\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx\) [568]

Optimal result

Integrand size = 29, antiderivative size = 91 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^{1+n}(c+d x)}{a d (1+n)}-\frac {\sin ^{2+n}(c+d x)}{a d (2+n)}-\frac {\sin ^{3+n}(c+d x)}{a d (3+n)}+\frac {\sin ^{4+n}(c+d x)}{a d (4+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 91, 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} \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^{n+1}(c+d x)}{a d (n+1)}-\frac {\sin ^{n+2}(c+d x)}{a d (n+2)}-\frac {\sin ^{n+3}(c+d x)}{a d (n+3)}+\frac {\sin ^{n+4}(c+d x)}{a d (n+4)} \]

[In]

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

[Out]

Sin[c + d*x]^(1 + n)/(a*d*(1 + n)) - Sin[c + d*x]^(2 + n)/(a*d*(2 + n)) - Sin[c + d*x]^(3 + n)/(a*d*(3 + n)) +
Sin[c + d*x]^(4 + n)/(a*d*(4 + 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)^2 \left (\frac {x}{a}\right )^n (a+x) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (a^3 \left (\frac {x}{a}\right )^n-a^3 \left (\frac {x}{a}\right )^{1+n}-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^5 d} \\ & = \frac {\sin ^{1+n}(c+d x)}{a d (1+n)}-\frac {\sin ^{2+n}(c+d x)}{a d (2+n)}-\frac {\sin ^{3+n}(c+d x)}{a d (3+n)}+\frac {\sin ^{4+n}(c+d x)}{a d (4+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^{1+n}(c+d x) \left (\frac {4+n}{1+n}-\frac {(4+n) \sin (c+d x)}{2+n}-\frac {(4+n) \sin ^2(c+d x)}{3+n}+\sin ^3(c+d x)\right )}{a d (4+n)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (1+n \right )}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (4+n \right )}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (2+n \right )}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (3+n \right )}\)	\(122\)
default	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (1+n \right )}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (4+n \right )}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (2+n \right )}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d a \left (3+n \right )}\)	\(122\)
parallelrisch	\(\frac {\left (\sin ^{n}\left (d x +c \right )\right ) \left (-30+\left (n^{3}+6 n^{2}+11 n +6\right ) \cos \left (4 d x +4 c \right )+2 \left (n^{3}+7 n^{2}+14 n +8\right ) \sin \left (3 d x +3 c \right )+8 \left (n^{2}+4 n +3\right ) \cos \left (2 d x +2 c \right )+2 \left (n^{3}+15 n^{2}+62 n +72\right ) \sin \left (d x +c \right )-n^{3}-14 n^{2}-43 n \right )}{8 a d \left (n^{2}+4 n +3\right ) \left (n^{2}+6 n +8\right )}\)	\(139\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.47 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} \cos \left (d x + c\right )^{4} - {\left (n^{3} + 4 \, n^{2} + 3 \, n\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} + {\left ({\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \cos \left (d x + c\right )^{2} + 2 \, n^{2} + 12 \, n + 16\right )} \sin \left (d x + c\right ) - 8 \, n - 6\right )} \sin \left (d x + c\right )^{n}}{a d n^{4} + 10 \, a d n^{3} + 35 \, a d n^{2} + 50 \, a d n + 24 \, a d} \]

[In]

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

[Out]

((n^3 + 6*n^2 + 11*n + 6)*cos(d*x + c)^4 - (n^3 + 4*n^2 + 3*n)*cos(d*x + c)^2 - 2*n^2 + ((n^3 + 7*n^2 + 14*n +
8)*cos(d*x + c)^2 + 2*n^2 + 12*n + 16)*sin(d*x + c) - 8*n - 6)*sin(d*x + c)^n/(a*d*n^4 + 10*a*d*n^3 + 35*a*d*
n^2 + 50*a*d*n + 24*a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} \sin \left (d x + c\right )^{4} - {\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \sin \left (d x + c\right )^{3} - {\left (n^{3} + 8 \, n^{2} + 19 \, n + 12\right )} \sin \left (d x + c\right )^{2} + {\left (n^{3} + 9 \, n^{2} + 26 \, n + 24\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} a d} \]

[In]

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

[Out]

((n^3 + 6*n^2 + 11*n + 6)*sin(d*x + c)^4 - (n^3 + 7*n^2 + 14*n + 8)*sin(d*x + c)^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*d)

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c)),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]%%%}+%%%{-1,[0,1,2,0,0]%%%}+%%%{-1,[0,1,1,0
,0]%%%}+%%%

Mupad [B] (verification not implemented)

Time = 12.15 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^n\,\left (144\,\sin \left (c+d\,x\right )-43\,n+24\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+16\,\sin \left (3\,c+3\,d\,x\right )+124\,n\,\sin \left (c+d\,x\right )+32\,n\,\cos \left (2\,c+2\,d\,x\right )+11\,n\,\cos \left (4\,c+4\,d\,x\right )+28\,n\,\sin \left (3\,c+3\,d\,x\right )+30\,n^2\,\sin \left (c+d\,x\right )+2\,n^3\,\sin \left (c+d\,x\right )-14\,n^2-n^3+8\,n^2\,\cos \left (2\,c+2\,d\,x\right )+6\,n^2\,\cos \left (4\,c+4\,d\,x\right )+n^3\,\cos \left (4\,c+4\,d\,x\right )+14\,n^2\,\sin \left (3\,c+3\,d\,x\right )+2\,n^3\,\sin \left (3\,c+3\,d\,x\right )-30\right )}{8\,a\,d\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )} \]

[In]

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

[Out]

(sin(c + d*x)^n*(144*sin(c + d*x) - 43*n + 24*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + 16*sin(3*c + 3*d*x) + 12
4*n*sin(c + d*x) + 32*n*cos(2*c + 2*d*x) + 11*n*cos(4*c + 4*d*x) + 28*n*sin(3*c + 3*d*x) + 30*n^2*sin(c + d*x)
+ 2*n^3*sin(c + d*x) - 14*n^2 - n^3 + 8*n^2*cos(2*c + 2*d*x) + 6*n^2*cos(4*c + 4*d*x) + n^3*cos(4*c + 4*d*x)
+ 14*n^2*sin(3*c + 3*d*x) + 2*n^3*sin(3*c + 3*d*x) - 30))/(8*a*d*(50*n + 35*n^2 + 10*n^3 + n^4 + 24))