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

Optimal. Leaf size=152 \[ \frac{(2 A-7 B+27 C) \sin (c+d x)}{15 a^3 d}-\frac{(B-3 C) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (B-3 C)}{a^3}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.47639, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3041, 2977, 2968, 3023, 12, 2735, 2648} \[ \frac{(2 A-7 B+27 C) \sin (c+d x)}{15 a^3 d}-\frac{(B-3 C) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (B-3 C)}{a^3}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(A+4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 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 ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx &=-\frac{(A-B+C) \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) (a (2 A+3 B-3 C)+a (A-B+6 C) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A+4 B-9 C) \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 (A+4 B-9 C)+a^2 (2 A-7 B+27 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A+4 B-9 C) \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 (A+4 B-9 C) \cos (c+d x)+a^2 (2 A-7 B+27 C) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{(2 A-7 B+27 C) \sin (c+d x)}{15 a^3 d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A+4 B-9 C) \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 (B-3 C) \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5}\\ &=\frac{(2 A-7 B+27 C) \sin (c+d x)}{15 a^3 d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A+4 B-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(B-3 C) \int \frac{\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(B-3 C) x}{a^3}+\frac{(2 A-7 B+27 C) \sin (c+d x)}{15 a^3 d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A+4 B-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(B-3 C) \int \frac{1}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(B-3 C) x}{a^3}+\frac{(2 A-7 B+27 C) \sin (c+d x)}{15 a^3 d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A+4 B-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(B-3 C) \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.00057, size = 423, normalized size = 2.78 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-120 A \sin \left (c+\frac{d x}{2}\right )+80 A \sin \left (c+\frac{3 d x}{2}\right )-60 A \sin \left (2 c+\frac{3 d x}{2}\right )+28 A \sin \left (2 c+\frac{5 d x}{2}\right )+160 A \sin \left (\frac{d x}{2}\right )+300 d x (B-3 C) \cos \left (c+\frac{d x}{2}\right )+540 B \sin \left (c+\frac{d x}{2}\right )-460 B \sin \left (c+\frac{3 d x}{2}\right )+180 B \sin \left (2 c+\frac{3 d x}{2}\right )-128 B \sin \left (2 c+\frac{5 d x}{2}\right )+150 B d x \cos \left (c+\frac{3 d x}{2}\right )+150 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+30 B d x \cos \left (2 c+\frac{5 d x}{2}\right )+30 B d x \cos \left (3 c+\frac{5 d x}{2}\right )+300 d x (B-3 C) \cos \left (\frac{d x}{2}\right )-740 B \sin \left (\frac{d x}{2}\right )-1125 C \sin \left (c+\frac{d x}{2}\right )+1215 C \sin \left (c+\frac{3 d x}{2}\right )-225 C \sin \left (2 c+\frac{3 d x}{2}\right )+363 C \sin \left (2 c+\frac{5 d x}{2}\right )+75 C \sin \left (3 c+\frac{5 d x}{2}\right )+15 C \sin \left (3 c+\frac{7 d x}{2}\right )+15 C \sin \left (4 c+\frac{7 d x}{2}\right )-450 C d x \cos \left (c+\frac{3 d x}{2}\right )-450 C d x \cos \left (2 c+\frac{3 d x}{2}\right )-90 C d x \cos \left (2 c+\frac{5 d x}{2}\right )-90 C d x \cos \left (3 c+\frac{5 d x}{2}\right )+1755 C \sin \left (\frac{d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(300*(B - 3*C)*d*x*Cos[(d*x)/2] + 300*(B - 3*C)*d*x*Cos[c + (d*x)/2] + 150*B*d*x*Co
s[c + (3*d*x)/2] - 450*C*d*x*Cos[c + (3*d*x)/2] + 150*B*d*x*Cos[2*c + (3*d*x)/2] - 450*C*d*x*Cos[2*c + (3*d*x)
/2] + 30*B*d*x*Cos[2*c + (5*d*x)/2] - 90*C*d*x*Cos[2*c + (5*d*x)/2] + 30*B*d*x*Cos[3*c + (5*d*x)/2] - 90*C*d*x
*Cos[3*c + (5*d*x)/2] + 160*A*Sin[(d*x)/2] - 740*B*Sin[(d*x)/2] + 1755*C*Sin[(d*x)/2] - 120*A*Sin[c + (d*x)/2]
+ 540*B*Sin[c + (d*x)/2] - 1125*C*Sin[c + (d*x)/2] + 80*A*Sin[c + (3*d*x)/2] - 460*B*Sin[c + (3*d*x)/2] + 121
5*C*Sin[c + (3*d*x)/2] - 60*A*Sin[2*c + (3*d*x)/2] + 180*B*Sin[2*c + (3*d*x)/2] - 225*C*Sin[2*c + (3*d*x)/2] +
28*A*Sin[2*c + (5*d*x)/2] - 128*B*Sin[2*c + (5*d*x)/2] + 363*C*Sin[2*c + (5*d*x)/2] + 75*C*Sin[3*c + (5*d*x)/
