\(\int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx\) [1034]

Optimal result

Integrand size = 38, antiderivative size = 157 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=-\frac {B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{1-p}}{2 f g}+\frac {2^{\frac {1}{2}-\frac {p}{2}} (2 A-B (1-p)) (g \cos (e+f x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+p),\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (-3+p)} (c-c \sin (e+f x))^{1-p}}{f g (1+p)} \] Output:

-1/2*B*(g*cos(f*x+e))^(p+1)*(c-c*sin(f*x+e))^(1-p)/f/g+2^(1/2-1/2*p)*(2*A- 
B*(1-p))*(g*cos(f*x+e))^(p+1)*hypergeom([-1/2+1/2*p, 1/2*p+1/2],[3/2+1/2*p 
],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(-3/2+1/2*p)*(c-c*sin(f*x+e))^(1-p)/f 
/g/(p+1)
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.96 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\frac {2^{\frac {1}{2} (-3-p)} c \cos (e+f x) (g \cos (e+f x))^p \left (-4 (2 A+B (-1+p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+p),\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1+p}{2}}+2^{\frac {1+p}{2}} B (1+p) (-1+\sin (e+f x))^2\right ) (c-c \sin (e+f x))^{-p}}{f (1+p) (-1+\sin (e+f x))} \] Input:

Integrate[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(1 
- p),x]
 

Output:

(2^((-3 - p)/2)*c*Cos[e + f*x]*(g*Cos[e + f*x])^p*(-4*(2*A + B*(-1 + p))*H 
ypergeometric2F1[(-1 + p)/2, (1 + p)/2, (3 + p)/2, (1 + Sin[e + f*x])/2]*( 
1 - Sin[e + f*x])^((1 + p)/2) + 2^((1 + p)/2)*B*(1 + p)*(-1 + Sin[e + f*x] 
)^2))/(f*(1 + p)*(-1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^p)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3339, 3042, 3168, 80, 79}

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.

\(\displaystyle \int (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^pdx\)

\(\Big \downarrow \) 3339

\(\displaystyle \frac {1}{2} (2 A-B (1-p)) \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{1-p}dx-\frac {B (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^{p+1}}{2 f g}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (2 A-B (1-p)) \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{1-p}dx-\frac {B (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^{p+1}}{2 f g}\)

\(\Big \downarrow \) 3168

\(\displaystyle \frac {c^2 (2 A-B (1-p)) (c-c \sin (e+f x))^{\frac {1}{2} (-p-1)} (c \sin (e+f x)+c)^{\frac {1}{2} (-p-1)} (g \cos (e+f x))^{p+1} \int (c-c \sin (e+f x))^{\frac {1-p}{2}} (\sin (e+f x) c+c)^{\frac {p-1}{2}}d\sin (e+f x)}{2 f g}-\frac {B (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^{p+1}}{2 f g}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {c^2 2^{-\frac {p}{2}-\frac {1}{2}} (2 A-B (1-p)) (1-\sin (e+f x))^{\frac {p-1}{2}} (c-c \sin (e+f x))^{\frac {1}{2} (-p-1)+\frac {1-p}{2}} (c \sin (e+f x)+c)^{\frac {1}{2} (-p-1)} (g \cos (e+f x))^{p+1} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1-p}{2}} (\sin (e+f x) c+c)^{\frac {p-1}{2}}d\sin (e+f x)}{f g}-\frac {B (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^{p+1}}{2 f g}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {c 2^{\frac {1}{2}-\frac {p}{2}} (2 A-B (1-p)) (1-\sin (e+f x))^{\frac {p-1}{2}} (c-c \sin (e+f x))^{\frac {1}{2} (-p-1)+\frac {1-p}{2}} (c \sin (e+f x)+c)^{\frac {1}{2} (-p-1)+\frac {p+1}{2}} (g \cos (e+f x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {p-1}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (p+1)}-\frac {B (c-c \sin (e+f x))^{1-p} (g \cos (e+f x))^{p+1}}{2 f g}\)

Input:

Int[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(1 - p),x 
]
 

Output:

-1/2*(B*(g*Cos[e + f*x])^(1 + p)*(c - c*Sin[e + f*x])^(1 - p))/(f*g) + (2^ 
(1/2 - p/2)*c*(2*A - B*(1 - p))*(g*Cos[e + f*x])^(1 + p)*Hypergeometric2F1 
[(-1 + p)/2, (1 + p)/2, (3 + p)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x] 
)^((-1 + p)/2)*(c - c*Sin[e + f*x])^((-1 - p)/2 + (1 - p)/2)*(c + c*Sin[e 
+ f*x])^((-1 - p)/2 + (1 + p)/2))/(f*g*(1 + p))
 

Defintions of rubi rules used

rule 79

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

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

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

rule 3168

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin 
[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)))   Subst[Int[(a + 
b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre 
eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

rule 3339

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

Maple [F]

\[\int \left (g \cos \left (f x +e \right )\right )^{p} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{1-p}d x\]

Input:

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1-p),x)
 

Output:

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1-p),x)
 

Fricas [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 1} \,d x } \] Input:

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1-p),x, algo 
rithm="fricas")
 

Output:

integral((B*sin(f*x + e) + A)*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(-p 
 + 1), x)
 

Sympy [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{1 - p} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \] Input:

integrate((g*cos(f*x+e))**p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1-p),x)
 

Output:

Integral((g*cos(e + f*x))**p*(-c*(sin(e + f*x) - 1))**(1 - p)*(A + B*sin(e 
 + f*x)), x)
 

Maxima [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 1} \,d x } \] Input:

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1-p),x, algo 
rithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(- 
p + 1), x)
 

Giac [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 1} \,d x } \] Input:

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1-p),x, algo 
rithm="giac")
 

Output:

integrate((B*sin(f*x + e) + A)*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(- 
p + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{1-p} \,d x \] Input:

int((g*cos(e + f*x))^p*(A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(1 - p),x 
)
 

Output:

int((g*cos(e + f*x))^p*(A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(1 - p), 
x)
 

Reduce [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{1-p} \, dx=g^{p} c \left (\left (\int \frac {\cos \left (f x +e \right )^{p}}{\left (-\sin \left (f x +e \right ) c +c \right )^{p}}d x \right ) a -\left (\int \frac {\cos \left (f x +e \right )^{p} \sin \left (f x +e \right )^{2}}{\left (-\sin \left (f x +e \right ) c +c \right )^{p}}d x \right ) b -\left (\int \frac {\cos \left (f x +e \right )^{p} \sin \left (f x +e \right )}{\left (-\sin \left (f x +e \right ) c +c \right )^{p}}d x \right ) a +\left (\int \frac {\cos \left (f x +e \right )^{p} \sin \left (f x +e \right )}{\left (-\sin \left (f x +e \right ) c +c \right )^{p}}d x \right ) b \right ) \] Input:

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1-p),x)
 

Output:

g**p*c*(int(cos(e + f*x)**p/( - sin(e + f*x)*c + c)**p,x)*a - int((cos(e + 
 f*x)**p*sin(e + f*x)**2)/( - sin(e + f*x)*c + c)**p,x)*b - int((cos(e + f 
*x)**p*sin(e + f*x))/( - sin(e + f*x)*c + c)**p,x)*a + int((cos(e + f*x)** 
p*sin(e + f*x))/( - sin(e + f*x)*c + c)**p,x)*b)