\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx\) [331]

Optimal result

Integrand size = 35, antiderivative size = 223 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=-\frac {B \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{\sqrt {2} a^2 f \sqrt {1+\sin (e+f x)}}-\frac {(A-B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{2 \sqrt {2} a^2 f \sqrt {1+\sin (e+f x)}} \] Output:

-1/2*B*AppellF1(1/2,-n,3/2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))* 
cos(f*x+e)*(c+d*sin(f*x+e))^n*2^(1/2)/a^2/f/(1+sin(f*x+e))^(1/2)/(((c+d*si 
n(f*x+e))/(c+d))^n)-1/4*(A-B)*AppellF1(1/2,-n,5/2,3/2,d*(1-sin(f*x+e))/(c+ 
d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^n*2^(1/2)/a^2/f/(1+sin( 
f*x+e))^(1/2)/(((c+d*sin(f*x+e))/(c+d))^n)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 3466, 3042, 3263, 156, 155}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3466

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3263

\(\displaystyle \frac {(A-B) \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{\sqrt {1-\sin (e+f x)} (\sin (e+f x)+1)^{5/2}}d\sin (e+f x)}{a^2 f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}+\frac {B \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{\sqrt {1-\sin (e+f x)} (\sin (e+f x)+1)^{3/2}}d\sin (e+f x)}{a^2 f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle -\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{2 \sqrt {2} a^2 f \sqrt {\sin (e+f x)+1}}-\frac {B \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{\sqrt {2} a^2 f \sqrt {\sin (e+f x)+1}}\)

Input:

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

Output:

-((B*AppellF1[1/2, 3/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x 
]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(Sqrt[2]*a^2*f*Sqrt[1 + 
Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)) - ((A - B)*AppellF1[1/2, 
5/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e 
+ f*x]*(c + d*Sin[e + f*x])^n)/(2*Sqrt[2]*a^2*f*Sqrt[1 + Sin[e + f*x]]*((c 
 + d*Sin[e + f*x])/(c + d))^n)
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

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

rule 3263

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

rule 3466

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*b - a*B)/b   Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x 
] + Simp[B/b   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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