\(\int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\) [47]

Optimal result

Integrand size = 31, antiderivative size = 170 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {(7 A-10 B) x}{2 a^2}-\frac {4 (2 A-3 B) \sin (c+d x)}{a^2 d}+\frac {(7 A-10 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {4 (2 A-3 B) \sin ^3(c+d x)}{3 a^2 d} \]

[Out]

1/2*(7*A-10*B)*x/a^2-4*(2*A-3*B)*sin(d*x+c)/a^2/d+1/2*(7*A-10*B)*cos(d*x+c)*sin(d*x+c)/a^2/d+1/3*(7*A-10*B)*co
s(d*x+c)^3*sin(d*x+c)/a^2/d/(1+cos(d*x+c))+1/3*(A-B)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^2+4/3*(2*A-3*B
)*sin(d*x+c)^3/a^2/d

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3056, 2827, 2715, 8, 2713} \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {4 (2 A-3 B) \sin ^3(c+d x)}{3 a^2 d}-\frac {4 (2 A-3 B) \sin (c+d x)}{a^2 d}+\frac {(7 A-10 B) \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {(7 A-10 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {x (7 A-10 B)}{2 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

((7*A - 10*B)*x)/(2*a^2) - (4*(2*A - 3*B)*Sin[c + d*x])/(a^2*d) + ((7*A - 10*B)*Cos[c + d*x]*Sin[c + d*x])/(2*
a^2*d) + ((7*A - 10*B)*Cos[c + d*x]^3*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + ((A - B)*Cos[c + d*x]^4*Sin
[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (4*(2*A - 3*B)*Sin[c + d*x]^3)/(3*a^2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2827

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 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)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^3(c+d x) (4 a (A-B)-3 a (A-2 B) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2} \\ & = \frac {(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \cos ^2(c+d x) \left (3 a^2 (7 A-10 B)-12 a^2 (2 A-3 B) \cos (c+d x)\right ) \, dx}{3 a^4} \\ & = \frac {(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(7 A-10 B) \int \cos ^2(c+d x) \, dx}{a^2}-\frac {(4 (2 A-3 B)) \int \cos ^3(c+d x) \, dx}{a^2} \\ & = \frac {(7 A-10 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(7 A-10 B) \int 1 \, dx}{2 a^2}+\frac {(4 (2 A-3 B)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d} \\ & = \frac {(7 A-10 B) x}{2 a^2}-\frac {4 (2 A-3 B) \sin (c+d x)}{a^2 d}+\frac {(7 A-10 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {4 (2 A-3 B) \sin ^3(c+d x)}{3 a^2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(369\) vs. \(2(170)=340\).

