Integrand size = 23, antiderivative size = 236 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=-\frac {4 e^5 (e \sin (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {7 e^3 (e \sin (c+d x))^{-3+m}}{a^3 d (3-m)}-\frac {3 e (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)} \]
-4*e^5*(e*sin(d*x+c))^(-5+m)/a^3/d/(5-m)+7*e^3*(e*sin(d*x+c))^(-3+m)/a^3/d /(3-m)-3*e*(e*sin(d*x+c))^(-1+m)/a^3/d/(1-m)+e^5*cos(d*x+c)*hypergeom([-5/ 2, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^(-5+m)/a^3/d/(5-m )/(cos(d*x+c)^2)^(1/2)+3*e^5*cos(d*x+c)*hypergeom([-3/2, -5/2+1/2*m],[-3/2 +1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^(-5+m)/a^3/d/(5-m)/(cos(d*x+c)^2)^(1/ 2)
Result contains complex when optimal does not.
Time = 6.09 (sec) , antiderivative size = 995, normalized size of antiderivative = 4.22 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=-\frac {i 2^{4-m} \left (-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )^m \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\left (-1+e^{2 i (c+d x)}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )}{2 m}+\frac {3 e^{i (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (\left (6-5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},2-m,\frac {3-m}{2},e^{2 i (c+d x)}\right )+e^{i (c+d x)} (-1+m) \left (e^{i (c+d x)} (-2+m) \operatorname {Hypergeometric2F1}\left (2-m,\frac {3-m}{2},\frac {5-m}{2},e^{2 i (c+d x)}\right )-2 (-3+m) \operatorname {Hypergeometric2F1}\left (2-m,1-\frac {m}{2},2-\frac {m}{2},e^{2 i (c+d x)}\right )\right )\right )}{(-3+m) (-2+m) (-1+m)}+6 e^{2 i (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (\frac {4 e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},4-m,\frac {5-m}{2},e^{2 i (c+d x)}\right )}{-3+m}+\frac {4 e^{3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (4-m,\frac {5-m}{2},\frac {7-m}{2},e^{2 i (c+d x)}\right )}{-5+m}-\frac {\operatorname {Hypergeometric2F1}\left (4-m,1-\frac {m}{2},2-\frac {m}{2},e^{2 i (c+d x)}\right )}{-2+m}-\frac {6 e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (4-m,2-\frac {m}{2},3-\frac {m}{2},e^{2 i (c+d x)}\right )}{-4+m}-\frac {e^{4 i (c+d x)} \operatorname {Hypergeometric2F1}\left (4-m,3-\frac {m}{2},4-\frac {m}{2},e^{2 i (c+d x)}\right )}{-6+m}\right )-4 e^{3 i (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (-\frac {\operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},6-m,\frac {5-m}{2},e^{2 i (c+d x)}\right )}{-3+m}-\frac {15 e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {5-m}{2},6-m,\frac {7-m}{2},e^{2 i (c+d x)}\right )}{-5+m}-\frac {15 e^{4 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,\frac {7-m}{2},\frac {9-m}{2},e^{2 i (c+d x)}\right )}{-7+m}-\frac {e^{6 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,\frac {9-m}{2},\frac {11-m}{2},e^{2 i (c+d x)}\right )}{-9+m}+\frac {6 e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,2-\frac {m}{2},3-\frac {m}{2},e^{2 i (c+d x)}\right )}{-4+m}+\frac {20 e^{3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,3-\frac {m}{2},4-\frac {m}{2},e^{2 i (c+d x)}\right )}{-6+m}+\frac {6 e^{5 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,4-\frac {m}{2},5-\frac {m}{2},e^{2 i (c+d x)}\right )}{-8+m}\right )\right ) \sec ^3(c+d x) \sin ^{-m}(c+d x) (e \sin (c+d x))^m}{a^3 d (1+\sec (c+d x))^3} \]
