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

Optimal. Leaf size=191 \[ \frac {8 (A+2 C) \sin ^3(c+d x)}{3 a^2 d}-\frac {8 (A+2 C) \sin (c+d x)}{a^2 d}-\frac {2 (A+2 C) \sin (c+d x) \cos ^4(c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac {(28 A+55 C) \sin (c+d x) \cos ^3(c+d x)}{12 a^2 d}+\frac {(28 A+55 C) \sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac {x (28 A+55 C)}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^5(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

1/8*(28*A+55*C)*x/a^2-8*(A+2*C)*sin(d*x+c)/a^2/d+1/8*(28*A+55*C)*cos(d*x+c)*sin(d*x+c)/a^2/d+1/12*(28*A+55*C)*
cos(d*x+c)^3*sin(d*x+c)/a^2/d-2*(A+2*C)*cos(d*x+c)^4*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A+C)*cos(d*x+c)^5*si
n(d*x+c)/d/(a+a*cos(d*x+c))^2+8/3*(A+2*C)*sin(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.34, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 2977, 2748, 2633, 2635, 8} \[ \frac {8 (A+2 C) \sin ^3(c+d x)}{3 a^2 d}-\frac {8 (A+2 C) \sin (c+d x)}{a^2 d}-\frac {2 (A+2 C) \sin (c+d x) \cos ^4(c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac {(28 A+55 C) \sin (c+d x) \cos ^3(c+d x)}{12 a^2 d}+\frac {(28 A+55 C) \sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac {x (28 A+55 C)}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^5(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((28*A + 55*C)*x)/(8*a^2) - (8*(A + 2*C)*Sin[c + d*x])/(a^2*d) + ((28*A + 55*C)*Cos[c + d*x]*Sin[c + d*x])/(8*
a^2*d) + ((28*A + 55*C)*Cos[c + d*x]^3*Sin[c + d*x])/(12*a^2*d) - (2*(A + 2*C)*Cos[c + d*x]^4*Sin[c + d*x])/(a
^2*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^5*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (8*(A + 2*C)*S
in[c + d*x]^3)/(3*a^2*d)

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]

