Integrand size = 27, antiderivative size = 38 \[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\operatorname {Hypergeometric2F1}(4,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^4 d (1+n)} \] Output:
hypergeom([4, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^(1+n)/a^4/d/(1+n)
Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\operatorname {Hypergeometric2F1}(4,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^4 d (1+n)} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^4,x]
Output:
(Hypergeometric2F1[4, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/( a^4*d*(1 + n))
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3312, 74}
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.
| \(\displaystyle \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x) \sin (c+d x)^n}{(a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {\int \frac {\sin ^n(c+d x)}{(\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 74 |
| \(\displaystyle \frac {\sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(4,n+1,n+2,-\sin (c+d x))}{a^4 d (n+1)}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^4,x]
Output:
(Hypergeometric2F1[4, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/( a^4*d*(1 + n))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
\[\int \frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )^{n}}{\left (a +a \sin \left (d x +c \right )\right )^{4}}d x\]
Input:
int(cos(d*x+c)*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)
Output:
int(cos(d*x+c)*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)
\[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
integral(sin(d*x + c)^n*cos(d*x + c)/(a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*cos(d*x + c)^2 - 2*a^4)*sin(d*x + c)), x)
Timed out. \[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**n/(a+a*sin(d*x+c))**4,x)
Output:
Timed out
\[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
integrate(sin(d*x + c)^n*cos(d*x + c)/(a*sin(d*x + c) + a)^4, x)
\[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
integrate(sin(d*x + c)^n*cos(d*x + c)/(a*sin(d*x + c) + a)^4, x)
Timed out. \[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \] Input:
int((cos(c + d*x)*sin(c + d*x)^n)/(a + a*sin(c + d*x))^4,x)
Output:
int((cos(c + d*x)*sin(c + d*x)^n)/(a + a*sin(c + d*x))^4, x)
\[ \int \frac {\cos (c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)
Output:
(sin(c + d*x)**n - int((sin(c + d*x)**n*cos(c + d*x))/(sin(c + d*x)**5*n - 3*sin(c + d*x)**5 + 4*sin(c + d*x)**4*n - 12*sin(c + d*x)**4 + 6*sin(c + d*x)**3*n - 18*sin(c + d*x)**3 + 4*sin(c + d*x)**2*n - 12*sin(c + d*x)**2 + sin(c + d*x)*n - 3*sin(c + d*x)),x)*sin(c + d*x)**3*d*n**2 + 3*int((sin( c + d*x)**n*cos(c + d*x))/(sin(c + d*x)**5*n - 3*sin(c + d*x)**5 + 4*sin(c + d*x)**4*n - 12*sin(c + d*x)**4 + 6*sin(c + d*x)**3*n - 18*sin(c + d*x)* *3 + 4*sin(c + d*x)**2*n - 12*sin(c + d*x)**2 + sin(c + d*x)*n - 3*sin(c + d*x)),x)*sin(c + d*x)**3*d*n - 3*int((sin(c + d*x)**n*cos(c + d*x))/(sin( c + d*x)**5*n - 3*sin(c + d*x)**5 + 4*sin(c + d*x)**4*n - 12*sin(c + d*x)* *4 + 6*sin(c + d*x)**3*n - 18*sin(c + d*x)**3 + 4*sin(c + d*x)**2*n - 12*s in(c + d*x)**2 + sin(c + d*x)*n - 3*sin(c + d*x)),x)*sin(c + d*x)**2*d*n** 2 + 9*int((sin(c + d*x)**n*cos(c + d*x))/(sin(c + d*x)**5*n - 3*sin(c + d* x)**5 + 4*sin(c + d*x)**4*n - 12*sin(c + d*x)**4 + 6*sin(c + d*x)**3*n - 1 8*sin(c + d*x)**3 + 4*sin(c + d*x)**2*n - 12*sin(c + d*x)**2 + sin(c + d*x )*n - 3*sin(c + d*x)),x)*sin(c + d*x)**2*d*n - 3*int((sin(c + d*x)**n*cos( c + d*x))/(sin(c + d*x)**5*n - 3*sin(c + d*x)**5 + 4*sin(c + d*x)**4*n - 1 2*sin(c + d*x)**4 + 6*sin(c + d*x)**3*n - 18*sin(c + d*x)**3 + 4*sin(c + d *x)**2*n - 12*sin(c + d*x)**2 + sin(c + d*x)*n - 3*sin(c + d*x)),x)*sin(c + d*x)*d*n**2 + 9*int((sin(c + d*x)**n*cos(c + d*x))/(sin(c + d*x)**5*n - 3*sin(c + d*x)**5 + 4*sin(c + d*x)**4*n - 12*sin(c + d*x)**4 + 6*sin(c ...