Integrand size = 35, antiderivative size = 336 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=-\frac {2 a^3 \left (2 B \left (115+203 n+104 n^2+16 n^3\right )+A \left (301+478 n+224 n^2+32 n^3\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f (3+2 n) (5+2 n) (7+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^3 \left (2 B \left (35+23 n+4 n^2\right )+A \left (77+50 n+8 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) (7+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (2 B (5+n)+A (7+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n) (7+2 n)}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^{3/2}}{d f (7+2 n)} \]
-2*a*B*cos(f*x+e)*(d*sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(3/2)/d/f/(7+2*n)- 2*a^3*(2*B*(16*n^3+104*n^2+203*n+115)+A*(32*n^3+224*n^2+478*n+301))*cos(f* x+e)*hypergeom([1/2, -n],[3/2],1-sin(f*x+e))*(d*sin(f*x+e))^n/f/(3+2*n)/(5 +2*n)/(7+2*n)/(sin(f*x+e)^n)/(a+a*sin(f*x+e))^(1/2)-2*a^3*(2*B*(4*n^2+23*n +35)+A*(8*n^2+50*n+77))*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(3+2*n)/(5+2*n )/(7+2*n)/(a+a*sin(f*x+e))^(1/2)-2*a^2*(2*B*(5+n)+A*(7+2*n))*cos(f*x+e)*(d *sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(1/2)/d/f/(5+2*n)/(7+2*n)
Time = 25.42 (sec) , antiderivative size = 596, normalized size of antiderivative = 1.77 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\frac {2^{1+n} \sec \left (\frac {1}{2} (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n (a (1+\sin (e+f x)))^{5/2} \tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}\right )^n \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^n \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {9}{2}+n,\frac {3+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{1+n}+\frac {A \operatorname {Hypergeometric2F1}\left (4+\frac {n}{2},\frac {9}{2}+n,5+\frac {n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^7\left (\frac {1}{2} (e+f x)\right )}{8+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(5 A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {2+n}{2},\frac {9}{2}+n,\frac {4+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{2+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(11 A+10 B) \operatorname {Hypergeometric2F1}\left (\frac {3+n}{2},\frac {9}{2}+n,\frac {5+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{3+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {5 (3 A+4 B) \operatorname {Hypergeometric2F1}\left (\frac {4+n}{2},\frac {9}{2}+n,\frac {6+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{4+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {5 (3 A+4 B) \operatorname {Hypergeometric2F1}\left (\frac {9}{2}+n,\frac {5+n}{2},\frac {7+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{5+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(11 A+10 B) \operatorname {Hypergeometric2F1}\left (\frac {9}{2}+n,\frac {6+n}{2},\frac {8+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{6+n}+\frac {(5 A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {9}{2}+n,\frac {7+n}{2},\frac {9+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{7+n}\right )\right )\right )\right )\right )\right )}{f \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
(2^(1 + n)*Sec[(e + f*x)/2]*(d*Sin[e + f*x])^n*(a*(1 + Sin[e + f*x]))^(5/2 )*Tan[(e + f*x)/2]*(Tan[(e + f*x)/2]/(1 + Tan[(e + f*x)/2]^2))^n*(1 + Tan[ (e + f*x)/2]^2)^n*((A*Hypergeometric2F1[(1 + n)/2, 9/2 + n, (3 + n)/2, -Ta n[(e + f*x)/2]^2])/(1 + n) + (A*Hypergeometric2F1[4 + n/2, 9/2 + n, 5 + n/ 2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^7)/(8 + n) + Tan[(e + f*x)/2]*((( 5*A + 2*B)*Hypergeometric2F1[(2 + n)/2, 9/2 + n, (4 + n)/2, -Tan[(e + f*x) /2]^2])/(2 + n) + Tan[(e + f*x)/2]*(((11*A + 10*B)*Hypergeometric2F1[(3 + n)/2, 9/2 + n, (5 + n)/2, -Tan[(e + f*x)/2]^2])/(3 + n) + Tan[(e + f*x)/2] *((5*(3*A + 4*B)*Hypergeometric2F1[(4 + n)/2, 9/2 + n, (6 + n)/2, -Tan[(e + f*x)/2]^2])/(4 + n) + Tan[(e + f*x)/2]*((5*(3*A + 4*B)*Hypergeometric2F1 [9/2 + n, (5 + n)/2, (7 + n)/2, -Tan[(e + f*x)/2]^2])/(5 + n) + Tan[(e + f *x)/2]*(((11*A + 10*B)*Hypergeometric2F1[9/2 + n, (6 + n)/2, (8 + n)/2, -T an[(e + f*x)/2]^2])/(6 + n) + ((5*A + 2*B)*Hypergeometric2F1[9/2 + n, (7 + n)/2, (9 + n)/2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2])/(7 + n))))))))/(f *Sqrt[Sec[(e + f*x)/2]^2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*Sin[e + f*x]^n)
Time = 1.47 (sec) , antiderivative size = 323, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3455, 27, 3042, 3455, 27, 3042, 3460, 3042, 3255, 77, 75}
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.