Time = 1.81 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (36 (7 A-10 B) d x \cos \left (\frac {d x}{2}\right )+36 (7 A-10 B) d x \cos \left (c+\frac {d x}{2}\right )+84 A d x \cos \left (c+\frac {3 d x}{2}\right )-120 B d x \cos \left (c+\frac {3 d x}{2}\right )+84 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-120 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 A \sin \left (\frac {d x}{2}\right )+516 B \sin \left (\frac {d x}{2}\right )+147 A \sin \left (c+\frac {d x}{2}\right )-156 B \sin \left (c+\frac {d x}{2}\right )-239 A \sin \left (c+\frac {3 d x}{2}\right )+342 B \sin \left (c+\frac {3 d x}{2}\right )-63 A \sin \left (2 c+\frac {3 d x}{2}\right )+118 B \sin \left (2 c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {5 d x}{2}\right )+30 B \sin \left (2 c+\frac {5 d x}{2}\right )-15 A \sin \left (3 c+\frac {5 d x}{2}\right )+30 B \sin \left (3 c+\frac {5 d x}{2}\right )+3 A \sin \left (3 c+\frac {7 d x}{2}\right )-3 B \sin \left (3 c+\frac {7 d x}{2}\right )+3 A \sin \left (4 c+\frac {7 d x}{2}\right )-3 B \sin \left (4 c+\frac {7 d x}{2}\right )+B \sin \left (4 c+\frac {9 d x}{2}\right )+B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(36*(7*A - 10*B)*d*x*Cos[(d*x)/2] + 36*(7*A - 10*B)*d*x*Cos[c + (d*x)/2] + 84*A*d*x
*Cos[c + (3*d*x)/2] - 120*B*d*x*Cos[c + (3*d*x)/2] + 84*A*d*x*Cos[2*c + (3*d*x)/2] - 120*B*d*x*Cos[2*c + (3*d*
x)/2] - 381*A*Sin[(d*x)/2] + 516*B*Sin[(d*x)/2] + 147*A*Sin[c + (d*x)/2] - 156*B*Sin[c + (d*x)/2] - 239*A*Sin[
c + (3*d*x)/2] + 342*B*Sin[c + (3*d*x)/2] - 63*A*Sin[2*c + (3*d*x)/2] + 118*B*Sin[2*c + (3*d*x)/2] - 15*A*Sin[
2*c + (5*d*x)/2] + 30*B*Sin[2*c + (5*d*x)/2] - 15*A*Sin[3*c + (5*d*x)/2] + 30*B*Sin[3*c + (5*d*x)/2] + 3*A*Sin
[3*c + (7*d*x)/2] - 3*B*Sin[3*c + (7*d*x)/2] + 3*A*Sin[4*c + (7*d*x)/2] - 3*B*Sin[4*c + (7*d*x)/2] + B*Sin[4*c
+ (9*d*x)/2] + B*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(\frac {-163 \left (\frac {4 \left (3 A -7 B \right ) \cos \left (2 d x +2 c \right )}{163}+\frac {\left (-3 A +2 B \right ) \cos \left (3 d x +3 c \right )}{163}-\frac {B \cos \left (4 d x +4 c \right )}{163}+\left (A -\frac {258 B}{163}\right ) \cos \left (d x +c \right )+\frac {140 A}{163}-\frac {219 B}{163}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 \left (A -\frac {10 B}{7}\right ) x d}{48 a^{2} d}\)	\(108\)
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+9 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 \left (-\frac {5 A}{4}+\frac {5 B}{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (-2 A +\frac {10 B}{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (-\frac {3 A}{4}+\frac {3 B}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+2 \left (7 A -10 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(154\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+9 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 \left (-\frac {5 A}{4}+\frac {5 B}{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (-2 A +\frac {10 B}{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (-\frac {3 A}{4}+\frac {3 B}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+2 \left (7 A -10 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(154\)
risch	\(\frac {7 x A}{2 a^{2}}-\frac {5 B x}{a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} A}{8 a^{2} d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B}{4 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{a^{2} d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )} B}{8 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{a^{2} d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} B}{8 a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{8 a^{2} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a^{2} d}-\frac {2 i \left (12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-15 B \,{\mathrm e}^{2 i \left (d x +c \right )}+21 A \,{\mathrm e}^{i \left (d x +c \right )}-27 B \,{\mathrm e}^{i \left (d x +c \right )}+11 A -14 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {B \sin \left (3 d x +3 c \right )}{12 a^{2} d}\)	\(263\)
norman	\(\frac {\frac {\left (7 A -10 B \right ) x}{2 a}+\frac {\left (A -B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {5 \left (7 A -10 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {5 \left (7 A -10 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {5 \left (7 A -10 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {5 \left (7 A -10 B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {\left (7 A -10 B \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {\left (8 A -11 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (13 A -21 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {5 \left (25 A -37 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {2 \left (77 A -115 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (94 A -143 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (349 A -521 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} a}\)	\(332\)

[In]

int(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^2,x,method=_RETURNVERBOSE)

[Out]

