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

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

[Out]

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

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Rubi [A]  time = 0.0329195, 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} \[ \frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 b^2 d}+\frac{2 \sin (c+d x)}{3 b d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(-3/2),x]

[Out]

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

Mathematica [A]  time = 0.0492983, size = 59, normalized size = 0.82 \[ \frac{\sec ^2(c+d x) \left (2 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (2 (c+d x))\right )}{3 d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(-3/2),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(d*x + c))^(-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 (d x + c\right )}}{b^{2} \sec \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sec(c + d*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 (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(d*x + c))^(-3/2), x)