3.1.13 \(\int (d \sin (e+f x))^n (a-a \sin (e+f x)) (a+a \sin (e+f x))^m \, dx\) [13]

3.1.13.1 Optimal result

Integrand size = 34, antiderivative size = 114 \[ \int (d \sin (e+f x))^n (a-a \sin (e+f x)) (a+a \sin (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (1+n,-\frac {1}{2},\frac {1}{2}-m,2+n,\sin (e+f x),-\sin (e+f x)\right ) \sec (e+f x) (d \sin (e+f x))^{1+n} (1+\sin (e+f x))^{\frac {1}{2}-m} (a-a \sin (e+f x)) (a+a \sin (e+f x))^m}{d f (1+n) \sqrt {1-\sin (e+f x)}} \]

output

AppellF1(1+n,1/2-m,-1/2,2+n,-sin(f*x+e),sin(f*x+e))*sec(f*x+e)*(d*sin(f*x+ 
e))^(1+n)*(1+sin(f*x+e))^(1/2-m)*(a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m/d/f/( 
1+n)/(1-sin(f*x+e))^(1/2)
 

3.1.13.2 Mathematica [F]

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

input

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

output

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

3.1.13.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3042, 3487, 152, 152, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3487

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

input

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

output

(AppellF1[1 + n, -1/2, 1/2 - m, 2 + n, Sin[e + f*x], -Sin[e + f*x]]*Sec[e 
+ f*x]*(d*Sin[e + f*x])^(1 + n)*(1 + Sin[e + f*x])^(1/2 - m)*(a - a*Sin[e 
+ f*x])*(a + a*Sin[e + f*x])^m)/(d*f*(1 + n)*Sqrt[1 - Sin[e + f*x]])
 

3.1.13.3.1 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 152

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

rule 3042

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

rule 3487

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

3.1.13.4 Maple [F]

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

input

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

output

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

3.1.13.5 Fricas [F]

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

input

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

output

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

3.1.13.6 Sympy [F]

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

input

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

output

-a*(Integral(-(d*sin(e + f*x))**n*(a*sin(e + f*x) + a)**m, x) + Integral(( 
d*sin(e + f*x))**n*(a*sin(e + f*x) + a)**m*sin(e + f*x), x))
 

3.1.13.7 Maxima [F]

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

input

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

output

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

3.1.13.8 Giac [F]

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

input

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

output

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

3.1.13.9 Mupad [F(-1)]

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

input

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

output

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