\(\int \csc (e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [125]

Optimal result

Integrand size = 21, antiderivative size = 83 \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\cos ^2(e+f x),\frac {b \cos ^2(e+f x)}{a+b}\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p}}{f} \] Output:

-AppellF1(1/2,1,-p,3/2,cos(f*x+e)^2,b*cos(f*x+e)^2/(a+b))*cos(f*x+e)*(a+b- 
b*cos(f*x+e)^2)^p/f/((1-b*cos(f*x+e)^2/(a+b))^p)
 

Mathematica [F]

\[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx \] Input:

Integrate[Csc[e + f*x]*(a + b*Sin[e + f*x]^2)^p,x]
 

Output:

Integrate[Csc[e + f*x]*(a + b*Sin[e + f*x]^2)^p, x]
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3665, 334, 333}

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.

\(\displaystyle \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3665

\(\displaystyle -\frac {\int \frac {\left (-b \cos ^2(e+f x)+a+b\right )^p}{1-\cos ^2(e+f x)}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 334

\(\displaystyle -\frac {\left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \int \frac {\left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^p}{1-\cos ^2(e+f x)}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 333

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

Input:

Int[Csc[e + f*x]*(a + b*Sin[e + f*x]^2)^p,x]
 

Output:

-((AppellF1[1/2, 1, -p, 3/2, Cos[e + f*x]^2, (b*Cos[e + f*x]^2)/(a + b)]*C 
os[e + f*x]*(a + b - b*Cos[e + f*x]^2)^p)/(f*(1 - (b*Cos[e + f*x]^2)/(a + 
b))^p))
 

Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 3042

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

rule 3665

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + 
 f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [F]

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

Input:

int(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)
 

Output:

int(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)
 

Fricas [F]

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

integrate(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((-b*cos(f*x + e)^2 + a + b)^p*csc(f*x + e), x)
 

Sympy [F]

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

integrate(csc(f*x+e)*(a+b*sin(f*x+e)**2)**p,x)
 

Output:

Integral((a + b*sin(e + f*x)**2)**p*csc(e + f*x), x)
 

Maxima [F]

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

integrate(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e)^2 + a)^p*csc(f*x + e), x)
 

Giac [F]

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

integrate(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*sin(f*x + e)^2 + a)^p*csc(f*x + e), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)
 

Output:

int((sin(e + f*x)**2*b + a)**p*csc(e + f*x),x)