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

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

[Out]

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

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Rubi [A]  time = 0.0378077, 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} \[ \frac{6 E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b c^2 \sqrt{\cos (a+b x)} \sqrt{c \sec (a+b x)}}+\frac{2 \sin (a+b x)}{5 b c (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^(-5/2),x]

[Out]

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

Mathematica [A]  time = 0.0679626, size = 60, normalized size = 0.83 \[ \frac{\sqrt{c \sec (a+b x)} \left (\sin (a+b x)+\sin (3 (a+b x))+12 \sqrt{\cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )\right )}{10 b c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sec[a + b*x])^(-5/2),x]

[Out]

(Sqrt[c*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*c^3)

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Maple [C]  time = 0.2, size = 323, normalized size = 4.5 \begin{align*}{\frac{2}{5\,b \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sin \left ( bx+a \right ) } \left ( 3\,i\sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \cos \left ( bx+a \right ) \sin \left ( bx+a \right ) -3\,i{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \cos \left ( bx+a \right ) \sin \left ( bx+a \right ) \sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}+3\,i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}\sin \left ( bx+a \right ) -3\,i{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \sin \left ( bx+a \right ) \sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}- \left ( \cos \left ( bx+a \right ) \right ) ^{4}-2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+3\,\cos \left ( bx+a \right ) \right ) \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sec(b*x+a))^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*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{\sqrt{c \sec \left (b x + a\right )}}{c^{3} \sec \left (b x + a\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*sec(b*x + a))/(c^3*sec(b*x + a)^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec{\left (a + b x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sec(b*x+a))**(5/2),x)

[Out]

Integral((c*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}{\left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sec(b*x + a))^(-5/2), x)