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

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

[Out]

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

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Rubi [A]  time = 0.0337027, 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} \[ \frac{6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(-5/2),x]

[Out]

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

Mathematica [A]  time = 0.0180601, size = 60, normalized size = 0.83 \[ \frac{\sqrt{b \sec (c+d x)} \left (\sin (c+d x)+\sin (3 (c+d x))+12 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{10 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(-5/2),x]

[Out]

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

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Maple [C]  time = 0.199, size = 321, normalized size = 4.5 \begin{align*} -{\frac{2}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) } \left ( 3\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) -3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +3\,i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) -3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) + \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-3\,\cos \left ( dx+c \right ) \right ) \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(d*x + c))^(-5/2), x)