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

Optimal. Leaf size=170 \[ \frac{4 (2 B-3 C) \sin ^3(c+d x)}{3 a^2 d}-\frac{4 (2 B-3 C) \sin (c+d x)}{a^2 d}+\frac{(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{(7 B-10 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{x (7 B-10 C)}{2 a^2}+\frac{(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

((7*B - 10*C)*x)/(2*a^2) - (4*(2*B - 3*C)*Sin[c + d*x])/(a^2*d) + ((7*B - 10*C)*Cos[c + d*x]*Sin[c + d*x])/(2*
a^2*d) + ((7*B - 10*C)*Cos[c + d*x]^3*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + ((B - C)*Cos[c + d*x]^4*Sin
[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (4*(2*B - 3*C)*Sin[c + d*x]^3)/(3*a^2*d)

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Rubi [A]  time = 0.396788, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3029, 2977, 2748, 2635, 8, 2633} \[ \frac{4 (2 B-3 C) \sin ^3(c+d x)}{3 a^2 d}-\frac{4 (2 B-3 C) \sin (c+d x)}{a^2 d}+\frac{(7 B-10 C) \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{(7 B-10 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{x (7 B-10 C)}{2 a^2}+\frac{(B-C) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*B - 10*C)*x)/(2*a^2) - (4*(2*B - 3*C)*Sin[c + d*x])/(a^2*d) + ((7*B - 10*C)*Cos[c + d*x]*Sin[c + d*x])/(2*
a^2*d) + ((7*B - 10*C)*Cos[c + d*x]^3*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + ((B - C)*Cos[c + d*x]^4*Sin
[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (4*(2*B - 3*C)*Sin[c + d*x]^3)/(3*a^2*d)

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]

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 (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx &=\int \frac{\cos ^4(c+d x) (B+C \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\\ &=\frac{(B-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) (4 a (B-C)-3 a (B-2 C) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac{(7 B-10 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(B-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 (3 a^2 (7 B-10 C)-12 a^2 (2 B-3 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac{(7 B-10 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(B-C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(7 B-10 C) \int \cos ^2(c+d x) \, dx}{a^2}-\frac{(4 (2 B-3 C)) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=\frac{(7 B-10 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{(7 B-10 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(B-C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(7 B-10 C) \int 1 \, dx}{2 a^2}+\frac{(4 (2 B-3 C)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=\frac{(7 B-10 C) x}{2 a^2}-\frac{4 (2 B-3 C) \sin (c+d x)}{a^2 d}+\frac{(7 B-10 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{(7 B-10 C) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(B-C) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{4 (2 B-3 C) \sin ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [B]  time = 0.607838, size = 369, normalized size = 2.17 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (36 d x (7 B-10 C) \cos \left (c+\frac{d x}{2}\right )+147 B \sin \left (c+\frac{d x}{2}\right )-239 B \sin \left (c+\frac{3 d x}{2}\right )-63 B \sin \left (2 c+\frac{3 d x}{2}\right )-15 B \sin \left (2 c+\frac{5 d x}{2}\right )-15 B \sin \left (3 c+\frac{5 d x}{2}\right )+3 B \sin \left (3 c+\frac{7 d x}{2}\right )+3 B \sin \left (4 c+\frac{7 d x}{2}\right )+84 B d x \cos \left (c+\frac{3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+36 d x (7 B-10 C) \cos \left (\frac{d x}{2}\right )-381 B \sin \left (\frac{d x}{2}\right )-156 C \sin \left (c+\frac{d x}{2}\right )+342 C \sin \left (c+\frac{3 d x}{2}\right )+118 C \sin \left (2 c+\frac{3 d x}{2}\right )+30 C \sin \left (2 c+\frac{5 d x}{2}\right )+30 C \sin \left (3 c+\frac{5 d x}{2}\right )-3 C \sin \left (3 c+\frac{7 d x}{2}\right )-3 C \sin \left (4 c+\frac{7 d x}{2}\right )+C \sin \left (4 c+\frac{9 d x}{2}\right )+C \sin \left (5 c+\frac{9 d x}{2}\right )-120 C d x \cos \left (c+\frac{3 d x}{2}\right )-120 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+516 C \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(36*(7*B - 10*C)*d*x*Cos[(d*x)/2] + 36*(7*B - 10*C)*d*x*Cos[c + (d*x)/2] + 84*B*d*x
*Cos[c + (3*d*x)/2] - 120*C*d*x*Cos[c + (3*d*x)/2] + 84*B*d*x*Cos[2*c + (3*d*x)/2] - 120*C*d*x*Cos[2*c + (3*d*
x)/2] - 381*B*Sin[(d*x)/2] + 516*C*Sin[(d*x)/2] + 147*B*Sin[c + (d*x)/2] - 156*C*Sin[c + (d*x)/2] - 239*B*Sin[
c + (3*d*x)/2] + 342*C*Sin[c + (3*d*x)/2] - 63*B*Sin[2*c + (3*d*x)/2] + 118*C*Sin[2*c + (3*d*x)/2] - 15*B*Sin[
2*c + (5*d*x)/2] + 30*C*Sin[2*c + (5*d*x)/2] - 15*B*Sin[3*c + (5*d*x)/2] + 30*C*Sin[3*c + (5*d*x)/2] + 3*B*Sin
[3*c + (7*d*x)/2] - 3*C*Sin[3*c + (7*d*x)/2] + 3*B*Sin[4*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 = 1.9 \begin{align*}{\frac{B}{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{7\,B}{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 ) }-5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}B}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+10\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-8\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{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}}-3\,{\frac{B\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}}}+7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{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*(B*cos(d*x+c)+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*B-1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3-7/2/d/a^2*B*tan(1/2*d*x+1/2*c)+9/2/d/a^2*C*t
an(1/2*d*x+1/2*c)-5/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*B+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-8/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*B+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)^3-3/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^3*B*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)+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B-10/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [B]  time = 2.0116, size = 502, normalized size = 2.95 \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 )} - B{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \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{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(B*cos(d*x+c)+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) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(c
os(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arct
an(sin(d*x + c)/(cos(d*x + c) + 1))/a^2))/d

