3.2.23 \(\int \frac {\sec ^5(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [123]

Optimal. Leaf size=72 \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 b^3 d}+\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^4 d} \]

[Out]

2/3*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/b^4/d+2/3*(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*sec(d*x+c))^(1/2)/b^3/d

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 3853, 3856, 2720} \begin {gather*} \frac {2 \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 b^4 d}+\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)}}{3 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(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]])/(3*b^3*d) + (2*(b*Sec[c + d*x])^(3/2)*Si
n[c + d*x])/(3*b^4*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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 {\sec ^5(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {\int (b \sec (c+d x))^{5/2} \, dx}{b^5}\\ &=\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^4 d}+\frac {\int \sqrt {b \sec (c+d x)} \, dx}{3 b^3}\\ &=\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^4 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 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)}}{3 b^3 d}+\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 51, normalized size = 0.71 \begin {gather*} \frac {2 \sqrt {b \sec (c+d x)} \left (\sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\tan (c+d x)\right )}{3 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 52.88, size = 130, normalized size = 1.81

method	result	size
default	\(-\frac {2 \left (\cos \left (d x +c \right )-1\right ) \left (i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right )-\cos \left (d x +c \right )+1\right ) \left (\cos \left (d x +c \right )+1\right )^{2}}{3 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.21, size = 101, normalized size = 1.40 \begin {gather*} \frac {-i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, b^{3} d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-I*sqrt(2)*sqrt(b)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*sqr
t(b)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(b/cos(d*x + c))*sin(d*x +
c))/(b^3*d*cos(d*x + c))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________