3.67 \(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=152 \[ \frac {(16 A-215 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(8 A-55 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {C x}{a^4}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac {2 (2 A-5 C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

C*x/a^4-1/105*(8*A-55*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2+1/105*(16*A-215*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))-
1/7*(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+2/35*(2*A-5*C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x
+c))^3

________________________________________________________________________________________

Rubi [A]  time = 0.44, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 2977, 2968, 3019, 2735, 2648} \[ \frac {(16 A-215 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(8 A-55 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {C x}{a^4}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac {2 (2 A-5 C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^4,x]

[Out]

(C*x)/a^4 - ((8*A - 55*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((16*A - 215*C)*Sin[c + d*x])/(105*
a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + (2*(2*A - 5*C
)*Cos[c + d*x]^2*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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]

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 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 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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 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))^4} \, dx &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {\cos ^2(c+d x) (a (4 A-3 C)+7 a C \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos (c+d x) \left (4 a^2 (2 A-5 C)+35 a^2 C \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {4 a^2 (2 A-5 C) \cos (c+d x)+35 a^2 C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(8 A-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {\int \frac {-2 a^3 (8 A-55 C)-105 a^3 C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac {C x}{a^4}-\frac {(8 A-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(16 A-215 C) \int \frac {1}{a+a \cos (c+d x)} \, dx}{105 a^3}\\ &=\frac {C x}{a^4}-\frac {(8 A-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(16 A-215 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.74, size = 315, normalized size = 2.07 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (-350 A \sin \left (c+\frac {d x}{2}\right )+336 A \sin \left (c+\frac {3 d x}{2}\right )-210 A \sin \left (2 c+\frac {3 d x}{2}\right )+182 A \sin \left (2 c+\frac {5 d x}{2}\right )+26 A \sin \left (3 c+\frac {7 d x}{2}\right )+560 A \sin \left (\frac {d x}{2}\right )+8260 C \sin \left (c+\frac {d x}{2}\right )-7140 C \sin \left (c+\frac {3 d x}{2}\right )+3780 C \sin \left (2 c+\frac {3 d x}{2}\right )-2800 C \sin \left (2 c+\frac {5 d x}{2}\right )+840 C \sin \left (3 c+\frac {5 d x}{2}\right )-520 C \sin \left (3 c+\frac {7 d x}{2}\right )+3675 C d x \cos \left (c+\frac {d x}{2}\right )+2205 C d x \cos \left (c+\frac {3 d x}{2}\right )+2205 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+735 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+735 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+105 C d x \cos \left (3 c+\frac {7 d x}{2}\right )+105 C d x \cos \left (4 c+\frac {7 d x}{2}\right )-9940 C \sin \left (\frac {d x}{2}\right )+3675 C d x \cos \left (\frac {d x}{2}\right )\right )}{13440 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^4,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(3675*C*d*x*Cos[(d*x)/2] + 3675*C*d*x*Cos[c + (d*x)/2] + 2205*C*d*x*Cos[c + (3*d*
x)/2] + 2205*C*d*x*Cos[2*c + (3*d*x)/2] + 735*C*d*x*Cos[2*c + (5*d*x)/2] + 735*C*d*x*Cos[3*c + (5*d*x)/2] + 10
5*C*d*x*Cos[3*c + (7*d*x)/2] + 105*C*d*x*Cos[4*c + (7*d*x)/2] + 560*A*Sin[(d*x)/2] - 9940*C*Sin[(d*x)/2] - 350
*A*Sin[c + (d*x)/2] + 8260*C*Sin[c + (d*x)/2] + 336*A*Sin[c + (3*d*x)/2] - 7140*C*Sin[c + (3*d*x)/2] - 210*A*S
in[2*c + (3*d*x)/2] + 3780*C*Sin[2*c + (3*d*x)/2] + 182*A*Sin[2*c + (5*d*x)/2] - 2800*C*Sin[2*c + (5*d*x)/2] +
840*C*Sin[3*c + (5*d*x)/2] + 26*A*Sin[3*c + (7*d*x)/2] - 520*C*Sin[3*c + (7*d*x)/2]))/(13440*a^4*d)

