\(\int \frac {\csc ^p(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\) [281]

Optimal result

Integrand size = 21, antiderivative size = 76 \[ \int \frac {\csc ^p(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {d \cos ^2(a+b x)^{3/4} \csc ^{-1+p}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {1-p}{2},\frac {3-p}{2},\sin ^2(a+b x)\right )}{b (1-p) (d \cos (a+b x))^{3/2}} \] Output:

d*(cos(b*x+a)^2)^(3/4)*csc(b*x+a)^(-1+p)*hypergeom([3/4, 1/2-1/2*p],[3/2-1 
/2*p],sin(b*x+a)^2)/b/(1-p)/(d*cos(b*x+a))^(3/2)
 

Mathematica [A] (verified)

Time = 10.58 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^p(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {2 \sqrt {d \cos (a+b x)} \csc ^{1+p}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1+p}{2},\frac {5}{4},\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac {1+p}{2}}}{b d} \] Input:

Integrate[Csc[a + b*x]^p/Sqrt[d*Cos[a + b*x]],x]
 

Output:

(-2*Sqrt[d*Cos[a + b*x]]*Csc[a + b*x]^(1 + p)*Hypergeometric2F1[1/4, (1 + 
p)/2, 5/4, Cos[a + b*x]^2]*(Sin[a + b*x]^2)^((1 + p)/2))/(b*d)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3067, 3042, 3057}

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.

\(\displaystyle \int \frac {\csc ^p(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-\sec \left (a+b x+\frac {\pi }{2}\right )\right )^p}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3067

\(\displaystyle \sin ^p(a+b x) \csc ^p(a+b x) \int \frac {\sin ^{-p}(a+b x)}{\sqrt {d \cos (a+b x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sin ^p(a+b x) \csc ^p(a+b x) \int \frac {\sin (a+b x)^{-p}}{\sqrt {d \cos (a+b x)}}dx\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {d \cos ^2(a+b x)^{3/4} \csc ^{p-1}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {1-p}{2},\frac {3-p}{2},\sin ^2(a+b x)\right )}{b (1-p) (d \cos (a+b x))^{3/2}}\)

Input:

Int[Csc[a + b*x]^p/Sqrt[d*Cos[a + b*x]],x]
 

Output:

(d*(Cos[a + b*x]^2)^(3/4)*Csc[a + b*x]^(-1 + p)*Hypergeometric2F1[3/4, (1 
- p)/2, (3 - p)/2, Sin[a + b*x]^2])/(b*(1 - p)*(d*Cos[a + b*x])^(3/2))
 

Defintions of rubi rules used

rule 3042

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

rule 3057

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac 
Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr 
acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ 
e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
 

rule 3067

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[b^2*(b*Cos[e + f*x])^(n - 1)*(b*Sec[e + f*x])^(n - 1) 
   Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, 
m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
 

Maple [F]

\[\int \frac {\csc \left (b x +a \right )^{p}}{\sqrt {\cos \left (b x +a \right ) d}}d x\]

Input:

int(csc(b*x+a)^p/(cos(b*x+a)*d)^(1/2),x)
 

Output:

int(csc(b*x+a)^p/(cos(b*x+a)*d)^(1/2),x)
 

Fricas [F]

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

integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(d*cos(b*x + a))*csc(b*x + a)^p/(d*cos(b*x + a)), x)
 

Sympy [F]

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

integrate(csc(b*x+a)**p/(d*cos(b*x+a))**(1/2),x)
 

Output:

Integral(csc(a + b*x)**p/sqrt(d*cos(a + b*x)), x)
 

Maxima [F]

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

integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

integrate(csc(b*x + a)^p/sqrt(d*cos(b*x + a)), x)
 

Giac [F]

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

integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

integrate(csc(b*x + a)^p/sqrt(d*cos(b*x + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^p(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int \frac {{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^p}{\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \] Input:

int((1/sin(a + b*x))^p/(d*cos(a + b*x))^(1/2),x)
 

Output:

int((1/sin(a + b*x))^p/(d*cos(a + b*x))^(1/2), x)
 

Reduce [F]

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

int(csc(b*x+a)^p/(d*cos(b*x+a))^(1/2),x)
 

Output:

(sqrt(d)*int((csc(a + b*x)**p*sqrt(cos(a + b*x)))/cos(a + b*x),x))/d