\(\int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [701]

Optimal result

Integrand size = 29, antiderivative size = 92 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^2 d (1+n)}-\frac {2 \sin ^{2+n}(c+d x)}{a^2 d (2+n)}+\frac {2 \sin ^{4+n}(c+d x)}{a^2 d (4+n)}-\frac {\sin ^{5+n}(c+d x)}{a^2 d (5+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2915, 76} \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{n+1}(c+d x)}{a^2 d (n+1)}-\frac {2 \sin ^{n+2}(c+d x)}{a^2 d (n+2)}+\frac {2 \sin ^{n+4}(c+d x)}{a^2 d (n+4)}-\frac {\sin ^{n+5}(c+d x)}{a^2 d (n+5)} \]

[In]

Int[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]

[Out]

Sin[c + d*x]^(1 + n)/(a^2*d*(1 + n)) - (2*Sin[c + d*x]^(2 + n))/(a^2*d*(2 + n)) + (2*Sin[c + d*x]^(4 + n))/(a^
2*d*(4 + n)) - Sin[c + d*x]^(5 + n)/(a^2*d*(5 + n))

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 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)^3 \left (\frac {x}{a}\right )^n (a+x) \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {\text {Subst}\left (\int \left (a^4 \left (\frac {x}{a}\right )^n-2 a^4 \left (\frac {x}{a}\right )^{1+n}+2 a^4 \left (\frac {x}{a}\right )^{3+n}-a^4 \left (\frac {x}{a}\right )^{4+n}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {\sin ^{1+n}(c+d x)}{a^2 d (1+n)}-\frac {2 \sin ^{2+n}(c+d x)}{a^2 d (2+n)}+\frac {2 \sin ^{4+n}(c+d x)}{a^2 d (4+n)}-\frac {\sin ^{5+n}(c+d x)}{a^2 d (5+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x) \left (40+38 n+11 n^2+n^3-2 \left (20+29 n+10 n^2+n^3\right ) \sin (c+d x)+2 \left (10+17 n+8 n^2+n^3\right ) \sin ^3(c+d x)-\left (8+14 n+7 n^2+n^3\right ) \sin ^4(c+d x)\right )}{a^2 d (1+n) (2+n) (4+n) (5+n)} \]

[In]

Integrate[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]

[Out]

(Sin[c + d*x]^(1 + n)*(40 + 38*n + 11*n^2 + n^3 - 2*(20 + 29*n + 10*n^2 + n^3)*Sin[c + d*x] + 2*(10 + 17*n + 8
*n^2 + n^3)*Sin[c + d*x]^3 - (8 + 14*n + 7*n^2 + n^3)*Sin[c + d*x]^4))/(a^2*d*(1 + n)*(2 + n)*(4 + n)*(5 + n))

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}-\frac {2 \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}+\frac {2 \left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (4+n \right )}-\frac {\left (\sin ^{5}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (5+n \right )}\)	\(123\)
default	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}-\frac {2 \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}+\frac {2 \left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (4+n \right )}-\frac {\left (\sin ^{5}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (5+n \right )}\)	\(123\)
parallelrisch	\(-\frac {\left (\sin ^{n}\left (d x +c \right )\right ) \left (\left (-4 n^{3}-32 n^{2}-68 n -40\right ) \cos \left (4 d x +4 c \right )+\left (-5 n^{3}-35 n^{2}-70 n -40\right ) \sin \left (3 d x +3 c \right )+\left (n^{3}+7 n^{2}+14 n +8\right ) \sin \left (5 d x +5 c \right )+\left (-32 n^{2}-192 n -160\right ) \cos \left (2 d x +2 c \right )+\left (-6 n^{3}-106 n^{2}-468 n -560\right ) \sin \left (d x +c \right )+4 n^{3}+64 n^{2}+260 n +200\right )}{16 a^{2} d \left (n^{2}+6 n +5\right ) \left (n^{2}+6 n +8\right )}\)	\(167\)

[In]

int(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/a^2/d/(1+n)*sin(d*x+c)*exp(n*ln(sin(d*x+c)))-2/a^2/d/(2+n)*sin(d*x+c)^2*exp(n*ln(sin(d*x+c)))+2/a^2/d/(4+n)*
sin(d*x+c)^4*exp(n*ln(sin(d*x+c)))-1/a^2/d/(5+n)*sin(d*x+c)^5*exp(n*ln(sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (2 \, {\left (n^{3} + 8 \, n^{2} + 17 \, n + 10\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (n^{3} + 6 \, n^{2} + 5 \, n\right )} \cos \left (d x + c\right )^{2} - 4 \, n^{2} - {\left ({\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \cos \left (d x + c\right )^{2} - 4 \, n^{2} - 24 \, n - 32\right )} \sin \left (d x + c\right ) - 24 \, n - 20\right )} \sin \left (d x + c\right )^{n}}{a^{2} d n^{4} + 12 \, a^{2} d n^{3} + 49 \, a^{2} d n^{2} + 78 \, a^{2} d n + 40 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

