Integrand size = 21, antiderivative size = 58 \[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{3},\frac {2}{3},\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}} \] Output:
-3/2*cos(f*x+e)*hypergeom([-1/2, -1/3],[2/3],sin(f*x+e)^2)/b/f/(cos(f*x+e) ^2)^(1/2)/(b*sin(f*x+e))^(2/3)
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{3},\frac {2}{3},\sin ^2(e+f x)\right ) \tan (e+f x)}{2 f (b \sin (e+f x))^{5/3}} \] Input:
Integrate[Cos[e + f*x]^2/(b*Sin[e + f*x])^(5/3),x]
Output:
(-3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[-1/2, -1/3, 2/3, Sin[e + f*x]^2 ]*Tan[e + f*x])/(2*f*(b*Sin[e + f*x])^(5/3))
Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {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.
\(\displaystyle \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)^2}{(b \sin (e+f x))^{5/3}}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle -\frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{3},\frac {2}{3},\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}}\) |
Input:
Int[Cos[e + f*x]^2/(b*Sin[e + f*x])^(5/3),x]
Output:
(-3*Cos[e + f*x]*Hypergeometric2F1[-1/2, -1/3, 2/3, Sin[e + f*x]^2])/(2*b* f*Sqrt[Cos[e + f*x]^2]*(b*Sin[e + f*x])^(2/3))
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]
\[\int \frac {\cos \left (f x +e \right )^{2}}{\left (b \sin \left (f x +e \right )\right )^{\frac {5}{3}}}d x\]
Input:
int(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x)
Output:
int(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x)
\[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="fricas")
Output:
integral(-(b*sin(f*x + e))^(1/3)*cos(f*x + e)^2/(b^2*cos(f*x + e)^2 - b^2) , x)
\[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int \frac {\cos ^{2}{\left (e + f x \right )}}{\left (b \sin {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx \] Input:
integrate(cos(f*x+e)**2/(b*sin(f*x+e))**(5/3),x)
Output:
Integral(cos(e + f*x)**2/(b*sin(e + f*x))**(5/3), x)
\[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="maxima")
Output:
integrate(cos(f*x + e)^2/(b*sin(f*x + e))^(5/3), x)
\[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="giac")
Output:
integrate(cos(f*x + e)^2/(b*sin(f*x + e))^(5/3), x)
Timed out. \[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2}{{\left (b\,\sin \left (e+f\,x\right )\right )}^{5/3}} \,d x \] Input:
int(cos(e + f*x)^2/(b*sin(e + f*x))^(5/3),x)
Output:
int(cos(e + f*x)^2/(b*sin(e + f*x))^(5/3), x)
\[ \int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\frac {\int \frac {\cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{\frac {5}{3}}}d x}{b^{\frac {5}{3}}} \] Input:
int(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x)
Output:
int(cos(e + f*x)**2/(sin(e + f*x)**(2/3)*sin(e + f*x)),x)/(b**(2/3)*b)