\(\int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) [427]

Optimal result

Integrand size = 19, antiderivative size = 20 \[ \int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2702, 30} \[ \int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

(-2*b)/(5*f*(b*Sec[e + f*x])^(5/2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

(-2*b)/(5*f*(b*Sec[e + f*x])^(5/2))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(-\frac {2 b}{5 f \left (b \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\)	\(17\)
default	\(-\frac {2 b}{5 f \left (b \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\)	\(17\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{3}}{5 \, b^{2} f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, \cos \left (f x + e\right )}{5 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75 \[ \int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, \sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )^{2}}{5 \, b^{2} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]

[In]

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

[Out]

-2/5*sqrt(b*cos(f*x + e))*cos(f*x + e)^2/(b^2*f*sgn(cos(f*x + e)))

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {\sin (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2\,{\cos \left (e+f\,x\right )}^3\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{5\,b^2\,f} \]

[In]

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

[Out]

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