3.58 \(\int \frac{\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=147 \[ -\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}-\frac{(A-3 B) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (A-3 B)}{a^3}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(4 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

((A - 3*B)*x)/a^3 - ((7*A - 27*B)*Sin[c + d*x])/(15*a^3*d) + ((A - B)*Cos[c + d*x]^3*Sin[c + d*x])/(5*d*(a + a
*Cos[c + d*x])^3) + ((4*A - 9*B)*Cos[c + d*x]^2*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((A - 3*B)*Sin
[c + d*x])/(d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.457031, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2977, 2968, 3023, 12, 2735, 2648} \[ -\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}-\frac{(A-3 B) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (A-3 B)}{a^3}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(4 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - 3*B)*x)/a^3 - ((7*A - 27*B)*Sin[c + d*x])/(15*a^3*d) + ((A - B)*Cos[c + d*x]^3*Sin[c + d*x])/(5*d*(a + a
*Cos[c + d*x])^3) + ((4*A - 9*B)*Cos[c + d*x]^2*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((A - 3*B)*Sin
[c + d*x])/(d*(a^3 + a^3*Cos[c + d*x]))

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 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 3023

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos ^2(c+d x) (3 a (A-B)-a (A-6 B) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos (c+d x) \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{2 a^2 (4 A-9 B) \cos (c+d x)-a^2 (7 A-27 B) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{15 a^3 (A-3 B) \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5}\\ &=-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(A-3 B) \int \frac{\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(A-3 B) x}{a^3}-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(A-3 B) \int \frac{1}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(A-3 B) x}{a^3}-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(A-3 B) \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.832577, size = 361, normalized size = 2.46 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (300 d x (A-3 B) \cos \left (c+\frac{d x}{2}\right )+300 d x (A-3 B) \cos \left (\frac{d x}{2}\right )+540 A \sin \left (c+\frac{d x}{2}\right )-460 A \sin \left (c+\frac{3 d x}{2}\right )+180 A \sin \left (2 c+\frac{3 d x}{2}\right )-128 A \sin \left (2 c+\frac{5 d x}{2}\right )+150 A d x \cos \left (c+\frac{3 d x}{2}\right )+150 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+30 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+30 A d x \cos \left (3 c+\frac{5 d x}{2}\right )-740 A \sin \left (\frac{d x}{2}\right )-1125 B \sin \left (c+\frac{d x}{2}\right )+1215 B \sin \left (c+\frac{3 d x}{2}\right )-225 B \sin \left (2 c+\frac{3 d x}{2}\right )+363 B \sin \left (2 c+\frac{5 d x}{2}\right )+75 B \sin \left (3 c+\frac{5 d x}{2}\right )+15 B \sin \left (3 c+\frac{7 d x}{2}\right )+15 B \sin \left (4 c+\frac{7 d x}{2}\right )-450 B d x \cos \left (c+\frac{3 d x}{2}\right )-450 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-90 B d x \cos \left (2 c+\frac{5 d x}{2}\right )-90 B d x \cos \left (3 c+\frac{5 d x}{2}\right )+1755 B \sin \left (\frac{d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(300*(A - 3*B)*d*x*Cos[(d*x)/2] + 300*(A - 3*B)*d*x*Cos[c + (d*x)/2] + 150*A*d*x*Co
s[c + (3*d*x)/2] - 450*B*d*x*Cos[c + (3*d*x)/2] + 150*A*d*x*Cos[2*c + (3*d*x)/2] - 450*B*d*x*Cos[2*c + (3*d*x)
/2] + 30*A*d*x*Cos[2*c + (5*d*x)/2] - 90*B*d*x*Cos[2*c + (5*d*x)/2] + 30*A*d*x*Cos[3*c + (5*d*x)/2] - 90*B*d*x
*Cos[3*c + (5*d*x)/2] - 740*A*Sin[(d*x)/2] + 1755*B*Sin[(d*x)/2] + 540*A*Sin[c + (d*x)/2] - 1125*B*Sin[c + (d*
x)/2] - 460*A*Sin[c + (3*d*x)/2] + 1215*B*Sin[c + (3*d*x)/2] + 180*A*Sin[2*c + (3*d*x)/2] - 225*B*Sin[2*c + (3
*d*x)/2] - 128*A*Sin[2*c + (5*d*x)/2] + 363*B*Sin[2*c + (5*d*x)/2] + 75*B*Sin[3*c + (5*d*x)/2] + 15*B*Sin[3*c
+ (7*d*x)/2] + 15*B*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.065, size = 189, normalized size = 1.3 \begin{align*} -{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{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{B}{2\,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{17\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{d{a}^{3}}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5+1/3/d/a^3*tan(1/2*d*x+1/2*c)^3*A-1/2/d/a^
3*B*tan(1/2*d*x+1/2*c)^3-7/4/d/a^3*A*tan(1/2*d*x+1/2*c)+17/4/d/a^3*B*tan(1/2*d*x+1/2*c)+2/d/a^3*B*tan(1/2*d*x+
1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*A-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B

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Maxima [A]  time = 1.56786, size = 312, normalized size = 2.12 \begin{align*} \frac{3 \, B{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac{a^{3} \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 )}} + \frac{\frac{85 \, \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{\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 )} - 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+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(3*B*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^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) - 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 = 1.36359, size = 429, normalized size = 2.92 \begin{align*} \frac{15 \,{\left (A - 3 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (A - 3 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (A - 3 \, B\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (A - 3 \, B\right )} d x +{\left (15 \, B \cos \left (d x + c\right )^{3} -{\left (32 \, A - 117 \, B\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (17 \, A - 57 \, B\right )} \cos \left (d x + c\right ) - 22 \, A + 72 \, B\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)^3*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*(A - 3*B)*d*x*cos(d*x + c)^3 + 45*(A - 3*B)*d*x*cos(d*x + c)^2 + 45*(A - 3*B)*d*x*cos(d*x + c) + 15*(
A - 3*B)*d*x + (15*B*cos(d*x + c)^3 - (32*A - 117*B)*cos(d*x + c)^2 - 3*(17*A - 57*B)*cos(d*x + c) - 22*A + 72
*B)*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 = 16.5648, size = 496, normalized size = 3.37 \begin{align*} \begin{cases} \frac{60 A d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{60 A d x}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{3 A \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{17 A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{85 A \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{105 A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{180 B d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{180 B d x}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{3 B \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{27 B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{225 B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{375 B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \left (A + B \cos{\left (c \right )}\right ) \cos ^{3}{\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)**3*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((60*A*d*x*tan(c/2 + d*x/2)**2/(60*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)**2 + 60*a**3*d) - 3*A*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 17*A*tan(
c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 85*A*tan(c/2 + d*x/2)**3/(60*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)**2 + 60*a**3*d) - 180*B*d*x*tan(c/2
+ d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 180*B*d*x/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d
) + 3*B*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 27*B*tan(c/2 + d*x/2)**5/(60*a**3*d*
tan(c/2 + d*x/2)**2 + 60*a**3*d) + 225*B*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 375
*B*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**3/(a*cos
(c) + a)**3, True))

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Giac [A]  time = 1.2116, size = 209, normalized size = 1.42 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )}{\left (A - 3 \, B\right )}}{a^{3}} + \frac{120 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B 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} + 30 \, B 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 ) - 255 \, B 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+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)*(A - 3*B)/a^3 + 120*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) - (3*A*a^12*t
an(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^12*tan(1/2
*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d