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

Optimal. Leaf size=223 \[ -\frac {32 (5 A+54 C) \sin (c+d x)}{105 a^4 d}-\frac {(10 A+129 C) \sin (c+d x) \cos ^3(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {16 (5 A+54 C) \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)}+\frac {(2 A+21 C) \sin (c+d x) \cos (c+d x)}{2 a^4 d}+\frac {x (2 A+21 C)}{2 a^4}-\frac {(A+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {2 C \sin (c+d x) \cos ^4(c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

[Out]

1/2*(2*A+21*C)*x/a^4-32/105*(5*A+54*C)*sin(d*x+c)/a^4/d+1/2*(2*A+21*C)*cos(d*x+c)*sin(d*x+c)/a^4/d-1/105*(10*A
+129*C)*cos(d*x+c)^3*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-16/105*(5*A+54*C)*cos(d*x+c)^2*sin(d*x+c)/a^4/d/(1+cos(
d*x+c))-1/7*(A+C)*cos(d*x+c)^5*sin(d*x+c)/d/(a+a*cos(d*x+c))^4-2/5*C*cos(d*x+c)^4*sin(d*x+c)/a/d/(a+a*cos(d*x+
c))^3

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Rubi [A]  time = 0.61, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 2977, 2734} \[ -\frac {32 (5 A+54 C) \sin (c+d x)}{105 a^4 d}-\frac {(10 A+129 C) \sin (c+d x) \cos ^3(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {16 (5 A+54 C) \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)}+\frac {(2 A+21 C) \sin (c+d x) \cos (c+d x)}{2 a^4 d}+\frac {x (2 A+21 C)}{2 a^4}-\frac {(A+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {2 C \sin (c+d x) \cos ^4(c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A + 21*C)*x)/(2*a^4) - (32*(5*A + 54*C)*Sin[c + d*x])/(105*a^4*d) + ((2*A + 21*C)*Cos[c + d*x]*Sin[c + d*x
])/(2*a^4*d) - ((10*A + 129*C)*Cos[c + d*x]^3*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) - (16*(5*A + 54*C
)*Cos[c + d*x]^2*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^5*Sin[c + d*x])/(7*d*(a
+ a*Cos[c + d*x])^4) - (2*C*Cos[c + d*x]^4*Sin[c + d*x])/(5*a*d*(a + a*Cos[c + d*x])^3)