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 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 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 ^4(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx &=-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^4(c+d x) (-a (2 A+5 C)+a (4 A+7 C) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 (A+2 C) \cos ^4(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \cos ^3(c+d x) \left (-24 a^2 (A+2 C)+a^2 (28 A+55 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 (A+2 C) \cos ^4(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(8 (A+2 C)) \int \cos ^3(c+d x) \, dx}{a^2}+\frac {(28 A+55 C) \int \cos ^4(c+d x) \, dx}{3 a^2}\\ &=\frac {(28 A+55 C) \cos ^3(c+d x) \sin (c+d x)}{12 a^2 d}-\frac {2 (A+2 C) \cos ^4(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(28 A+55 C) \int \cos ^2(c+d x) \, dx}{4 a^2}+\frac {(8 (A+2 C)) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {8 (A+2 C) \sin (c+d x)}{a^2 d}+\frac {(28 A+55 C) \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {(28 A+55 C) \cos ^3(c+d x) \sin (c+d x)}{12 a^2 d}-\frac {2 (A+2 C) \cos ^4(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {8 (A+2 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {(28 A+55 C) \int 1 \, dx}{8 a^2}\\ &=\frac {(28 A+55 C) x}{8 a^2}-\frac {8 (A+2 C) \sin (c+d x)}{a^2 d}+\frac {(28 A+55 C) \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {(28 A+55 C) \cos ^3(c+d x) \sin (c+d x)}{12 a^2 d}-\frac {2 (A+2 C) \cos ^4(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {8 (A+2 C) \sin ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 0.88, size = 399, normalized size = 2.09 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (72 d x (28 A+55 C) \cos \left (c+\frac {d x}{2}\right )+1176 A \sin \left (c+\frac {d x}{2}\right )-1912 A \sin \left (c+\frac {3 d x}{2}\right )-504 A \sin \left (2 c+\frac {3 d x}{2}\right )-120 A \sin \left (2 c+\frac {5 d x}{2}\right )-120 A \sin \left (3 c+\frac {5 d x}{2}\right )+24 A \sin \left (3 c+\frac {7 d x}{2}\right )+24 A \sin \left (4 c+\frac {7 d x}{2}\right )+672 A d x \cos \left (c+\frac {3 d x}{2}\right )+672 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+72 d x (28 A+55 C) \cos \left (\frac {d x}{2}\right )-3048 A \sin \left (\frac {d x}{2}\right )+1344 C \sin \left (c+\frac {d x}{2}\right )-3488 C \sin \left (c+\frac {3 d x}{2}\right )-1312 C \sin \left (2 c+\frac {3 d x}{2}\right )-285 C \sin \left (2 c+\frac {5 d x}{2}\right )-285 C \sin \left (3 c+\frac {5 d x}{2}\right )+57 C \sin \left (3 c+\frac {7 d x}{2}\right )+57 C \sin \left (4 c+\frac {7 d x}{2}\right )-7 C \sin \left (4 c+\frac {9 d x}{2}\right )-7 C \sin \left (5 c+\frac {9 d x}{2}\right )+3 C \sin \left (5 c+\frac {11 d x}{2}\right )+3 C \sin \left (6 c+\frac {11 d x}{2}\right )+1320 C d x \cos \left (c+\frac {3 d x}{2}\right )+1320 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-5184 C \sin \left (\frac {d x}{2}\right )\right )}{384 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(72*(28*A + 55*C)*d*x*Cos[(d*x)/2] + 72*(28*A + 55*C)*d*x*Cos[c + (d*x)/2] + 672*A*
d*x*Cos[c + (3*d*x)/2] + 1320*C*d*x*Cos[c + (3*d*x)/2] + 672*A*d*x*Cos[2*c + (3*d*x)/2] + 1320*C*d*x*Cos[2*c +
(3*d*x)/2] - 3048*A*Sin[(d*x)/2] - 5184*C*Sin[(d*x)/2] + 1176*A*Sin[c + (d*x)/2] + 1344*C*Sin[c + (d*x)/2] -
1912*A*Sin[c + (3*d*x)/2] - 3488*C*Sin[c + (3*d*x)/2] - 504*A*Sin[2*c + (3*d*x)/2] - 1312*C*Sin[2*c + (3*d*x)/
2] - 120*A*Sin[2*c + (5*d*x)/2] - 285*C*Sin[2*c + (5*d*x)/2] - 120*A*Sin[3*c + (5*d*x)/2] - 285*C*Sin[3*c + (5
*d*x)/2] + 24*A*Sin[3*c + (7*d*x)/2] + 57*C*Sin[3*c + (7*d*x)/2] + 24*A*Sin[4*c + (7*d*x)/2] + 57*C*Sin[4*c +
(7*d*x)/2] - 7*C*Sin[4*c + (9*d*x)/2] - 7*C*Sin[5*c + (9*d*x)/2] + 3*C*Sin[5*c + (11*d*x)/2] + 3*C*Sin[6*c + (
11*d*x)/2]))/(384*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.62, size = 167, normalized size = 0.87 \[ \frac {3 \, {\left (28 \, A + 55 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (28 \, A + 55 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (28 \, A + 55 \, C\right )} d x + {\left (6 \, C \cos \left (d x + c\right )^{5} - 4 \, C \cos \left (d x + c\right )^{4} + {\left (12 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (4 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (172 \, A + 347 \, C\right )} \cos \left (d x + c\right ) - 128 \, A - 256 \, C\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(3*(28*A + 55*C)*d*x*cos(d*x + c)^2 + 6*(28*A + 55*C)*d*x*cos(d*x + c) + 3*(28*A + 55*C)*d*x + (6*C*cos(d
*x + c)^5 - 4*C*cos(d*x + c)^4 + (12*A + 19*C)*cos(d*x + c)^3 - 6*(4*A + 9*C)*cos(d*x + c)^2 - (172*A + 347*C)
*cos(d*x + c) - 128*A - 256*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 = 0.50, size = 220, normalized size = 1.15 \[ \frac {\frac {3 \, {\left (d x + c\right )} {\left (28 \, A + 55 \, C\right )}}{a^{2}} + \frac {4 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 33 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{6}} - \frac {2 \, {\left (60 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 195 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 156 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 395 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 132 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 341 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 93 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(3*(d*x + c)*(28*A + 55*C)/a^2 + 4*(A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 21*A*a^
4*tan(1/2*d*x + 1/2*c) - 33*C*a^4*tan(1/2*d*x + 1/2*c))/a^6 - 2*(60*A*tan(1/2*d*x + 1/2*c)^7 + 195*C*tan(1/2*d
*x + 1/2*c)^7 + 156*A*tan(1/2*d*x + 1/2*c)^5 + 395*C*tan(1/2*d*x + 1/2*c)^5 + 132*A*tan(1/2*d*x + 1/2*c)^3 + 3
41*C*tan(1/2*d*x + 1/2*c)^3 + 36*A*tan(1/2*d*x + 1/2*c) + 93*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2
+ 1)^4*a^2))/d

