\(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [208]

Optimal result

Integrand size = 38, antiderivative size = 118 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a (1+\sin (e+f x)))^m \left (2 A (3+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )+B \left (-6-4 m+(1+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+m,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )\right )}{2 f (1+2 m) (3+2 m) \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/Sqrt[c - c*Sin[e + 
 f*x]],x]
 

Output:

(Cos[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(2*A*(3 + 2*m)*Hypergeometric2F1[1, 
 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2] + B*(-6 - 4*m + (1 + 2*m)*Hyperge 
ometric2F1[1, 3/2 + m, 5/2 + m, (1 + Sin[e + f*x])/2]*(1 + Sin[e + f*x]))) 
)/(2*f*(1 + 2*m)*(3 + 2*m)*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3042, 3452, 3042, 3224, 3042, 3146, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3452

\(\displaystyle (A+B) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A+B) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}\)

\(\Big \downarrow \) 3224

\(\displaystyle \frac {(A+B) \cos (e+f x) \int \sec (e+f x) (\sin (e+f x) a+a)^{m+\frac {1}{2}}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+B) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\cos (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}\)

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 78

\(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {\sin (e+f x) a+a}{2 a}\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}\)

Input:

Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/Sqrt[c - c*Sin[e + f*x]] 
,x]
 

Output:

(-2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*Sin[e + 
 f*x]]) + ((A + B)*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (a 
+ a*Sin[e + f*x])/(2*a)]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*S 
in[e + f*x]])
 

Defintions of rubi rules used

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 3042

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

rule 3146

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x 
)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

rule 3224

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e 
+ f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*FracP 
art[m]))   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F 
reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
 && (FractionQ[m] ||  !FractionQ[n])
 

rule 3452

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1))   Int[( 
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, 
e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, 
 -2^(-1)] && NeQ[m + n + 1, 0]
 

Maple [F]

\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\sqrt {c -c \sin \left (f x +e \right )}}d x\]

Input:

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)
 

Output:

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)
 

Fricas [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="fricas")
 

Output:

integral(-(B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + 
 a)^m/(c*sin(f*x + e) - c), x)
 

Sympy [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right )}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \] Input:

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x))/sqrt(-c*(sin(e + f 
*x) - 1)), x)
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/sqrt(-c*sin(f*x + e) 
 + c), x)
 

Giac [F(-1)]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(1/ 
2),x)
 

Output:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(1/ 
2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {c}\, \left (\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )-1}d x \right ) b +\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )-1}d x \right ) a \right )}{c} \] Input:

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)
 

Output:

( - sqrt(c)*(int(((sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + 1)*sin(e 
+ f*x))/(sin(e + f*x) - 1),x)*b + int(((sin(e + f*x)*a + a)**m*sqrt( - sin 
(e + f*x) + 1))/(sin(e + f*x) - 1),x)*a))/c