\(\int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx\) [498]

Optimal result

Integrand size = 19, antiderivative size = 73 \[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=-\frac {b \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sin (e+f x)}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2712, 2656} \[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=-\frac {b \sin (e+f x) (b \sec (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]

[In]

Int[(b*Sec[e + f*x])^n*Sin[e + f*x]^4,x]

[Out]

-((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]))

Rule 2656

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^(2*IntPar
t[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*FracPart[(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, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rule 2712

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(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*Co
s[e + f*x])^m*(b*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \left (b^2 (b \cos (e+f x))^{-1+n} (b \sec (e+f x))^{-1+n}\right ) \int (b \cos (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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{2},3-\frac {n}{2},\frac {7}{2},-\tan ^2(e+f x)\right ) (b \sec (e+f x))^n \sec ^2(e+f x)^{-n/2} \tan ^5(e+f x)}{5 f} \]

[In]

Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x]^4,x]

[Out]

(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))

Maple [F]

\[\int \left (b \sec \left (f x +e \right )\right )^{n} \left (\sin ^{4}\left (f x +e \right )\right )d x\]

[In]

int((b*sec(f*x+e))^n*sin(f*x+e)^4,x)

[Out]

int((b*sec(f*x+e))^n*sin(f*x+e)^4,x)

Fricas [F]

\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(b*sec(f*x + e))^n, x)

Sympy [F]

\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int \left (b \sec {\left (e + f x \right )}\right )^{n} \sin ^{4}{\left (e + f x \right )}\, dx \]

[In]

integrate((b*sec(f*x+e))**n*sin(f*x+e)**4,x)

[Out]

Integral((b*sec(e + f*x))**n*sin(e + f*x)**4, x)

Maxima [F]

\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^n*sin(f*x + e)^4, x)

Giac [F]

\[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n*sin(f*x + e)^4, x)

Mupad [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[In]

int(sin(e + f*x)^4*(b/cos(e + f*x))^n,x)

[Out]

int(sin(e + f*x)^4*(b/cos(e + f*x))^n, x)