\(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x)) (d \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x)) (d \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3455 |
\(\displaystyle \frac {2 \int \frac {1}{2} (d \sin (e+f x))^n (\sin (e+f x) a+a)^{3/2} \left (a d \left (2 B (n+1)+2 A \left (n+\frac {7}{2}\right )\right )+a d (2 B (n+5)+A (2 n+7)) \sin (e+f x)\right )dx}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (d \sin (e+f x))^n (\sin (e+f x) a+a)^{3/2} (a d (2 B (n+1)+A (2 n+7))+a d (2 B (n+5)+A (2 n+7)) \sin (e+f x))dx}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (d \sin (e+f x))^n (\sin (e+f x) a+a)^{3/2} (a d (2 B (n+1)+A (2 n+7))+a d (2 B (n+5)+A (2 n+7)) \sin (e+f x))dx}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 3455 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{2} (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a} \left (a^2 \left (2 B \left (4 n^2+19 n+15\right )+A \left (8 n^2+42 n+49\right )\right ) d^2+a^2 \left (2 B \left (4 n^2+23 n+35\right )+A \left (8 n^2+50 n+77\right )\right ) \sin (e+f x) d^2\right )dx}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a} \left (a^2 \left (2 B \left (4 n^2+19 n+15\right )+A \left (8 n^2+42 n+49\right )\right ) d^2+a^2 \left (2 B \left (4 n^2+23 n+35\right )+A \left (8 n^2+50 n+77\right )\right ) \sin (e+f x) d^2\right )dx}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a} \left (a^2 \left (2 B \left (4 n^2+19 n+15\right )+A \left (8 n^2+42 n+49\right )\right ) d^2+a^2 \left (2 B \left (4 n^2+23 n+35\right )+A \left (8 n^2+50 n+77\right )\right ) \sin (e+f x) d^2\right )dx}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \frac {\frac {\frac {a^2 d^2 \left (A \left (32 n^3+224 n^2+478 n+301\right )+2 B \left (16 n^3+104 n^2+203 n+115\right )\right ) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx}{2 n+3}-\frac {2 a^3 d \left (A \left (8 n^2+50 n+77\right )+2 B \left (4 n^2+23 n+35\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {a^2 d^2 \left (A \left (32 n^3+224 n^2+478 n+301\right )+2 B \left (16 n^3+104 n^2+203 n+115\right )\right ) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx}{2 n+3}-\frac {2 a^3 d \left (A \left (8 n^2+50 n+77\right )+2 B \left (4 n^2+23 n+35\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 3255 |
\(\displaystyle \frac {\frac {\frac {a^4 d^2 \left (A \left (32 n^3+224 n^2+478 n+301\right )+2 B \left (16 n^3+104 n^2+203 n+115\right )\right ) \cos (e+f x) \int \frac {(d \sin (e+f x))^n}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f (2 n+3) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {2 a^3 d \left (A \left (8 n^2+50 n+77\right )+2 B \left (4 n^2+23 n+35\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 77 |
\(\displaystyle \frac {\frac {\frac {a^4 d^2 \left (A \left (32 n^3+224 n^2+478 n+301\right )+2 B \left (16 n^3+104 n^2+203 n+115\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f (2 n+3) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {2 a^3 d \left (A \left (8 n^2+50 n+77\right )+2 B \left (4 n^2+23 n+35\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {\frac {-\frac {2 a^3 d^2 \left (A \left (32 n^3+224 n^2+478 n+301\right )+2 B \left (16 n^3+104 n^2+203 n+115\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^3 d \left (A \left (8 n^2+50 n+77\right )+2 B \left (4 n^2+23 n+35\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 (A (2 n+7)+2 B (n+5)) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{f (2 n+5)}}{d (2 n+7)}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (d \sin (e+f x))^{n+1}}{d f (2 n+7)}\) |
(-2*a*B*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n)*(a + a*Sin[e + f*x])^(3/2))/ (d*f*(7 + 2*n)) + ((-2*a^2*(2*B*(5 + n) + A*(7 + 2*n))*Cos[e + f*x]*(d*Sin [e + f*x])^(1 + n)*Sqrt[a + a*Sin[e + f*x]])/(f*(5 + 2*n)) + ((-2*a^3*d^2* (2*B*(115 + 203*n + 104*n^2 + 16*n^3) + A*(301 + 478*n + 224*n^2 + 32*n^3) )*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[e + f*x]]*(d*Sin[e + f*x])^n)/(f*(3 + 2*n)*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]]) - (2*a^3* d*(2*B*(35 + 23*n + 4*n^2) + A*(77 + 50*n + 8*n^2))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(f*(3 + 2*n)*Sqrt[a + a*Sin[e + f*x]]))/(d*(5 + 2*n)))/(d *(7 + 2*n))
3.1.7.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-b)*(c/ d))^IntPart[m]*((b*x)^FracPart[m]/((-d)*(x/c))^FracPart[m]) Int[((-d)*(x/ c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(c + d*x)^n/Sqrt[a - b*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[2*n]
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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]
\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (A +B \sin \left (f x +e \right )\right )d x\]
\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
integral(-((A + 2*B)*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2 + (B*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e))^n, x)
Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]