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 ^4(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx &=-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {\cos ^4(c+d x) (a (2 A-5 C)+a (2 A+9 C) \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^3(c+d x) \left (-56 a^2 C+a^2 (10 A+73 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(10 A+129 C) \cos ^3(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) \left (-3 a^3 (10 A+129 C)+a^3 (50 A+477 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(10 A+129 C) \cos ^3(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {16 (5 A+54 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \cos (c+d x) \left (-32 a^4 (5 A+54 C)+105 a^4 (2 A+21 C) \cos (c+d x)\right ) \, dx}{105 a^8}\\ &=\frac {(2 A+21 C) x}{2 a^4}-\frac {32 (5 A+54 C) \sin (c+d x)}{105 a^4 d}+\frac {(2 A+21 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(10 A+129 C) \cos ^3(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \cos ^5(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 C \cos ^4(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {16 (5 A+54 C) \cos ^2(c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.10, size = 513, normalized size = 2.30 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (14700 d x (2 A+21 C) \cos \left (c+\frac {d x}{2}\right )+66080 A \sin \left (c+\frac {d x}{2}\right )-57120 A \sin \left (c+\frac {3 d x}{2}\right )+30240 A \sin \left (2 c+\frac {3 d x}{2}\right )-22400 A \sin \left (2 c+\frac {5 d x}{2}\right )+6720 A \sin \left (3 c+\frac {5 d x}{2}\right )-4160 A \sin \left (3 c+\frac {7 d x}{2}\right )+17640 A d x \cos \left (c+\frac {3 d x}{2}\right )+17640 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+5880 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+5880 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+840 A d x \cos \left (3 c+\frac {7 d x}{2}\right )+840 A d x \cos \left (4 c+\frac {7 d x}{2}\right )+14700 d x (2 A+21 C) \cos \left (\frac {d x}{2}\right )-79520 A \sin \left (\frac {d x}{2}\right )+386190 C \sin \left (c+\frac {d x}{2}\right )-422478 C \sin \left (c+\frac {3 d x}{2}\right )+132930 C \sin \left (2 c+\frac {3 d x}{2}\right )-181461 C \sin \left (2 c+\frac {5 d x}{2}\right )+3675 C \sin \left (3 c+\frac {5 d x}{2}\right )-36003 C \sin \left (3 c+\frac {7 d x}{2}\right )-9555 C \sin \left (4 c+\frac {7 d x}{2}\right )-945 C \sin \left (4 c+\frac {9 d x}{2}\right )-945 C \sin \left (5 c+\frac {9 d x}{2}\right )+105 C \sin \left (5 c+\frac {11 d x}{2}\right )+105 C \sin \left (6 c+\frac {11 d x}{2}\right )+185220 C d x \cos \left (c+\frac {3 d x}{2}\right )+185220 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+61740 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+61740 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+8820 C d x \cos \left (3 c+\frac {7 d x}{2}\right )+8820 C d x \cos \left (4 c+\frac {7 d x}{2}\right )-539490 C \sin \left (\frac {d x}{2}\right )\right )}{6720 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(14700*(2*A + 21*C)*d*x*Cos[(d*x)/2] + 14700*(2*A + 21*C)*d*x*Cos[c + (d*x)/2] + 17
640*A*d*x*Cos[c + (3*d*x)/2] + 185220*C*d*x*Cos[c + (3*d*x)/2] + 17640*A*d*x*Cos[2*c + (3*d*x)/2] + 185220*C*d
*x*Cos[2*c + (3*d*x)/2] + 5880*A*d*x*Cos[2*c + (5*d*x)/2] + 61740*C*d*x*Cos[2*c + (5*d*x)/2] + 5880*A*d*x*Cos[
3*c + (5*d*x)/2] + 61740*C*d*x*Cos[3*c + (5*d*x)/2] + 840*A*d*x*Cos[3*c + (7*d*x)/2] + 8820*C*d*x*Cos[3*c + (7
*d*x)/2] + 840*A*d*x*Cos[4*c + (7*d*x)/2] + 8820*C*d*x*Cos[4*c + (7*d*x)/2] - 79520*A*Sin[(d*x)/2] - 539490*C*
Sin[(d*x)/2] + 66080*A*Sin[c + (d*x)/2] + 386190*C*Sin[c + (d*x)/2] - 57120*A*Sin[c + (3*d*x)/2] - 422478*C*Si
n[c + (3*d*x)/2] + 30240*A*Sin[2*c + (3*d*x)/2] + 132930*C*Sin[2*c + (3*d*x)/2] - 22400*A*Sin[2*c + (5*d*x)/2]
- 181461*C*Sin[2*c + (5*d*x)/2] + 6720*A*Sin[3*c + (5*d*x)/2] + 3675*C*Sin[3*c + (5*d*x)/2] - 4160*A*Sin[3*c
+ (7*d*x)/2] - 36003*C*Sin[3*c + (7*d*x)/2] - 9555*C*Sin[4*c + (7*d*x)/2] - 945*C*Sin[4*c + (9*d*x)/2] - 945*C
*Sin[5*c + (9*d*x)/2] + 105*C*Sin[5*c + (11*d*x)/2] + 105*C*Sin[6*c + (11*d*x)/2]))/(6720*a^4*d*(1 + Cos[c + d
*x])^4)

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fricas [A]  time = 0.85, size = 234, normalized size = 1.05 \[ \frac {105 \, {\left (2 \, A + 21 \, C\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (2 \, A + 21 \, C\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (2 \, A + 21 \, C\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (2 \, A + 21 \, C\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (2 \, A + 21 \, C\right )} d x + {\left (105 \, C \cos \left (d x + c\right )^{5} - 420 \, C \cos \left (d x + c\right )^{4} - 4 \, {\left (130 \, A + 1509 \, C\right )} \cos \left (d x + c\right )^{3} - 4 \, {\left (310 \, A + 3411 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (1070 \, A + 11619 \, C\right )} \cos \left (d x + c\right ) - 320 \, A - 3456 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*(2*A + 21*C)*d*x*cos(d*x + c)^4 + 420*(2*A + 21*C)*d*x*cos(d*x + c)^3 + 630*(2*A + 21*C)*d*x*cos(d*
x + c)^2 + 420*(2*A + 21*C)*d*x*cos(d*x + c) + 105*(2*A + 21*C)*d*x + (105*C*cos(d*x + c)^5 - 420*C*cos(d*x +
c)^4 - 4*(130*A + 1509*C)*cos(d*x + c)^3 - 4*(310*A + 3411*C)*cos(d*x + c)^2 - (1070*A + 11619*C)*cos(d*x + c)
- 320*A - 3456*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)

