3.56 \(\int \frac{\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=218 \[ \frac{4 (19 A-34 B) \sin ^3(c+d x)}{15 a^3 d}-\frac{4 (19 A-34 B) \sin (c+d x)}{5 a^3 d}+\frac{(13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{3 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(13 A-23 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{x (13 A-23 B)}{2 a^3}+\frac{(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

((13*A - 23*B)*x)/(2*a^3) - (4*(19*A - 34*B)*Sin[c + d*x])/(5*a^3*d) + ((13*A - 23*B)*Cos[c + d*x]*Sin[c + d*x
])/(2*a^3*d) + ((A - B)*Cos[c + d*x]^5*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((8*A - 13*B)*Cos[c + d*x]
^4*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + ((13*A - 23*B)*Cos[c + d*x]^3*Sin[c + d*x])/(3*d*(a^3 + a^3
*Cos[c + d*x])) + (4*(19*A - 34*B)*Sin[c + d*x]^3)/(15*a^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.515435, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2977, 2748, 2635, 8, 2633} \[ \frac{4 (19 A-34 B) \sin ^3(c+d x)}{15 a^3 d}-\frac{4 (19 A-34 B) \sin (c+d x)}{5 a^3 d}+\frac{(13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{3 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(13 A-23 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{x (13 A-23 B)}{2 a^3}+\frac{(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

((13*A - 23*B)*x)/(2*a^3) - (4*(19*A - 34*B)*Sin[c + d*x])/(5*a^3*d) + ((13*A - 23*B)*Cos[c + d*x]*Sin[c + d*x
])/(2*a^3*d) + ((A - B)*Cos[c + d*x]^5*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((8*A - 13*B)*Cos[c + d*x]
^4*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + ((13*A - 23*B)*Cos[c + d*x]^3*Sin[c + d*x])/(3*d*(a^3 + a^3
*Cos[c + d*x])) + (4*(19*A - 34*B)*Sin[c + d*x]^3)/(15*a^3*d)

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac{(A-B) \cos ^5(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos ^4(c+d x) (5 a (A-B)-a (3 A-8 B) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \cos ^5(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(8 A-13 B) \cos ^4(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^3(c+d x) \left (4 a^2 (8 A-13 B)-3 a^2 (11 A-21 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \cos ^5(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(8 A-13 B) \cos ^4(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(13 A-23 B) \cos ^3(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \cos ^2(c+d x) \left (15 a^3 (13 A-23 B)-12 a^3 (19 A-34 B) \cos (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(A-B) \cos ^5(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(8 A-13 B) \cos ^4(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(13 A-23 B) \cos ^3(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(4 (19 A-34 B)) \int \cos ^3(c+d x) \, dx}{5 a^3}+\frac{(13 A-23 B) \int \cos ^2(c+d x) \, dx}{a^3}\\ &=\frac{(13 A-23 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac{(A-B) \cos ^5(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(8 A-13 B) \cos ^4(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(13 A-23 B) \cos ^3(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(13 A-23 B) \int 1 \, dx}{2 a^3}+\frac{(4 (19 A-34 B)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 a^3 d}\\ &=\frac{(13 A-23 B) x}{2 a^3}-\frac{4 (19 A-34 B) \sin (c+d x)}{5 a^3 d}+\frac{(13 A-23 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac{(A-B) \cos ^5(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(8 A-13 B) \cos ^4(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(13 A-23 B) \cos ^3(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{4 (19 A-34 B) \sin ^3(c+d x)}{15 a^3 d}\\ \end{align*}

