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

Optimal. Leaf size=163 \[ -\frac{(5 A+12 C) \sin ^3(c+d x)}{3 a^2 d}+\frac{(5 A+12 C) \sin (c+d x)}{a^2 d}-\frac{2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(2 A+5 C) \sin (c+d x) \cos (c+d x)}{a^2 d}-\frac{x (2 A+5 C)}{a^2}-\frac{(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-(((2*A + 5*C)*x)/a^2) + ((5*A + 12*C)*Sin[c + d*x])/(a^2*d) - ((2*A + 5*C)*Cos[c + d*x]*Sin[c + d*x])/(a^2*d)
- (2*(2*A + 5*C)*Cos[c + d*x]^3*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^4*Sin[c +
d*x])/(3*d*(a + a*Cos[c + d*x])^2) - ((5*A + 12*C)*Sin[c + d*x]^3)/(3*a^2*d)

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Rubi [A]  time = 0.327101, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3042, 2977, 2748, 2635, 8, 2633} \[ -\frac{(5 A+12 C) \sin ^3(c+d x)}{3 a^2 d}+\frac{(5 A+12 C) \sin (c+d x)}{a^2 d}-\frac{2 (2 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(2 A+5 C) \sin (c+d x) \cos (c+d x)}{a^2 d}-\frac{x (2 A+5 C)}{a^2}-\frac{(A+C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*A + 5*C)*x)/a^2) + ((5*A + 12*C)*Sin[c + d*x])/(a^2*d) - ((2*A + 5*C)*Cos[c + d*x]*Sin[c + d*x])/(a^2*d)
- (2*(2*A + 5*C)*Cos[c + d*x]^3*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^4*Sin[c +
d*x])/(3*d*(a + a*Cos[c + d*x])^2) - ((5*A + 12*C)*Sin[c + d*x]^3)/(3*a^2*d)

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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2635

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 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))^2} \, dx &=-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^3(c+d x) (-a (A+4 C)+3 a (A+2 C) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{2 (2 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \cos ^2(c+d x) \left (-6 a^2 (2 A+5 C)+3 a^2 (5 A+12 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{2 (2 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(2 (2 A+5 C)) \int \cos ^2(c+d x) \, dx}{a^2}+\frac{(5 A+12 C) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac{(2 A+5 C) \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac{2 (2 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(2 A+5 C) \int 1 \, dx}{a^2}-\frac{(5 A+12 C) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac{(2 A+5 C) x}{a^2}+\frac{(5 A+12 C) \sin (c+d x)}{a^2 d}-\frac{(2 A+5 C) \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac{2 (2 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(5 A+12 C) \sin ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [B]  time = 0.651771, size = 341, normalized size = 2.09 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-72 d x (2 A+5 C) \cos \left (c+\frac{d x}{2}\right )-120 A \sin \left (c+\frac{d x}{2}\right )+164 A \sin \left (c+\frac{3 d x}{2}\right )+36 A \sin \left (2 c+\frac{3 d x}{2}\right )+12 A \sin \left (2 c+\frac{5 d x}{2}\right )+12 A \sin \left (3 c+\frac{5 d x}{2}\right )-48 A d x \cos \left (c+\frac{3 d x}{2}\right )-48 A d x \cos \left (2 c+\frac{3 d x}{2}\right )-72 d x (2 A+5 C) \cos \left (\frac{d x}{2}\right )+264 A \sin \left (\frac{d x}{2}\right )-156 C \sin \left (c+\frac{d x}{2}\right )+342 C \sin \left (c+\frac{3 d x}{2}\right )+118 C \sin \left (2 c+\frac{3 d x}{2}\right )+30 C \sin \left (2 c+\frac{5 d x}{2}\right )+30 C \sin \left (3 c+\frac{5 d x}{2}\right )-3 C \sin \left (3 c+\frac{7 d x}{2}\right )-3 C \sin \left (4 c+\frac{7 d x}{2}\right )+C \sin \left (4 c+\frac{9 d x}{2}\right )+C \sin \left (5 c+\frac{9 d x}{2}\right )-120 C d x \cos \left (c+\frac{3 d x}{2}\right )-120 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+516 C \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-72*(2*A + 5*C)*d*x*Cos[(d*x)/2] - 72*(2*A + 5*C)*d*x*Cos[c + (d*x)/2] - 48*A*d*x*
Cos[c + (3*d*x)/2] - 120*C*d*x*Cos[c + (3*d*x)/2] - 48*A*d*x*Cos[2*c + (3*d*x)/2] - 120*C*d*x*Cos[2*c + (3*d*x
)/2] + 264*A*Sin[(d*x)/2] + 516*C*Sin[(d*x)/2] - 120*A*Sin[c + (d*x)/2] - 156*C*Sin[c + (d*x)/2] + 164*A*Sin[c
+ (3*d*x)/2] + 342*C*Sin[c + (3*d*x)/2] + 36*A*Sin[2*c + (3*d*x)/2] + 118*C*Sin[2*c + (3*d*x)/2] + 12*A*Sin[2
*c + (5*d*x)/2] + 30*C*Sin[2*c + (5*d*x)/2] + 12*A*Sin[3*c + (5*d*x)/2] + 30*C*Sin[3*c + (5*d*x)/2] - 3*C*Sin[
3*c + (7*d*x)/2] - 3*C*Sin[4*c + (7*d*x)/2] + C*Sin[4*c + (9*d*x)/2] + C*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 +
Cos[c + d*x])^2)

