\(\int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [702]

Optimal result

Integrand size = 29, antiderivative size = 92 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^3 d (1+n)}-\frac {3 \sin ^{2+n}(c+d x)}{a^3 d (2+n)}+\frac {3 \sin ^{3+n}(c+d x)}{a^3 d (3+n)}-\frac {\sin ^{4+n}(c+d x)}{a^3 d (4+n)} \]

[Out]

sin(d*x+c)^(1+n)/a^3/d/(1+n)-3*sin(d*x+c)^(2+n)/a^3/d/(2+n)+3*sin(d*x+c)^(3+n)/a^3/d/(3+n)-sin(d*x+c)^(4+n)/a^
3/d/(4+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, 45} \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^{n+1}(c+d x)}{a^3 d (n+1)}-\frac {3 \sin ^{n+2}(c+d x)}{a^3 d (n+2)}+\frac {3 \sin ^{n+3}(c+d x)}{a^3 d (n+3)}-\frac {\sin ^{n+4}(c+d x)}{a^3 d (n+4)} \]

[In]

Int[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^{1+n}(c+d x) \left (\frac {1}{1+n}-\frac {3 \sin (c+d x)}{2+n}+\frac {3 \sin ^2(c+d x)}{3+n}-\frac {\sin ^3(c+d x)}{4+n}\right )}{a^3 d} \]

[In]

Integrate[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Maple [A] (verified)

Time = 1.66 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (1+n \right )}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (2+n \right )}+\frac {3 \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (3+n \right )}-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (4+n \right )}\)	\(123\)
default	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (1+n \right )}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (2+n \right )}+\frac {3 \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (3+n \right )}-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{3} d \left (4+n \right )}\)	\(123\)
parallelrisch	\(\frac {2 \left (\sin ^{n}\left (d x +c \right )\right ) \left (\left (n^{3}+\frac {15}{2} n^{2}+17 n +\frac {21}{2}\right ) \cos \left (2 d x +2 c \right )-\frac {\left (3+n \right ) \left (2+n \right ) \left (1+n \right ) \cos \left (4 d x +4 c \right )}{16}+\left (-\frac {3}{8} n^{3}-\frac {21}{8} n^{2}-\frac {21}{4} n -3\right ) \sin \left (3 d x +3 c \right )+\left (21+\frac {99}{8} n^{2}+\frac {13}{8} n^{3}+\frac {115}{4} n \right ) \sin \left (d x +c \right )-\frac {15 \left (1+n \right ) \left (n +\frac {18}{5}\right ) \left (3+n \right )}{16}\right )}{a^{3} d \left (n^{2}+4 n +3\right ) \left (n^{2}+6 n +8\right )}\)	\(139\)

[In]

