Integrand size = 40, antiderivative size = 81 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {(g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (1,-m+n,1-m+n,\frac {1}{2} (1-\sin (e+f x))\right ) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{2 f g (m-n)} \]
1/2*hypergeom([1, -m+n],[1-m+n],1/2-1/2*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c- c*sin(f*x+e))^n/f/g/(m-n)/((g*cos(f*x+e))^(2*m))
\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \]
Time = 0.47 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3332, 27, 3042, 3146, 78}
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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-2 m-1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-2 m-1}dx\) |
\(\Big \downarrow \) 3332 |
\(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {\sec (e+f x) (c-c \sin (e+f x))^{n-m}}{g}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \sec (e+f x) (c-c \sin (e+f x))^{n-m}dx}{g}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {(c-c \sin (e+f x))^{n-m}}{\cos (e+f x)}dx}{g}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle -\frac {c (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {(c-c \sin (e+f x))^{-m+n-1}}{\sin (e+f x) c+c}d(-c \sin (e+f x))}{f g}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-2 m} \operatorname {Hypergeometric2F1}\left (1,n-m,-m+n+1,\frac {c-c \sin (e+f x)}{2 c}\right )}{2 f g (m-n)}\) |
(Hypergeometric2F1[1, -m + n, 1 - m + n, (c - c*Sin[e + f*x])/(2*c)]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n)/(2*f*g*(m - n)*(g*Cos[e + f*x])^ (2*m))
3.2.78.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^ IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f* x])^FracPart[m]/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[m]))) Int[ (g*Cos[e + f*x])^(2*m + p)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a , b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
\[\int \left (g \cos \left (f x +e \right )\right )^{-1-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]
\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="fricas")
Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\text {Timed out} \]
\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="maxima")
\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+1}} \,d x \]