3.2.25 \(\int \frac {\sec ^3(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [125]

Optimal. Leaf size=41 \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{b^3 d} \]

[Out]

2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(b*se
c(d*x+c))^(1/2)/b^3/d

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Rubi [A]
time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {16, 3856, 2720} \begin {gather*} \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(b*Sec[c + d*x])^(5/2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(b^3*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2720

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

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 {\sec ^3(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {\int \sqrt {b \sec (c+d x)} \, dx}{b^3}\\ &=\frac {\left (\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{b^3}\\ &=\frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{b^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 41, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(b*Sec[c + d*x])^(5/2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(b^3*d)

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Maple [C] Result contains complex when optimal does not.
time = 33.81, size = 98, normalized size = 2.39

method	result	size
default	\(-\frac {2 i \left (\frac {1}{\cos \left (d x +c \right )+1}\right )^{\frac {7}{2}} \left (\cos \left (d x +c \right )-1\right ) \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos \left (d x +c \right )+1\right )^{2}}{d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{2}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2*I/d*(1/(cos(d*x+c)+1))^(7/2)*(cos(d*x+c)-1)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(cos(d*x+c)+1)^2/(b/co
s(d*x+c))^(5/2)/(cos(d*x+c)/(cos(d*x+c)+1))^(5/2)/sin(d*x+c)^2

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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(sec(d*x+c)^3/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^3/(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 = 0.76, size = 60, normalized size = 1.46 \begin {gather*} \frac {-i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{b^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

(-I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*sqrt(b)*weierstrassP
Inverse(-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 {\sec ^{3}{\left (c + d x \right )}}{\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(sec(d*x+c)**3/(b*sec(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**3/(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(sec(d*x+c)^3/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/(b*sec(d*x + c))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\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/(cos(c + d*x)^3*(b/cos(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(b/cos(c + d*x))^(5/2)), x)

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