\(\int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx\) [1558]

Optimal result

Integrand size = 35, antiderivative size = 508 \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=-\frac {b g \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {a+b}{a+b \sin (e+f x)},\frac {a-b}{a+b \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac {b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {b (1+\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}}}{(b c-a d)^2 f (1-p)}+\frac {b g \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {c+d}{c+d \sin (e+f x)},\frac {c-d}{c+d \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac {d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {d (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}}}{(b c-a d)^2 f (1-p)}+\frac {g \operatorname {AppellF1}\left (2-p,\frac {1-p}{2},\frac {1-p}{2},3-p,\frac {c+d}{c+d \sin (e+f x)},\frac {c-d}{c+d \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac {d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {d (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}}}{(b c-a d) f (2-p) (c+d \sin (e+f x))} \] Output:

-b*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)^2/f 
/(1-p)+b*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)^2/f/(1-p)+g*AppellF1(2-p,1/2-1/2*p,1/2-1/2*p,3-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/(2-p)/(c+d*sin(f*x+e))
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(12568\) vs. \(2(508)=1016\).

Time = 58.88 (sec) , antiderivative size = 12568, normalized size of antiderivative = 24.74 \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\text {Result too large to show} \] Input:

Integrate[(g*Cos[e + f*x])^p/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^2) 
,x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 508, 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.

\(\displaystyle \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2}dx\)

\(\Big \downarrow \) 3403

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

\(\Big \downarrow \) 2009

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

Input:

Int[(g*Cos[e + f*x])^p/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^2),x]
 

Output:

-((b*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 - 
 Sin[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)^2*f*(1 - p))) + (b*g*Appe 
llF1[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)^2*f*(1 - p)) + (g*AppellF1[2 - p, (1 
 - p)/2, (1 - p)/2, 3 - p, (c + d)/(c + d*Sin[e + f*x]), (c - d)/(c + d*Si 
n[e + f*x])]*(g*Cos[e + f*x])^(-1 + p)*(-((d*(1 - Sin[e + f*x]))/(c + d*Si 
n[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*(2 - p)*(c + d*Sin[e + f*x]))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((g*cos(f*x+e))^p/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorith 
m="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

integrate((g*cos(f*x+e))^p/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((g*cos(f*x+e))^p/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=g^{p} \left (\int \frac {\cos \left (f x +e \right )^{p}}{\sin \left (f x +e \right )^{3} b \,d^{2}+\sin \left (f x +e \right )^{2} a \,d^{2}+2 \sin \left (f x +e \right )^{2} b c d +2 \sin \left (f x +e \right ) a c d +\sin \left (f x +e \right ) b \,c^{2}+a \,c^{2}}d x \right ) \] Input:

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

Output:

g**p*int(cos(e + f*x)**p/(sin(e + f*x)**3*b*d**2 + sin(e + f*x)**2*a*d**2 
+ 2*sin(e + f*x)**2*b*c*d + 2*sin(e + f*x)*a*c*d + sin(e + f*x)*b*c**2 + a 
*c**2),x)