(2*(n^3 + 8*n^2 + 17*n + 10)*cos(d*x + c)^4 - 2*(n^3 + 6*n^2 + 5*n)*cos(d*x + c)^2 - 4*n^2 - ((n^3 + 7*n^2 + 1
4*n + 8)*cos(d*x + c)^4 - 2*(n^3 + 7*n^2 + 14*n + 8)*cos(d*x + c)^2 - 4*n^2 - 24*n - 32)*sin(d*x + c) - 24*n -
20)*sin(d*x + c)^n/(a^2*d*n^4 + 12*a^2*d*n^3 + 49*a^2*d*n^2 + 78*a^2*d*n + 40*a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**7*sin(d*x+c)**n/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {{\left ({\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \sin \left (d x + c\right )^{5} - 2 \, {\left (n^{3} + 8 \, n^{2} + 17 \, n + 10\right )} \sin \left (d x + c\right )^{4} + 2 \, {\left (n^{3} + 10 \, n^{2} + 29 \, n + 20\right )} \sin \left (d x + c\right )^{2} - {\left (n^{3} + 11 \, n^{2} + 38 \, n + 40\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{4} + 12 \, n^{3} + 49 \, n^{2} + 78 \, n + 40\right )} a^{2} d} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-((n^3 + 7*n^2 + 14*n + 8)*sin(d*x + c)^5 - 2*(n^3 + 8*n^2 + 17*n + 10)*sin(d*x + c)^4 + 2*(n^3 + 10*n^2 + 29*
n + 20)*sin(d*x + c)^2 - (n^3 + 11*n^2 + 38*n + 40)*sin(d*x + c))*sin(d*x + c)^n/((n^4 + 12*n^3 + 49*n^2 + 78*
n + 40)*a^2*d)

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-1,[0,1,4,0,0]%%%}+%%%{2,[0,1,3,0,0]%%%}+%%%{-2,[0,1,1,0
,0]%%%}+%%%

Mupad [B] (verification not implemented)

Time = 12.30 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^n\,\left (560\,\sin \left (c+d\,x\right )-260\,n+160\,\cos \left (2\,c+2\,d\,x\right )+40\,\cos \left (4\,c+4\,d\,x\right )+40\,\sin \left (3\,c+3\,d\,x\right )-8\,\sin \left (5\,c+5\,d\,x\right )+468\,n\,\sin \left (c+d\,x\right )+192\,n\,\cos \left (2\,c+2\,d\,x\right )+68\,n\,\cos \left (4\,c+4\,d\,x\right )+70\,n\,\sin \left (3\,c+3\,d\,x\right )-14\,n\,\sin \left (5\,c+5\,d\,x\right )+106\,n^2\,\sin \left (c+d\,x\right )+6\,n^3\,\sin \left (c+d\,x\right )-64\,n^2-4\,n^3+32\,n^2\,\cos \left (2\,c+2\,d\,x\right )+32\,n^2\,\cos \left (4\,c+4\,d\,x\right )+4\,n^3\,\cos \left (4\,c+4\,d\,x\right )+35\,n^2\,\sin \left (3\,c+3\,d\,x\right )+5\,n^3\,\sin \left (3\,c+3\,d\,x\right )-7\,n^2\,\sin \left (5\,c+5\,d\,x\right )-n^3\,\sin \left (5\,c+5\,d\,x\right )-200\right )}{16\,a^2\,d\,\left (n^4+12\,n^3+49\,n^2+78\,n+40\right )} \]

[In]

int((cos(c + d*x)^7*sin(c + d*x)^n)/(a + a*sin(c + d*x))^2,x)

[Out]

(sin(c + d*x)^n*(560*sin(c + d*x) - 260*n + 160*cos(2*c + 2*d*x) + 40*cos(4*c + 4*d*x) + 40*sin(3*c + 3*d*x) -
8*sin(5*c + 5*d*x) + 468*n*sin(c + d*x) + 192*n*cos(2*c + 2*d*x) + 68*n*cos(4*c + 4*d*x) + 70*n*sin(3*c + 3*d
*x) - 14*n*sin(5*c + 5*d*x) + 106*n^2*sin(c + d*x) + 6*n^3*sin(c + d*x) - 64*n^2 - 4*n^3 + 32*n^2*cos(2*c + 2*
d*x) + 32*n^2*cos(4*c + 4*d*x) + 4*n^3*cos(4*c + 4*d*x) + 35*n^2*sin(3*c + 3*d*x) + 5*n^3*sin(3*c + 3*d*x) - 7
*n^2*sin(5*c + 5*d*x) - n^3*sin(5*c + 5*d*x) - 200))/(16*a^2*d*(78*n + 49*n^2 + 12*n^3 + n^4 + 40))