Integrand size = 19, antiderivative size = 73 \[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=-\frac {b \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sin (e+f x)}{f (1-n) \sqrt {\sin ^2(e+f x)}} \] Output:
-b*hypergeom([-3/2, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*sec(f*x+e))^(- 1+n)*sin(f*x+e)/f/(1-n)/(sin(f*x+e)^2)^(1/2)
Time = 0.49 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{2},3-\frac {n}{2},\frac {7}{2},-\tan ^2(e+f x)\right ) (b \sec (e+f x))^n \sec ^2(e+f x)^{-n/2} \tan ^5(e+f x)}{5 f} \] Input:
Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x]^4,x]
Output:
(Hypergeometric2F1[5/2, 3 - n/2, 7/2, -Tan[e + f*x]^2]*(b*Sec[e + f*x])^n* Tan[e + f*x]^5)/(5*f*(Sec[e + f*x]^2)^(n/2))
Time = 0.51 (sec) , antiderivative size = 73, 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.
\(\displaystyle \int \sin ^4(e+f x) (b \sec (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \sec (e+f x))^n}{\csc (e+f x)^4}dx\) |
\(\Big \downarrow \) 3112 |
\(\displaystyle b^2 (b \cos (e+f x))^{n-1} (b \sec (e+f x))^{n-1} \int (b \cos (e+f x))^{-n} \sin ^4(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 (b \cos (e+f x))^{n-1} (b \sec (e+f x))^{n-1} \int (b \cos (e+f x))^{-n} \sin (e+f x)^4dx\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle -\frac {b \sin (e+f x) (b \sec (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}}\) |
Input:
Int[(b*Sec[e + f*x])^n*Sin[e + f*x]^4,x]
Output:
-((b*Hypergeometric2F1[-3/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Sec[ e + f*x])^(-1 + n)*Sin[e + f*x])/(f*(1 - n)*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]
\[\int \left (b \sec \left (f x +e \right )\right )^{n} \sin \left (f x +e \right )^{4}d x\]
Input:
int((b*sec(f*x+e))^n*sin(f*x+e)^4,x)
Output:
int((b*sec(f*x+e))^n*sin(f*x+e)^4,x)
\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4} \,d x } \] Input:
integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="fricas")
Output:
integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(b*sec(f*x + e))^n, x)
\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int \left (b \sec {\left (e + f x \right )}\right )^{n} \sin ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate((b*sec(f*x+e))**n*sin(f*x+e)**4,x)
Output:
Integral((b*sec(e + f*x))**n*sin(e + f*x)**4, x)
\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4} \,d x } \] Input:
integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="maxima")
Output:
integrate((b*sec(f*x + e))^n*sin(f*x + e)^4, x)
\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4} \,d x } \] Input:
integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="giac")
Output:
integrate((b*sec(f*x + e))^n*sin(f*x + e)^4, x)
Timed out. \[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n \,d x \] Input:
int(sin(e + f*x)^4*(b/cos(e + f*x))^n,x)
Output:
int(sin(e + f*x)^4*(b/cos(e + f*x))^n, x)
\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=b^{n} \left (\int \sec \left (f x +e \right )^{n} \sin \left (f x +e \right )^{4}d x \right ) \] Input:
int((b*sec(f*x+e))^n*sin(f*x+e)^4,x)
Output:
b**n*int(sec(e + f*x)**n*sin(e + f*x)**4,x)