\(\int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx\) [470]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F(-1)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=-\frac {2 b}{11 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}-\frac {16 b}{77 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}-\frac {64 b}{231 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \]

[Out]

-2/11*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(11/2)-16/77*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(7/2)-64/231*b/f/(b
*sec(f*x+e))^(3/2)/sin(f*x+e)^(3/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2664, 2658} \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=-\frac {64 b}{231 f \sin ^{\frac {3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {16 b}{77 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

[In]

Int[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(13/2)),x]

[Out]

(-2*b)/(11*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(11/2)) - (16*b)/(77*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7
/2)) - (64*b)/(231*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}{11 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}+\frac {8}{11} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)} \, dx \\ & = -\frac {2 b}{11 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}-\frac {16 b}{77 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}+\frac {32}{77} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {5}{2}}(e+f x)} \, dx \\ & = -\frac {2 b}{11 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}-\frac {16 b}{77 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}-\frac {64 b}{231 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\frac {2 b (-45+28 \cos (2 (e+f x))-4 \cos (4 (e+f x)))}{231 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)} \]

[In]

Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(13/2)),x]

[Out]

(2*b*(-45 + 28*Cos[2*(e + f*x)] - 4*Cos[4*(e + f*x)]))/(231*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(11/2))

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58

method result size
default \(-\frac {2 \left (32 \left (\cos ^{5}\left (f x +e \right )\right )-88 \left (\cos ^{3}\left (f x +e \right )\right )+77 \cos \left (f x +e \right )\right )}{231 f \sin \left (f x +e \right )^{\frac {11}{2}} \sqrt {b \sec \left (f x +e \right )}}\) \(53\)

[In]

int(1/sin(f*x+e)^(13/2)/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/231/f/sin(f*x+e)^(11/2)/(b*sec(f*x+e))^(1/2)*(32*cos(f*x+e)^5-88*cos(f*x+e)^3+77*cos(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\frac {2 \, {\left (32 \, \cos \left (f x + e\right )^{6} - 88 \, \cos \left (f x + e\right )^{4} + 77 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{231 \, {\left (b f \cos \left (f x + e\right )^{6} - 3 \, b f \cos \left (f x + e\right )^{4} + 3 \, b f \cos \left (f x + e\right )^{2} - b f\right )}} \]

[In]

integrate(1/sin(f*x+e)^(13/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/231*(32*cos(f*x + e)^6 - 88*cos(f*x + e)^4 + 77*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))*sqrt(sin(f*x + e))/(b*f
*cos(f*x + e)^6 - 3*b*f*cos(f*x + e)^4 + 3*b*f*cos(f*x + e)^2 - b*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(1/sin(f*x+e)**(13/2)/(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\int { \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {13}{2}}} \,d x } \]

[In]

integrate(1/sin(f*x+e)^(13/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(13/2)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(1/sin(f*x+e)^(13/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.79 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\frac {{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\sqrt {\frac {b}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,992{}\mathrm {i}}{231\,b\,f}+\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,608{}\mathrm {i}}{231\,b\,f}-\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,320{}\mathrm {i}}{231\,b\,f}+\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,64{}\mathrm {i}}{231\,b\,f}\right )\,1{}\mathrm {i}}{32\,{\sin \left (e+f\,x\right )}^{11/2}} \]

[In]

int(1/(sin(e + f*x)^(13/2)*(b/cos(e + f*x))^(1/2)),x)

[Out]

(exp(- e*6i - f*x*6i)*(b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((exp(e*6i + f*x*6i)*992i)/(23
1*b*f) + (exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*608i)/(231*b*f) - (exp(e*6i + f*x*6i)*cos(4*e + 4*f*x)*320i)/(23
1*b*f) + (exp(e*6i + f*x*6i)*cos(6*e + 6*f*x)*64i)/(231*b*f))*1i)/(32*sin(e + f*x)^(11/2))