\(\int \sin (e+f x) (a+b \sin (e+f x))^n \, dx\) [224]

Optimal result

Integrand size = 19, antiderivative size = 221 \[ \int \sin (e+f x) (a+b \sin (e+f x))^n \, dx=-\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^{1+n} \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-1-n}}{b f \sqrt {1+\sin (e+f x)}}+\frac {\sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n}}{b f \sqrt {1+\sin (e+f x)}} \] Output:

-2^(1/2)*AppellF1(1/2,-1-n,1/2,3/2,b*(1-sin(f*x+e))/(a+b),1/2-1/2*sin(f*x+ 
e))*cos(f*x+e)*(a+b*sin(f*x+e))^(1+n)*((a+b*sin(f*x+e))/(a+b))^(-1-n)/b/f/ 
(1+sin(f*x+e))^(1/2)+2^(1/2)*a*AppellF1(1/2,-n,1/2,3/2,b*(1-sin(f*x+e))/(a 
+b),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^n/b/f/(1+sin(f*x+e))^( 
1/2)/(((a+b*sin(f*x+e))/(a+b))^n)
 

Mathematica [A] (warning: unable to verify)

Time = 0.61 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.87 \[ \int \sin (e+f x) (a+b \sin (e+f x))^n \, dx=\frac {\sec (e+f x) \sqrt {-\frac {b (-1+\sin (e+f x))}{a+b}} \sqrt {\frac {b (1+\sin (e+f x))}{-a+b}} (a+b \sin (e+f x))^{1+n} \left (-a (2+n) \operatorname {AppellF1}\left (1+n,\frac {1}{2},\frac {1}{2},2+n,\frac {a+b \sin (e+f x)}{a-b},\frac {a+b \sin (e+f x)}{a+b}\right )+(1+n) \operatorname {AppellF1}\left (2+n,\frac {1}{2},\frac {1}{2},3+n,\frac {a+b \sin (e+f x)}{a-b},\frac {a+b \sin (e+f x)}{a+b}\right ) (a+b \sin (e+f x))\right )}{b^2 f (1+n) (2+n)} \] Input:

Integrate[Sin[e + f*x]*(a + b*Sin[e + f*x])^n,x]
 

Output:

(Sec[e + f*x]*Sqrt[-((b*(-1 + Sin[e + f*x]))/(a + b))]*Sqrt[(b*(1 + Sin[e 
+ f*x]))/(-a + b)]*(a + b*Sin[e + f*x])^(1 + n)*(-(a*(2 + n)*AppellF1[1 + 
n, 1/2, 1/2, 2 + n, (a + b*Sin[e + f*x])/(a - b), (a + b*Sin[e + f*x])/(a 
+ b)]) + (1 + n)*AppellF1[2 + n, 1/2, 1/2, 3 + n, (a + b*Sin[e + f*x])/(a 
- b), (a + b*Sin[e + f*x])/(a + b)]*(a + b*Sin[e + f*x])))/(b^2*f*(1 + n)* 
(2 + n))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3235, 3042, 3144, 156, 155}

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.

\(\displaystyle \int \sin (e+f x) (a+b \sin (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x) (a+b \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3235

\(\displaystyle \frac {\int (a+b \sin (e+f x))^{n+1}dx}{b}-\frac {a \int (a+b \sin (e+f x))^ndx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (a+b \sin (e+f x))^{n+1}dx}{b}-\frac {a \int (a+b \sin (e+f x))^ndx}{b}\)

\(\Big \downarrow \) 3144

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

\(\Big \downarrow \) 156

\(\displaystyle \frac {(a+b) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^{n+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {a \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^n}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 155

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

Input:

Int[Sin[e + f*x]*(a + b*Sin[e + f*x])^n,x]
 

Output:

-((Sqrt[2]*(a + b)*AppellF1[1/2, 1/2, -1 - n, 3/2, (1 - Sin[e + f*x])/2, ( 
b*(1 - Sin[e + f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^n)/(b*f*S 
qrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^n)) + (Sqrt[2]*a*Appe 
llF1[1/2, 1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Sin[e + f*x]))/(a + 
b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^n)/(b*f*Sqrt[1 + Sin[e + f*x]]*((a + 
 b*Sin[e + f*x])/(a + b))^n)
 

Defintions of rubi rules used

rule 155

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

rule 156

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

rule 3042

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

rule 3144

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + 
d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]])   Subst[Int[(a + b*x 
)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, 
d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]
 

rule 3235

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b   Int[(a + b*Sin[e + f*x])^m, 
 x], x] + Simp[d/b   Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, 
b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

Maple [F]

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

Input:

int(sin(f*x+e)*(a+b*sin(f*x+e))^n,x)
 

Output:

int(sin(f*x+e)*(a+b*sin(f*x+e))^n,x)
 

Fricas [F]

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

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^n,x, algorithm="fricas")
 

Output:

integral((b*sin(f*x + e) + a)^n*sin(f*x + e), x)
 

Sympy [F]

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

integrate(sin(f*x+e)*(a+b*sin(f*x+e))**n,x)
 

Output:

Integral((a + b*sin(e + f*x))**n*sin(e + f*x), x)
 

Maxima [F]

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

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^n,x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e) + a)^n*sin(f*x + e), x)
 

Giac [F]

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

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^n,x, algorithm="giac")
 

Output:

integrate((b*sin(f*x + e) + a)^n*sin(f*x + e), x)
 

Mupad [F(-1)]

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

int(sin(e + f*x)*(a + b*sin(e + f*x))^n,x)
 

Output:

int(sin(e + f*x)*(a + b*sin(e + f*x))^n, x)
 

Reduce [F]

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

int(sin(f*x+e)*(a+b*sin(f*x+e))^n,x)
 

Output:

int((sin(e + f*x)*b + a)**n*sin(e + f*x),x)