\(\int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx\) [5]

Optimal result

Integrand size = 33, antiderivative size = 279 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=-\frac {n (A-2 A n+2 B (1+n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+n) \sqrt {\cos ^2(e+f x)}}+\frac {(1+n) (B+2 A (1-n)+2 B n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{3 a^2 d^2 f (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2} \] Output:

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

Mathematica [A] (verified)

Time = 4.36 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.79 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\frac {(d \sin (e+f x))^n \left ((A-B) \sin (2 (e+f x))-\frac {2 (1+\sin (e+f x)) \left ((1+n) (2+n) (2 A (-1+n)-B (1+2 n)) \cos ^2(e+f x)+\sqrt {\cos ^2(e+f x)} (1+\sin (e+f x)) \left (n (2+n) (A-2 A n+2 B (1+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right )+(1+n)^2 (2 A (-1+n)-B (1+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right ) \tan (e+f x)}{(1+n) (2+n)}\right )}{6 a^2 f (1+\sin (e+f x))^2} \] Input:

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

Output:

((d*Sin[e + f*x])^n*((A - B)*Sin[2*(e + f*x)] - (2*(1 + Sin[e + f*x])*((1 
+ n)*(2 + n)*(2*A*(-1 + n) - B*(1 + 2*n))*Cos[e + f*x]^2 + Sqrt[Cos[e + f* 
x]^2]*(1 + Sin[e + f*x])*(n*(2 + n)*(A - 2*A*n + 2*B*(1 + n))*Hypergeometr 
ic2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2] + (1 + n)^2*(2*A*(-1 + n) 
 - B*(1 + 2*n))*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^ 
2]*Sin[e + f*x]))*Tan[e + f*x])/((1 + n)*(2 + n))))/(6*a^2*f*(1 + Sin[e + 
f*x])^2)
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3457, 3042, 3457, 25, 3042, 3227, 3042, 3122}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {(2 A (1-n)+B (2 n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (\sin (e+f x)+1)}-\frac {\frac {a^2 d n (-2 A n+A+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{f (n+1) \sqrt {\cos ^2(e+f x)}}-\frac {a^2 (n+1) (2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(e+f x)\right )}{f (n+2) \sqrt {\cos ^2(e+f x)}}}{a^2 d}}{3 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 d f (a \sin (e+f x)+a)^2}\)

Input:

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

Output:

((A - B)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(3*d*f*(a + a*Sin[e + f*x] 
)^2) + (((2*A*(1 - n) + B*(1 + 2*n))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n) 
)/(f*(1 + Sin[e + f*x])) - ((a^2*d*n*(A - 2*A*n + 2*B*(1 + n))*Cos[e + f*x 
]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2]*(d*Sin[e + 
f*x])^(1 + n))/(f*(1 + n)*Sqrt[Cos[e + f*x]^2]) - (a^2*(1 + n)*(B + 2*A*(1 
 - n) + 2*B*n)*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, S 
in[e + f*x]^2]*(d*Sin[e + f*x])^(2 + n))/(f*(2 + n)*Sqrt[Cos[e + f*x]^2])) 
/(a^2*d))/(3*a^2*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3457

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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