Integrand size = 29, antiderivative size = 68 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^2 d (1+n)}-\frac {2 \sin ^{2+n}(c+d x)}{a^2 d (2+n)}+\frac {\sin ^{3+n}(c+d x)}{a^2 d (3+n)} \] Output:
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)
Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x) \left (\frac {1}{1+n}-\frac {2 \sin (c+d x)}{2+n}+\frac {\sin ^2(c+d x)}{3+n}\right )}{a^2 d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]
Output:
(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)
Time = 0.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 3315, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 \sin (c+d x)^n}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^n(c+d x) (a-a \sin (c+d x))^2d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (a^2 \sin ^n(c+d x)-2 a^2 \sin ^{n+1}(c+d x)+a^2 \sin ^{n+2}(c+d x)\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^3 \sin ^{n+1}(c+d x)}{n+1}-\frac {2 a^3 \sin ^{n+2}(c+d x)}{n+2}+\frac {a^3 \sin ^{n+3}(c+d x)}{n+3}}{a^5 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]
Output:
((a^3*Sin[c + d*x]^(1 + n))/(1 + n) - (2*a^3*Sin[c + d*x]^(2 + n))/(2 + n) + (a^3*Sin[c + d*x]^(3 + n))/(3 + n))/(a^5*d)
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] && 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])
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 1.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}+\frac {\sin \left (d x +c \right )^{3} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (3+n \right )}-\frac {2 \sin \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}\) | \(91\) |
default | \(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}+\frac {\sin \left (d x +c \right )^{3} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (3+n \right )}-\frac {2 \sin \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}\) | \(91\) |
parallelrisch | \(\frac {\sin \left (d x +c \right )^{n} \left (\left (n^{2}+4 n +3\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {1}{4} n^{2}-\frac {3}{4} n -\frac {1}{2}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {7}{4} n^{2}+\frac {29}{4} n +\frac {15}{2}\right ) \sin \left (d x +c \right )-n^{2}-4 n -3\right )}{a^{2} \left (3+n \right ) \left (1+n \right ) d \left (2+n \right )}\) | \(96\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
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)))
Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.54 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (2 \, {\left (n^{2} + 4 \, n + 3\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} - {\left ({\left (n^{2} + 3 \, n + 2\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} - 8 \, n - 8\right )} \sin \left (d x + c\right ) - 8 \, n - 6\right )} \sin \left (d x + c\right )^{n}}{a^{2} d n^{3} + 6 \, a^{2} d n^{2} + 11 \, a^{2} d n + 6 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
(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)
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**n/(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} \sin \left (d x + c\right )^{3} - 2 \, {\left (n^{2} + 4 \, n + 3\right )} \sin \left (d x + c\right )^{2} + {\left (n^{2} + 5 \, n + 6\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} a^{2} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
((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)
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}}{n + 3} - \frac {2 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}}{n + 2} + \frac {\sin \left (d x + c\right )^{n + 1}}{n + 1}}{a^{2} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
(sin(d*x + c)^n*sin(d*x + c)^3/(n + 3) - 2*sin(d*x + c)^n*sin(d*x + c)^2/( n + 2) + sin(d*x + c)^(n + 1)/(n + 1))/(a^2*d)
Time = 34.77 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.15 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^n\,\left (24\,{\sin \left (c+d\,x\right )}^2-30\,\sin \left (c+d\,x\right )+2\,\sin \left (3\,c+3\,d\,x\right )\right )}{4}+\frac {n\,{\sin \left (c+d\,x\right )}^n\,\left (32\,{\sin \left (c+d\,x\right )}^2-29\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )\right )}{4}+\frac {n^2\,{\sin \left (c+d\,x\right )}^n\,\left (8\,{\sin \left (c+d\,x\right )}^2-7\,\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\right )}{4}}{a^2\,d\,\left (n^3+6\,n^2+11\,n+6\right )} \] Input:
int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^2,x)
Output:
-((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))
\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1}d x}{a^{2}} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x)
Output:
int((sin(c + d*x)**n*cos(c + d*x)**5)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1),x)/a**2