3.481 \(\int \frac{1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=137 \[ -\frac{2 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{21 a^4 b^2 f \sqrt{a \sin (e+f x)}}+\frac{2}{21 a^3 b f (a \sin (e+f x))^{3/2} \sqrt{b \sec (e+f x)}}-\frac{2}{7 a b f (a \sin (e+f x))^{7/2} \sqrt{b \sec (e+f x)}} \]

[Out]

-2/(7*a*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(7/2)) + 2/(21*a^3*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])
^(3/2)) - (2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(21*a^4*b^2*f*Sqrt[a*Si
n[e + f*x]])

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Rubi [A]  time = 0.21898, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2581, 2584, 2585, 2573, 2641} \[ -\frac{2 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{21 a^4 b^2 f \sqrt{a \sin (e+f x)}}+\frac{2}{21 a^3 b f (a \sin (e+f x))^{3/2} \sqrt{b \sec (e+f x)}}-\frac{2}{7 a b f (a \sin (e+f x))^{7/2} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(9/2)),x]

[Out]

-2/(7*a*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(7/2)) + 2/(21*a^3*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])
^(3/2)) - (2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(21*a^4*b^2*f*Sqrt[a*Si
n[e + f*x]])

Rule 2581

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

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{1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{9/2}} \, dx &=-\frac{2}{7 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}-\frac{\int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx}{7 a^2 b^2}\\ &=-\frac{2}{7 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{2}{21 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{2 \int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx}{21 a^4 b^2}\\ &=-\frac{2}{7 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{2}{21 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{\left (2 \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}{21 a^4 b^2}\\ &=-\frac{2}{7 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{2}{21 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{\left (2 \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}{21 a^4 b^2 \sqrt{a \sin (e+f x)}}\\ &=-\frac{2}{7 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{2}{21 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{2 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)}}{21 a^4 b^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.867706, size = 119, normalized size = 0.87 \[ \frac{\cos (2 (e+f x)) \csc ^4(e+f x) \sqrt{a \sin (e+f x)} \left ((\cos (2 (e+f x))+5) \sec ^2(e+f x)-2 \left (-\tan ^2(e+f x)\right )^{7/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\sec ^2(e+f x)\right )\right )}{21 a^5 b f \left (\sec ^2(e+f x)-2\right ) \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(9/2)),x]

[Out]

(Cos[2*(e + f*x)]*Csc[e + f*x]^4*Sqrt[a*Sin[e + f*x]]*((5 + Cos[2*(e + f*x)])*Sec[e + f*x]^2 - 2*Hypergeometri
c2F1[1/2, 3/4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(7/4)))/(21*a^5*b*f*Sqrt[b*Sec[e + f*x]]*(-2 + Sec[e + f
*x]^2))

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Maple [B]  time = 0.122, size = 540, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(9/2),x)

[Out]

1/21/f*2^(1/2)*(2*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))+2*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))-2*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))-2*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^(1/2)*cos(f*x+e)^3-2*2^(1/2)*cos(f*x+e))*sin(f*x+e)/cos(f*x+e)^2/(b/cos(f*x+e))^(3/2)/(a*sin(f*x+e))^(9/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \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(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(f*x + e))^(3/2)*(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} b^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{5} b^{2} \cos \left (f x + e\right )^{2} + a^{5} b^{2}\right )} \sec \left (f x + e\right )^{2} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2)/(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*b^2*cos(f*x + e)^4 - 2*a^5*b^2*cos(f*x + e)^2 + a^5*b
^2)*sec(f*x + e)^2*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(1/(b*sec(f*x+e))**(3/2)/(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{1}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \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(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(f*x + e))^(3/2)*(a*sin(f*x + e))^(9/2)), x)