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

Optimal. Leaf size=94 \[ \frac{\sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{2 b^2 f \sqrt{a \sin (e+f x)}}+\frac{\sqrt{a \sin (e+f x)}}{a b f \sqrt{b \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.152337, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2582, 2585, 2573, 2641} \[ \frac{\sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{2 b^2 f \sqrt{a \sin (e+f x)}}+\frac{\sqrt{a \sin (e+f x)}}{a b f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2582

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 - n)), x] - Dist[(n + 1)/(b^2*(m - n)), Int[(a*Sin[e + f*x])^
m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m - n, 0] && 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} \sqrt{a \sin (e+f x)}} \, dx &=\frac{\sqrt{a \sin (e+f x)}}{a b f \sqrt{b \sec (e+f x)}}+\frac{\int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx}{2 b^2}\\ &=\frac{\sqrt{a \sin (e+f x)}}{a b f \sqrt{b \sec (e+f x)}}+\frac{\left (\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}{2 b^2}\\ &=\frac{\sqrt{a \sin (e+f x)}}{a b f \sqrt{b \sec (e+f x)}}+\frac{\left (\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}{2 b^2 \sqrt{a \sin (e+f x)}}\\ &=\frac{\sqrt{a \sin (e+f x)}}{a b f \sqrt{b \sec (e+f x)}}+\frac{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)}}{2 b^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.53812, size = 84, normalized size = 0.89 \[ -\frac{\cot (e+f x) \sqrt{b \sec (e+f x)} \left (-\left (-\tan ^2(e+f x)\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\sec ^2(e+f x)\right )+\cos (2 (e+f x))-1\right )}{2 b^2 f \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.118, size = 190, normalized size = 2. \begin{align*} -{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{2\,f \left ( -1+\cos \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( \sin \left ( fx+e \right ) \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ) -\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+\sqrt{2}\cos \left ( fx+e \right ) \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a\sin \left ( fx+e \right ) }}}} \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))^(1/2),x)

[Out]

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

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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}} \sqrt{a \sin \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(f*x + e))^(3/2)*sqrt(a*sin(f*x + e))), 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 )}}{a b^{2} \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))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))/(a*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))**(1/2),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}} \sqrt{a \sin \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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