\(\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx\) [617]

Optimal result

Integrand size = 23, antiderivative size = 170 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=-\frac {e \operatorname {AppellF1}\left (2-p,\frac {1-p}{2},\frac {1-p}{2},3-p,\frac {a+b}{a+b \sin (c+d x)},\frac {a-b}{a+b \sin (c+d x)}\right ) (e \cos (c+d x))^{-1+p} \left (-\frac {b (1-\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}} \left (\frac {b (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1-p}{2}}}{b d (2-p) (a+b \sin (c+d x))} \] Output:

-e*AppellF1(2-p,1/2-1/2*p,1/2-1/2*p,3-p,(a-b)/(a+b*sin(d*x+c)),(a+b)/(a+b* 
sin(d*x+c)))*(e*cos(d*x+c))^(-1+p)*(-b*(1-sin(d*x+c))/(a+b*sin(d*x+c)))^(1 
/2-1/2*p)*(b*(1+sin(d*x+c))/(a+b*sin(d*x+c)))^(1/2-1/2*p)/b/d/(2-p)/(a+b*s 
in(d*x+c))
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4727\) vs. \(2(170)=340\).

Time = 27.08 (sec) , antiderivative size = 4727, normalized size of antiderivative = 27.81 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

((e*Cos[c + d*x])^p*Tan[c + d*x]*(b*(a^2 - b^2)*AppellF1[1, (-1 + p)/2, 2, 
 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Tan[c + d*x] + (3*a^5* 
((-2*a^2*b^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Ta 
n[c + d*x]^2])/((-3*a^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, (-1 + 
b^2/a^2)*Tan[c + d*x]^2] + (4*(a^2 - b^2)*AppellF1[3/2, p/2, 3, 5/2, -Tan[ 
c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2 
, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2) 
*(b^2*Tan[c + d*x]^2 - a^2*(1 + Tan[c + d*x]^2))^2) + ((a^2 + b^2)*AppellF 
1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])/((-3* 
a^2*AppellF1[1/2, p/2, 1, 3/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x 
]^2] + (2*(a^2 - b^2)*AppellF1[3/2, p/2, 2, 5/2, -Tan[c + d*x]^2, (-1 + b^ 
2/a^2)*Tan[c + d*x]^2] + a^2*p*AppellF1[3/2, (2 + p)/2, 1, 5/2, -Tan[c + d 
*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2])*Tan[c + d*x]^2)*(-(b^2*Tan[c + d*x] 
^2) + a^2*(1 + Tan[c + d*x]^2)))))/(1 + Tan[c + d*x]^2)^(p/2)))/(a^3*(-a^2 
 + b^2)*d*(a + b*Sin[c + d*x])^2*((Sec[c + d*x]^2*(b*(a^2 - b^2)*AppellF1[ 
1, (-1 + p)/2, 2, 2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2]*Tan[c 
 + d*x] + (3*a^5*((-2*a^2*b^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c + d*x]^2, 
(-1 + b^2/a^2)*Tan[c + d*x]^2])/((-3*a^2*AppellF1[1/2, p/2, 2, 3/2, -Tan[c 
 + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + (4*(a^2 - b^2)*AppellF1[3/2, p 
/2, 3, 5/2, -Tan[c + d*x]^2, (-1 + b^2/a^2)*Tan[c + d*x]^2] + a^2*p*App...
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3042, 3182}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3182

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

Input:

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

Output:

-((e*AppellF1[2 - p, (1 - p)/2, (1 - p)/2, 3 - p, (a + b)/(a + b*Sin[c + d 
*x]), (a - b)/(a + b*Sin[c + d*x])]*(e*Cos[c + d*x])^(-1 + p)*(-((b*(1 - S 
in[c + d*x]))/(a + b*Sin[c + d*x])))^((1 - p)/2)*((b*(1 + Sin[c + d*x]))/( 
a + b*Sin[c + d*x]))^((1 - p)/2))/(b*d*(2 - p)*(a + b*Sin[c + d*x])))
 

Defintions of rubi rules used

rule 3042

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

rule 3182

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + p)*((-b)*((1 - Sin[e + f*x])/(a + b*Sin[e + f*x])))^(( 
p - 1)/2)*(b*((1 + Sin[e + f*x])/(a + b*Sin[e + f*x])))^((p - 1)/2)))*Appel 
lF1[-p - m, (1 - p)/2, (1 - p)/2, 1 - p - m, (a + b)/(a + b*Sin[e + f*x]), 
(a - b)/(a + b*Sin[e + f*x])], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^ 
2 - b^2, 0] && ILtQ[m, 0] &&  !IGtQ[m + p + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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