\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [569]

Optimal result

Integrand size = 29, antiderivative size = 68 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^2 d (1+n)}-\frac {2 \sin ^{2+n}(c+d x)}{a^2 d (2+n)}+\frac {\sin ^{3+n}(c+d x)}{a^2 d (3+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 68, 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} \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{n+1}(c+d x)}{a^2 d (n+1)}-\frac {2 \sin ^{n+2}(c+d x)}{a^2 d (n+2)}+\frac {\sin ^{n+3}(c+d x)}{a^2 d (n+3)} \]

[In]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x) \left (\frac {1}{1+n}-\frac {2 \sin (c+d x)}{2+n}+\frac {\sin ^2(c+d x)}{3+n}\right )}{a^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}+\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (3+n \right )}-\frac {2 \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}\)	\(91\)
default	\(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}+\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (3+n \right )}-\frac {2 \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}\)	\(91\)
parallelrisch	\(\frac {\left (\left (n^{2}+4 n +3\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {1}{4} n^{2}-\frac {3}{4} n -\frac {1}{2}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {7}{4} n^{2}+\frac {29}{4} n +\frac {15}{2}\right ) \sin \left (d x +c \right )-n^{2}-4 n -3\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a^{2} d \left (2+n \right ) \left (3+n \right ) \left (1+n \right )}\)	\(96\)

[In]

int(cos(d*x+c)^5*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)))+1/a^2/d/(3+n)*sin(d*x+c)^3*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)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.54 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (2 \, {\left (n^{2} + 4 \, n + 3\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} - {\left ({\left (n^{2} + 3 \, n + 2\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} - 8 \, n - 8\right )} \sin \left (d x + c\right ) - 8 \, n - 6\right )} \sin \left (d x + c\right )^{n}}{a^{2} d n^{3} + 6 \, a^{2} d n^{2} + 11 \, a^{2} d n + 6 \, a^{2} d} \]

[In]

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

[Out]

(2*(n^2 + 4*n + 3)*cos(d*x + c)^2 - 2*n^2 - ((n^2 + 3*n + 2)*cos(d*x + c)^2 - 2*n^2 - 8*n - 8)*sin(d*x + c) -
8*n - 6)*sin(d*x + c)^n/(a^2*d*n^3 + 6*a^2*d*n^2 + 11*a^2*d*n + 6*a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*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 = 81, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} \sin \left (d x + c\right )^{3} - 2 \, {\left (n^{2} + 4 \, n + 3\right )} \sin \left (d x + c\right )^{2} + {\left (n^{2} + 5 \, n + 6\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} a^{2} d} \]

[In]

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

[Out]

((n^2 + 3*n + 2)*sin(d*x + c)^3 - 2*(n^2 + 4*n + 3)*sin(d*x + c)^2 + (n^2 + 5*n + 6)*sin(d*x + c))*sin(d*x + c
)^n/((n^3 + 6*n^2 + 11*n + 6)*a^2*d)

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 11.65 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.15 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^n\,\left (24\,{\sin \left (c+d\,x\right )}^2-30\,\sin \left (c+d\,x\right )+2\,\sin \left (3\,c+3\,d\,x\right )\right )}{4}+\frac {n\,{\sin \left (c+d\,x\right )}^n\,\left (32\,{\sin \left (c+d\,x\right )}^2-29\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )\right )}{4}+\frac {n^2\,{\sin \left (c+d\,x\right )}^n\,\left (8\,{\sin \left (c+d\,x\right )}^2-7\,\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\right )}{4}}{a^2\,d\,\left (n^3+6\,n^2+11\,n+6\right )} \]

[In]

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

[Out]

-((sin(c + d*x)^n*(2*sin(3*c + 3*d*x) - 30*sin(c + d*x) + 24*sin(c + d*x)^2))/4 + (n*sin(c + d*x)^n*(3*sin(3*c
+ 3*d*x) - 29*sin(c + d*x) + 32*sin(c + d*x)^2))/4 + (n^2*sin(c + d*x)^n*(sin(3*c + 3*d*x) - 7*sin(c + d*x) +
8*sin(c + d*x)^2))/4)/(a^2*d*(11*n + 6*n^2 + n^3 + 6))