\(\int (d \sin (e+f x))^n (1+\sin (e+f x))^m \, dx\) [134]

Optimal result

Integrand size = 21, antiderivative size = 91 \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^m \, dx=-\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f \sqrt {1+\sin (e+f x)}} \] Output:

-2^(1/2+m)*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/f/(sin(f*x+e)^n)/(1+sin(f*x+e))^(1/2)
 

Mathematica [F]

\[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^m \, dx=\int (d \sin (e+f x))^n (1+\sin (e+f x))^m \, dx \] Input:

Integrate[(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^m,x]
 

Output:

Integrate[(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^m, x]
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {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.

\(\displaystyle \int (\sin (e+f x)+1)^m (d \sin (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (\sin (e+f x)+1)^m (d \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3265

\(\displaystyle \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \sin ^n(e+f x) (\sin (e+f x)+1)^mdx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \sin (e+f x)^n (\sin (e+f x)+1)^mdx\)

\(\Big \downarrow \) 3264

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

\(\Big \downarrow \) 150

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

Input:

Int[(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^m,x]
 

Output:

-((2^(1/2 + m)*AppellF1[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)/(f*Sin[e + f*x]^n*Sqrt[1 + S 
in[e + f*x]]))
 

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]
 

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (1+\sin \left (f x +e \right )\right )^{m}d x\]

Input:

int((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x)
 

Output:

int((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x)
 

Fricas [F]

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

integrate((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x, algorithm="fricas")
 

Output:

integral((d*sin(f*x + e))^n*(sin(f*x + e) + 1)^m, x)
 

Sympy [F]

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

integrate((d*sin(f*x+e))**n*(1+sin(f*x+e))**m,x)
 

Output:

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

Maxima [F]

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

integrate((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x, algorithm="maxima")
 

Output:

integrate((d*sin(f*x + e))^n*(sin(f*x + e) + 1)^m, x)
 

Giac [F]

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

integrate((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((d*sin(f*x + e))^n*(sin(f*x + e) + 1)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^m \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,{\left (\sin \left (e+f\,x\right )+1\right )}^m \,d x \] Input:

int((d*sin(e + f*x))^n*(sin(e + f*x) + 1)^m,x)
 

Output:

int((d*sin(e + f*x))^n*(sin(e + f*x) + 1)^m, x)
 

Reduce [F]

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

int((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x)
 

Output:

d**n*int(sin(e + f*x)**n*(sin(e + f*x) + 1)**m,x)