Mathematica [B]  time = 0.908189, size = 491, normalized size = 2.25 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (600 d x (13 A-23 B) \cos \left (c+\frac{d x}{2}\right )+600 d x (13 A-23 B) \cos \left (\frac{d x}{2}\right )+7560 A \sin \left (c+\frac{d x}{2}\right )-9230 A \sin \left (c+\frac{3 d x}{2}\right )+930 A \sin \left (2 c+\frac{3 d x}{2}\right )-2782 A \sin \left (2 c+\frac{5 d x}{2}\right )-750 A \sin \left (3 c+\frac{5 d x}{2}\right )-105 A \sin \left (3 c+\frac{7 d x}{2}\right )-105 A \sin \left (4 c+\frac{7 d x}{2}\right )+15 A \sin \left (4 c+\frac{9 d x}{2}\right )+15 A \sin \left (5 c+\frac{9 d x}{2}\right )+3900 A d x \cos \left (c+\frac{3 d x}{2}\right )+3900 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+780 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+780 A d x \cos \left (3 c+\frac{5 d x}{2}\right )-12760 A \sin \left (\frac{d x}{2}\right )-11110 B \sin \left (c+\frac{d x}{2}\right )+15380 B \sin \left (c+\frac{3 d x}{2}\right )-380 B \sin \left (2 c+\frac{3 d x}{2}\right )+4777 B \sin \left (2 c+\frac{5 d x}{2}\right )+1625 B \sin \left (3 c+\frac{5 d x}{2}\right )+230 B \sin \left (3 c+\frac{7 d x}{2}\right )+230 B \sin \left (4 c+\frac{7 d x}{2}\right )-20 B \sin \left (4 c+\frac{9 d x}{2}\right )-20 B \sin \left (5 c+\frac{9 d x}{2}\right )+5 B \sin \left (5 c+\frac{11 d x}{2}\right )+5 B \sin \left (6 c+\frac{11 d x}{2}\right )-6900 B d x \cos \left (c+\frac{3 d x}{2}\right )-6900 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-1380 B d x \cos \left (2 c+\frac{5 d x}{2}\right )-1380 B d x \cos \left (3 c+\frac{5 d x}{2}\right )+20410 B \sin \left (\frac{d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(600*(13*A - 23*B)*d*x*Cos[(d*x)/2] + 600*(13*A - 23*B)*d*x*Cos[c + (d*x)/2] + 3900
*A*d*x*Cos[c + (3*d*x)/2] - 6900*B*d*x*Cos[c + (3*d*x)/2] + 3900*A*d*x*Cos[2*c + (3*d*x)/2] - 6900*B*d*x*Cos[2
*c + (3*d*x)/2] + 780*A*d*x*Cos[2*c + (5*d*x)/2] - 1380*B*d*x*Cos[2*c + (5*d*x)/2] + 780*A*d*x*Cos[3*c + (5*d*
x)/2] - 1380*B*d*x*Cos[3*c + (5*d*x)/2] - 12760*A*Sin[(d*x)/2] + 20410*B*Sin[(d*x)/2] + 7560*A*Sin[c + (d*x)/2
] - 11110*B*Sin[c + (d*x)/2] - 9230*A*Sin[c + (3*d*x)/2] + 15380*B*Sin[c + (3*d*x)/2] + 930*A*Sin[2*c + (3*d*x
)/2] - 380*B*Sin[2*c + (3*d*x)/2] - 2782*A*Sin[2*c + (5*d*x)/2] + 4777*B*Sin[2*c + (5*d*x)/2] - 750*A*Sin[3*c
+ (5*d*x)/2] + 1625*B*Sin[3*c + (5*d*x)/2] - 105*A*Sin[3*c + (7*d*x)/2] + 230*B*Sin[3*c + (7*d*x)/2] - 105*A*S
in[4*c + (7*d*x)/2] + 230*B*Sin[4*c + (7*d*x)/2] + 15*A*Sin[4*c + (9*d*x)/2] - 20*B*Sin[4*c + (9*d*x)/2] + 15*
A*Sin[5*c + (9*d*x)/2] - 20*B*Sin[5*c + (9*d*x)/2] + 5*B*Sin[5*c + (11*d*x)/2] + 5*B*Sin[6*c + (11*d*x)/2]))/(
480*a^3*d*(1 + Cos[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.065, size = 362, normalized size = 1.7 \begin{align*} -{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{2\,A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{5\,B}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{31\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{49\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-7\,{\frac{A \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+17\,{\frac{B \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-12\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}A}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{76\,B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-5\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+11\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+13\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{d{a}^{3}}}-23\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3,x)

[Out]

-1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5+2/3/d/a^3*tan(1/2*d*x+1/2*c)^3*A-5/6/d/a^
3*B*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*A*tan(1/2*d*x+1/2*c)+49/4/d/a^3*B*tan(1/2*d*x+1/2*c)-7/d/a^3/(1+tan(1/2*d*
x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)^5+17/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)^5-12/d/a^3/(1+ta
n(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A+76/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)^3-5/d/
a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)+11/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)+1
3/d/a^3*arctan(tan(1/2*d*x+1/2*c))*A-23/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B