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Fricas [A]  time = 1.70942, size = 387, normalized size = 2.28 \begin{align*} \frac{3 \,{\left (7 \, B - 10 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (7 \, B - 10 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (7 \, B - 10 \, C\right )} d x +{\left (2 \, C \cos \left (d x + c\right )^{4} +{\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{3} - 6 \,{\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (43 \, B - 66 \, C\right )} \cos \left (d x + c\right ) - 32 \, B + 48 \, C\right )} \sin \left (d x + c\right )}{6 \,{\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*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*(7*B - 10*C)*d*x*cos(d*x + c)^2 + 6*(7*B - 10*C)*d*x*cos(d*x + c) + 3*(7*B - 10*C)*d*x + (2*C*cos(d*x +
c)^4 + (3*B - 2*C)*cos(d*x + c)^3 - 6*(B - 2*C)*cos(d*x + c)^2 - (43*B - 66*C)*cos(d*x + c) - 32*B + 48*C)*si
n(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 = 28.4945, size = 1430, normalized size = 8.41 \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*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((21*B*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) + 63*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*B*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) + 21*B*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) + B*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) - 18*B*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*ta
n(c/2 + d*x/2)**2 + 6*a**2*d) - 90*B*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) - 110*B*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) - 39*B*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*(B*cos(c) + C*cos(c
)**2)*cos(c)**3/(a*cos(c) + a)**2, True))

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Giac [A]  time = 1.4393, size = 259, normalized size = 1.52 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}{\left (7 \, B - 10 \, C\right )}}{a^{2}} - \frac{2 \,{\left (15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 30 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 18 \, 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{B 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} - 21 \, B 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*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(3*(d*x + c)*(7*B - 10*C)/a^2 - 2*(15*B*tan(1/2*d*x + 1/2*c)^5 - 30*C*tan(1/2*d*x + 1/2*c)^5 + 24*B*tan(1/
2*d*x + 1/2*c)^3 - 40*C*tan(1/2*d*x + 1/2*c)^3 + 9*B*tan(1/2*d*x + 1/2*c) - 18*C*tan(1/2*d*x + 1/2*c))/((tan(1
/2*d*x + 1/2*c)^2 + 1)^3*a^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3 - 21*B*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