Integrand size = 21, antiderivative size = 78 \[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\cos ^2(a+b x)^{3/4} \csc ^{-1+p}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {1-p}{2},\frac {3-p}{2},\sin ^2(a+b x)\right )}{b d (1-p) (d \cos (a+b x))^{3/2}} \] Output:
(cos(b*x+a)^2)^(3/4)*csc(b*x+a)^(-1+p)*hypergeom([7/4, 1/2-1/2*p],[3/2-1/2 *p],sin(b*x+a)^2)/b/d/(1-p)/(d*cos(b*x+a))^(3/2)
Time = 10.73 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 \csc ^{-1+p}(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1+p}{2},\frac {1}{4},\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac {1}{2} (-1+p)}}{3 b d (d \cos (a+b x))^{3/2}} \] Input:
Integrate[Csc[a + b*x]^p/(d*Cos[a + b*x])^(5/2),x]
Output:
(2*Csc[a + b*x]^(-1 + p)*Hypergeometric2F1[-3/4, (1 + p)/2, 1/4, Cos[a + b *x]^2]*(Sin[a + b*x]^2)^((-1 + p)/2))/(3*b*d*(d*Cos[a + b*x])^(3/2))
Time = 0.37 (sec) , antiderivative size = 78, 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.
\(\displaystyle \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (-\sec \left (a+b x+\frac {\pi }{2}\right )\right )^p}{\left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3067 |
\(\displaystyle \sin ^p(a+b x) \csc ^p(a+b x) \int \frac {\sin ^{-p}(a+b x)}{(d \cos (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^p(a+b x) \csc ^p(a+b x) \int \frac {\sin (a+b x)^{-p}}{(d \cos (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {\cos ^2(a+b x)^{3/4} \csc ^{p-1}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {1-p}{2},\frac {3-p}{2},\sin ^2(a+b x)\right )}{b d (1-p) (d \cos (a+b x))^{3/2}}\) |
Input:
Int[Csc[a + b*x]^p/(d*Cos[a + b*x])^(5/2),x]
Output:
((Cos[a + b*x]^2)^(3/4)*Csc[a + b*x]^(-1 + p)*Hypergeometric2F1[7/4, (1 - p)/2, (3 - p)/2, Sin[a + b*x]^2])/(b*d*(1 - p)*(d*Cos[a + b*x])^(3/2))
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]
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]
\[\int \frac {\csc \left (b x +a \right )^{p}}{\left (\cos \left (b x +a \right ) d \right )^{\frac {5}{2}}}d x\]
Input:
int(csc(b*x+a)^p/(cos(b*x+a)*d)^(5/2),x)
Output:
int(csc(b*x+a)^p/(cos(b*x+a)*d)^(5/2),x)
\[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{p}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")
Output:
integral(sqrt(d*cos(b*x + a))*csc(b*x + a)^p/(d^3*cos(b*x + a)^3), x)
Timed out. \[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)**p/(d*cos(b*x+a))**(5/2),x)
Output:
Timed out
\[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{p}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate(csc(b*x + a)^p/(d*cos(b*x + a))^(5/2), x)
\[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{p}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)^p/(d*cos(b*x+a))^(5/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)^p/(d*cos(b*x + a))^(5/2), x)
Timed out. \[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^p}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \] Input:
int((1/sin(a + b*x))^p/(d*cos(a + b*x))^(5/2),x)
Output:
int((1/sin(a + b*x))^p/(d*cos(a + b*x))^(5/2), x)
\[ \int \frac {\csc ^p(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\csc \left (b x +a \right )^{p} \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right )^{3}}d x \right )}{d^{3}} \] Input:
int(csc(b*x+a)^p/(d*cos(b*x+a))^(5/2),x)
Output:
(sqrt(d)*int((csc(a + b*x)**p*sqrt(cos(a + b*x)))/cos(a + b*x)**3,x))/d**3