Integrand size = 30, antiderivative size = 101 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c f \left (3+8 m+4 m^2\right )} \] Output:
cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/f/(3+2*m)+cos(f*x+e) *(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1-m)/c/f/(4*m^2+8*m+3)
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {\sec (e+f x) (a (1+\sin (e+f x)))^{1+m} (-2 (1+m)+\sin (e+f x)) (c-c \sin (e+f x))^{-m}}{a c^2 f (1+2 m) (3+2 m) (-1+\sin (e+f x))} \] Input:
Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m),x]
Output:
(Sec[e + f*x]*(a*(1 + Sin[e + f*x]))^(1 + m)*(-2*(1 + m) + Sin[e + f*x]))/ (a*c^2*f*(1 + 2*m)*(3 + 2*m)*(-1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^m)
Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3222, 3042, 3221}
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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}dx\) |
\(\Big \downarrow \) 3222 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^{-m-1}dx}{c (2 m+3)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^{-m-1}dx}{c (2 m+3)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}\) |
\(\Big \downarrow \) 3221 |
\(\displaystyle \frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c f (2 m+1) (2 m+3)}\) |
Input:
Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m),x]
Output:
(Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(f*(3 + 2*m)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c*f*(1 + 2*m)*(3 + 2*m))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne Q[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) ) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
\[\int \left (a +\sin \left (f x +e \right ) a \right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-2-m}d x\]
Input:
int((a+sin(f*x+e)*a)^m*(c-c*sin(f*x+e))^(-2-m),x)
Output:
int((a+sin(f*x+e)*a)^m*(c-c*sin(f*x+e))^(-2-m),x)
Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.71 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {{\left (2 \, {\left (m + 1\right )} \cos \left (f x + e\right ) - \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}}{4 \, f m^{2} + 8 \, f m + 3 \, f} \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m),x, algorithm="fricas" )
Output:
(2*(m + 1)*cos(f*x + e) - cos(f*x + e)*sin(f*x + e))*(a*sin(f*x + e) + a)^ m*(-c*sin(f*x + e) + c)^(-m - 2)/(4*f*m^2 + 8*f*m + 3*f)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 2}\, dx \] Input:
integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-2-m),x)
Output:
Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 2), x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m),x, algorithm="maxima" )
Output:
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 2), x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m),x, algorithm="giac")
Output:
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 2), x)
Time = 0.53 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.10 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (\sin \left (2\,e+2\,f\,x\right )+8\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+4\,m\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )-4\right )}{c^2\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (4\,m^2+8\,m+3\right )\,\left (2\,{\sin \left (e+f\,x\right )}^2-4\,\sin \left (e+f\,x\right )+2\right )} \] Input:
int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x))^(m + 2),x)
Output:
-((a*(sin(e + f*x) + 1))^m*(sin(2*e + 2*f*x) + 8*sin(e/2 + (f*x)/2)^2 + 4* m*(2*sin(e/2 + (f*x)/2)^2 - 1) - 4))/(c^2*f*(-c*(sin(e + f*x) - 1))^m*(8*m + 4*m^2 + 3)*(2*sin(e + f*x)^2 - 4*sin(e + f*x) + 2))
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )^{2}-2 \left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )+\left (-\sin \left (f x +e \right ) c +c \right )^{m}}d x}{c^{2}} \] Input:
int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m),x)
Output:
int((sin(e + f*x)*a + a)**m/(( - sin(e + f*x)*c + c)**m*sin(e + f*x)**2 - 2*( - sin(e + f*x)*c + c)**m*sin(e + f*x) + ( - sin(e + f*x)*c + c)**m),x) /c**2