3.22 \(\int \frac{1}{(c \sec (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right ) \sqrt{c \sec (a+b x)}}{3 b c^2}+\frac{2 \sin (a+b x)}{3 b c \sqrt{c \sec (a+b x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0461409, 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} \[ \frac{2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \sec (a+b x)}}{3 b c^2}+\frac{2 \sin (a+b x)}{3 b c \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^(-3/2),x]

[Out]

(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[c*Sec[a + b*x]])/(3*b*c^2) + (2*Sin[a + b*x])/(3*b*c*Sqrt
[c*Sec[a + b*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}{(c \sec (a+b x))^{3/2}} \, dx &=\frac{2 \sin (a+b x)}{3 b c \sqrt{c \sec (a+b x)}}+\frac{\int \sqrt{c \sec (a+b x)} \, dx}{3 c^2}\\ &=\frac{2 \sin (a+b x)}{3 b c \sqrt{c \sec (a+b x)}}+\frac{\left (\sqrt{\cos (a+b x)} \sqrt{c \sec (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{3 c^2}\\ &=\frac{2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \sec (a+b x)}}{3 b c^2}+\frac{2 \sin (a+b x)}{3 b c \sqrt{c \sec (a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0579416, size = 59, normalized size = 0.82 \[ \frac{\sec ^2(a+b x) \left (2 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )+\sin (2 (a+b x))\right )}{3 b (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sec[a + b*x])^(-3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.155, size = 131, normalized size = 1.8 \begin{align*} -{\frac{2\, \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( bx+a \right ) \right ) }{3\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{3}} \left ( i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}\sin \left ( bx+a \right ) - \left ( \cos \left ( bx+a \right ) \right ) ^{2}+\cos \left ( bx+a \right ) \right ) \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sec(b*x+a))^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sec(b*x + a))^(-3/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c \sec \left (b x + a\right )}}{c^{2} \sec \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*sec(b*x + a))/(c^2*sec(b*x + a)^2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sec(b*x+a))**(3/2),x)

[Out]

Integral((c*sec(a + b*x))**(-3/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sec(b*x + a))^(-3/2), x)