\(\int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1021]

Optimal result

Integrand size = 31, antiderivative size = 104 \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a (A (2-m)-B m) \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-1+m}}{4 f (1-m)}+\frac {a^3 (A+B) (a+a \sin (e+f x))^{-1+m}}{2 f \left (a^2-a^2 \sin (e+f x)\right )} \] Output:

-1/4*a*(A*(2-m)-B*m)*hypergeom([1, -1+m],[m],1/2+1/2*sin(f*x+e))*(a+a*sin( 
f*x+e))^(-1+m)/f/(1-m)+1/2*a^3*(A+B)*(a+a*sin(f*x+e))^(-1+m)/f/(a^2-a^2*si 
n(f*x+e))
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79 \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a \left (2 (A+B) (-1+m)+(A (-2+m)+B m) \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {1}{2} (1+\sin (e+f x))\right ) (-1+\sin (e+f x))\right ) (a (1+\sin (e+f x)))^{-1+m}}{4 f (-1+m) (-1+\sin (e+f x))} \] Input:

Integrate[Sec[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01, 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.

\(\displaystyle \int \sec ^3(e+f x) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{\cos (e+f x)^3}dx\)

\(\Big \downarrow \) 3315

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 78

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

Input:

Int[Sec[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
 

Output:

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

\[\int \sec \left (f x +e \right )^{3} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]

Input:

int(sec(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
 

Output:

int(sec(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
 

Fricas [F]

\[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="f 
ricas")
 

Output:

integral((B*sec(f*x + e)^3*sin(f*x + e) + A*sec(f*x + e)^3)*(a*sin(f*x + e 
) + a)^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**3*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(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)^3, x)
 

Giac [F]

\[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(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)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^3} \,d x \] Input:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^3,x)
 

Output:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^3, x)
 

Reduce [F]

\[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sec \left (f x +e \right )^{3} \sin \left (f x +e \right )d x \right ) b +\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sec \left (f x +e \right )^{3}d x \right ) a \] Input:

int(sec(f*x+e)^3*(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)**3*sin(e + f*x),x)*b + int((sin(e 
 + f*x)*a + a)**m*sec(e + f*x)**3,x)*a