int(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/a^3/d/(1+n)*sin(d*x+c)*exp(n*ln(sin(d*x+c)))-3/a^3/d/(2+n)*sin(d*x+c)^2*exp(n*ln(sin(d*x+c)))+3/a^3/d/(3+n)*
sin(d*x+c)^3*exp(n*ln(sin(d*x+c)))-1/a^3/d/(4+n)*sin(d*x+c)^4*exp(n*ln(sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.74 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} \cos \left (d x + c\right )^{4} + 4 \, n^{3} - {\left (5 \, n^{3} + 36 \, n^{2} + 79 \, n + 48\right )} \cos \left (d x + c\right )^{2} + 30 \, n^{2} - {\left (4 \, n^{3} - 3 \, {\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \cos \left (d x + c\right )^{2} + 30 \, n^{2} + 68 \, n + 48\right )} \sin \left (d x + c\right ) + 68 \, n + 42\right )} \sin \left (d x + c\right )^{n}}{a^{3} d n^{4} + 10 \, a^{3} d n^{3} + 35 \, a^{3} d n^{2} + 50 \, a^{3} d n + 24 \, a^{3} d} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-((n^3 + 6*n^2 + 11*n + 6)*cos(d*x + c)^4 + 4*n^3 - (5*n^3 + 36*n^2 + 79*n + 48)*cos(d*x + c)^2 + 30*n^2 - (4*
n^3 - 3*(n^3 + 7*n^2 + 14*n + 8)*cos(d*x + c)^2 + 30*n^2 + 68*n + 48)*sin(d*x + c) + 68*n + 42)*sin(d*x + c)^n
/(a^3*d*n^4 + 10*a^3*d*n^3 + 35*a^3*d*n^2 + 50*a^3*d*n + 24*a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**7*sin(d*x+c)**n/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} \sin \left (d x + c\right )^{4} - 3 \, {\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \sin \left (d x + c\right )^{3} + 3 \, {\left (n^{3} + 8 \, n^{2} + 19 \, n + 12\right )} \sin \left (d x + c\right )^{2} - {\left (n^{3} + 9 \, n^{2} + 26 \, n + 24\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} a^{3} d} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-((n^3 + 6*n^2 + 11*n + 6)*sin(d*x + c)^4 - 3*(n^3 + 7*n^2 + 14*n + 8)*sin(d*x + c)^3 + 3*(n^3 + 8*n^2 + 19*n
+ 12)*sin(d*x + c)^2 - (n^3 + 9*n^2 + 26*n + 24)*sin(d*x + c))*sin(d*x + c)^n/((n^4 + 10*n^3 + 35*n^2 + 50*n +
24)*a^3*d)

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,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,[0,1,3,0,0]%%%}+%%%{3,[0,1,2,0,0]%%%}+%%%{-3,[0,1,1,0
,0]%%%}+%%%

Mupad [B] (verification not implemented)

Time = 11.53 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.63 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {{\sin \left (c+d\,x\right )}^n\,\left (261\,n-336\,\sin \left (c+d\,x\right )-168\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+48\,\sin \left (3\,c+3\,d\,x\right )-460\,n\,\sin \left (c+d\,x\right )-272\,n\,\cos \left (2\,c+2\,d\,x\right )+11\,n\,\cos \left (4\,c+4\,d\,x\right )+84\,n\,\sin \left (3\,c+3\,d\,x\right )-198\,n^2\,\sin \left (c+d\,x\right )-26\,n^3\,\sin \left (c+d\,x\right )+114\,n^2+15\,n^3-120\,n^2\,\cos \left (2\,c+2\,d\,x\right )-16\,n^3\,\cos \left (2\,c+2\,d\,x\right )+6\,n^2\,\cos \left (4\,c+4\,d\,x\right )+n^3\,\cos \left (4\,c+4\,d\,x\right )+42\,n^2\,\sin \left (3\,c+3\,d\,x\right )+6\,n^3\,\sin \left (3\,c+3\,d\,x\right )+162\right )}{8\,a^3\,d\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )} \]

[In]

int((cos(c + d*x)^7*sin(c + d*x)^n)/(a + a*sin(c + d*x))^3,x)

[Out]

-(sin(c + d*x)^n*(261*n - 336*sin(c + d*x) - 168*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + 48*sin(3*c + 3*d*x) -
460*n*sin(c + d*x) - 272*n*cos(2*c + 2*d*x) + 11*n*cos(4*c + 4*d*x) + 84*n*sin(3*c + 3*d*x) - 198*n^2*sin(c +
d*x) - 26*n^3*sin(c + d*x) + 114*n^2 + 15*n^3 - 120*n^2*cos(2*c + 2*d*x) - 16*n^3*cos(2*c + 2*d*x) + 6*n^2*co
s(4*c + 4*d*x) + n^3*cos(4*c + 4*d*x) + 42*n^2*sin(3*c + 3*d*x) + 6*n^3*sin(3*c + 3*d*x) + 162))/(8*a^3*d*(50*
n + 35*n^2 + 10*n^3 + n^4 + 24))