3.1.12 \(\int (d \sin (e+f x))^n (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [12]

3.1.12.1 Optimal result

Integrand size = 33, antiderivative size = 221 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {2^{\frac {3}{2}+m} B \operatorname {AppellF1}\left (\frac {1}{2},-n,-\frac {1}{2}-m,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f}-\frac {2^{\frac {1}{2}+m} (A-B) \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f} \]

output

-2^(3/2+m)*B*AppellF1(1/2,-n,-1/2-m,3/2,1-sin(f*x+e),1/2-1/2*sin(f*x+e))*c 
os(f*x+e)*(d*sin(f*x+e))^n*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^m/f/(s 
in(f*x+e)^n)-2^(1/2+m)*(A-B)*AppellF1(1/2,-n,1/2-m,3/2,1-sin(f*x+e),1/2-1/ 
2*sin(f*x+e))*cos(f*x+e)*(d*sin(f*x+e))^n*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin 
(f*x+e))^m/f/(sin(f*x+e)^n)
 

3.1.12.2 Mathematica [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int (d \sin (e+f x))^n (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx \]

input

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

output

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

3.1.12.3 Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3466, 3042, 3266, 3042, 3265, 3042, 3264, 150}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (d \sin (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (d \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3466

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3266

\(\displaystyle (A-B) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (d \sin (e+f x))^n (\sin (e+f x)+1)^mdx+B (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (d \sin (e+f x))^n (\sin (e+f x)+1)^{m+1}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (d \sin (e+f x))^n (\sin (e+f x)+1)^mdx+B (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (d \sin (e+f x))^n (\sin (e+f x)+1)^{m+1}dx\)

\(\Big \downarrow \) 3265

\(\displaystyle (A-B) (\sin (e+f x)+1)^{-m} \sin ^{-n}(e+f x) (a \sin (e+f x)+a)^m (d \sin (e+f x))^n \int \sin ^n(e+f x) (\sin (e+f x)+1)^mdx+B (\sin (e+f x)+1)^{-m} \sin ^{-n}(e+f x) (a \sin (e+f x)+a)^m (d \sin (e+f x))^n \int \sin ^n(e+f x) (\sin (e+f x)+1)^{m+1}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) (\sin (e+f x)+1)^{-m} \sin ^{-n}(e+f x) (a \sin (e+f x)+a)^m (d \sin (e+f x))^n \int \sin (e+f x)^n (\sin (e+f x)+1)^mdx+B (\sin (e+f x)+1)^{-m} \sin ^{-n}(e+f x) (a \sin (e+f x)+a)^m (d \sin (e+f x))^n \int \sin (e+f x)^n (\sin (e+f x)+1)^{m+1}dx\)

\(\Big \downarrow \) 3264

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

\(\Big \downarrow \) 150

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

input

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

output

-((2^(3/2 + m)*B*AppellF1[1/2, -n, -1/2 - m, 3/2, 1 - Sin[e + f*x], (1 - S 
in[e + f*x])/2]*Cos[e + f*x]*(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^(-1/2 - 
 m)*(a + a*Sin[e + f*x])^m)/(f*Sin[e + f*x]^n)) - (2^(1/2 + m)*(A - B)*App 
ellF1[1/2, -n, 1/2 - m, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e 
 + f*x]*(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f* 
x])^m)/(f*Sin[e + f*x]^n)
 

3.1.12.3.1 Defintions of rubi rules used

rule 150

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

rule 3042

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

rule 3264

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

rule 3265

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_), x_Symbol] :> Simp[(d/b)^IntPart[n]*((d*Sin[e + f*x])^FracPart[n 
]/(b*Sin[e + f*x])^FracPart[n])   Int[(a + b*Sin[e + f*x])^m*(b*Sin[e + f*x 
])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !I 
ntegerQ[m] && GtQ[a, 0] &&  !GtQ[d/b, 0]
 

rule 3266

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

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]
 

3.1.12.4 Maple [F]

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

input

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

output

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

3.1.12.5 Fricas [F]

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

input

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

output

integral((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^n, x 
)
 

3.1.12.6 Sympy [F]

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

input

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

output

Integral((a*(sin(e + f*x) + 1))**m*(d*sin(e + f*x))**n*(A + B*sin(e + f*x) 
), x)
 

3.1.12.7 Maxima [F]

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

input

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

output

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

3.1.12.8 Giac [F]

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

input

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

output

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

3.1.12.9 Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]

input

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

output

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