Integrand size = 26, antiderivative size = 83 \[ \int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (3+3 \sin (e+f x))^{1+m}}{9 c^2 f} \]
1/3*2^(1/2+m)*hypergeom([-3/2, 1/2-m],[-1/2],1/2-1/2*sin(f*x+e))*sec(f*x+e )^3*(1+sin(f*x+e))^(1/2-m)*(a+a*sin(f*x+e))^(1+m)/a/c^2/f
\[ \int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx \]
Time = 0.35 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3215, 3042, 3168, 80, 79}
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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m}{(c-c \sin (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m}{(c-c \sin (e+f x))^2}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle \frac {\int \sec ^4(e+f x) (\sin (e+f x) a+a)^{m+2}dx}{a^2 c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{m+2}}{\cos (e+f x)^4}dx}{a^2 c^2}\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {\sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{3/2} \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{(a-a \sin (e+f x))^{5/2}}d\sin (e+f x)}{c^2 f}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {2^{m-\frac {1}{2}} \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+1} \int \frac {\left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m-\frac {1}{2}}}{(a-a \sin (e+f x))^{5/2}}d\sin (e+f x)}{c^2 f}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {2^{m+\frac {1}{2}} \sec ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f}\) |
(2^(1/2 + m)*Hypergeometric2F1[-3/2, 1/2 - m, -1/2, (1 - Sin[e + f*x])/2]* Sec[e + f*x]^3*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^(1 + m))/ (3*a*c^2*f)
3.5.9.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin [e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))) Subst[Int[(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{2}}d x\]
\[ \int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
\[ \int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {\int \frac {\left (a \sin {\left (e + f x \right )} + a\right )^{m}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx}{c^{2}} \]
\[ \int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
\[ \int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]