Integrand size = 31, antiderivative size = 123 \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {B \sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m)}+\frac {2^{-\frac {1}{2}+m} (A (1-m)-B m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1-m)} \]
B*sec(f*x+e)*(a+a*sin(f*x+e))^m/f/(1-m)+2^(-1/2+m)*(A*(1-m)-B*m)*hypergeom ([-1/2, 3/2-m],[1/2],1/2-1/2*sin(f*x+e))*sec(f*x+e)*(1+sin(f*x+e))^(1/2-m) *(a+a*sin(f*x+e))^m/f/(1-m)
Time = 0.75 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94 \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {\sec (e+f x) (1+\sin (e+f x))^{-m} (a (1+\sin (e+f x)))^m \left (2^m (A (-1+m)+B m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sqrt {1+\sin (e+f x)}-\sqrt {2} B (1+\sin (e+f x))^m\right )}{\sqrt {2} f (-1+m)} \]
(Sec[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(2^m*(A*(-1 + m) + B*m)*Hypergeomet ric2F1[-1/2, 3/2 - m, 1/2, (1 - Sin[e + f*x])/2]*Sqrt[1 + Sin[e + f*x]] - Sqrt[2]*B*(1 + Sin[e + f*x])^m))/(Sqrt[2]*f*(-1 + m)*(1 + Sin[e + f*x])^m)
Time = 0.45 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 3339, 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.
\(\displaystyle \int \sec ^2(e+f x) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{\cos (e+f x)^2}dx\) |
\(\Big \downarrow \) 3339 |
\(\displaystyle \left (A-\frac {B m}{1-m}\right ) \int \sec ^2(e+f x) (\sin (e+f x) a+a)^mdx+\frac {B \sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (A-\frac {B m}{1-m}\right ) \int \frac {(\sin (e+f x) a+a)^m}{\cos (e+f x)^2}dx+\frac {B \sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m)}\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^2 \left (A-\frac {B m}{1-m}\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {(\sin (e+f x) a+a)^{m-\frac {3}{2}}}{(a-a \sin (e+f x))^{3/2}}d\sin (e+f x)}{f}+\frac {B \sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m)}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a 2^{m-\frac {3}{2}} \left (A-\frac {B m}{1-m}\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \int \frac {\left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m-\frac {3}{2}}}{(a-a \sin (e+f x))^{3/2}}d\sin (e+f x)}{f}+\frac {B \sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m)}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {2^{m-\frac {1}{2}} \left (A-\frac {B m}{1-m}\right ) \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f}+\frac {B \sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m)}\) |
(B*Sec[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 - m)) + (2^(-1/2 + m)*(A - ( B*m)/(1 - m))*Hypergeometric2F1[-1/2, 3/2 - m, 1/2, (1 - Sin[e + f*x])/2]* Sec[e + f*x]*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^m)/f
3.11.30.3.1 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && NeQ[m + p + 1, 0]
\[\int \left (\sec ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]
\[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
Timed out. \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
\[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
\[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
Timed out. \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^2} \,d x \]