \frac{15 \, C d x \cos \left (d x + c\right )^{3} + 45 \, C d x \cos \left (d x + c\right )^{2} + 45 \, C d x \cos \left (d x + c\right ) + 15 \, C d x +{\left ({\left (3 \, A - 32 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, A - 17 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 22 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*C*d*x*cos(d*x + c)^3 + 45*C*d*x*cos(d*x + c)^2 + 45*C*d*x*cos(d*x + c) + 15*C*d*x + ((3*A - 32*C)*cos
(d*x + c)^2 + 3*(3*A - 17*C)*cos(d*x + c) + 3*A - 22*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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Sympy [A]  time = 9.56269, size = 128, normalized size = 1.12 \begin{align*} \begin{cases} - \frac{A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} + \frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} + \frac{C x}{a^{3}} - \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 )}}{3 a^{3} d} - \frac{7 C \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-A*tan(c/2 + d*x/2)**5/(20*a**3*d) + A*tan(c/2 + d*x/2)/(4*a**3*d) + C*x/a**3 - C*tan(c/2 + d*x/2)*
*5/(20*a**3*d) + C*tan(c/2 + d*x/2)**3/(3*a**3*d) - 7*C*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(A + C*cos(
c)**2)*cos(c)/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.25333, size = 140, normalized size = 1.23 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )} C}{a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*(d*x + c)*C/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*C*a^12*tan(
1/2*d*x + 1/2*c)^3 - 15*A*a^12*tan(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d