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

Optimal. Leaf size=141 \[ -\frac {4 (A+4 C) \sin (c+d x)}{3 a^2 d}-\frac {2 (A+4 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {(2 A+7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {x (2 A+7 C)}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

1/2*(2*A+7*C)*x/a^2-4/3*(A+4*C)*sin(d*x+c)/a^2/d+1/2*(2*A+7*C)*cos(d*x+c)*sin(d*x+c)/a^2/d-2/3*(A+4*C)*cos(d*x
+c)^2*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^2

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Rubi [A]  time = 0.26, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 2977, 2734} \[ -\frac {4 (A+4 C) \sin (c+d x)}{3 a^2 d}-\frac {2 (A+4 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {(2 A+7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {x (2 A+7 C)}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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)]

Rubi steps

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

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Mathematica [A]  time = 0.65, size = 273, normalized size = 1.94 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (36 d x (2 A+7 C) \cos \left (c+\frac {d x}{2}\right )+96 A \sin \left (c+\frac {d x}{2}\right )-80 A \sin \left (c+\frac {3 d x}{2}\right )+24 A d x \cos \left (c+\frac {3 d x}{2}\right )+24 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+36 d x (2 A+7 C) \cos \left (\frac {d x}{2}\right )-144 A \sin \left (\frac {d x}{2}\right )+147 C \sin \left (c+\frac {d x}{2}\right )-239 C \sin \left (c+\frac {3 d x}{2}\right )-63 C \sin \left (2 c+\frac {3 d x}{2}\right )-15 C \sin \left (2 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 C \sin \left (3 c+\frac {7 d x}{2}\right )+3 C \sin \left (4 c+\frac {7 d x}{2}\right )+84 C d x \cos \left (c+\frac {3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 C \sin \left (\frac {d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(36*(2*A + 7*C)*d*x*Cos[(d*x)/2] + 36*(2*A + 7*C)*d*x*Cos[c + (d*x)/2] + 24*A*d*x*C
os[c + (3*d*x)/2] + 84*C*d*x*Cos[c + (3*d*x)/2] + 24*A*d*x*Cos[2*c + (3*d*x)/2] + 84*C*d*x*Cos[2*c + (3*d*x)/2
] - 144*A*Sin[(d*x)/2] - 381*C*Sin[(d*x)/2] + 96*A*Sin[c + (d*x)/2] + 147*C*Sin[c + (d*x)/2] - 80*A*Sin[c + (3
*d*x)/2] - 239*C*Sin[c + (3*d*x)/2] - 63*C*Sin[2*c + (3*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] - 15*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]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.71, size = 134, normalized size = 0.95 \[ \frac {3 \, {\left (2 \, A + 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, A + 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (2 \, A + 7 \, C\right )} d x + {\left (3 \, C \cos \left (d x + c\right )^{3} - 6 \, C \cos \left (d x + c\right )^{2} - {\left (10 \, A + 43 \, C\right )} \cos \left (d x + c\right ) - 8 \, A - 32 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*A + 7*C)*d*x*cos(d*x + c)^2 + 6*(2*A + 7*C)*d*x*cos(d*x + c) + 3*(2*A + 7*C)*d*x + (3*C*cos(d*x + c)
^3 - 6*C*cos(d*x + c)^2 - (10*A + 43*C)*cos(d*x + c) - 8*A - 32*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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giac [A]  time = 0.54, size = 137, normalized size = 0.97 \[ \frac {\frac {3 \, {\left (d x + c\right )} {\left (2 \, A + 7 \, C\right )}}{a^{2}} - \frac {6 \, {\left (5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(d*x + c)*(2*A + 7*C)/a^2 - 6*(5*C*tan(1/2*d*x + 1/2*c)^3 + 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1
/2*c)^2 + 1)^2*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 - 9*A*a^4*tan(1/2*d*x + 1/2
*c) - 21*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.12, size = 184, normalized size = 1.30 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {5 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(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)^3-3/2/d/a^2*A*tan(1/2*d*x+1/2*c)-7/2/d/a^2*C*t
an(1/2*d*x+1/2*c)-5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)^3-3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2
*C*tan(1/2*d*x+1/2*c)+2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*A+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [A]  time = 0.45, size = 236, normalized size = 1.67 \[ -\frac {C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} + A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(C*(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/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) + A*((9*sin(d*x + c)
/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^
2))/d

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mupad [B]  time = 0.96, size = 134, normalized size = 0.95 \[ \frac {x\,\left (2\,A+7\,C\right )}{2\,a^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^2}+\frac {2\,C}{a^2}\right )}{d}-\frac {5\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^2,x)

[Out]

(x*(2*A + 7*C))/(2*a^2) - (tan(c/2 + (d*x)/2)*((3*(A + C))/(2*a^2) + (2*C)/a^2))/d - (3*C*tan(c/2 + (d*x)/2) +
5*C*tan(c/2 + (d*x)/2)^3)/(d*(2*a^2*tan(c/2 + (d*x)/2)^2 + a^2*tan(c/2 + (d*x)/2)^4 + a^2)) + (tan(c/2 + (d*x
)/2)^3*(A + C))/(6*a^2*d)

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sympy [A]  time = 7.01, size = 845, normalized size = 5.99 \[ \begin {cases} \frac {6 A d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {12 A d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {6 A d x}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {7 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {17 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {9 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {21 C d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {42 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {21 C d x}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {19 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {71 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {39 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\relax (c )}\right ) \cos ^{2}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((6*A*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*
d) + 12*A*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) +
6*A*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + A*tan(c/2 + d*x/2)**7/(6*a
**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 7*A*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(
c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 17*A*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x
/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 9*A*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a
**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d
*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 42*C*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(
c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*
d) + C*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 19*C*ta
n(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 71*C*tan(c/2 + d
*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 39*C*tan(c/2 + d*x/2)/(6*
a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)
**2/(a*cos(c) + a)**2, True))

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