Integrand size = 19, antiderivative size = 53 \[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\frac {3 \operatorname {Hypergeometric2F1}\left (-\frac {13}{6},-\frac {3}{2},-\frac {7}{6},\cos ^2(e+f x)\right ) \sec ^{\frac {13}{3}}(e+f x) \sin (e+f x)}{13 f \sqrt {\sin ^2(e+f x)}} \] Output:
3/13*hypergeom([-13/6, -3/2],[-7/6],cos(f*x+e)^2)*sec(f*x+e)^(13/3)*sin(f* x+e)/f/(sin(f*x+e)^2)^(1/2)
Time = 0.91 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.68 \[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\frac {3 \sqrt [3]{\sec (e+f x)} \left (27 \sin (e+f x)-18 \sqrt [6]{\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\sin ^2(e+f x)\right ) \sin (e+f x)+\sec (e+f x) \left (-16+7 \sec ^2(e+f x)\right ) \tan (e+f x)\right )}{91 f} \] Input:
Integrate[Sec[e + f*x]^(16/3)*Sin[e + f*x]^4,x]
Output:
(3*Sec[e + f*x]^(1/3)*(27*Sin[e + f*x] - 18*(Cos[e + f*x]^2)^(1/6)*Hyperge ometric2F1[1/6, 1/2, 3/2, Sin[e + f*x]^2]*Sin[e + f*x] + Sec[e + f*x]*(-16 + 7*Sec[e + f*x]^2)*Tan[e + f*x]))/(91*f)
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3112, 3042, 3056}
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.
\(\displaystyle \int \sin ^4(e+f x) \sec ^{\frac {16}{3}}(e+f x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (e+f x)^{16/3}}{\csc (e+f x)^4}dx\) |
\(\Big \downarrow \) 3112 |
\(\displaystyle \sqrt [3]{\cos (e+f x)} \sqrt [3]{\sec (e+f x)} \int \frac {\sin ^4(e+f x)}{\cos ^{\frac {16}{3}}(e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt [3]{\cos (e+f x)} \sqrt [3]{\sec (e+f x)} \int \frac {\sin (e+f x)^4}{\cos (e+f x)^{16/3}}dx\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle \frac {3 \sin (e+f x) \sec ^{\frac {13}{3}}(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {13}{6},-\frac {3}{2},-\frac {7}{6},\cos ^2(e+f x)\right )}{13 f \sqrt {\sin ^2(e+f x)}}\) |
Input:
Int[Sec[e + f*x]^(16/3)*Sin[e + f*x]^4,x]
Output:
(3*Hypergeometric2F1[-13/6, -3/2, -7/6, Cos[e + f*x]^2]*Sec[e + f*x]^(13/3 )*Sin[e + f*x])/(13*f*Sqrt[Sin[e + f*x]^2])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-b^(2*IntPart[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*F racPart[(n - 1)/2])*((a*Cos[e + f*x])^(m + 1)/(a*f*(m + 1)*(Sin[e + f*x]^2) ^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, C os[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && SimplerQ[n, m]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(a^2/b^2)*(a*Sec[e + f*x])^(m - 1)*(b*Csc[e + f*x])^( n + 1)*(a*Cos[e + f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1) Int[1/((a*Cos[e + f*x])^m*(b*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x]
\[\int \sec \left (f x +e \right )^{\frac {16}{3}} \sin \left (f x +e \right )^{4}d x\]
Input:
int(sec(f*x+e)^(16/3)*sin(f*x+e)^4,x)
Output:
int(sec(f*x+e)^(16/3)*sin(f*x+e)^4,x)
\[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\int { \sec \left (f x + e\right )^{\frac {16}{3}} \sin \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^(16/3)*sin(f*x+e)^4,x, algorithm="fricas")
Output:
integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sec(f*x + e)^(16/3), x)
Timed out. \[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\text {Timed out} \] Input:
integrate(sec(f*x+e)**(16/3)*sin(f*x+e)**4,x)
Output:
Timed out
\[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\int { \sec \left (f x + e\right )^{\frac {16}{3}} \sin \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^(16/3)*sin(f*x+e)^4,x, algorithm="maxima")
Output:
integrate(sec(f*x + e)^(16/3)*sin(f*x + e)^4, x)
\[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\int { \sec \left (f x + e\right )^{\frac {16}{3}} \sin \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^(16/3)*sin(f*x+e)^4,x, algorithm="giac")
Output:
integrate(sec(f*x + e)^(16/3)*sin(f*x + e)^4, x)
Timed out. \[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^{16/3} \,d x \] Input:
int(sin(e + f*x)^4*(1/cos(e + f*x))^(16/3),x)
Output:
int(sin(e + f*x)^4*(1/cos(e + f*x))^(16/3), x)
\[ \int \sec ^{\frac {16}{3}}(e+f x) \sin ^4(e+f x) \, dx=\int \sec \left (f x +e \right )^{\frac {16}{3}} \sin \left (f x +e \right )^{4}d x \] Input:
int(sec(f*x+e)^(16/3)*sin(f*x+e)^4,x)
Output:
int(sec(e + f*x)**(1/3)*sec(e + f*x)**5*sin(e + f*x)**4,x)