\(\int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [57]

Optimal result

Integrand size = 31, antiderivative size = 193 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {(6 A-13 B) x}{2 a^3}+\frac {8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {4 (9 A-19 B) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

-1/2*(6*A-13*B)*x/a^3+8/15*(9*A-19*B)*sin(d*x+c)/a^3/d-1/2*(6*A-13*B)*cos(d*x+c)*sin(d*x+c)/a^3/d+1/5*(A-B)*co
s(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(6*A-11*B)*cos(d*x+c)^3*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+4/15
*(9*A-19*B)*cos(d*x+c)^2*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {3056, 2813} \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}+\frac {4 (9 A-19 B) \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(6 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {x (6 A-13 B)}{2 a^3}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(6 A-11 B) \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[In]

Int[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/2*((6*A - 13*B)*x)/a^3 + (8*(9*A - 19*B)*Sin[c + d*x])/(15*a^3*d) - ((6*A - 13*B)*Cos[c + d*x]*Sin[c + d*x]
)/(2*a^3*d) + ((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((6*A - 11*B)*Cos[c + d*x]^
3*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + (4*(9*A - 19*B)*Cos[c + d*x]^2*Sin[c + d*x])/(15*d*(a^3 + a^
3*Cos[c + d*x]))

Rule 2813

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 3056

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])

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^3(c+d x) (4 a (A-B)-a (2 A-7 B) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = \frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^2(c+d x) \left (3 a^2 (6 A-11 B)-a^2 (18 A-43 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = \frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {4 (9 A-19 B) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \cos (c+d x) \left (8 a^3 (9 A-19 B)-15 a^3 (6 A-13 B) \cos (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(6 A-13 B) x}{2 a^3}+\frac {8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {4 (9 A-19 B) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(435\) vs. \(2(193)=386\).

Time = 2.17 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.25 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-600 (6 A-13 B) d x \cos \left (\frac {d x}{2}\right )-600 (6 A-13 B) d x \cos \left (c+\frac {d x}{2}\right )-1800 A d x \cos \left (c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (c+\frac {3 d x}{2}\right )-1800 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-360 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (2 c+\frac {5 d x}{2}\right )-360 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac {5 d x}{2}\right )+7020 A \sin \left (\frac {d x}{2}\right )-12760 B \sin \left (\frac {d x}{2}\right )-4500 A \sin \left (c+\frac {d x}{2}\right )+7560 B \sin \left (c+\frac {d x}{2}\right )+4860 A \sin \left (c+\frac {3 d x}{2}\right )-9230 B \sin \left (c+\frac {3 d x}{2}\right )-900 A \sin \left (2 c+\frac {3 d x}{2}\right )+930 B \sin \left (2 c+\frac {3 d x}{2}\right )+1452 A \sin \left (2 c+\frac {5 d x}{2}\right )-2782 B \sin \left (2 c+\frac {5 d x}{2}\right )+300 A \sin \left (3 c+\frac {5 d x}{2}\right )-750 B \sin \left (3 c+\frac {5 d x}{2}\right )+60 A \sin \left (3 c+\frac {7 d x}{2}\right )-105 B \sin \left (3 c+\frac {7 d x}{2}\right )+60 A \sin \left (4 c+\frac {7 d x}{2}\right )-105 B \sin \left (4 c+\frac {7 d x}{2}\right )+15 B \sin \left (4 c+\frac {9 d x}{2}\right )+15 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \]

[In]

Integrate[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-600*(6*A - 13*B)*d*x*Cos[(d*x)/2] - 600*(6*A - 13*B)*d*x*Cos[c + (d*x)/2] - 1800*
A*d*x*Cos[c + (3*d*x)/2] + 3900*B*d*x*Cos[c + (3*d*x)/2] - 1800*A*d*x*Cos[2*c + (3*d*x)/2] + 3900*B*d*x*Cos[2*
c + (3*d*x)/2] - 360*A*d*x*Cos[2*c + (5*d*x)/2] + 780*B*d*x*Cos[2*c + (5*d*x)/2] - 360*A*d*x*Cos[3*c + (5*d*x)
/2] + 780*B*d*x*Cos[3*c + (5*d*x)/2] + 7020*A*Sin[(d*x)/2] - 12760*B*Sin[(d*x)/2] - 4500*A*Sin[c + (d*x)/2] +
7560*B*Sin[c + (d*x)/2] + 4860*A*Sin[c + (3*d*x)/2] - 9230*B*Sin[c + (3*d*x)/2] - 900*A*Sin[2*c + (3*d*x)/2] +
930*B*Sin[2*c + (3*d*x)/2] + 1452*A*Sin[2*c + (5*d*x)/2] - 2782*B*Sin[2*c + (5*d*x)/2] + 300*A*Sin[3*c + (5*d
*x)/2] - 750*B*Sin[3*c + (5*d*x)/2] + 60*A*Sin[3*c + (7*d*x)/2] - 105*B*Sin[3*c + (7*d*x)/2] + 60*A*Sin[4*c +
(7*d*x)/2] - 105*B*Sin[4*c + (7*d*x)/2] + 15*B*Sin[4*c + (9*d*x)/2] + 15*B*Sin[5*c + (9*d*x)/2]))/(480*a^3*d*(
1 + Cos[c + d*x])^3)

