Integrand size = 29, antiderivative size = 360 \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (3 a^2-9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^3 d (1+n)}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^3 d (1+n)}-\frac {b^6 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a \left (a^2-b^2\right )^3 d (1+n)}+\frac {(3 a-5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^2 d (1+n)}+\frac {(3 a+5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^2 d (1+n)}+\frac {\operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a-b) d (1+n)}+\frac {\operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a+b) d (1+n)} \]
1/16*(3*a^2-9*a*b+8*b^2)*hypergeom([1, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^ (1+n)/(a-b)^3/d/(1+n)+1/16*(3*a^2+9*a*b+8*b^2)*hypergeom([1, 1+n],[2+n],si n(d*x+c))*sin(d*x+c)^(1+n)/(a+b)^3/d/(1+n)-b^6*hypergeom([1, 1+n],[2+n],-b *sin(d*x+c)/a)*sin(d*x+c)^(1+n)/a/(a^2-b^2)^3/d/(1+n)+1/16*(3*a-5*b)*hyper geom([2, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^(1+n)/(a-b)^2/d/(1+n)+1/16*(3* a+5*b)*hypergeom([2, 1+n],[2+n],sin(d*x+c))*sin(d*x+c)^(1+n)/(a+b)^2/d/(1+ n)+1/8*hypergeom([3, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^(1+n)/(a-b)/d/(1+n )+1/8*hypergeom([3, 1+n],[2+n],sin(d*x+c))*sin(d*x+c)^(1+n)/(a+b)/d/(1+n)
Time = 0.31 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (\frac {\left (3 a^2-9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x))}{(a-b)^3}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,\sin (c+d x))}{(a+b)^3}-\frac {16 b^6 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{a (a-b)^3 (a+b)^3}+\frac {(3 a-5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x))}{(a-b)^2}+\frac {(3 a+5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x))}{(a+b)^2}+\frac {2 \operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x))}{a-b}+\frac {2 \operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x))}{a+b}\right ) \sin ^{1+n}(c+d x)}{16 d (1+n)} \]
((((3*a^2 - 9*a*b + 8*b^2)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x ]])/(a - b)^3 + ((3*a^2 + 9*a*b + 8*b^2)*Hypergeometric2F1[1, 1 + n, 2 + n , Sin[c + d*x]])/(a + b)^3 - (16*b^6*Hypergeometric2F1[1, 1 + n, 2 + n, -( (b*Sin[c + d*x])/a)])/(a*(a - b)^3*(a + b)^3) + ((3*a - 5*b)*Hypergeometri c2F1[2, 1 + n, 2 + n, -Sin[c + d*x]])/(a - b)^2 + ((3*a + 5*b)*Hypergeomet ric2F1[2, 1 + n, 2 + n, Sin[c + d*x]])/(a + b)^2 + (2*Hypergeometric2F1[3, 1 + n, 2 + n, -Sin[c + d*x]])/(a - b) + (2*Hypergeometric2F1[3, 1 + n, 2 + n, Sin[c + d*x]])/(a + b))*Sin[c + d*x]^(1 + n))/(16*d*(1 + n))
Time = 0.74 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 3316, 615, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^n}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^n(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^5 \int \left (\frac {\left (3 a^2+9 b a+8 b^2\right ) \sin ^n(c+d x)}{16 b^5 (a+b)^3 (b-b \sin (c+d x))}-\frac {\sin ^n(c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+\frac {\left (3 a^2-9 b a+8 b^2\right ) \sin ^n(c+d x)}{16 (a-b)^3 b^5 (\sin (c+d x) b+b)}+\frac {(3 a+5 b) \sin ^n(c+d x)}{16 b^4 (a+b)^2 (b-b \sin (c+d x))^2}+\frac {(3 a-5 b) \sin ^n(c+d x)}{16 (a-b)^2 b^4 (\sin (c+d x) b+b)^2}+\frac {\sin ^n(c+d x)}{8 b^3 (a+b) (b-b \sin (c+d x))^3}-\frac {\sin ^n(c+d x)}{8 b^3 (b-a) (\sin (c+d x) b+b)^3}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^5 \left (-\frac {b \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {b \sin (c+d x)}{a}\right )}{a (n+1) \left (a^2-b^2\right )^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,-\sin (c+d x))}{16 b^5 (n+1) (a-b)^3}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,\sin (c+d x))}{16 b^5 (n+1) (a+b)^3}+\frac {(3 a-5 b) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(2,n+1,n+2,-\sin (c+d x))}{16 b^5 (n+1) (a-b)^2}+\frac {(3 a+5 b) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(2,n+1,n+2,\sin (c+d x))}{16 b^5 (n+1) (a+b)^2}+\frac {\sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(3,n+1,n+2,-\sin (c+d x))}{8 b^5 (n+1) (a-b)}+\frac {\sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(3,n+1,n+2,\sin (c+d x))}{8 b^5 (n+1) (a+b)}\right )}{d}\) |
(b^5*(((3*a^2 - 9*a*b + 8*b^2)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(16*(a - b)^3*b^5*(1 + n)) + ((3*a^2 + 9*a*b + 8*b^2)*Hypergeometric2F1[1, 1 + n, 2 + n, Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(16*b^5*(a + b)^3*(1 + n)) - (b*Hypergeometric2F1[1, 1 + n, 2 + n, - ((b*Sin[c + d*x])/a)]*Sin[c + d*x]^(1 + n))/(a*(a^2 - b^2)^3*(1 + n)) + (( 3*a - 5*b)*Hypergeometric2F1[2, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^ (1 + n))/(16*(a - b)^2*b^5*(1 + n)) + ((3*a + 5*b)*Hypergeometric2F1[2, 1 + n, 2 + n, Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(16*b^5*(a + b)^2*(1 + n)) + (Hypergeometric2F1[3, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n) )/(8*(a - b)*b^5*(1 + n)) + (Hypergeometric2F1[3, 1 + n, 2 + n, Sin[c + d* x]]*Sin[c + d*x]^(1 + n))/(8*b^5*(a + b)*(1 + n))))/d
3.16.13.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
\[\int \frac {\left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]
\[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n}{{\cos \left (c+d\,x\right )}^5\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \,d x \]