\(\int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx\) [223]

Optimal result

Integrand size = 21, antiderivative size = 275 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=-\frac {\cos (e+f x) (a+b \sin (e+f x))^{1+n}}{b f (2+n)}+\frac {\sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^{1+n} \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-1-n}}{b^2 f (2+n) \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} \left (a^2+b^2 (1+n)\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n}}{b^2 f (2+n) \sqrt {1+\sin (e+f x)}} \] Output:

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

Mathematica [F]

\[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=\int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx \] Input:

Integrate[Sin[e + f*x]^2*(a + b*Sin[e + f*x])^n,x]
 

Output:

Integrate[Sin[e + f*x]^2*(a + b*Sin[e + f*x])^n, x]
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3270, 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.

\(\displaystyle \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^2 (a+b \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3270

\(\displaystyle \frac {\int (b (n+1)-a \sin (e+f x)) (a+b \sin (e+f x))^ndx}{b (n+2)}-\frac {\cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (b (n+1)-a \sin (e+f x)) (a+b \sin (e+f x))^ndx}{b (n+2)}-\frac {\cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}\)

\(\Big \downarrow \) 3235

\(\displaystyle \frac {\frac {\left (a^2+b^2 (n+1)\right ) \int (a+b \sin (e+f x))^ndx}{b}-\frac {a \int (a+b \sin (e+f x))^{n+1}dx}{b}}{b (n+2)}-\frac {\cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (a^2+b^2 (n+1)\right ) \int (a+b \sin (e+f x))^ndx}{b}-\frac {a \int (a+b \sin (e+f x))^{n+1}dx}{b}}{b (n+2)}-\frac {\cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}\)

\(\Big \downarrow \) 3144

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

Input:

Int[Sin[e + f*x]^2*(a + b*Sin[e + f*x])^n,x]
 

Output:

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

rule 3270

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

Maple [F]

\[\int \sin \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )\right )^{n}d x\]

Input:

int(sin(f*x+e)^2*(a+b*sin(f*x+e))^n,x)
 

Output:

int(sin(f*x+e)^2*(a+b*sin(f*x+e))^n,x)
 

Fricas [F]

\[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{n} \sin \left (f x + e\right )^{2} \,d x } \] Input:

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^n,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**2*(a+b*sin(f*x+e))**n,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{n} \sin \left (f x + e\right )^{2} \,d x } \] Input:

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^n,x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e) + a)^n*sin(f*x + e)^2, x)
 

Giac [F]

\[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{n} \sin \left (f x + e\right )^{2} \,d x } \] Input:

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^n,x, algorithm="giac")
 

Output:

integrate((b*sin(f*x + e) + a)^n*sin(f*x + e)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=\int {\sin \left (e+f\,x\right )}^2\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^n \,d x \] Input:

int(sin(e + f*x)^2*(a + b*sin(e + f*x))^n,x)
 

Output:

int(sin(e + f*x)^2*(a + b*sin(e + f*x))^n, x)
 

Reduce [F]

\[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^n \, dx=\int \left (\sin \left (f x +e \right ) b +a \right )^{n} \sin \left (f x +e \right )^{2}d x \] Input:

int(sin(f*x+e)^2*(a+b*sin(f*x+e))^n,x)
                                                                                    
                                                                                    
 

Output:

int((sin(e + f*x)*b + a)**n*sin(e + f*x)**2,x)