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

Optimal. Leaf size=80 \[ -\frac{b^5 (b \sec (e+f x))^{n-5}}{f (5-n)}+\frac{2 b^3 (b \sec (e+f x))^{n-3}}{f (3-n)}-\frac{b (b \sec (e+f x))^{n-1}}{f (1-n)} \]

[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))

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Rubi [A]  time = 0.0732299, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2622, 270} \[ -\frac{b^5 (b \sec (e+f x))^{n-5}}{f (5-n)}+\frac{2 b^3 (b \sec (e+f x))^{n-3}}{f (3-n)}-\frac{b (b \sec (e+f x))^{n-1}}{f (1-n)} \]

Antiderivative was successfully verified.

[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 2622

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])

Rule 270

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]

Rubi steps

\begin{align*} \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx &=\frac{b^5 \operatorname{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 \operatorname{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]  time = 0.367233, size = 80, normalized size = 1. \[ \frac{b \left (-4 \left (n^2-8 n+7\right ) \cos (2 (e+f x))+\left (n^2-4 n+3\right ) \cos (4 (e+f x))+3 n^2-28 n+89\right ) (b \sec (e+f x))^{n-1}}{8 f (n-5) (n-3) (n-1)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 1.006, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( fx+e \right ) \right ) ^{n} \left ( \sin \left ( fx+e \right ) \right ) ^{5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03923, size = 115, normalized size = 1.44 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.80017, size = 208, normalized size = 2.6 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)