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

Optimal. Leaf size=91 \[ -\frac{64 b}{231 f \sin ^{\frac{3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{16 b}{77 f \sin ^{\frac{7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{2 b}{11 f \sin ^{\frac{11}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

[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))

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Rubi [A]  time = 0.121811, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2584, 2578} \[ -\frac{64 b}{231 f \sin ^{\frac{3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{16 b}{77 f \sin ^{\frac{7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{2 b}{11 f \sin ^{\frac{11}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[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 2584

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]

Rule 2578

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]

Rubi steps

\begin{align*} \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{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]  time = 0.201415, size = 52, normalized size = 0.57 \[ \frac{2 b (28 \cos (2 (e+f x))-4 \cos (4 (e+f x))-45)}{231 f \sin ^{\frac{11}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.129, size = 92, normalized size = 1. \begin{align*} -{\frac{128\,\cos \left ( fx+e \right ) \left ( 32\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-88\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+77 \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{6}}{231\,f \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{2}+ \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\cos \left ( fx+e \right ) +1 \right ) ^{6}} \left ( \sin \left ( fx+e \right ) \right ) ^{-{\frac{11}{2}}}{\frac{1}{\sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-128/231/f*cos(f*x+e)*(32*cos(f*x+e)^4-88*cos(f*x+e)^2+77)*(-1+cos(f*x+e))^6/sin(f*x+e)^(11/2)/(sin(f*x+e)^2+c
os(f*x+e)^2-2*cos(f*x+e)+1)^6/(b/cos(f*x+e))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.6834, size = 243, normalized size = 2.67 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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