\(\int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx\) [469]

Optimal result

Integrand size = 23, antiderivative size = 61 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=-\frac {2 b}{7 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}-\frac {8 b}{21 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \]

[Out]

-2/7*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(7/2)-8/21*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(3/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2664, 2658} \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=-\frac {8 b}{21 f \sin ^{\frac {3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {2 b}{7 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

[In]

Int[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(9/2)),x]

[Out]

(-2*b)/(7*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7/2)) - (8*b)/(21*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(3/2)
)

Rule 2658

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[b*(a*Sin[e
+ f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/(a*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m - n + 2,
0] && NeQ[m, -1]

Rule 2664

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(a*Sin[e +
f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/(a*f*(m + 1))), x] + Dist[(m - n + 2)/(a^2*(m + 1)), Int[(a*Sin[e + f*
x])^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b}{7 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}+\frac {4}{7} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {5}{2}}(e+f x)} \, dx \\ & = -\frac {2 b}{7 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}-\frac {8 b}{21 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=\frac {2 b (-5+2 \cos (2 (e+f x)))}{21 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)} \]

[In]

Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(9/2)),x]

[Out]

(2*b*(-5 + 2*Cos[2*(e + f*x)]))/(21*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7/2))

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {\frac {8 \left (\cos ^{3}\left (f x +e \right )\right )}{21}-\frac {2 \cos \left (f x +e \right )}{3}}{f \sin \left (f x +e \right )^{\frac {7}{2}} \sqrt {b \sec \left (f x +e \right )}}\)	\(43\)

[In]

int(1/sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/21/f/sin(f*x+e)^(7/2)/(b*sec(f*x+e))^(1/2)*(4*cos(f*x+e)^3-7*cos(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=\frac {2 \, {\left (4 \, \cos \left (f x + e\right )^{4} - 7 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{21 \, {\left (b f \cos \left (f x + e\right )^{4} - 2 \, b f \cos \left (f x + e\right )^{2} + b f\right )}} \]

[In]

integrate(1/sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/21*(4*cos(f*x + e)^4 - 7*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))*sqrt(sin(f*x + e))/(b*f*cos(f*x + e)^4 - 2*b*f
*cos(f*x + e)^2 + b*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(1/sin(f*x+e)**(9/2)/(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=\int { \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate(1/sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(9/2)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(1/sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 2.41 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.69 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx=\frac {4\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}\,\left (11\,\sin \left (e+f\,x\right )+4\,\sin \left (3\,e+3\,f\,x\right )-6\,\sin \left (5\,e+5\,f\,x\right )+\sin \left (7\,e+7\,f\,x\right )\right )}{21\,b\,f\,\sqrt {\sin \left (e+f\,x\right )}\,\left (15\,\cos \left (2\,e+2\,f\,x\right )-6\,\cos \left (4\,e+4\,f\,x\right )+\cos \left (6\,e+6\,f\,x\right )-10\right )} \]

[In]

int(1/(sin(e + f*x)^(9/2)*(b/cos(e + f*x))^(1/2)),x)

[Out]

(4*(b/cos(e + f*x))^(1/2)*(11*sin(e + f*x) + 4*sin(3*e + 3*f*x) - 6*sin(5*e + 5*f*x) + sin(7*e + 7*f*x)))/(21*
b*f*sin(e + f*x)^(1/2)*(15*cos(2*e + 2*f*x) - 6*cos(4*e + 4*f*x) + cos(6*e + 6*f*x) - 10))