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 179, normalized size = 1.18 \[ \frac {105 \, C d x \cos \left (d x + c\right )^{4} + 420 \, C d x \cos \left (d x + c\right )^{3} + 630 \, C d x \cos \left (d x + c\right )^{2} + 420 \, C d x \cos \left (d x + c\right ) + 105 \, C d x + {\left (13 \, {\left (A - 20 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (13 \, A - 155 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A - 535 \, C\right )} \cos \left (d x + c\right ) + 8 \, A - 160 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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))^4,x, algorithm="fricas")

[Out]

1/105*(105*C*d*x*cos(d*x + c)^4 + 420*C*d*x*cos(d*x + c)^3 + 630*C*d*x*cos(d*x + c)^2 + 420*C*d*x*cos(d*x + c)
+ 105*C*d*x + (13*(A - 20*C)*cos(d*x + c)^3 + 4*(13*A - 155*C)*cos(d*x + c)^2 + (32*A - 535*C)*cos(d*x + c) +
8*A - 160*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*
cos(d*x + c) + a^4*d)

________________________________________________________________________________________

giac [A]  time = 0.41, size = 154, normalized size = 1.01 \[ \frac {\frac {840 \, {\left (d x + c\right )} C}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

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))^4,x, algorithm="giac")

[Out]

1/840*(840*(d*x + c)*C/a^4 + (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 21*A*a^24*
tan(1/2*d*x + 1/2*c)^5 - 105*C*a^24*tan(1/2*d*x + 1/2*c)^5 - 35*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan
(1/2*d*x + 1/2*c)^3 + 105*A*a^24*tan(1/2*d*x + 1/2*c) - 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

________________________________________________________________________________________

maple [A]  time = 0.12, size = 177, normalized size = 1.16 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}+\frac {11 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{4}} \]

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))^4,x)

[Out]

1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A+1/56/d/a^4*C*tan(1/2*d*x+1/2*c)^7-1/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5-1/8/d/a^
4*C*tan(1/2*d*x+1/2*c)^5-1/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A+11/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3+1/8/d/a^4*A*tan(
1/2*d*x+1/2*c)-15/8/d/a^4*C*tan(1/2*d*x+1/2*c)+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*C

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 201, normalized size = 1.32 \[ -\frac {5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

-1/840*(5*C*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^4) - A*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/
(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

________________________________________________________________________________________

mupad [B]  time = 0.97, size = 162, normalized size = 1.07 \[ \frac {C\,x}{a^4}+\frac {\left (\frac {13\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}-\frac {52\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {13\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{210}+\frac {16\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {11\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{140}-\frac {5\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{28}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

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))^4,x)

[Out]

(C*x)/a^4 + ((A*sin(c/2 + (d*x)/2))/56 + (C*sin(c/2 + (d*x)/2))/56 - cos(c/2 + (d*x)/2)^2*((11*A*sin(c/2 + (d*
x)/2))/140 + (5*C*sin(c/2 + (d*x)/2))/28) + cos(c/2 + (d*x)/2)^6*((13*A*sin(c/2 + (d*x)/2))/105 - (52*C*sin(c/
2 + (d*x)/2))/21) + cos(c/2 + (d*x)/2)^4*((13*A*sin(c/2 + (d*x)/2))/210 + (16*C*sin(c/2 + (d*x)/2))/21))/(a^4*
d*cos(c/2 + (d*x)/2)^7)

________________________________________________________________________________________

sympy [A]  time = 13.43, size = 192, normalized size = 1.26 \[ \begin {cases} \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {C x}{a^{4}} + \frac {C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {11 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} - \frac {15 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\relax (c )}\right ) \cos ^{2}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

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))**4,x)

[Out]

Piecewise((A*tan(c/2 + d*x/2)**7/(56*a**4*d) - A*tan(c/2 + d*x/2)**5/(40*a**4*d) - A*tan(c/2 + d*x/2)**3/(24*a
**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d) + C*x/a**4 + C*tan(c/2 + d*x/2)**7/(56*a**4*d) - C*tan(c/2 + d*x/2)**5/
(8*a**4*d) + 11*C*tan(c/2 + d*x/2)**3/(24*a**4*d) - 15*C*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + C*cos
(c)**2)*cos(c)**2/(a*cos(c) + a)**4, True))

________________________________________________________________________________________