Integrand size = 21, antiderivative size = 58 \[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{2},\frac {4}{3},\sin ^2(e+f x)\right ) \sec (e+f x) (b \sin (e+f x))^{2/3}}{2 b f} \] Output:
3/2*(cos(f*x+e)^2)^(1/2)*hypergeom([1/3, 5/2],[4/3],sin(f*x+e)^2)*sec(f*x+ e)*(b*sin(f*x+e))^(2/3)/b/f
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{2},\frac {4}{3},\sin ^2(e+f x)\right ) \tan (e+f x)}{2 f \sqrt [3]{b \sin (e+f x)}} \] Input:
Integrate[Sec[e + f*x]^4/(b*Sin[e + f*x])^(1/3),x]
Output:
(3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[1/3, 5/2, 4/3, Sin[e + f*x]^2]*T an[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.
\(\displaystyle \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (e+f x)^4 \sqrt [3]{b \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {3 \sqrt {\cos ^2(e+f x)} \sec (e+f x) (b \sin (e+f x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{2},\frac {4}{3},\sin ^2(e+f x)\right )}{2 b f}\) |
Input:
Int[Sec[e + f*x]^4/(b*Sin[e + f*x])^(1/3),x]
Output:
(3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[1/3, 5/2, 4/3, Sin[e + f*x]^2]*S ec[e + f*x]*(b*Sin[e + f*x])^(2/3))/(2*b*f)
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]
\[\int \frac {\sec \left (f x +e \right )^{4}}{\left (b \sin \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]
Input:
int(sec(f*x+e)^4/(b*sin(f*x+e))^(1/3),x)
Output:
int(sec(f*x+e)^4/(b*sin(f*x+e))^(1/3),x)
\[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(sec(f*x+e)^4/(b*sin(f*x+e))^(1/3),x, algorithm="fricas")
Output:
integral((b*sin(f*x + e))^(2/3)*sec(f*x + e)^4/(b*sin(f*x + e)), x)
\[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int \frac {\sec ^{4}{\left (e + f x \right )}}{\sqrt [3]{b \sin {\left (e + f x \right )}}}\, dx \] Input:
integrate(sec(f*x+e)**4/(b*sin(f*x+e))**(1/3),x)
Output:
Integral(sec(e + f*x)**4/(b*sin(e + f*x))**(1/3), x)
\[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(sec(f*x+e)^4/(b*sin(f*x+e))^(1/3),x, algorithm="maxima")
Output:
integrate(sec(f*x + e)^4/(b*sin(f*x + e))^(1/3), x)
\[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(sec(f*x+e)^4/(b*sin(f*x+e))^(1/3),x, algorithm="giac")
Output:
integrate(sec(f*x + e)^4/(b*sin(f*x + e))^(1/3), x)
Timed out. \[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^4\,{\left (b\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \] Input:
int(1/(cos(e + f*x)^4*(b*sin(e + f*x))^(1/3)),x)
Output:
int(1/(cos(e + f*x)^4*(b*sin(e + f*x))^(1/3)), x)
\[ \int \frac {\sec ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {\int \frac {\sec \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{\frac {1}{3}}}d x}{b^{\frac {1}{3}}} \] Input:
int(sec(f*x+e)^4/(b*sin(f*x+e))^(1/3),x)
Output:
int(sec(e + f*x)**4/sin(e + f*x)**(1/3),x)/b**(1/3)