3.455 \(\int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac{64 b}{45 a^5 f \sqrt{a \sin (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{16 b}{45 a^3 f (a \sin (e+f x))^{5/2} \sqrt{b \sec (e+f x)}}-\frac{2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt{b \sec (e+f x)}} \]

[Out]

(-2*b)/(9*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(9/2)) - (16*b)/(45*a^3*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e +
f*x])^(5/2)) - (64*b)/(45*a^5*f*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])

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Rubi [A]  time = 0.165228, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2584, 2578} \[ -\frac{64 b}{45 a^5 f \sqrt{a \sin (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{16 b}{45 a^3 f (a \sin (e+f x))^{5/2} \sqrt{b \sec (e+f x)}}-\frac{2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Sec[e + f*x]]/(a*Sin[e + f*x])^(11/2),x]

[Out]

(-2*b)/(9*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(9/2)) - (16*b)/(45*a^3*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e +
f*x])^(5/2)) - (64*b)/(45*a^5*f*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])

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{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx &=-\frac{2 b}{9 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{9/2}}+\frac{8 \int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{7/2}} \, dx}{9 a^2}\\ &=-\frac{2 b}{9 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{9/2}}-\frac{16 b}{45 a^3 f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{5/2}}+\frac{32 \int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx}{45 a^4}\\ &=-\frac{2 b}{9 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{9/2}}-\frac{16 b}{45 a^3 f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{5/2}}-\frac{64 b}{45 a^5 f \sqrt{b \sec (e+f x)} \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.214393, size = 65, normalized size = 0.61 \[ \frac{2 b (20 \cos (2 (e+f x))-4 \cos (4 (e+f x))-21) \csc ^5(e+f x) \sqrt{a \sin (e+f x)}}{45 a^6 f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Sec[e + f*x]]/(a*Sin[e + f*x])^(11/2),x]

[Out]

(2*b*(-21 + 20*Cos[2*(e + f*x)] - 4*Cos[4*(e + f*x)])*Csc[e + f*x]^5*Sqrt[a*Sin[e + f*x]])/(45*a^6*f*Sqrt[b*Se
c[e + f*x]])

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Maple [A]  time = 0.133, size = 62, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 64\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-144\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+90 \right ) \cos \left ( fx+e \right ) \sin \left ( fx+e \right ) }{45\,f}\sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}} \left ( a\sin \left ( fx+e \right ) \right ) ^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x)

[Out]

-2/45/f*(32*cos(f*x+e)^4-72*cos(f*x+e)^2+45)*cos(f*x+e)*(b/cos(f*x+e))^(1/2)*sin(f*x+e)/(a*sin(f*x+e))^(11/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(11/2), x)

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Fricas [A]  time = 4.38894, size = 240, normalized size = 2.26 \begin{align*} -\frac{2 \,{\left (32 \, \cos \left (f x + e\right )^{5} - 72 \, \cos \left (f x + e\right )^{3} + 45 \, \cos \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{45 \,{\left (a^{6} f \cos \left (f x + e\right )^{4} - 2 \, a^{6} f \cos \left (f x + e\right )^{2} + a^{6} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-2/45*(32*cos(f*x + e)^5 - 72*cos(f*x + e)^3 + 45*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e))/((a^
6*f*cos(f*x + e)^4 - 2*a^6*f*cos(f*x + e)^2 + a^6*f)*sin(f*x + e))

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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((b*sec(f*x+e))**(1/2)/(a*sin(f*x+e))**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(11/2), x)