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

Optimal. Leaf size=189 \[ -\frac{2 (11 A+76 C) \sin (c+d x)}{15 a^3 d}-\frac{(11 A+76 C) \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(2 A+13 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{x (2 A+13 C)}{2 a^3}-\frac{(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(A+11 C) \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

((2*A + 13*C)*x)/(2*a^3) - (2*(11*A + 76*C)*Sin[c + d*x])/(15*a^3*d) + ((2*A + 13*C)*Cos[c + d*x]*Sin[c + d*x]
)/(2*a^3*d) - ((A + C)*Cos[c + d*x]^4*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((A + 11*C)*Cos[c + d*x]^3*
Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((11*A + 76*C)*Cos[c + d*x]^2*Sin[c + d*x])/(15*d*(a^3 + a^3*C
os[c + d*x]))

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Rubi [A]  time = 0.46004, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3042, 2977, 2734} \[ -\frac{2 (11 A+76 C) \sin (c+d x)}{15 a^3 d}-\frac{(11 A+76 C) \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(2 A+13 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{x (2 A+13 C)}{2 a^3}-\frac{(A+C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(A+11 C) \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A + 13*C)*x)/(2*a^3) - (2*(11*A + 76*C)*Sin[c + d*x])/(15*a^3*d) + ((2*A + 13*C)*Cos[c + d*x]*Sin[c + d*x]
)/(2*a^3*d) - ((A + C)*Cos[c + d*x]^4*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((A + 11*C)*Cos[c + d*x]^3*
Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((11*A + 76*C)*Cos[c + d*x]^2*Sin[c + d*x])/(15*d*(a^3 + a^3*C
os[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 2977

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

Rule 2734

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

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx &=-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos ^3(c+d x) (a (A-4 C)+a (2 A+7 C) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(A+11 C) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^2(c+d x) \left (-3 a^2 (A+11 C)+a^2 (8 A+43 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(A+11 C) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(11 A+76 C) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \cos (c+d x) \left (-2 a^3 (11 A+76 C)+15 a^3 (2 A+13 C) \cos (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(2 A+13 C) x}{2 a^3}-\frac{2 (11 A+76 C) \sin (c+d x)}{15 a^3 d}+\frac{(2 A+13 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(A+11 C) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(11 A+76 C) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.659566, size = 393, normalized size = 2.08 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (600 d x (2 A+13 C) \cos \left (c+\frac{d x}{2}\right )+2160 A \sin \left (c+\frac{d x}{2}\right )-1840 A \sin \left (c+\frac{3 d x}{2}\right )+720 A \sin \left (2 c+\frac{3 d x}{2}\right )-512 A \sin \left (2 c+\frac{5 d x}{2}\right )+600 A d x \cos \left (c+\frac{3 d x}{2}\right )+600 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+120 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+120 A d x \cos \left (3 c+\frac{5 d x}{2}\right )+600 d x (2 A+13 C) \cos \left (\frac{d x}{2}\right )-2960 A \sin \left (\frac{d x}{2}\right )+7560 C \sin \left (c+\frac{d x}{2}\right )-9230 C \sin \left (c+\frac{3 d x}{2}\right )+930 C \sin \left (2 c+\frac{3 d x}{2}\right )-2782 C \sin \left (2 c+\frac{5 d x}{2}\right )-750 C \sin \left (3 c+\frac{5 d x}{2}\right )-105 C \sin \left (3 c+\frac{7 d x}{2}\right )-105 C \sin \left (4 c+\frac{7 d x}{2}\right )+15 C \sin \left (4 c+\frac{9 d x}{2}\right )+15 C \sin \left (5 c+\frac{9 d x}{2}\right )+3900 C d x \cos \left (c+\frac{3 d x}{2}\right )+3900 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+780 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+780 C d x \cos \left (3 c+\frac{5 d x}{2}\right )-12760 C \sin \left (\frac{d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(600*(2*A + 13*C)*d*x*Cos[(d*x)/2] + 600*(2*A + 13*C)*d*x*Cos[c + (d*x)/2] + 600*A*
d*x*Cos[c + (3*d*x)/2] + 3900*C*d*x*Cos[c + (3*d*x)/2] + 600*A*d*x*Cos[2*c + (3*d*x)/2] + 3900*C*d*x*Cos[2*c +
(3*d*x)/2] + 120*A*d*x*Cos[2*c + (5*d*x)/2] + 780*C*d*x*Cos[2*c + (5*d*x)/2] + 120*A*d*x*Cos[3*c + (5*d*x)/2]
+ 780*C*d*x*Cos[3*c + (5*d*x)/2] - 2960*A*Sin[(d*x)/2] - 12760*C*Sin[(d*x)/2] + 2160*A*Sin[c + (d*x)/2] + 756
0*C*Sin[c + (d*x)/2] - 1840*A*Sin[c + (3*d*x)/2] - 9230*C*Sin[c + (3*d*x)/2] + 720*A*Sin[2*c + (3*d*x)/2] + 93
0*C*Sin[2*c + (3*d*x)/2] - 512*A*Sin[2*c + (5*d*x)/2] - 2782*C*Sin[2*c + (5*d*x)/2] - 750*C*Sin[3*c + (5*d*x)/
2] - 105*C*Sin[3*c + (7*d*x)/2] - 105*C*Sin[4*c + (7*d*x)/2] + 15*C*Sin[4*c + (9*d*x)/2] + 15*C*Sin[5*c + (9*d
*x)/2]))/(480*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.034, size = 224, normalized size = 1.2 \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{A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{2\,C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{31\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-7\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-5\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{d{a}^{3}}}+13\,{\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)^3*(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*tan(1/2*d*x+1/2*c)^3*A+2/3/d/a^
3*C*tan(1/2*d*x+1/2*c)^3-7/4/d/a^3*A*tan(1/2*d*x+1/2*c)-31/4/d/a^3*C*tan(1/2*d*x+1/2*c)-7/d/a^3/(tan(1/2*d*x+1
/2*c)^2+1)^2*C*tan(1/2*d*x+1/2*c)^3-5/d/a^3/(tan(1/2*d*x+1/2*c)^2+1)^2*C*tan(1/2*d*x+1/2*c)+2/d/a^3*arctan(tan
(1/2*d*x+1/2*c))*A+13/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.55897, size = 373, normalized size = 1.97 \begin{align*} -\frac{C{\left (\frac{60 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac{2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{40 \, \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{780 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + A{\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 )}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
- 40*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 - 780*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^3) + A*((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/(cos(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d

