Integrand size = 29, antiderivative size = 80 \[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {(A-B) (a+a \sin (e+f x))^m}{2 f m}+\frac {(A+B) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{1+m}}{4 a f (1+m)} \]
1/2*(A-B)*(a+a*sin(f*x+e))^m/f/m+1/4*(A+B)*hypergeom([1, 1+m],[2+m],1/2+1/ 2*sin(f*x+e))*(a+a*sin(f*x+e))^(1+m)/a/f/(1+m)
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {(a (1+\sin (e+f x)))^m \left (2 (A-B) (1+m)+(A+B) m \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )}{4 f m (1+m)} \]
((a*(1 + Sin[e + f*x]))^m*(2*(A - B)*(1 + m) + (A + B)*m*Hypergeometric2F1 [1, 1 + m, 2 + m, (1 + Sin[e + f*x])/2]*(1 + Sin[e + f*x])))/(4*f*m*(1 + m ))
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 88, 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.
\(\displaystyle \int \sec (e+f x) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{\cos (e+f x)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a \int \frac {(\sin (e+f x) a+a)^{m-1} (a A+a B \sin (e+f x))}{a (a-a \sin (e+f x))}d(a \sin (e+f x))}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{m-1} (a A+a B \sin (e+f x))}{a-a \sin (e+f x)}d(a \sin (e+f x))}{f}\) |
\(\Big \downarrow \) 88 |
\(\displaystyle \frac {\frac {1}{2} (A+B) \int \frac {(\sin (e+f x) a+a)^m}{a-a \sin (e+f x)}d(a \sin (e+f x))+\frac {(A-B) (a \sin (e+f x)+a)^m}{2 m}}{f}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {\frac {(A+B) (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {\sin (e+f x) a+a}{2 a}\right )}{4 a (m+1)}+\frac {(A-B) (a \sin (e+f x)+a)^m}{2 m}}{f}\) |
(((A - B)*(a + a*Sin[e + f*x])^m)/(2*m) + ((A + B)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + a*Sin[e + f*x])/(2*a)]*(a + a*Sin[e + f*x])^(1 + m))/(4*a *(1 + m)))/f
3.11.24.3.1 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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && !RationalQ[p] && SumSimpl erQ[p, 1]
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]
\[\int \sec \left (f x +e \right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]
\[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \]
\[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right ) \sec {\left (e + f x \right )}\, dx \]
\[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \]
\[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \]
Timed out. \[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\cos \left (e+f\,x\right )} \,d x \]