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

Optimal. Leaf size=62 \[ \frac{2 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )}{3 b}+\frac{2 \sin (a+b x)}{3 b \sqrt{\sec (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0280829, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3769, 3771, 2641} \[ \frac{2 \sin (a+b x)}{3 b \sqrt{\sec (a+b x)}}+\frac{2 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^(-3/2),x]

[Out]

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

Mathematica [A]  time = 0.0442057, size = 49, normalized size = 0.79 \[ \frac{\sqrt{\sec (a+b x)} \left (2 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )+\sin (2 (a+b x))\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^(-3/2),x]

[Out]

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

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Maple [B]  time = 1.485, size = 179, normalized size = 2.9 \begin{align*} -{\frac{2}{3\,b}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}} \left ( 4\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}\cos \left ( 1/2\,bx+a/2 \right ) +\sqrt{ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) ,\sqrt{2} \right ) -2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}\cos \left ( 1/2\,bx+a/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(4*sin(1/2*b*x+1/2*a)^4*cos(1/2*b*x+1/2*a)+(sin(1
/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))-2*sin(1/2*b*x+1/
2*a)^2*cos(1/2*b*x+1/2*a))/(-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1/2)/sin(1/2*b*x+1/2*a)/(2*cos(1/2*
b*x+1/2*a)^2-1)^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sec(b*x + a)^(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(a + b*x)**(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(-3/2), x)