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Fricas [A]  time = 1.68491, size = 478, normalized size = 2.53 \begin{align*} \frac{15 \,{\left (2 \, A + 13 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (2 \, A + 13 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (2 \, A + 13 \, C\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (2 \, A + 13 \, C\right )} d x +{\left (15 \, C \cos \left (d x + c\right )^{4} - 45 \, C \cos \left (d x + c\right )^{3} -{\left (64 \, A + 479 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (34 \, A + 239 \, C\right )} \cos \left (d x + c\right ) - 44 \, A - 304 \, C\right )} \sin \left (d x + c\right )}{30 \,{\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)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/30*(15*(2*A + 13*C)*d*x*cos(d*x + c)^3 + 45*(2*A + 13*C)*d*x*cos(d*x + c)^2 + 45*(2*A + 13*C)*d*x*cos(d*x +
c) + 15*(2*A + 13*C)*d*x + (15*C*cos(d*x + c)^4 - 45*C*cos(d*x + c)^3 - (64*A + 479*C)*cos(d*x + c)^2 - 3*(34*
A + 239*C)*cos(d*x + c) - 44*A - 304*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 = 29.4935, size = 967, normalized size = 5.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((60*A*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a
**3*d) + 120*A*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a*
*3*d) + 60*A*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 3*A*tan(c/2 +
d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 14*A*tan(c/2 + d*x/2)
**7/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 68*A*tan(c/2 + d*x/2)**5/(6
0*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 190*A*tan(c/2 + d*x/2)**3/(60*a**
3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 105*A*tan(c/2 + d*x/2)/(60*a**3*d*tan(
c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*C*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c
/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 780*C*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/
2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*C*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a
**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 3*C*tan(c/2 + d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*t
an(c/2 + d*x/2)**2 + 60*a**3*d) + 34*C*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2
+ d*x/2)**2 + 60*a**3*d) - 388*C*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*
x/2)**2 + 60*a**3*d) - 1310*C*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)
**2 + 60*a**3*d) - 765*C*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60
*a**3*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)**3/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.30967, size = 235, normalized size = 1.24 \begin{align*} \frac{\frac{30 \,{\left (d x + c\right )}{\left (2 \, A + 13 \, C\right )}}{a^{3}} - \frac{60 \,{\left (7 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} 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 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 465 \, 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)^3*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(d*x + c)*(2*A + 13*C)/a^3 - 60*(7*C*tan(1/2*d*x + 1/2*c)^3 + 5*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x
+ 1/2*c)^2 + 1)^2*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*A*a^12*tan(1
/2*d*x + 1/2*c)^3 - 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) + 465*C*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d