\(\int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\) [1012]

Optimal result

Integrand size = 31, antiderivative size = 51 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {(A+B) (a-a \sin (c+d x))^3}{3 a^5 d}+\frac {B (a-a \sin (c+d x))^4}{4 a^6 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, 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, 45} \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {B (a-a \sin (c+d x))^4}{4 a^6 d}-\frac {(A+B) (a-a \sin (c+d x))^3}{3 a^5 d} \]

[In]

Int[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/3*((A + B)*(a - a*Sin[c + d*x])^3)/(a^5*d) + (B*(a - a*Sin[c + d*x])^4)/(4*a^6*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {(-1+\sin (c+d x))^3 (4 A+B+3 B \sin (c+d x))}{12 a^2 d} \]

[In]

Integrate[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

((-1 + Sin[c + d*x])^3*(4*A + B + 3*B*Sin[c + d*x]))/(12*a^2*d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right ) B}{4}+\frac {\left (A -2 B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (B -2 A \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )}{a^{2} d}\)	\(58\)
default	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right ) B}{4}+\frac {\left (A -2 B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (B -2 A \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )}{a^{2} d}\)	\(58\)
parallelrisch	\(\frac {\left (48 A -36 B \right ) \cos \left (2 d x +2 c \right )+\left (-8 A +16 B \right ) \sin \left (3 d x +3 c \right )+3 B \cos \left (4 d x +4 c \right )+\left (120 A -48 B \right ) \sin \left (d x +c \right )-48 A +33 B}{96 a^{2} d}\)	\(76\)
risch	\(\frac {5 \sin \left (d x +c \right ) A}{4 a^{2} d}-\frac {\sin \left (d x +c \right ) B}{2 a^{2} d}+\frac {\cos \left (4 d x +4 c \right ) B}{32 a^{2} d}-\frac {\sin \left (3 d x +3 c \right ) A}{12 a^{2} d}+\frac {\sin \left (3 d x +3 c \right ) B}{6 a^{2} d}+\frac {\cos \left (2 d x +2 c \right ) A}{2 a^{2} d}-\frac {3 \cos \left (2 d x +2 c \right ) B}{8 a^{2} d}\)	\(122\)
norman	\(\frac {\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 A \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (2 A +2 B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (2 A +2 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (12 A +2 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (12 A +2 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (14 A +6 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (14 A +6 B \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (24 A +8 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (24 A +8 B \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (20 A +2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (20 A +2 B \right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (74 A +8 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (74 A +8 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(376\)

[In]

int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, B \cos \left (d x + c\right )^{4} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, A + 2 \, B\right )} \sin \left (d x + c\right )}{12 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(3*B*cos(d*x + c)^4 + 12*(A - B)*cos(d*x + c)^2 - 4*((A - 2*B)*cos(d*x + c)^2 - 4*A + 2*B)*sin(d*x + c))/
(a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1182 vs. \(2 (44) = 88\).

Time = 20.66 (sec) , antiderivative size = 1182, normalized size of antiderivative = 23.18 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)**5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((6*A*tan(c/2 + d*x/2)**7/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 12*A*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 +
d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*
a**2*d) + 26*A*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 24*A*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 +
d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*
a**2*d) + 26*A*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 12*A*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 +
d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*
a**2*d) + 6*A*tan(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c
/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 6*B*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2
)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*
d) - 16*B*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/
2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 24*B*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2
)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*
d) - 16*B*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/
2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 6*B*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)
**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d
), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)**5/(a*sin(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, B \sin \left (d x + c\right )^{4} + 4 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )^{3} - 6 \, {\left (2 \, A - B\right )} \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(3*B*sin(d*x + c)^4 + 4*(A - 2*B)*sin(d*x + c)^3 - 6*(2*A - B)*sin(d*x + c)^2 + 12*A*sin(d*x + c))/(a^2*d
)

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, B \sin \left (d x + c\right )^{4} + 4 \, A \sin \left (d x + c\right )^{3} - 8 \, B \sin \left (d x + c\right )^{3} - 12 \, A \sin \left (d x + c\right )^{2} + 6 \, B \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(3*B*sin(d*x + c)^4 + 4*A*sin(d*x + c)^3 - 8*B*sin(d*x + c)^3 - 12*A*sin(d*x + c)^2 + 6*B*sin(d*x + c)^2
+ 12*A*sin(d*x + c))/(a^2*d)

Mupad [B] (verification not implemented)

Time = 9.71 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^3\,\left (A-2\,B\right )}{3\,a^2}+\frac {B\,{\sin \left (c+d\,x\right )}^4}{4\,a^2}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,A-B\right )}{2\,a^2}+\frac {A\,\sin \left (c+d\,x\right )}{a^2}}{d} \]

[In]

int((cos(c + d*x)^5*(A + B*sin(c + d*x)))/(a + a*sin(c + d*x))^2,x)

[Out]

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