\(\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx\) [1510]

Optimal result

Integrand size = 29, antiderivative size = 487 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\frac {3 \operatorname {AppellF1}\left (1+n,-p,1,2+n,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 \operatorname {AppellF1}\left (1+n,-p,1,2+n,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 \operatorname {AppellF1}\left (1+n,-p,2,2+n,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 \operatorname {AppellF1}\left (1+n,-p,2,2+n,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {\operatorname {AppellF1}\left (1+n,-p,3,2+n,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac {\operatorname {AppellF1}\left (1+n,-p,3,2+n,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)} \] Output:

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)
 

Mathematica [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx \] Input:

Integrate[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^p,x]
 

Output:

Integrate[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^p, x]
 

Rubi [A] (verified)

Time = 0.77 (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

\(\displaystyle \frac {b^5 \left (\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,1,n+2,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,1,n+2,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,2,n+2,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,2,n+2,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 b^5 (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,3,n+2,-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{8 b^5 (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,3,n+2,-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{8 b^5 (n+1)}\right )}{d}\)

Input:

Int[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^p,x]
 

Output:

(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
 

Defintions of rubi rules used

rule 615

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316

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]
 

Maple [F]

\[\int \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )^{n} \left (a +b \sin \left (d x +c \right )\right )^{p}d x\]

Input:

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^p,x)
 

Output:

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^p,x)
 

Fricas [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \] Input:

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^p,x, algorithm="frica 
s")
 

Output:

integral((b*sin(d*x + c) + a)^p*sin(d*x + c)^n*sec(d*x + c)^5, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**5*sin(d*x+c)**n*(a+b*sin(d*x+c))**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \] Input:

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^p,x, algorithm="maxim 
a")
 

Output:

integrate((b*sin(d*x + c) + a)^p*sin(d*x + c)^n*sec(d*x + c)^5, x)
 

Giac [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \] Input:

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^p,x, algorithm="giac" 
)
 

Output:

integrate((b*sin(d*x + c) + a)^p*sin(d*x + c)^n*sec(d*x + c)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^p}{{\cos \left (c+d\,x\right )}^5} \,d x \] Input:

int((sin(c + d*x)^n*(a + b*sin(c + d*x))^p)/cos(c + d*x)^5,x)
 

Output:

int((sin(c + d*x)^n*(a + b*sin(c + d*x))^p)/cos(c + d*x)^5, x)
 

Reduce [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx=\int \sin \left (d x +c \right )^{n} \left (\sin \left (d x +c \right ) b +a \right )^{p} \sec \left (d x +c \right )^{5}d x \] Input:

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^p,x)
 

Output:

int(sin(c + d*x)**n*(sin(c + d*x)*b + a)**p*sec(c + d*x)**5,x)