3.460 \(\int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=130 \[ -\frac{4 b}{7 a^3 f (a \sin (e+f x))^{3/2} \sqrt{b \sec (e+f x)}}+\frac{4 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{7 a^4 f \sqrt{a \sin (e+f x)}}-\frac{2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt{b \sec (e+f x)}} \]

[Out]

(-2*b)/(7*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(7/2)) - (4*b)/(7*a^3*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*
x])^(3/2)) + (4*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(7*a^4*f*Sqrt[a*Sin[
e + f*x]])

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Rubi [A]  time = 0.213507, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2584, 2585, 2573, 2641} \[ -\frac{4 b}{7 a^3 f (a \sin (e+f x))^{3/2} \sqrt{b \sec (e+f x)}}+\frac{4 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{7 a^4 f \sqrt{a \sin (e+f x)}}-\frac{2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b)/(7*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(7/2)) - (4*b)/(7*a^3*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*
x])^(3/2)) + (4*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(7*a^4*f*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 2585

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

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{9/2}} \, dx &=-\frac{2 b}{7 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{6 \int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx}{7 a^2}\\ &=-\frac{2 b}{7 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}-\frac{4 b}{7 a^3 f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac{4 \int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx}{7 a^4}\\ &=-\frac{2 b}{7 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}-\frac{4 b}{7 a^3 f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac{\left (4 \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{b \cos (e+f x)} \sqrt{a \sin (e+f x)}} \, dx}{7 a^4}\\ &=-\frac{2 b}{7 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}-\frac{4 b}{7 a^3 f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac{\left (4 \sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{7 a^4 \sqrt{a \sin (e+f x)}}\\ &=-\frac{2 b}{7 a f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}-\frac{4 b}{7 a^3 f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac{4 F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}}{7 a^4 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.01634, size = 111, normalized size = 0.85 \[ -\frac{2 \cos (2 (e+f x)) (b \sec (e+f x))^{3/2} \left (2 \left (-\tan ^2(e+f x)\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\sec ^2(e+f x)\right )+(\cos (2 (e+f x))-2) \csc ^2(e+f x)\right )}{7 a^3 b f \left (\sec ^2(e+f x)-2\right ) (a \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[2*(e + f*x)]*(b*Sec[e + f*x])^(3/2)*((-2 + Cos[2*(e + f*x)])*Csc[e + f*x]^2 + 2*Hypergeometric2F1[1/2,
3/4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(7*a^3*b*f*(-2 + Sec[e + f*x]^2)*(a*Sin[e + f*x])^(3/2))

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Maple [B]  time = 0.154, size = 532, normalized size = 4.1 \begin{align*} \text{result too large to display} \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))^(9/2),x)

[Out]

-1/7/f*2^(1/2)*(4*sin(f*x+e)*cos(f*x+e)^3*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x
+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/
2),1/2*2^(1/2))+4*sin(f*x+e)*cos(f*x+e)^2*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x
+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/
2),1/2*2^(1/2))-4*sin(f*x+e)*cos(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e
))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)
,1/2*2^(1/2))-4*sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e)
)^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))
-2*2^(1/2)*cos(f*x+e)^3+3*2^(1/2)*cos(f*x+e))*(b/cos(f*x+e))^(1/2)*sin(f*x+e)/(a*sin(f*x+e))^(9/2)

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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{9}{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))^(9/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (f x + e\right )} \sqrt{a \sin \left (f x + e\right )}}{{\left (a^{5} \cos \left (f x + e\right )^{4} - 2 \, a^{5} \cos \left (f x + e\right )^{2} + a^{5}\right )} \sin \left (f x + e\right )}, x\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))^(9/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))/((a^5*cos(f*x + e)^4 - 2*a^5*cos(f*x + e)^2 + a^5)*sin(f*x
+ e)), x)

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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))**(9/2),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{9}{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))^(9/2),x, algorithm="giac")

[Out]

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