3.15 \(\int \frac{1}{\sec ^{\frac{5}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=62 \[ \frac{2 \sin (a+b x)}{5 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{6 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b} \]

[Out]

(6*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(5*b) + (2*Sin[a + b*x])/(5*b*Sec[a + b*x]
^(3/2))

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Rubi [A]  time = 0.0286673, 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, 2639} \[ \frac{2 \sin (a+b x)}{5 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{6 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^(-5/2),x]

[Out]

(6*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(5*b) + (2*Sin[a + b*x])/(5*b*Sec[a + b*x]
^(3/2))

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{\sec ^{\frac{5}{2}}(a+b x)} \, dx &=\frac{2 \sin (a+b x)}{5 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{3}{5} \int \frac{1}{\sqrt{\sec (a+b x)}} \, dx\\ &=\frac{2 \sin (a+b x)}{5 b \sec ^{\frac{3}{2}}(a+b x)}+\frac{1}{5} \left (3 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx\\ &=\frac{6 \sqrt{\cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{\sec (a+b x)}}{5 b}+\frac{2 \sin (a+b x)}{5 b \sec ^{\frac{3}{2}}(a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0693045, size = 55, normalized size = 0.89 \[ \frac{\sqrt{\sec (a+b x)} \left (\sin (a+b x)+\sin (3 (a+b x))+12 \sqrt{\cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )\right )}{10 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^(-5/2),x]

[Out]

(Sqrt[Sec[a + b*x]]*(12*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2] + Sin[a + b*x] + Sin[3*(a + b*x)]))/(10*b
)

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Maple [B]  time = 1.212, size = 202, normalized size = 3.3 \begin{align*} -{\frac{2}{5\,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 ( -8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}\cos \left ( 1/2\,bx+a/2 \right ) +8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}\cos \left ( 1/2\,bx+a/2 \right ) -3\,\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+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)^(5/2),x)

[Out]

-2/5*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(-8*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)+8*sin
(1/2*b*x+1/2*a)^4*cos(1/2*b*x+1/2*a)-3*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*EllipticE
(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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(b*x + a)^(-5/2), 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{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sec(b*x + a)^(-5/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sec ^{\frac{5}{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)**(5/2),x)

[Out]

Integral(sec(a + b*x)**(-5/2), 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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(sec(b*x + a)^(-5/2), x)