Integrand size = 23, antiderivative size = 195 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx=\frac {a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {2 a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {a^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)} \]
2*a^2*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^(1 +m)/d/e/(1+m)+a^2*cos(d*x+c)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(d* x+c)^2)*(e*sin(d*x+c))^(1+m)/d/e/(1+m)/(cos(d*x+c)^2)^(1/2)+a^2*hypergeom( [3/2, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*sec(d*x+c)*(e*sin(d*x+c))^(1+m) *(cos(d*x+c)^2)^(1/2)/d/e/(1+m)
Time = 0.42 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.65 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx=\frac {a^2 (e \sin (c+d x))^m \left (2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin (c+d x)+\sqrt {\cos ^2(c+d x)} \left (\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )+\operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )\right ) \tan (c+d x)\right )}{d (1+m)} \]
(a^2*(e*Sin[c + d*x])^m*(2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[ c + d*x]^2]*Sin[c + d*x] + Sqrt[Cos[c + d*x]^2]*(Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2] + Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2])*Tan[c + d*x]))/(d*(1 + m))
Time = 0.58 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4360, 3042, 3352, 2009}
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.
| \(\displaystyle \int (a \sec (c+d x)+a)^2 (e \sin (c+d x))^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^mdx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \sec ^2(c+d x) (a (-\cos (c+d x))-a)^2 (e \sin (c+d x))^mdx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^m}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \int \left (a^2 (e \sin (c+d x))^m+a^2 \sec ^2(c+d x) (e \sin (c+d x))^m+2 a^2 \sec (c+d x) (e \sin (c+d x))^m\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^2 (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {a^2 \cos (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {a^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}\) |
(a^2*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x ]^2]*(e*Sin[c + d*x])^(1 + m))/(d*e*(1 + m)*Sqrt[Cos[c + d*x]^2]) + (2*a^2 *Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x ])^(1 + m))/(d*e*(1 + m)) + (a^2*Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[3/ 2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Sin[c + d*x])^(1 + m))/(d*e*(1 + m))
3.2.35.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
\[\int \left (a +a \sec \left (d x +c \right )\right )^{2} \left (e \sin \left (d x +c \right )\right )^{m}d x\]
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx=a^{2} \left (\int \left (e \sin {\left (c + d x \right )}\right )^{m}\, dx + \int 2 \left (e \sin {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\, dx + \int \left (e \sin {\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral((e*sin(c + d*x))**m, x) + Integral(2*(e*sin(c + d*x))**m*se c(c + d*x), x) + Integral((e*sin(c + d*x))**m*sec(c + d*x)**2, x))
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^m\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]