\(\int (b \sec (e+f x))^{5/2} \sin (e+f x) \, dx\) [400]

Optimal result

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

[Out]

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

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

[In]

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

[Out]

(2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

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 \sqrt {x} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {2 b (b \sec (e+f x))^{3/2}}{3 f} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {2 b \left (b \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}\)	\(17\)
default	\(\frac {2 b \left (b \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}\)	\(17\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int (b \sec (e+f x))^{5/2} \sin (e+f x) \, dx=\frac {2 \, b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{3 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int (b \sec (e+f x))^{5/2} \sin (e+f x) \, dx=\frac {2 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {5}{2}} \cos \left (f x + e\right )}{3 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).

Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int (b \sec (e+f x))^{5/2} \sin (e+f x) \, dx=\frac {2 \, b^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{3 \, \sqrt {b \cos \left (f x + e\right )} f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95 \[ \int (b \sec (e+f x))^{5/2} \sin (e+f x) \, dx=\frac {4\,b^2\,\cos \left (e+f\,x\right )\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{3\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

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