3.446 \(\int \frac{1}{(b \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{6 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.034779, 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, 2639} \[ \frac{6 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(-5/2),x]

[Out]

(6*EllipticE[(e + f*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (2*Sin[e + f*x])/(5*b*f*(b*S
ec[e + f*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}{(b \sec (e+f x))^{5/2}} \, dx &=\frac{2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac{3 \int \frac{1}{\sqrt{b \sec (e+f x)}} \, dx}{5 b^2}\\ &=\frac{2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac{3 \int \sqrt{\cos (e+f x)} \, dx}{5 b^2 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=\frac{6 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{2 \sin (e+f x)}{5 b f (b \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0849257, size = 60, normalized size = 0.83 \[ \frac{\sqrt{b \sec (e+f x)} \left (\sin (e+f x)+\sin (3 (e+f x))+12 \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{10 b^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(-5/2),x]

[Out]

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

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Maple [C]  time = 0.15, size = 323, normalized size = 4.5 \begin{align*}{\frac{2}{5\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) } \left ( 3\,i\cos \left ( fx+e \right ){\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 ) -3\,i{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}+3\,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 ) -3\,i{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}- \left ( \cos \left ( fx+e \right ) \right ) ^{4}-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3\,\cos \left ( fx+e \right ) \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sec(f*x+e))^(5/2),x)

[Out]

2/5/f*(3*I*cos(f*x+e)*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)-3*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*(1/(cos(f*x+e)+1))
^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+3*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)-3*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*(1/(co
s(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)^4-2*cos(f*x+e)^2+3*cos(f*x+e))/(b/cos(f*x+e))^
(5/2)/cos(f*x+e)^3/sin(f*x+e)

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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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(f*x + e))^(-5/2), 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^{3} \sec \left (f x + e\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(f*x + e))/(b^3*sec(f*x + e)^3), 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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))**(5/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(f*x + e))^(-5/2), x)