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]

2/5*sin(f*x+e)/b/f/(b*sec(f*x+e))^(3/2)+6/5*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*
f*x+1/2*e),2^(1/2))/b^2/f/cos(f*x+e)^(1/2)/(b*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right )}}{b^{3} \sec \left (f x + e\right )^{3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [C]  time = 0.22, size = 321, normalized size = 4.46 \[ -\frac {2 \left (3 i \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )-3 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+3 i \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )-3 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )+\cos ^{4}\left (f x +e \right )+2 \left (\cos ^{2}\left (f x +e \right )\right )-3 \cos \left (f x +e \right )\right )}{5 f \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{3} \sin \left (f x +e \right )} \]

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*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*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-1+cos(f*
x+e))/sin(f*x+e),I)*sin(f*x+e)*cos(f*x+e)+3*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*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)*sin(f*x+e)*(1/(c
os(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/cos(e + f*x))^(5/2),x)

[Out]

int(1/(b/cos(e + f*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

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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