________________________________________________________________________________________

Maxima [B]  time = 1.51081, size = 556, normalized size = 2.55 \begin{align*} \frac{B{\left (\frac{20 \,{\left (\frac{33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{76 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{51 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{735 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{50 \, \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{1380 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - A{\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 )}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(B*(20*(33*sin(d*x + c)/(cos(d*x + c) + 1) + 76*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 51*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5)/(a^3 + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 + a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (735*sin(d*x + c)/(cos(d*x + c) + 1) - 50*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 - 1380*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^3) - A*(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))/d

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Fricas [A]  time = 1.40411, size = 539, normalized size = 2.47 \begin{align*} \frac{15 \,{\left (13 \, A - 23 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (13 \, A - 23 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (13 \, A - 23 \, B\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (13 \, A - 23 \, B\right )} d x +{\left (10 \, B \cos \left (d x + c\right )^{5} + 15 \,{\left (A - B\right )} \cos \left (d x + c\right )^{4} - 5 \,{\left (9 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{3} -{\left (479 \, A - 869 \, B\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (239 \, A - 429 \, B\right )} \cos \left (d x + c\right ) - 304 \, A + 544 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/30*(15*(13*A - 23*B)*d*x*cos(d*x + c)^3 + 45*(13*A - 23*B)*d*x*cos(d*x + c)^2 + 45*(13*A - 23*B)*d*x*cos(d*x
+ c) + 15*(13*A - 23*B)*d*x + (10*B*cos(d*x + c)^5 + 15*(A - B)*cos(d*x + c)^4 - 5*(9*A - 19*B)*cos(d*x + c)^
3 - (479*A - 869*B)*cos(d*x + c)^2 - 3*(239*A - 429*B)*cos(d*x + c) - 304*A + 544*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)

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Sympy [A]  time = 45.6154, size = 1584, normalized size = 7.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((390*A*d*x*tan(c/2 + d*x/2)**6/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180
*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 1170*A*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180
*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 1170*A*d*x*tan(c/2 + d*x/2)**2/(60
*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3
90*A*d*x/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60
*a**3*d) - 3*A*tan(c/2 + d*x/2)**11/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3
*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 31*A*tan(c/2 + d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan
(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 354*A*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2
+ d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 1698*A*tan(c/2 +
d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 6
0*a**3*d) - 2075*A*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a
**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 765*A*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*ta
n(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 690*B*d*x*tan(c/2 + d*x/2)**6/(60*a**3*d*tan
(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 2070*B*d*x*t
an(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/
2)**2 + 60*a**3*d) - 2070*B*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/
2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 690*B*d*x/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*ta
n(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3*B*tan(c/2 + d*x/2)**11/(60*a**3*d*tan(c/2
+ d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 41*B*tan(c/2 + d*
x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*
a**3*d) + 594*B*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3
*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3078*B*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*t
an(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3675*B*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c
/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 1395*B*tan(c/2
+ d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 6
0*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**5/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.21456, size = 308, normalized size = 1.41 \begin{align*} \frac{\frac{30 \,{\left (d x + c\right )}{\left (13 \, A - 23 \, B\right )}}{a^{3}} - \frac{20 \,{\left (21 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 51 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 76 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 33 \, 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^{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} - 40 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 50 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 735 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(d*x + c)*(13*A - 23*B)/a^3 - 20*(21*A*tan(1/2*d*x + 1/2*c)^5 - 51*B*tan(1/2*d*x + 1/2*c)^5 + 36*A*ta
n(1/2*d*x + 1/2*c)^3 - 76*B*tan(1/2*d*x + 1/2*c)^3 + 15*A*tan(1/2*d*x + 1/2*c) - 33*B*tan(1/2*d*x + 1/2*c))/((
tan(1/2*d*x + 1/2*c)^2 + 1)^3*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 - 40*A
*a^12*tan(1/2*d*x + 1/2*c)^3 + 50*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) - 735*B*a^12
*tan(1/2*d*x + 1/2*c))/a^15)/d