\(\int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx\) [1557]

Optimal result

Integrand size = 35, antiderivative size = 330 \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=-\frac {g \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {a+b}{a+b \sin (e+f x)},\frac {a-b}{a+b \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac {b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {b (1+\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}}}{(b c-a d) f (1-p)}+\frac {g \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {c+d}{c+d \sin (e+f x)},\frac {c-d}{c+d \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac {d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {d (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}}}{(b c-a d) f (1-p)} \] Output:

-g*AppellF1(1-p,1/2-1/2*p,1/2-1/2*p,2-p,(a-b)/(a+b*sin(f*x+e)),(a+b)/(a+b* 
sin(f*x+e)))*(g*cos(f*x+e))^(-1+p)*(-b*(1-sin(f*x+e))/(a+b*sin(f*x+e)))^(1 
/2-1/2*p)*(b*(1+sin(f*x+e))/(a+b*sin(f*x+e)))^(1/2-1/2*p)/(-a*d+b*c)/f/(1- 
p)+g*AppellF1(1-p,1/2-1/2*p,1/2-1/2*p,2-p,(c-d)/(c+d*sin(f*x+e)),(c+d)/(c+ 
d*sin(f*x+e)))*(g*cos(f*x+e))^(-1+p)*(-d*(1-sin(f*x+e))/(c+d*sin(f*x+e)))^ 
(1/2-1/2*p)*(d*(1+sin(f*x+e))/(c+d*sin(f*x+e)))^(1/2-1/2*p)/(-a*d+b*c)/f/( 
1-p)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5085\) vs. \(2(330)=660\).

Time = 36.56 (sec) , antiderivative size = 5085, normalized size of antiderivative = 15.41 \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Result too large to show} \] Input:

Integrate[(g*Cos[e + f*x])^p/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])),x 
]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3042, 3403, 2009}

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.

\(\displaystyle \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))}dx\)

\(\Big \downarrow \) 3403

\(\displaystyle \int \left (\frac {b (g \cos (e+f x))^p}{(b c-a d) (a+b \sin (e+f x))}-\frac {d (g \cos (e+f x))^p}{(b c-a d) (c+d \sin (e+f x))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {g (g \cos (e+f x))^{p-1} \left (-\frac {d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {d (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {c+d}{c+d \sin (e+f x)},\frac {c-d}{c+d \sin (e+f x)}\right )}{f (1-p) (b c-a d)}-\frac {g (g \cos (e+f x))^{p-1} \left (-\frac {b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (e+f x)+1)}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {a+b}{a+b \sin (e+f x)},\frac {a-b}{a+b \sin (e+f x)}\right )}{f (1-p) (b c-a d)}\)

Input:

Int[(g*Cos[e + f*x])^p/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])),x]
 

Output:

-((g*AppellF1[1 - p, (1 - p)/2, (1 - p)/2, 2 - p, (a + b)/(a + b*Sin[e + f 
*x]), (a - b)/(a + b*Sin[e + f*x])]*(g*Cos[e + f*x])^(-1 + p)*(-((b*(1 - S 
in[e + f*x]))/(a + b*Sin[e + f*x])))^((1 - p)/2)*((b*(1 + Sin[e + f*x]))/( 
a + b*Sin[e + f*x]))^((1 - p)/2))/((b*c - a*d)*f*(1 - p))) + (g*AppellF1[1 
 - p, (1 - p)/2, (1 - p)/2, 2 - p, (c + d)/(c + d*Sin[e + f*x]), (c - d)/( 
c + d*Sin[e + f*x])]*(g*Cos[e + f*x])^(-1 + p)*(-((d*(1 - Sin[e + f*x]))/( 
c + d*Sin[e + f*x])))^((1 - p)/2)*((d*(1 + Sin[e + f*x]))/(c + d*Sin[e + f 
*x]))^((1 - p)/2))/((b*c - a*d)*f*(1 - p))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3403

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[Exp 
andTrig[(g*cos[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])^n, x 
], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Integer 
sQ[2*m, 2*n]
 

Maple [F]

\[\int \frac {\left (g \cos \left (f x +e \right )\right )^{p}}{\left (a +b \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((g*cos(f*x+e))^p/(a+b*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm= 
"fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**p/(a+b*sin(f*x+e))/(c+d*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((g*cos(f*x+e))^p/(a+b*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm= 
"maxima")
 

Output:

integrate((g*cos(f*x + e))^p/((b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)), 
x)
 

Giac [F]

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

integrate((g*cos(f*x+e))^p/(a+b*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm= 
"giac")
 

Output:

integrate((g*cos(f*x + e))^p/((b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)), 
x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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