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Maple [B]  time = 0.033, size = 322, normalized size = 2. \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{5\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{9\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}A}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+10\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}C}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}A}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{40\,C}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}+2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+6\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{d{a}^{2}}}-10\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{2}}} \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))^2,x)

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+5/2/d/a^2*A*tan(1/2*d*x+1/2*c)+9/2/d/a^2*C*
tan(1/2*d*x+1/2*c)+2/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*A+10/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)
^3*tan(1/2*d*x+1/2*c)^5*C+4/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*A+40/3/d/a^2/(tan(1/2*d*x+1/
2*c)^2+1)^3*C*tan(1/2*d*x+1/2*c)^3+2/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^3*A*tan(1/2*d*x+1/2*c)+6/d/a^2/(tan(1/2*d*
x+1/2*c)^2+1)^3*C*tan(1/2*d*x+1/2*c)-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*A-10/d/a^2*arctan(tan(1/2*d*x+1/2*c))*
C

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Maxima [B]  time = 1.56672, size = 439, normalized size = 2.69 \begin{align*} \frac{C{\left (\frac{4 \,{\left (\frac{9 \, \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{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \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}}}{a^{2}} - \frac{60 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} + A{\left (\frac{\frac{15 \, \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}}}{a^{2}} - \frac{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, 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))^2,x, algorithm="maxima")

[Out]

1/6*(C*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) + A*((15*sin(d*x + c)/(cos(d*x + c) + 1) - si
n(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^
2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 1.46484, size = 369, normalized size = 2.26 \begin{align*} -\frac{3 \,{\left (2 \, A + 5 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (2 \, A + 5 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (2 \, A + 5 \, C\right )} d x -{\left (C \cos \left (d x + c\right )^{4} - C \cos \left (d x + c\right )^{3} + 3 \,{\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (14 \, A + 33 \, C\right )} \cos \left (d x + c\right ) + 10 \, A + 24 \, 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 )}} \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))^2,x, algorithm="fricas")

[Out]

-1/3*(3*(2*A + 5*C)*d*x*cos(d*x + c)^2 + 6*(2*A + 5*C)*d*x*cos(d*x + c) + 3*(2*A + 5*C)*d*x - (C*cos(d*x + c)^
4 - C*cos(d*x + c)^3 + 3*(A + 2*C)*cos(d*x + c)^2 + (14*A + 33*C)*cos(d*x + c) + 10*A + 24*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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Sympy [A]  time = 19.7183, size = 1426, normalized size = 8.75 \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))**2,x)

[Out]

Piecewise((-12*A*d*x*tan(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a*
*2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 36*A*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*
tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 36*A*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c
/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*A*d*x/(6*a**2*
d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - A*tan(c/2
+ d*x/2)**9/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*
a**2*d) + 12*A*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*t
an(c/2 + d*x/2)**2 + 6*a**2*d) + 54*A*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 +
d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 68*A*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**6
+ 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*A*tan(c/2 + d*x/2)/(6*a**2*d
*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 30*C*d*x*ta
n(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**
2 + 6*a**2*d) - 90*C*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 1
8*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 90*C*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**
2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 30*C*d*x/(6*a**2*d*tan(c/2 + d*x/2)**6 +
18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - C*tan(c/2 + d*x/2)**9/(6*a**2*d*t
an(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 24*C*tan(c/2
+ d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*
a**2*d) + 138*C*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*
tan(c/2 + d*x/2)**2 + 6*a**2*d) + 160*C*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2
+ d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*C*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**6
+ 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*c
os(c)**3/(a*cos(c) + a)**2, True))

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Giac [A]  time = 1.31605, size = 258, normalized size = 1.58 \begin{align*} -\frac{\frac{6 \,{\left (d x + c\right )}{\left (2 \, A + 5 \, C\right )}}{a^{2}} - \frac{4 \,{\left (3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, 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 )}^{3} a^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 27 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

-1/6*(6*(d*x + c)*(2*A + 5*C)/a^2 - 4*(3*A*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2*c)^5 + 6*A*tan(1/2*
d*x + 1/2*c)^3 + 20*C*tan(1/2*d*x + 1/2*c)^3 + 3*A*tan(1/2*d*x + 1/2*c) + 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*
d*x + 1/2*c)^2 + 1)^3*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^4*tan(1/2*d
*x + 1/2*c) - 27*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d