\(\int (d \sin (e+f x))^n \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \, dx\) [9]

Optimal result

Integrand size = 35, antiderivative size = 161 \[ \int (d \sin (e+f x))^n \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \, dx=-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (2 B (1+n)+A (3+2 n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+n,2+n,\sin (e+f x)\right ) \sqrt {1-\sin (e+f x)} (d \sin (e+f x))^{1+n}}{d f (1+n) (3+2 n) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}} \] Output:

-2*a*B*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(3+2*n)/(a+a*sin(f*x+e))^(1/2)+ 
a^2*(2*B*(1+n)+A*(3+2*n))*cos(f*x+e)*hypergeom([1/2, 1+n],[2+n],sin(f*x+e) 
)*(1-sin(f*x+e))^(1/2)*(d*sin(f*x+e))^(1+n)/d/f/(1+n)/(3+2*n)/(a-a*sin(f*x 
+e))/(a+a*sin(f*x+e))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.84 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.54 \[ \int (d \sin (e+f x))^n \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) \, dx=-\frac {(1+i) 2^{-2-n} e^{-\frac {3 i e}{2}+i f n x} \left (1-e^{2 i (e+f x)}\right )^{-n} \left (-i e^{-i (e+f x)} \left (-1+e^{2 i (e+f x)}\right )\right )^n \left (\frac {2 B e^{-\frac {1}{2} i f (3+2 n) x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (-3-2 n),-n,\frac {1}{4} (1-2 n),e^{2 i (e+f x)}\right )}{f (3+2 n)}+2 e^{i e} \left (-\frac {i (2 A+B) e^{-\frac {1}{2} i f (1+2 n) x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (-1-2 n),-n,\frac {1}{4} (3-2 n),e^{2 i (e+f x)}\right )}{f+2 f n}+\frac {e^{\frac {1}{2} i (2 e+f (1-2 n) x)} \left (-\left ((2 A+B) (-3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (1-2 n),-n,\frac {1}{4} (5-2 n),e^{2 i (e+f x)}\right )\right )+i B e^{i (e+f x)} (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (3-2 n),-n,\frac {1}{4} (7-2 n),e^{2 i (e+f x)}\right )\right )}{f (-3+2 n) (-1+2 n)}\right )\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \sqrt {a (1+\sin (e+f x))}}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )} \] Input:

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

Output:

((-1 - I)*2^(-2 - n)*E^(((-3*I)/2)*e + I*f*n*x)*(((-I)*(-1 + E^((2*I)*(e + 
 f*x))))/E^(I*(e + f*x)))^n*((2*B*Hypergeometric2F1[(-3 - 2*n)/4, -n, (1 - 
 2*n)/4, E^((2*I)*(e + f*x))])/(E^((I/2)*f*(3 + 2*n)*x)*f*(3 + 2*n)) + 2*E 
^(I*e)*(((-I)*(2*A + B)*Hypergeometric2F1[(-1 - 2*n)/4, -n, (3 - 2*n)/4, E 
^((2*I)*(e + f*x))])/(E^((I/2)*f*(1 + 2*n)*x)*(f + 2*f*n)) + (E^((I/2)*(2* 
e + f*(1 - 2*n)*x))*(-((2*A + B)*(-3 + 2*n)*Hypergeometric2F1[(1 - 2*n)/4, 
 -n, (5 - 2*n)/4, E^((2*I)*(e + f*x))]) + I*B*E^(I*(e + f*x))*(-1 + 2*n)*H 
ypergeometric2F1[(3 - 2*n)/4, -n, (7 - 2*n)/4, E^((2*I)*(e + f*x))]))/(f*( 
-3 + 2*n)*(-1 + 2*n))))*(d*Sin[e + f*x])^n*Sqrt[a*(1 + Sin[e + f*x])])/((1 
 - E^((2*I)*(e + f*x)))^n*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sin[e + f* 
x]^n)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 3460, 3042, 3255, 77, 75}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

\(\displaystyle \left (A+\frac {2 B (n+1)}{2 n+3}\right ) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (A+\frac {2 B (n+1)}{2 n+3}\right ) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3255

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

\(\Big \downarrow \) 77

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

\(\Big \downarrow \) 75

\(\displaystyle -\frac {2 a \left (A+\frac {2 B (n+1)}{2 n+3}\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}\)

Input:

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

Output:

(-2*a*(A + (2*B*(1 + n))/(3 + 2*n))*Cos[e + f*x]*Hypergeometric2F1[1/2, -n 
, 3/2, 1 - Sin[e + f*x]]*(d*Sin[e + f*x])^n)/(f*Sin[e + f*x]^n*Sqrt[a + a* 
Sin[e + f*x]]) - (2*a*B*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(3 + 2 
*n)*Sqrt[a + a*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 77

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

rule 3042

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

rule 3255

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

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]
 

Maple [F]

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

Input:

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)),x)
 

Output:

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)),x)
 

Fricas [F]

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

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)),x, algo 
rithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((d*sin(f*x+e))**n*(a+a*sin(f*x+e))**(1/2)*(A+B*sin(f*x+e)),x)
 

Output:

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

Maxima [F]

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

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((d*sin(e + f*x))^n*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2),x)
 

Output:

int((d*sin(e + f*x))^n*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2), x)
 

Reduce [F]

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

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)),x)
 

Output:

d**n*sqrt(a)*(int(sin(e + f*x)**n*sqrt(sin(e + f*x) + 1)*sin(e + f*x),x)*b 
 + int(sin(e + f*x)**n*sqrt(sin(e + f*x) + 1),x)*a)