\(\int \frac {\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) [425]

Optimal result

Integrand size = 21, antiderivative size = 65 \[ \int \frac {\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac {4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[Out]

-2/13*b^5/f/(b*sec(f*x+e))^(13/2)+4/9*b^3/f/(b*sec(f*x+e))^(9/2)-2/5*b/f/(b*sec(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, 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, 276} \[ \int \frac {\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac {4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

(-2*b^5)/(13*f*(b*Sec[e + f*x])^(13/2)) + (4*b^3)/(9*f*(b*Sec[e + f*x])^(9/2)) - (2*b)/(5*f*(b*Sec[e + f*x])^(
5/2))

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 \frac {\left (-1+\frac {x^2}{b^2}\right )^2}{x^{15/2}} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^5 \text {Subst}\left (\int \left (\frac {1}{x^{15/2}}-\frac {2}{b^2 x^{11/2}}+\frac {1}{b^4 x^{7/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = -\frac {2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac {4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {b (-551+340 \cos (2 (e+f x))-45 \cos (4 (e+f x)))}{2340 f (b \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

(b*(-551 + 340*Cos[2*(e + f*x)] - 45*Cos[4*(e + f*x)]))/(2340*f*(b*Sec[e + f*x])^(5/2))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {2 \left (45 \left (\cos ^{6}\left (f x +e \right )\right )-130 \left (\cos ^{4}\left (f x +e \right )\right )+117 \left (\cos ^{2}\left (f x +e \right )\right )\right )}{585 f b \sqrt {b \sec \left (f x +e \right )}}\)	\(50\)

[In]

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

[Out]

-2/585/f/b/(b*sec(f*x+e))^(1/2)*(45*cos(f*x+e)^6-130*cos(f*x+e)^4+117*cos(f*x+e)^2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (45 \, \cos \left (f x + e\right )^{7} - 130 \, \cos \left (f x + e\right )^{5} + 117 \, \cos \left (f x + e\right )^{3}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{585 \, b^{2} f} \]

[In]

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

[Out]

-2/585*(45*cos(f*x + e)^7 - 130*cos(f*x + e)^5 + 117*cos(f*x + e)^3)*sqrt(b/cos(f*x + e))/(b^2*f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \frac {\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (45 \, b^{4} - \frac {130 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {117 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{585 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {13}{2}}} \]

[In]

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

[Out]

-2/585*(45*b^4 - 130*b^4/cos(f*x + e)^2 + 117*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(13/2))

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.34 \[ \int \frac {\sin ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (45 \, \sqrt {b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{6} - 130 \, \sqrt {b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{4} + 117 \, \sqrt {b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{2}\right )}}{585 \, b^{8} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]

[In]

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

[Out]

-2/585*(45*sqrt(b*cos(f*x + e))*b^6*cos(f*x + e)^6 - 130*sqrt(b*cos(f*x + e))*b^6*cos(f*x + e)^4 + 117*sqrt(b*
cos(f*x + e))*b^6*cos(f*x + e)^2)/(b^8*f*sgn(cos(f*x + e)))

Mupad [F(-1)]

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

[In]

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

[Out]

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