\(\int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [439]

Optimal result

Integrand size = 21, antiderivative size = 43 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[Out]

2/11*b^3/f/(b*sec(f*x+e))^(11/2)-2/7*b/f/(b*sec(f*x+e))^(7/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2702, 14} \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[In]

Int[Sin[e + f*x]^3/(b*Sec[e + f*x])^(5/2),x]

[Out]

(2*b^3)/(11*f*(b*Sec[e + f*x])^(11/2)) - (2*b)/(7*f*(b*Sec[e + f*x])^(7/2))

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 \frac {-1+\frac {x^2}{b^2}}{x^{13/2}} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^3 \text {Subst}\left (\int \left (-\frac {1}{x^{13/2}}+\frac {1}{b^2 x^{9/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {2 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\cos ^4(e+f x) (-15+7 \cos (2 (e+f x))) \sqrt {b \sec (e+f x)}}{77 b^3 f} \]

[In]

Integrate[Sin[e + f*x]^3/(b*Sec[e + f*x])^(5/2),x]

[Out]

(Cos[e + f*x]^4*(-15 + 7*Cos[2*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(77*b^3*f)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\frac {2 \left (\cos ^{5}\left (f x +e \right )\right )}{11}-\frac {2 \left (\cos ^{3}\left (f x +e \right )\right )}{7}}{f \,b^{2} \sqrt {b \sec \left (f x +e \right )}}\)	\(40\)

[In]

int(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/77/f/b^2/(b*sec(f*x+e))^(1/2)*(7*cos(f*x+e)^5-11*cos(f*x+e)^3)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (7 \, \cos \left (f x + e\right )^{6} - 11 \, \cos \left (f x + e\right )^{4}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{77 \, b^{3} f} \]

[In]

integrate(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/77*(7*cos(f*x + e)^6 - 11*cos(f*x + e)^4)*sqrt(b/cos(f*x + e))/(b^3*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**3/(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (7 \, b^{2} - \frac {11 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} b}{77 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {11}{2}}} \]

[In]

integrate(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/77*(7*b^2 - 11*b^2/cos(f*x + e)^2)*b/(f*(b/cos(f*x + e))^(11/2))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.49 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (7 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{5} - 11 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{3}\right )}}{77 \, b^{8} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]

[In]

integrate(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

2/77*(7*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^5 - 11*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^3)/(b^8*f*sgn(cos(f
*x + e)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(sin(e + f*x)^3/(b/cos(e + f*x))^(5/2),x)

[Out]

int(sin(e + f*x)^3/(b/cos(e + f*x))^(5/2), x)