2] + 15*C*Sin[3*c + (7*d*x)/2] + 15*C*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.033, size = 247, normalized size = 1.6 \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{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{A}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{2\,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\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\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\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{3}}}-6\,{\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)^2*(A+B*cos(d*x+c)+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*B*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-1/6/d/a^
3*tan(1/2*d*x+1/2*c)^3*A+1/3/d/a^3*tan(1/2*d*x+1/2*c)^3*B-1/2/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*B*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)+2/d/a^3*C*tan(1/2*d*x+1/2*c)/(tan(1
/2*d*x+1/2*c)^2+1)+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [B]  time = 1.55138, size = 398, normalized size = 2.62 \begin{align*} \frac{3 \, C{\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 )} - B{\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{A{\left (\frac{15 \, \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{3 \, \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)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(3*C*(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) - B*((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) + A*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(c
os(d*x + c) + 1)^5)/a^3)/d

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Fricas [A]  time = 1.91543, size = 454, normalized size = 2.99 \begin{align*} \frac{15 \,{\left (B - 3 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (B - 3 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (B - 3 \, C\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (B - 3 \, C\right )} d x +{\left (15 \, C \cos \left (d x + c\right )^{3} +{\left (7 \, A - 32 \, B + 117 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A - 17 \, B + 57 \, C\right )} \cos \left (d x + c\right ) + 2 \, A - 22 \, B + 72 \, 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)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*(B - 3*C)*d*x*cos(d*x + c)^3 + 45*(B - 3*C)*d*x*cos(d*x + c)^2 + 45*(B - 3*C)*d*x*cos(d*x + c) + 15*(
B - 3*C)*d*x + (15*C*cos(d*x + c)^3 + (7*A - 32*B + 117*C)*cos(d*x + c)^2 + 3*(2*A - 17*B + 57*C)*cos(d*x + c)
+ 2*A - 22*B + 72*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 = 24.1584, size = 665, normalized size = 4.38 \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)**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((3*A*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 7*A*tan(c/2 + d*x/2)**5/(60*a
**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 5*A*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) +
15*A*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 60*B*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*t
an(c/2 + d*x/2)**2 + 60*a**3*d) + 60*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) + 17*B*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*
a**3*d) - 85*B*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 105*B*tan(c/2 + d*x/2)/(60*a*
*3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 180*C*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*C*d*x/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3*C*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x
/2)**2 + 60*a**3*d) - 27*C*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 225*C*tan(c/2 + d
*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 375*C*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) + C*cos(c)**2)*cos(c)**2/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.22639, size = 274, normalized size = 1.8 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )}{\left (B - 3 \, C\right )}}{a^{3}} + \frac{120 \, C \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} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 30 \, 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 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 255 \, 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)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)*(B - 3*C)/a^3 + 120*C*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 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 10*A*a^12*tan(1/2*
d*x + 1/2*c)^3 + 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*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*B*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d