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giac [A]  time = 0.48, size = 207, normalized size = 0.93 \[ \frac {\frac {420 \, {\left (d x + c\right )} {\left (2 \, A + 21 \, C\right )}}{a^{4}} - \frac {840 \, {\left (9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 189 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1365 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11655 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(420*(d*x + c)*(2*A + 21*C)/a^4 - 840*(9*C*tan(1/2*d*x + 1/2*c)^3 + 7*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*
d*x + 1/2*c)^2 + 1)^2*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 - 105*A*a^24
*tan(1/2*d*x + 1/2*c)^5 - 189*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 1365*C*a^24*
tan(1/2*d*x + 1/2*c)^3 - 1575*A*a^24*tan(1/2*d*x + 1/2*c) - 11655*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.12, size = 264, normalized size = 1.18 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {9 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}+\frac {13 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {111 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {9 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{4}}+\frac {21 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(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)^7-1/8/d/a^4*A*tan(1/2*d*x+1/2*c)^5-9/40/d/a^
4*C*tan(1/2*d*x+1/2*c)^5+11/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A+13/8/d/a^4*C*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*A*tan
(1/2*d*x+1/2*c)-111/8/d/a^4*C*tan(1/2*d*x+1/2*c)-9/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)^3-7/d
/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*A+21/d/a^4*arctan(tan(
1/2*d*x+1/2*c))*C

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maxima [A]  time = 0.42, size = 318, normalized size = 1.43 \[ -\frac {3 \, C {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {5880 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} + 5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

-1/840*(3*C*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 + 2*a^4*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c)
+ 1) - 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7)/a^4 - 5880*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) + 5*A*((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))/d

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mupad [B]  time = 0.95, size = 245, normalized size = 1.10 \[ \frac {x\,\left (2\,A+21\,C\right )}{2\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A+C\right )}{4\,a^4}-\frac {3\,\left (A-15\,C\right )}{8\,a^4}+\frac {3\,\left (2\,A+6\,C\right )}{4\,a^4}-\frac {4\,A-20\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,\left (A+C\right )}{40\,a^4}+\frac {2\,A+6\,C}{40\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^4}-\frac {A-15\,C}{24\,a^4}+\frac {2\,A+6\,C}{8\,a^4}\right )}{d}-\frac {9\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+7\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

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))^4,x)

[Out]

(x*(2*A + 21*C))/(2*a^4) - (tan(c/2 + (d*x)/2)*((5*(A + C))/(4*a^4) - (3*(A - 15*C))/(8*a^4) + (3*(2*A + 6*C))
/(4*a^4) - (4*A - 20*C)/(8*a^4)))/d - (tan(c/2 + (d*x)/2)^5*((3*(A + C))/(40*a^4) + (2*A + 6*C)/(40*a^4)))/d +
(tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^4) - (A - 15*C)/(24*a^4) + (2*A + 6*C)/(8*a^4)))/d - (7*C*tan(c/2 + (d*x)
/2) + 9*C*tan(c/2 + (d*x)/2)^3)/(d*(2*a^4*tan(c/2 + (d*x)/2)^2 + a^4*tan(c/2 + (d*x)/2)^4 + a^4)) + (tan(c/2 +
(d*x)/2)^7*(A + C))/(56*a^4*d)

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sympy [A]  time = 33.13, size = 1086, normalized size = 4.87 \[ \text {result too large to display} \]

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))**4,x)

[Out]

Piecewise((840*A*d*x*tan(c/2 + d*x/2)**4/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 8
40*a**4*d) + 1680*A*d*x*tan(c/2 + d*x/2)**2/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2
+ 840*a**4*d) + 840*A*d*x/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 15
*A*tan(c/2 + d*x/2)**11/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 75*A
*tan(c/2 + d*x/2)**9/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 190*A*t
an(c/2 + d*x/2)**7/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 910*A*tan
(c/2 + d*x/2)**5/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 2765*A*tan(
c/2 + d*x/2)**3/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 1575*A*tan(c
/2 + d*x/2)/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 8820*C*d*x*tan(c
/2 + d*x/2)**4/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 17640*C*d*x*t
an(c/2 + d*x/2)**2/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 8820*C*d*
x/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 15*C*tan(c/2 + d*x/2)**11/
(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 159*C*tan(c/2 + d*x/2)**9/(8
40*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 1002*C*tan(c/2 + d*x/2)**7/(84
0*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 9114*C*tan(c/2 + d*x/2)**5/(840
*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 29505*C*tan(c/2 + d*x/2)**3/(840
*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 17535*C*tan(c/2 + d*x/2)/(840*a*
*4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(
c)**4/(a*cos(c) + a)**4, True))

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