\(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\) [201]

Optimal result

Integrand size = 36, antiderivative size = 148 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}+\frac {2^{\frac {1}{2}+m} (A (1-m)-B (2+m)) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{2+m}}{3 a^2 c^2 f (1-m)} \] Output:

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

Mathematica [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx \] Input:

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x 
])^2,x]
 

Output:

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x 
])^2, x]
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3446, 3042, 3339, 3042, 3168, 80, 79}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

\(\displaystyle \frac {\int \sec ^4(e+f x) (\sin (e+f x) a+a)^{m+2} (A+B \sin (e+f x))dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{m+2} (A+B \sin (e+f x))}{\cos (e+f x)^4}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3339

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

\(\displaystyle \frac {\frac {a^2 \left (A-\frac {B (m+2)}{1-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{3/2} \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{(a-a \sin (e+f x))^{5/2}}d\sin (e+f x)}{f}+\frac {B \sec ^3(e+f x) (a \sin (e+f x)+a)^{m+2}}{f (1-m)}}{a^2 c^2}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\frac {a^2 2^{m-\frac {1}{2}} \left (A-\frac {B (m+2)}{1-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+1} \int \frac {\left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m-\frac {1}{2}}}{(a-a \sin (e+f x))^{5/2}}d\sin (e+f x)}{f}+\frac {B \sec ^3(e+f x) (a \sin (e+f x)+a)^{m+2}}{f (1-m)}}{a^2 c^2}\)

\(\Big \downarrow \) 79

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

Input:

Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^2,x 
]
 

Output:

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

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

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

rule 3042

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

rule 3168

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

rule 3339

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* 
(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S 
imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1))   Int[(g*Cos[e + f*x])^p*(a + 
 b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ 
a^2 - b^2, 0] && NeQ[m + p + 1, 0]
 

rule 3446

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {\int \frac {A \left (a \sin {\left (e + f x \right )} + a\right )^{m}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx + \int \frac {B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx}{c^{2}} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[0,1,0]%%%} / %%%{1,[0,0,2]%%%} Error: Bad Argument Valu 
e
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) b}{c^{2}} \] Input:

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

Output:

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