\(\int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1022]

Optimal result

Integrand size = 31, antiderivative size = 104 \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a^2 (A (4-m)-B m) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-2+m}}{16 f (2-m)}+\frac {a^4 (A+B) (a+a \sin (e+f x))^{-2+m}}{4 f (a-a \sin (e+f x))^2} \] Output:

-1/16*a^2*(A*(4-m)-B*m)*hypergeom([2, -2+m],[-1+m],1/2+1/2*sin(f*x+e))*(a+ 
a*sin(f*x+e))^(-2+m)/f/(2-m)+1/4*a^4*(A+B)*(a+a*sin(f*x+e))^(-2+m)/f/(a-a* 
sin(f*x+e))^2
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.73 \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {a^2 \left (-\frac {(A (-4+m)+B m) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,\frac {1}{2} (1+\sin (e+f x))\right )}{-2+m}+\frac {4 (A+B)}{(-1+\sin (e+f x))^2}\right ) (a (1+\sin (e+f x)))^{-2+m}}{16 f} \] Input:

Integrate[Sec[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
 

Output:

(a^2*(-(((A*(-4 + m) + B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (1 + Sin[ 
e + f*x])/2])/(-2 + m)) + (4*(A + B))/(-1 + Sin[e + f*x])^2)*(a*(1 + Sin[e 
 + f*x]))^(-2 + m))/(16*f)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 87, 78}

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.

\(\displaystyle \int \sec ^5(e+f x) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{\cos (e+f x)^5}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {a^5 \int \frac {(\sin (e+f x) a+a)^{m-3} (a A+a B \sin (e+f x))}{a (a-a \sin (e+f x))^3}d(a \sin (e+f x))}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^4 \int \frac {(\sin (e+f x) a+a)^{m-3} (a A+a B \sin (e+f x))}{(a-a \sin (e+f x))^3}d(a \sin (e+f x))}{f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {a^4 \left (\frac {1}{4} (A (4-m)-B m) \int \frac {(\sin (e+f x) a+a)^{m-3}}{(a-a \sin (e+f x))^2}d(a \sin (e+f x))+\frac {(A+B) (a \sin (e+f x)+a)^{m-2}}{4 (a-a \sin (e+f x))^2}\right )}{f}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {a^4 \left (\frac {(A+B) (a \sin (e+f x)+a)^{m-2}}{4 (a-a \sin (e+f x))^2}-\frac {(A (4-m)-B m) (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (2,m-2,m-1,\frac {\sin (e+f x) a+a}{2 a}\right )}{16 a^2 (2-m)}\right )}{f}\)

Input:

Int[Sec[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
 

Output:

(a^4*(-1/16*((A*(4 - m) - B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (a + a 
*Sin[e + f*x])/(2*a)]*(a + a*Sin[e + f*x])^(-2 + m))/(a^2*(2 - m)) + ((A + 
 B)*(a + a*Sin[e + f*x])^(-2 + m))/(4*(a - a*Sin[e + f*x])^2)))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 3042

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

rule 3315

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, 
 x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege 
rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
 

Maple [F]

\[\int \sec \left (f x +e \right )^{5} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]

Input:

int(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
 

Output:

int(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
 

Fricas [F]

\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \] Input:

integrate(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="f 
ricas")
 

Output:

integral((B*sec(f*x + e)^5*sin(f*x + e) + A*sec(f*x + e)^5)*(a*sin(f*x + e 
) + a)^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**5*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \] Input:

integrate(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="m 
axima")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*sec(f*x + e)^5, x)
 

Giac [F]

\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \] Input:

integrate(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="g 
iac")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*sec(f*x + e)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^5} \,d x \] Input:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^5,x)
 

Output:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^5, x)
 

Reduce [F]

\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sec \left (f x +e \right )^{5} \sin \left (f x +e \right )d x \right ) b +\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sec \left (f x +e \right )^{5}d x \right ) a \] Input:

int(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
 

Output:

int((sin(e + f*x)*a + a)**m*sec(e + f*x)**5*sin(e + f*x),x)*b + int((sin(e 
 + f*x)*a + a)**m*sec(e + f*x)**5,x)*a