3.1.23 \(\int \frac {1}{(c \sec (a+b x))^{5/2}} \, dx\) [23]

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]

2/5*sin(b*x+a)/b/c/(c*sec(b*x+a))^(3/2)+6/5*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(1/2*
a+1/2*b*x),2^(1/2))/b/c^2/cos(b*x+a)^(1/2)/(c*sec(b*x+a))^(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} (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 {gather*}

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 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}{(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*}

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Mathematica [A]
time = 0.08, size = 60, normalized size = 0.83 \begin {gather*} \frac {\sqrt {c \sec (a+b x)} \left (12 \sqrt {\cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\sin (a+b x)+\sin (3 (a+b x))\right )}{10 b c^3} \end {gather*}

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] Result contains complex when optimal does not.
time = 28.65, size = 323, normalized size = 4.49

method	result	size
default	\(\frac {\frac {6 i \sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right )}{5}-\frac {6 i \EllipticE \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right ) \sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}}{5}+\frac {6 i \EllipticF \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right ) \sin \left (b x +a \right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}}{5}-\frac {6 i \EllipticE \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right ) \sin \left (b x +a \right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}}{5}-\frac {2 \left (\cos ^{4}\left (b x +a \right )\right )}{5}-\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{5}+\frac {6 \cos \left (b x +a \right )}{5}}{b \sin \left (b x +a \right ) \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \cos \left (b x +a \right )^{3}}\)	\(323\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*sin(b*x+a)*cos(b*x+a)*EllipticF(I*(-1+co
s(b*x+a))/sin(b*x+a),I)-3*I*(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*sin(b*x+a)*cos(b*x+a)*E
llipticE(I*(-1+cos(b*x+a))/sin(b*x+a),I)+3*I*(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*sin(b*
x+a)*EllipticF(I*(-1+cos(b*x+a))/sin(b*x+a),I)-3*I*(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*
sin(b*x+a)*EllipticE(I*(-1+cos(b*x+a))/sin(b*x+a),I)-cos(b*x+a)^4-2*cos(b*x+a)^2+3*cos(b*x+a))/sin(b*x+a)/(c/c
os(b*x+a))^(5/2)/cos(b*x+a)^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/(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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.02, size = 95, normalized size = 1.32 \begin {gather*} \frac {2 \, \sqrt {\frac {c}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + 3 i \, \sqrt {2} \sqrt {c} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) - 3 i \, \sqrt {2} \sqrt {c} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}{5 \, b c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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 = 0.00, 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/(c*sec(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((c*sec(b*x + a))^(-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 {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c/cos(a + b*x))^(5/2),x)

[Out]

int(1/(c/cos(a + b*x))^(5/2), x)

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