\(\int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx\) [493]

Optimal result

Integrand size = 19, antiderivative size = 52 \[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=\frac {b^3 (b \sec (e+f x))^{-3+n}}{f (3-n)}-\frac {b (b \sec (e+f x))^{-1+n}}{f (1-n)} \]

[Out]

b^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 = 0.04 (sec) , antiderivative size = 52, 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, 14} \[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=\frac {b^3 (b \sec (e+f x))^{n-3}}{f (3-n)}-\frac {b (b \sec (e+f x))^{n-1}}{f (1-n)} \]

[In]

Int[(b*Sec[e + f*x])^n*Sin[e + f*x]^3,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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^3 \text {Subst}\left (\int x^{-4+n} \left (-1+\frac {x^2}{b^2}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^3 \text {Subst}\left (\int \left (-x^{-4+n}+\frac {x^{-2+n}}{b^2}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {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 = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90 \[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=-\frac {b (5-n+(-1+n) \cos (2 (e+f x))) (b \sec (e+f x))^{-1+n}}{2 f (-3+n) (-1+n)} \]

[In]

Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x]^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {\left (\left (1-n \right ) \cos \left (3 f x +3 e \right )+\cos \left (f x +e \right ) \left (n -9\right )\right ) \left (\frac {b}{\cos \left (f x +e \right )}\right )^{n}}{4 \left (-3+n \right ) f \left (-1+n \right )}\)	\(54\)
default	\(\frac {\cos \left (f x +e \right ) {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-1+n \right )}-\frac {\left (\cos ^{3}\left (f x +e \right )\right ) {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-3+n \right )}\)	\(63\)

[In]

int((b*sec(f*x+e))^n*sin(f*x+e)^3,x,method=_RETURNVERBOSE)

[Out]

1/4*((1-n)*cos(3*f*x+3*e)+cos(f*x+e)*(n-9))*(b/cos(f*x+e))^n/(-3+n)/f/(-1+n)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=-\frac {{\left ({\left (n - 1\right )} \cos \left (f x + e\right )^{3} - {\left (n - 3\right )} \cos \left (f x + e\right )\right )} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{n}}{f n^{2} - 4 \, f n + 3 \, f} \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^3,x, algorithm="fricas")

[Out]

-((n - 1)*cos(f*x + e)^3 - (n - 3)*cos(f*x + e))*(b/cos(f*x + e))^n/(f*n^2 - 4*f*n + 3*f)

Sympy [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=\text {Timed out} \]

[In]

integrate((b*sec(f*x+e))**n*sin(f*x+e)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.13 \[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=-\frac {\frac {b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )^{3}}{n - 3} - \frac {b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )}{n - 1}}{f} \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^3,x, algorithm="maxima")

[Out]

-(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]

\[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{3} \,d x } \]

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^3,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n*sin(f*x + e)^3, x)

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.29 \[ \int (b \sec (e+f x))^n \sin ^3(e+f x) \, dx=-\frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n\,\left (9\,\cos \left (e+f\,x\right )-\cos \left (3\,e+3\,f\,x\right )-n\,\cos \left (e+f\,x\right )+n\,\cos \left (3\,e+3\,f\,x\right )\right )}{4\,f\,\left (n^2-4\,n+3\right )} \]

[In]

int(sin(e + f*x)^3*(b/cos(e + f*x))^n,x)

[Out]

-((b/cos(e + f*x))^n*(9*cos(e + f*x) - cos(3*e + 3*f*x) - n*cos(e + f*x) + n*cos(3*e + 3*f*x)))/(4*f*(n^2 - 4*
n + 3))