Integrand size = 29, antiderivative size = 487 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\frac {3 \operatorname {AppellF1}\left (1+n,-p,1,2+n,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 \operatorname {AppellF1}\left (1+n,-p,1,2+n,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 \operatorname {AppellF1}\left (1+n,-p,2,2+n,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 \operatorname {AppellF1}\left (1+n,-p,2,2+n,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {\operatorname {AppellF1}\left (1+n,-p,3,2+n,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac {\operatorname {AppellF1}\left (1+n,-p,3,2+n,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)} \]
3/16*AppellF1(1+n,-p,1,2+n,-b*sin(d*x+c)/a,-sin(d*x+c))*sin(d*x+c)^(1+n)*( a+b*sin(d*x+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)+3/16*AppellF1(1+n,-p,1,2+ n,-b*sin(d*x+c)/a,sin(d*x+c))*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n)/ ((1+b*sin(d*x+c)/a)^p)+3/16*AppellF1(1+n,-p,2,2+n,-b*sin(d*x+c)/a,-sin(d*x +c))*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)+3/ 16*AppellF1(1+n,-p,2,2+n,-b*sin(d*x+c)/a,sin(d*x+c))*sin(d*x+c)^(1+n)*(a+b *sin(d*x+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)+1/8*AppellF1(1+n,-p,3,2+n,-b *sin(d*x+c)/a,-sin(d*x+c))*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n)/((1 +b*sin(d*x+c)/a)^p)+1/8*AppellF1(1+n,-p,3,2+n,-b*sin(d*x+c)/a,sin(d*x+c))* sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx \]
Time = 0.75 (sec) , antiderivative size = 494, 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.
\(\displaystyle \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^n (a+b \sin (c+d x))^p}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b^5 \int \frac {\sin ^n(c+d x) (a+b \sin (c+d x))^p}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {b^5 \int \left (\frac {3 (a+b \sin (c+d x))^p \sin ^n(c+d x)}{16 b^4 (b-b \sin (c+d x))^2}+\frac {(a+b \sin (c+d x))^p \sin ^n(c+d x)}{8 b^3 (b-b \sin (c+d x))^3}+\frac {3 (a+b \sin (c+d x))^p \sin ^n(c+d x)}{8 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {3 (a+b \sin (c+d x))^p \sin ^n(c+d x)}{16 b^4 (\sin (c+d x) b+b)^2}+\frac {(a+b \sin (c+d x))^p \sin ^n(c+d x)}{8 b^3 (\sin (c+d x) b+b)^3}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^5 \left (\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,1,n+2,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,1,n+2,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,2,n+2,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,2,n+2,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,3,n+2,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{8 b^5 (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,3,n+2,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{8 b^5 (n+1)}\right )}{d}\) |
(b^5*((3*AppellF1[1 + n, -p, 1, 2 + n, -((b*Sin[c + d*x])/a), -Sin[c + d*x ]]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^p)/(16*b^5*(1 + n)*(1 + (b*Si n[c + d*x])/a)^p) + (3*AppellF1[1 + n, -p, 1, 2 + n, -((b*Sin[c + d*x])/a) , Sin[c + d*x]]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^p)/(16*b^5*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (3*AppellF1[1 + n, -p, 2, 2 + n, -((b*Sin [c + d*x])/a), -Sin[c + d*x]]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^p) /(16*b^5*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (3*AppellF1[1 + n, -p, 2, 2 + n, -((b*Sin[c + d*x])/a), Sin[c + d*x]]*Sin[c + d*x]^(1 + n)*(a + b*Sin [c + d*x])^p)/(16*b^5*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (AppellF1[1 + n, -p, 3, 2 + n, -((b*Sin[c + d*x])/a), -Sin[c + d*x]]*Sin[c + d*x]^(1 + n )*(a + b*Sin[c + d*x])^p)/(8*b^5*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (Ap pellF1[1 + n, -p, 3, 2 + n, -((b*Sin[c + d*x])/a), Sin[c + d*x]]*Sin[c + d *x]^(1 + n)*(a + b*Sin[c + d*x])^p)/(8*b^5*(1 + n)*(1 + (b*Sin[c + d*x])/a )^p)))/d
3.16.14.3.1 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]
\[\int \left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{p}d x\]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\text {Timed out} \]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^p}{{\cos \left (c+d\,x\right )}^5} \,d x \]