Integrand size = 21, antiderivative size = 58 \[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{3},\frac {4}{3},\sin ^2(e+f x)\right ) (b \sin (e+f x))^{2/3}}{2 b f \sqrt {\cos ^2(e+f x)}} \] Output:
3/2*cos(f*x+e)*hypergeom([-1/2, 1/3],[4/3],sin(f*x+e)^2)*(b*sin(f*x+e))^(2 /3)/b/f/(cos(f*x+e)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{3},\frac {4}{3},\sin ^2(e+f x)\right ) \tan (e+f x)}{2 f \sqrt [3]{b \sin (e+f x)}} \] Input:
Integrate[Cos[e + f*x]^2/(b*Sin[e + f*x])^(1/3),x]
Output:
(3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[-1/2, 1/3, 4/3, Sin[e + f*x]^2]* Tan[e + f*x])/(2*f*(b*Sin[e + f*x])^(1/3))
Time = 0.21 (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.
\(\displaystyle \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)^2}{\sqrt [3]{b \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {3 \cos (e+f x) (b \sin (e+f x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{3},\frac {4}{3},\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)}}\) |
Input:
Int[Cos[e + f*x]^2/(b*Sin[e + f*x])^(1/3),x]
Output:
(3*Cos[e + f*x]*Hypergeometric2F1[-1/2, 1/3, 4/3, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(2/3))/(2*b*f*Sqrt[Cos[e + f*x]^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]
\[\int \frac {\cos \left (f x +e \right )^{2}}{\left (b \sin \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]
Input:
int(cos(f*x+e)^2/(b*sin(f*x+e))^(1/3),x)
Output:
int(cos(f*x+e)^2/(b*sin(f*x+e))^(1/3),x)
\[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(1/3),x, algorithm="fricas")
Output:
integral((b*sin(f*x + e))^(2/3)*cos(f*x + e)^2/(b*sin(f*x + e)), x)
\[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int \frac {\cos ^{2}{\left (e + f x \right )}}{\sqrt [3]{b \sin {\left (e + f x \right )}}}\, dx \] Input:
integrate(cos(f*x+e)**2/(b*sin(f*x+e))**(1/3),x)
Output:
Integral(cos(e + f*x)**2/(b*sin(e + f*x))**(1/3), x)
\[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(1/3),x, algorithm="maxima")
Output:
integrate(cos(f*x + e)^2/(b*sin(f*x + e))^(1/3), x)
\[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(1/3),x, algorithm="giac")
Output:
integrate(cos(f*x + e)^2/(b*sin(f*x + e))^(1/3), x)
Timed out. \[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2}{{\left (b\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \] Input:
int(cos(e + f*x)^2/(b*sin(e + f*x))^(1/3),x)
Output:
int(cos(e + f*x)^2/(b*sin(e + f*x))^(1/3), x)
\[ \int \frac {\cos ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {\int \frac {\cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{\frac {1}{3}}}d x}{b^{\frac {1}{3}}} \] Input:
int(cos(f*x+e)^2/(b*sin(f*x+e))^(1/3),x)
Output:
int(cos(e + f*x)**2/sin(e + f*x)**(1/3),x)/b**(1/3)