((-I)*2^(4 - m)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/E^(I*(c + d*x)))^m*Cos[ (c + d*x)/2]^6*(((-1 + E^((2*I)*(c + d*x)))*Hypergeometric2F1[1, (2 + m)/2 , 1 - m/2, E^((2*I)*(c + d*x))])/(2*m) + (3*E^(I*(c + d*x))*((6 - 5*m + m^ 2)*Hypergeometric2F1[(1 - m)/2, 2 - m, (3 - m)/2, E^((2*I)*(c + d*x))] + E ^(I*(c + d*x))*(-1 + m)*(E^(I*(c + d*x))*(-2 + m)*Hypergeometric2F1[2 - m, (3 - m)/2, (5 - m)/2, E^((2*I)*(c + d*x))] - 2*(-3 + m)*Hypergeometric2F1 [2 - m, 1 - m/2, 2 - m/2, E^((2*I)*(c + d*x))])))/((1 - E^((2*I)*(c + d*x) ))^m*(-3 + m)*(-2 + m)*(-1 + m)) + (6*E^((2*I)*(c + d*x))*((4*E^(I*(c + d* x))*Hypergeometric2F1[(3 - m)/2, 4 - m, (5 - m)/2, E^((2*I)*(c + d*x))])/( -3 + m) + (4*E^((3*I)*(c + d*x))*Hypergeometric2F1[4 - m, (5 - m)/2, (7 - m)/2, E^((2*I)*(c + d*x))])/(-5 + m) - Hypergeometric2F1[4 - m, 1 - m/2, 2 - m/2, E^((2*I)*(c + d*x))]/(-2 + m) - (6*E^((2*I)*(c + d*x))*Hypergeomet ric2F1[4 - m, 2 - m/2, 3 - m/2, E^((2*I)*(c + d*x))])/(-4 + m) - (E^((4*I) *(c + d*x))*Hypergeometric2F1[4 - m, 3 - m/2, 4 - m/2, E^((2*I)*(c + d*x)) ])/(-6 + m)))/(1 - E^((2*I)*(c + d*x)))^m - (4*E^((3*I)*(c + d*x))*(-(Hype rgeometric2F1[(3 - m)/2, 6 - m, (5 - m)/2, E^((2*I)*(c + d*x))]/(-3 + m)) - (15*E^((2*I)*(c + d*x))*Hypergeometric2F1[(5 - m)/2, 6 - m, (7 - m)/2, E ^((2*I)*(c + d*x))])/(-5 + m) - (15*E^((4*I)*(c + d*x))*Hypergeometric2F1[ 6 - m, (7 - m)/2, (9 - m)/2, E^((2*I)*(c + d*x))])/(-7 + m) - (E^((6*I)*(c + d*x))*Hypergeometric2F1[6 - m, (9 - m)/2, (11 - m)/2, E^((2*I)*(c + ...
Time = 0.92 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4360, 25, 25, 3042, 3354, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^m}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^m}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cos ^3(c+d x) (e \sin (c+d x))^m}{(a (-\cos (c+d x))-a)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cos ^3(c+d x) (e \sin (c+d x))^m}{(\cos (c+d x) a+a)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cos ^3(c+d x) (e \sin (c+d x))^m}{(a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^m}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {e^6 \int \cos ^3(c+d x) (a-a \cos (c+d x))^3 (e \sin (c+d x))^{m-6}dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^6 \int \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{m-6} \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3dx}{a^6}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {e^6 \int \left (-a^3 \cos ^6(c+d x) (e \sin (c+d x))^{m-6}+3 a^3 \cos ^5(c+d x) (e \sin (c+d x))^{m-6}-3 a^3 \cos ^4(c+d x) (e \sin (c+d x))^{m-6}+a^3 \cos ^3(c+d x) (e \sin (c+d x))^{m-6}\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^6 \left (-\frac {3 a^3 (e \sin (c+d x))^{m-1}}{d e^5 (1-m)}+\frac {7 a^3 (e \sin (c+d x))^{m-3}}{d e^3 (3-m)}+\frac {a^3 \cos (c+d x) (e \sin (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {m-5}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{d e (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^3 \cos (c+d x) (e \sin (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m-5}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{d e (5-m) \sqrt {\cos ^2(c+d x)}}-\frac {4 a^3 (e \sin (c+d x))^{m-5}}{d e (5-m)}\right )}{a^6}\) |
(e^6*((-4*a^3*(e*Sin[c + d*x])^(-5 + m))/(d*e*(5 - m)) + (a^3*Cos[c + d*x] *Hypergeometric2F1[-5/2, (-5 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(d*e*(5 - m)*Sqrt[Cos[c + d*x]^2]) + (3*a^3*Cos[c + d*x] *Hypergeometric2F1[-3/2, (-5 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(d*e*(5 - m)*Sqrt[Cos[c + d*x]^2]) + (7*a^3*(e*Sin[c + d *x])^(-3 + m))/(d*e^3*(3 - m)) - (3*a^3*(e*Sin[c + d*x])^(-1 + m))/(d*e^5* (1 - m))))/a^6
3.2.39.3.1 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_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[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]
\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}d x\]
\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
integral((e*sin(d*x + c))^m/(a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3 *a^3*sec(d*x + c) + a^3), x)
\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Integral((e*sin(c + d*x))**m/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec( c + d*x) + 1), x)/a**3
\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \]