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

Optimal. Leaf size=72 \[ \frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 b^2 f}+\frac{2 \sin (e+f x)}{3 b f \sqrt{b \sec (e+f x)}} \]

[Out]

(2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*b^2*f) + (2*Sin[e + f*x])/(3*b*f*Sqrt
[b*Sec[e + f*x]])

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Rubi [A]  time = 0.035942, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3769, 3771, 2641} \[ \frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 b^2 f}+\frac{2 \sin (e+f x)}{3 b f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(-3/2),x]

[Out]

(2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*b^2*f) + (2*Sin[e + f*x])/(3*b*f*Sqrt
[b*Sec[e + f*x]])

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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}} \, dx &=\frac{2 \sin (e+f x)}{3 b f \sqrt{b \sec (e+f x)}}+\frac{\int \sqrt{b \sec (e+f x)} \, dx}{3 b^2}\\ &=\frac{2 \sin (e+f x)}{3 b f \sqrt{b \sec (e+f x)}}+\frac{\left (\sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx}{3 b^2}\\ &=\frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 b^2 f}+\frac{2 \sin (e+f x)}{3 b f \sqrt{b \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0598092, size = 59, normalized size = 0.82 \[ \frac{\sec ^2(e+f x) \left (\sin (2 (e+f x))+2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{3 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(-3/2),x]

[Out]

(Sec[e + f*x]^2*(2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] + Sin[2*(e + f*x)]))/(3*f*(b*Sec[e + f*x])^(3/
2))

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Maple [C]  time = 0.115, size = 131, normalized size = 1.8 \begin{align*} -{\frac{2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}+\cos \left ( fx+e \right ) \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(-3/2), 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 )}}{b^{2} \sec \left (f x + e\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))/(b^2*sec(f*x + e)^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))**(3/2),x)

[Out]

Integral((b*sec(e + f*x))**(-3/2), x)

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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}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(-3/2), x)