\(\int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\) [1004]

Optimal result

Integrand size = 31, antiderivative size = 79 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d}+\frac {(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac {B (a-a \sin (c+d x))^5}{5 a^6 d} \]

[Out]

-2/3*(A+B)*(a-a*sin(d*x+c))^3/a^4/d+1/4*(A+3*B)*(a-a*sin(d*x+c))^4/a^5/d-1/5*B*(a-a*sin(d*x+c))^5/a^6/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2915, 78} \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {B (a-a \sin (c+d x))^5}{5 a^6 d}+\frac {(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac {2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d} \]

[In]

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

[Out]

(-2*(A + B)*(a - a*Sin[c + d*x])^3)/(3*a^4*d) + ((A + 3*B)*(a - a*Sin[c + d*x])^4)/(4*a^5*d) - (B*(a - a*Sin[c
+ d*x])^5)/(5*a^6*d)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a-x)^2 (a+x) \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (2 a (A+B) (a-x)^2+(-A-3 B) (a-x)^3+\frac {B (a-x)^4}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = -\frac {2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d}+\frac {(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac {B (a-a \sin (c+d x))^5}{5 a^6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\sin (c+d x) \left (60 A-30 (A-B) \sin (c+d x)-20 (A+B) \sin ^2(c+d x)+15 (A-B) \sin ^3(c+d x)+12 B \sin ^4(c+d x)\right )}{60 a d} \]

[In]

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

[Out]

(Sin[c + d*x]*(60*A - 30*(A - B)*Sin[c + d*x] - 20*(A + B)*Sin[c + d*x]^2 + 15*(A - B)*Sin[c + d*x]^3 + 12*B*S
in[c + d*x]^4))/(60*a*d)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right ) B}{5}+\frac {\left (A -B \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-A -B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )}{d a}\)	\(75\)
default	\(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right ) B}{5}+\frac {\left (A -B \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-A -B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )}{d a}\)	\(75\)
parallelrisch	\(\frac {60 \cos \left (2 d x +2 c \right ) \left (A -B \right )+15 \left (A -B \right ) \cos \left (4 d x +4 c \right )+10 \left (4 A +B \right ) \sin \left (3 d x +3 c \right )+6 B \sin \left (5 d x +5 c \right )+60 \left (6 A -B \right ) \sin \left (d x +c \right )-75 A +75 B}{480 d a}\)	\(91\)
risch	\(\frac {3 A \sin \left (d x +c \right )}{4 a d}-\frac {B \sin \left (d x +c \right )}{8 a d}+\frac {\sin \left (5 d x +5 c \right ) B}{80 a d}+\frac {\cos \left (4 d x +4 c \right ) A}{32 a d}-\frac {\cos \left (4 d x +4 c \right ) B}{32 a d}+\frac {\sin \left (3 d x +3 c \right ) A}{12 a d}+\frac {\sin \left (3 d x +3 c \right ) B}{48 a d}+\frac {\cos \left (2 d x +2 c \right ) A}{8 a d}-\frac {\cos \left (2 d x +2 c \right ) B}{8 a d}\)	\(158\)
norman	\(\frac {\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 A \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 B \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \left (8 A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (8 A -B \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {4 \left (10 A +3 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}+\frac {4 \left (10 A +3 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}+\frac {\left (10 A +4 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (10 A +4 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (40 A +12 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}+\frac {\left (40 A +12 B \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(317\)

[In]

int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(1/5*sin(d*x+c)^5*B+1/4*(A-B)*sin(d*x+c)^4+1/3*(-A-B)*sin(d*x+c)^3+1/2*(-A+B)*sin(d*x+c)^2+A*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B \cos \left (d x + c\right )^{4} + {\left (5 \, A - B\right )} \cos \left (d x + c\right )^{2} + 10 \, A - 2 \, B\right )} \sin \left (d x + c\right )}{60 \, a d} \]

[In]

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

[Out]

1/60*(15*(A - B)*cos(d*x + c)^4 + 4*(3*B*cos(d*x + c)^4 + (5*A - B)*cos(d*x + c)^2 + 10*A - 2*B)*sin(d*x + c))
/(a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1703 vs. \(2 (73) = 146\).

Time = 11.68 (sec) , antiderivative size = 1703, normalized size of antiderivative = 21.56 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((30*A*tan(c/2 + d*x/2)**9/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/
2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*A*tan(c/2 + d*x/2)**8/
(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*
x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 80*A*tan(c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*
d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2
+ 15*a*d) - 30*A*tan(c/2 + d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(
c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 100*A*tan(c/2 + d*x/2)*
*5/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 +
d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*A*tan(c/2 + d*x/2)**4/(15*a*d*tan(c/2 + d*x/2)**10 + 75
*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)
**2 + 15*a*d) + 80*A*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*t
an(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*A*tan(c/2 + d*x/2
)**2/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2
+ d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*A*tan(c/2 + d*x/2)/(15*a*d*tan(c/2 + d*x/2)**10 + 75*
a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)*
*2 + 15*a*d) + 30*B*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*ta
n(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*B*tan(c/2 + d*x/2)
**7/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2
+ d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*B*tan(c/2 + d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**10 + 7
5*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2
)**2 + 15*a*d) + 16*B*tan(c/2 + d*x/2)**5/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*
tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*B*tan(c/2 + d*x/
2)**4/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/
2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*B*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**10 +
75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x
/2)**2 + 15*a*d) + 30*B*tan(c/2 + d*x/2)**2/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*
d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*(A +
B*sin(c))*cos(c)**5/(a*sin(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {12 \, B \sin \left (d x + c\right )^{5} + 15 \, {\left (A - B\right )} \sin \left (d x + c\right )^{4} - 20 \, {\left (A + B\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (A - B\right )} \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \]

[In]

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

[Out]

1/60*(12*B*sin(d*x + c)^5 + 15*(A - B)*sin(d*x + c)^4 - 20*(A + B)*sin(d*x + c)^3 - 30*(A - B)*sin(d*x + c)^2
+ 60*A*sin(d*x + c))/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {12 \, B \sin \left (d x + c\right )^{5} + 15 \, A \sin \left (d x + c\right )^{4} - 15 \, B \sin \left (d x + c\right )^{4} - 20 \, A \sin \left (d x + c\right )^{3} - 20 \, B \sin \left (d x + c\right )^{3} - 30 \, A \sin \left (d x + c\right )^{2} + 30 \, B \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \]

[In]

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

[Out]

1/60*(12*B*sin(d*x + c)^5 + 15*A*sin(d*x + c)^4 - 15*B*sin(d*x + c)^4 - 20*A*sin(d*x + c)^3 - 20*B*sin(d*x + c
)^3 - 30*A*sin(d*x + c)^2 + 30*B*sin(d*x + c)^2 + 60*A*sin(d*x + c))/(a*d)

Mupad [B] (verification not implemented)

Time = 10.00 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4\,\left (A-B\right )}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (A-B\right )}{2\,a}+\frac {B\,{\sin \left (c+d\,x\right )}^5}{5\,a}+\frac {A\,\sin \left (c+d\,x\right )}{a}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (A+B\right )}{3\,a}}{d} \]

[In]

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

[Out]

((sin(c + d*x)^4*(A - B))/(4*a) - (sin(c + d*x)^2*(A - B))/(2*a) + (B*sin(c + d*x)^5)/(5*a) + (A*sin(c + d*x))
/a - (sin(c + d*x)^3*(A + B))/(3*a))/d