3.2.28 \(\int \frac {1}{(b \sec (c+d x))^{5/2}} \, dx\) [128]

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]

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

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Rubi [A]
time = 0.02, 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 = {3854, 3856, 2719} \begin {gather*} \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 {gather*}

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 2719

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

Rule 3854

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 3856

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 (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*}

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Mathematica [A]
time = 0.02, size = 60, normalized size = 0.83 \begin {gather*} \frac {\sqrt {b \sec (c+d x)} \left (12 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sin (c+d x)+\sin (3 (c+d x))\right )}{10 b^3 d} \end {gather*}

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] Result contains complex when optimal does not.
time = 33.39, size = 321, normalized size = 4.46

method	result	size
default	\(-\frac {2 \left (-3 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right )+3 i \cos \left (d x +c \right ) \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )-3 i \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )+3 i \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\cos ^{4}\left (d x +c \right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-3 \cos \left (d x +c \right )\right )}{5 d \sin \left (d x +c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right )^{3}}\)	\(321\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.56, size = 95, normalized size = 1.32 \begin {gather*} \frac {2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + 3 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, b^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2*sqrt(b/cos(d*x + c))*cos(d*x + c)^2*sin(d*x + c) + 3*I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstra
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/cos(c + d*x))^(5/2),x)

[Out]

int(1/(b/cos(c + d*x))^(5/2), x)

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