\(\int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx\) [492]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=-\frac {b^5 (b \sec (e+f x))^{-5+n}}{f (5-n)}+\frac {2 b^3 (b \sec (e+f x))^{-3+n}}{f (3-n)}-\frac {b (b \sec (e+f x))^{-1+n}}{f (1-n)} \]

[Out]

-b^5*(b*sec(f*x+e))^(-5+n)/f/(5-n)+2*b^3*(b*sec(f*x+e))^(-3+n)/f/(3-n)-b*(b*sec(f*x+e))^(-1+n)/f/(1-n)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2702, 276} \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=-\frac {b^5 (b \sec (e+f x))^{n-5}}{f (5-n)}+\frac {2 b^3 (b \sec (e+f x))^{n-3}}{f (3-n)}-\frac {b (b \sec (e+f x))^{n-1}}{f (1-n)} \]

[In]

Int[(b*Sec[e + f*x])^n*Sin[e + f*x]^5,x]

[Out]

-((b^5*(b*Sec[e + f*x])^(-5 + n))/(f*(5 - n))) + (2*b^3*(b*Sec[e + f*x])^(-3 + n))/(f*(3 - n)) - (b*(b*Sec[e +
f*x])^(-1 + n))/(f*(1 - n))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int x^{-6+n} \left (-1+\frac {x^2}{b^2}\right )^2 \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^5 \text {Subst}\left (\int \left (x^{-6+n}-\frac {2 x^{-4+n}}{b^2}+\frac {x^{-2+n}}{b^4}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = -\frac {b^5 (b \sec (e+f x))^{-5+n}}{f (5-n)}+\frac {2 b^3 (b \sec (e+f x))^{-3+n}}{f (3-n)}-\frac {b (b \sec (e+f x))^{-1+n}}{f (1-n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {b \left (89-28 n+3 n^2-4 \left (7-8 n+n^2\right ) \cos (2 (e+f x))+\left (3-4 n+n^2\right ) \cos (4 (e+f x))\right ) (b \sec (e+f x))^{-1+n}}{8 f (-5+n) (-3+n) (-1+n)} \]

[In]

Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x]^5,x]

[Out]

(b*(89 - 28*n + 3*n^2 - 4*(7 - 8*n + n^2)*Cos[2*(e + f*x)] + (3 - 4*n + n^2)*Cos[4*(e + f*x)])*(b*Sec[e + f*x]
)^(-1 + n))/(8*f*(-5 + n)*(-3 + n)*(-1 + n))

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(\frac {\left (\left (-\frac {3}{2} n^{2}+14 n -\frac {25}{2}\right ) \cos \left (3 f x +3 e \right )+\left (\frac {1}{2} n^{2}-2 n +\frac {3}{2}\right ) \cos \left (5 f x +5 e \right )+\cos \left (f x +e \right ) \left (n^{2}-12 n +75\right )\right ) \left (\frac {b}{\cos \left (f x +e \right )}\right )^{n}}{8 \left (n^{3}-9 n^{2}+23 n -15\right ) f}\)	\(89\)
default	\(\frac {\cos \left (f x +e \right ) {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-1+n \right )}-\frac {2 \left (\cos ^{3}\left (f x +e \right )\right ) {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-3+n \right )}+\frac {\left (\cos ^{5}\left (f x +e \right )\right ) {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-5+n \right )}\)	\(94\)

[In]

int((b*sec(f*x+e))^n*sin(f*x+e)^5,x,method=_RETURNVERBOSE)

[Out]

1/8*((-3/2*n^2+14*n-25/2)*cos(3*f*x+3*e)+(1/2*n^2-2*n+3/2)*cos(5*f*x+5*e)+cos(f*x+e)*(n^2-12*n+75))*(b/cos(f*x
+e))^n/(n^3-9*n^2+23*n-15)/f

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {{\left ({\left (n^{2} - 4 \, n + 3\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (n^{2} - 6 \, n + 5\right )} \cos \left (f x + e\right )^{3} + {\left (n^{2} - 8 \, n + 15\right )} \cos \left (f x + e\right )\right )} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{n}}{f n^{3} - 9 \, f n^{2} + 23 \, f n - 15 \, f} \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^5,x, algorithm="fricas")

[Out]

((n^2 - 4*n + 3)*cos(f*x + e)^5 - 2*(n^2 - 6*n + 5)*cos(f*x + e)^3 + (n^2 - 8*n + 15)*cos(f*x + e))*(b/cos(f*x
+ e))^n/(f*n^3 - 9*f*n^2 + 23*f*n - 15*f)

Sympy [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\text {Timed out} \]

[In]

integrate((b*sec(f*x+e))**n*sin(f*x+e)**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {\frac {b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )^{5}}{n - 5} - \frac {2 \, b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )^{3}}{n - 3} + \frac {b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )}{n - 1}}{f} \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^5,x, algorithm="maxima")

[Out]

(b^n*cos(f*x + e)^(-n)*cos(f*x + e)^5/(n - 5) - 2*b^n*cos(f*x + e)^(-n)*cos(f*x + e)^3/(n - 3) + b^n*cos(f*x +
e)^(-n)*cos(f*x + e)/(n - 1))/f

Giac [F]

\[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{5} \,d x } \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n*sin(f*x + e)^5, x)

Mupad [B] (verification not implemented)

Time = 1.40 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.68 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n\,\left (150\,\cos \left (e+f\,x\right )-25\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )-24\,n\,\cos \left (e+f\,x\right )+28\,n\,\cos \left (3\,e+3\,f\,x\right )-4\,n\,\cos \left (5\,e+5\,f\,x\right )+2\,n^2\,\cos \left (e+f\,x\right )-3\,n^2\,\cos \left (3\,e+3\,f\,x\right )+n^2\,\cos \left (5\,e+5\,f\,x\right )\right )}{16\,f\,\left (n^3-9\,n^2+23\,n-15\right )} \]

[In]

int(sin(e + f*x)^5*(b/cos(e + f*x))^n,x)

[Out]

((b/cos(e + f*x))^n*(150*cos(e + f*x) - 25*cos(3*e + 3*f*x) + 3*cos(5*e + 5*f*x) - 24*n*cos(e + f*x) + 28*n*co
s(3*e + 3*f*x) - 4*n*cos(5*e + 5*f*x) + 2*n^2*cos(e + f*x) - 3*n^2*cos(3*e + 3*f*x) + n^2*cos(5*e + 5*f*x)))/(
16*f*(23*n - 9*n^2 + n^3 - 15))