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.55

method	result	size
parallelrisch	\(\frac {2916 \left (\left (\frac {26 A}{81}-\frac {464 B}{729}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {5 A}{243}-\frac {5 B}{162}\right ) \cos \left (3 d x +3 c \right )+\frac {5 B \cos \left (4 d x +4 c \right )}{972}+\left (A -\frac {1001 B}{486}\right ) \cos \left (d x +c \right )+\frac {58 A}{81}-\frac {4303 B}{2916}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2880 \left (A -\frac {13 B}{6}\right ) x d}{960 a^{3} d}\)	\(106\)
derivativedivides	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A +\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {16 \left (\left (-\frac {A}{2}+\frac {7 B}{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {A}{2}+\frac {5 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-4 \left (6 A -13 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(163\)
default	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A +\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {16 \left (\left (-\frac {A}{2}+\frac {7 B}{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {A}{2}+\frac {5 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-4 \left (6 A -13 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(163\)
risch	\(-\frac {3 A x}{a^{3}}+\frac {13 B x}{2 a^{3}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B}{8 a^{3} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 a^{3} d}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 a^{3} d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{8 a^{3} d}+\frac {2 i \left (90 A \,{\mathrm e}^{4 i \left (d x +c \right )}-150 B \,{\mathrm e}^{4 i \left (d x +c \right )}+300 A \,{\mathrm e}^{3 i \left (d x +c \right )}-525 B \,{\mathrm e}^{3 i \left (d x +c \right )}+420 A \,{\mathrm e}^{2 i \left (d x +c \right )}-745 B \,{\mathrm e}^{2 i \left (d x +c \right )}+270 A \,{\mathrm e}^{i \left (d x +c \right )}-485 B \,{\mathrm e}^{i \left (d x +c \right )}+72 A -127 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(255\)
norman	\(\frac {-\frac {\left (6 A -13 B \right ) x}{2 a}+\frac {\left (A -B \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (3 A -5 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}-\frac {5 \left (6 A -13 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {5 \left (6 A -13 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {5 \left (6 A -13 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {5 \left (6 A -13 B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {\left (6 A -13 B \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {25 \left (9 A -19 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (25 A -51 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (27 A -59 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (345 A -721 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (549 A -1165 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (3123 A -6613 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} a^{2}}\)	\(358\)

[In]

int(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/960*(2916*((26/81*A-464/729*B)*cos(2*d*x+2*c)+(5/243*A-5/162*B)*cos(3*d*x+3*c)+5/972*B*cos(4*d*x+4*c)+(A-100
1/486*B)*cos(d*x+c)+58/81*A-4303/2916*B)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4-2880*(A-13/6*B)*x*d)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {15 \, {\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (6 \, A - 13 \, B\right )} d x - {\left (15 \, B \cos \left (d x + c\right )^{4} + 15 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (234 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (114 \, A - 239 \, B\right )} \cos \left (d x + c\right ) + 144 \, A - 304 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/30*(15*(6*A - 13*B)*d*x*cos(d*x + c)^3 + 45*(6*A - 13*B)*d*x*cos(d*x + c)^2 + 45*(6*A - 13*B)*d*x*cos(d*x +
c) + 15*(6*A - 13*B)*d*x - (15*B*cos(d*x + c)^4 + 15*(2*A - 3*B)*cos(d*x + c)^3 + (234*A - 479*B)*cos(d*x + c
)^2 + 3*(114*A - 239*B)*cos(d*x + c) + 144*A - 304*B)*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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 966 vs. \(2 (178) = 356\).

Time = 3.90 (sec) , antiderivative size = 966, normalized size of antiderivative = 5.01 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)**4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((-180*A*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60
*a**3*d) - 360*A*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*
a**3*d) - 180*A*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3*A*tan(c/2
+ d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 24*A*tan(c/2 + d*x
/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 198*A*tan(c/2 + d*x/2)**
5/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 600*A*tan(c/2 + d*x/2)**3/(60
*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 375*A*tan(c/2 + d*x/2)/(60*a**3*d*
tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*B*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*t
an(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 780*B*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*ta
n(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*B*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 1
20*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 3*B*tan(c/2 + d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3
*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 34*B*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan
(c/2 + d*x/2)**2 + 60*a**3*d) - 388*B*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2
+ d*x/2)**2 + 60*a**3*d) - 1310*B*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*
x/2)**2 + 60*a**3*d) - 765*B*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2
+ 60*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**4/(a*cos(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - 3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
- 40*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 - 780*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^3) - 3*A*(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))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {30 \, {\left (d x + c\right )} {\left (6 \, A - 13 \, B\right )}}{a^{3}} - \frac {60 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(30*(d*x + c)*(6*A - 13*B)/a^3 - 60*(2*A*tan(1/2*d*x + 1/2*c)^3 - 7*B*tan(1/2*d*x + 1/2*c)^3 + 2*A*tan(1
/2*d*x + 1/2*c) - 5*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2
*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 40*B*a^12*tan(1/2*d*x + 1/2*c)^3
+ 255*A*a^12*tan(1/2*d*x + 1/2*c) - 465*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^3}+\frac {3\,\left (3\,A-5\,B\right )}{4\,a^3}+\frac {2\,A-10\,B}{4\,a^3}\right )}{d}-\frac {x\,\left (6\,A-13\,B\right )}{2\,a^3}+\frac {\left (2\,A-7\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A-5\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^3}+\frac {3\,A-5\,B}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]

[In]

int((cos(c + d*x)^4*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*((3*(A - B))/(2*a^3) + (3*(3*A - 5*B))/(4*a^3) + (2*A - 10*B)/(4*a^3)))/d - (x*(6*A - 13*B
))/(2*a^3) + (tan(c/2 + (d*x)/2)^3*(2*A - 7*B) + tan(c/2 + (d*x)/2)*(2*A - 5*B))/(d*(2*a^3*tan(c/2 + (d*x)/2)^
2 + a^3*tan(c/2 + (d*x)/2)^4 + a^3)) - (tan(c/2 + (d*x)/2)^3*((A - B)/(4*a^3) + (3*A - 5*B)/(12*a^3)))/d + (ta
n(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d)