1/48*(-163*(4/163*(3*A-7*B)*cos(2*d*x+2*c)+1/163*(-3*A+2*B)*cos(3*d*x+3*c)-1/163*B*cos(4*d*x+4*c)+(A-258/163*B
)*cos(d*x+c)+140/163*A-219/163*B)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2+168*(A-10/7*B)*x*d)/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (7 \, A - 10 \, B\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (7 \, A - 10 \, B\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (7 \, A - 10 \, B\right )} d x + {\left (2 \, B \cos \left (d x + c\right )^{4} + {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, A - 66 \, B\right )} \cos \left (d x + c\right ) - 32 \, A + 48 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/6*(3*(7*A - 10*B)*d*x*cos(d*x + c)^2 + 6*(7*A - 10*B)*d*x*cos(d*x + c) + 3*(7*A - 10*B)*d*x + (2*B*cos(d*x +
c)^4 + (3*A - 2*B)*cos(d*x + c)^3 - 6*(A - 2*B)*cos(d*x + c)^2 - (43*A - 66*B)*cos(d*x + c) - 32*A + 48*B)*si
n(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1425 vs. \(2 (155) = 310\).

Time = 3.18 (sec) , antiderivative size = 1425, normalized size of antiderivative = 8.38 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((21*A*d*x*tan(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**
2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*A*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*A*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/
2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*A*d*x/(6*a**2*d
*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + A*tan(c/2 +
d*x/2)**9/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a
**2*d) - 18*A*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*ta
n(c/2 + d*x/2)**2 + 6*a**2*d) - 90*A*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d
*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 110*A*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**6
+ 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 39*A*tan(c/2 + d*x/2)/(6*a**2*d
*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 30*B*d*x*ta
n(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**
2 + 6*a**2*d) - 90*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 1
8*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 90*B*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**
2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 30*B*d*x/(6*a**2*d*tan(c/2 + d*x/2)**6 +
18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - B*tan(c/2 + d*x/2)**9/(6*a**2*d*t
an(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 24*B*tan(c/2
+ d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*
a**2*d) + 138*B*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*
tan(c/2 + d*x/2)**2 + 6*a**2*d) + 160*B*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2
+ d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**6
+ 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(
c)**4/(a*cos(c) + a)**2, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (160) = 320\).

Time = 0.33 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {B {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

[In]

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

[Out]

1/6*(B*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^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/(c
os(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*arct
an(sin(d*x + c)/(cos(d*x + c) + 1))/a^2))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (d x + c\right )} {\left (7 \, A - 10 \, B\right )}}{a^{2}} - \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(d*x + c)*(7*A - 10*B)/a^2 - 2*(15*A*tan(1/2*d*x + 1/2*c)^5 - 30*B*tan(1/2*d*x + 1/2*c)^5 + 24*A*tan(1/
2*d*x + 1/2*c)^3 - 40*B*tan(1/2*d*x + 1/2*c)^3 + 9*A*tan(1/2*d*x + 1/2*c) - 18*B*tan(1/2*d*x + 1/2*c))/((tan(1
/2*d*x + 1/2*c)^2 + 1)^3*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 - 21*A*a^4*tan(1/
2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {x\,\left (7\,A-10\,B\right )}{2\,a^2}-\frac {\left (5\,A-10\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (8\,A-\frac {40\,B}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-6\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A-B\right )}{a^2}+\frac {3\,A-5\,B}{2\,a^2}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d} \]

[In]

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

[Out]

(x*(7*A - 10*B))/(2*a^2) - (tan(c/2 + (d*x)/2)^5*(5*A - 10*B) + tan(c/2 + (d*x)/2)^3*(8*A - (40*B)/3) + tan(c/
2 + (d*x)/2)*(3*A - 6*B))/(d*(3*a^2*tan(c/2 + (d*x)/2)^2 + 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)
^6 + a^2)) - (tan(c/2 + (d*x)/2)*((2*(A - B))/a^2 + (3*A - 5*B)/(2*a^2)))/d + (tan(c/2 + (d*x)/2)^3*(A - B))/(
6*a^2*d)