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maple [B]  time = 0.12, size = 392, normalized size = 2.05 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {11 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {65 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{4 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {13 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {395 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{12 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {341 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {31 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}}+\frac {55 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{4 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(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)^3-7/2/d/a^2*A*tan(1/2*d*x+1/2*c)-11/2/d/a^2*C*
tan(1/2*d*x+1/2*c)-5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*A-65/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^
2)^4*tan(1/2*d*x+1/2*c)^7*C-13/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*A-395/12/d/a^2/(1+tan(1/2
*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*C-11/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*A-341/12/d/a^
2/(1+tan(1/2*d*x+1/2*c)^2)^4*C*tan(1/2*d*x+1/2*c)^3-3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*A*tan(1/2*d*x+1/2*c)-31
/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*C*tan(1/2*d*x+1/2*c)+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*A+55/4/d/a^2*arcta
n(tan(1/2*d*x+1/2*c))*C

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maxima [B]  time = 0.45, size = 415, normalized size = 2.17 \[ -\frac {C {\left (\frac {\frac {93 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {341 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {395 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {195 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {2 \, {\left (\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {165 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} + 2 \, A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*(C*((93*sin(d*x + c)/(cos(d*x + c) + 1) + 341*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 395*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5 + 195*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^2 + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^
2 + 6*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^8
/(cos(d*x + c) + 1)^8) + 2*(33*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 16
5*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) + 2*A*(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) - sin(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))/d

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mupad [B]  time = 1.04, size = 220, normalized size = 1.15 \[ \frac {x\,\left (28\,A+55\,C\right )}{8\,a^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A+C\right )}{2\,a^2}+\frac {2\,A+6\,C}{2\,a^2}\right )}{d}-\frac {\left (5\,A+\frac {65\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (13\,A+\frac {395\,C}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (11\,A+\frac {341\,C}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A+\frac {31\,C}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^2,x)

[Out]

(x*(28*A + 55*C))/(8*a^2) - (tan(c/2 + (d*x)/2)*((5*(A + C))/(2*a^2) + (2*A + 6*C)/(2*a^2)))/d - (tan(c/2 + (d
*x)/2)^7*(5*A + (65*C)/4) + tan(c/2 + (d*x)/2)^3*(11*A + (341*C)/12) + tan(c/2 + (d*x)/2)^5*(13*A + (395*C)/12
) + tan(c/2 + (d*x)/2)*(3*A + (31*C)/4))/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 6*a^2*tan(c/2 + (d*x)/2)^4 + 4*a^2*t
an(c/2 + (d*x)/2)^6 + a^2*tan(c/2 + (d*x)/2)^8 + a^2)) + (tan(c/2 + (d*x)/2)^3*(A + C))/(6*a^2*d)

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sympy [A]  time = 17.38, size = 2161, normalized size = 11.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((84*A*d*x*tan(c/2 + d*x/2)**8/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a
**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 336*A*d*x*tan(c/2 + d*x/2)**6/(24*a**
2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 +
d*x/2)**2 + 24*a**2*d) + 504*A*d*x*tan(c/2 + d*x/2)**4/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d
*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 336*A*d*x*tan(c/2 + d
*x/2)**2/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*
a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 84*A*d*x/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)
**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 4*A*tan(c/2 + d*x/2)**11/(
24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan
(c/2 + d*x/2)**2 + 24*a**2*d) - 68*A*tan(c/2 + d*x/2)**9/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 +
d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 432*A*tan(c/2 + d*x/
2)**7/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**
2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 800*A*tan(c/2 + d*x/2)**5/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*ta
n(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 596*A*tan(c/
2 + d*x/2)**3/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4
+ 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 156*A*tan(c/2 + d*x/2)/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2
*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 165*C*d
*x*tan(c/2 + d*x/2)**8/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d
*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 660*C*d*x*tan(c/2 + d*x/2)**6/(24*a**2*d*tan(c/2 + d*x
/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a
**2*d) + 990*C*d*x*tan(c/2 + d*x/2)**4/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a*
*2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 660*C*d*x*tan(c/2 + d*x/2)**2/(24*a**2
*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 +
d*x/2)**2 + 24*a**2*d) + 165*C*d*x/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d
*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 4*C*tan(c/2 + d*x/2)**11/(24*a**2*d*tan(c/
2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2
+ 24*a**2*d) - 116*C*tan(c/2 + d*x/2)**9/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144
*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 894*C*tan(c/2 + d*x/2)**7/(24*a**2*
d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d
*x/2)**2 + 24*a**2*d) - 1566*C*tan(c/2 + d*x/2)**5/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)
**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 1206*C*tan(c/2 + d*x/2)**3
/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*t
an(c/2 + d*x/2)**2 + 24*a**2*d) - 318*C*tan(c/2 + d*x/2)/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 +
d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d), Ne(d, 0)), (x*(A + C*
cos(c)**2)*cos(c)**4/(a*cos(c) + a)**2, True))

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