3.2.21 \(\int \frac {\sec ^7(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [121]

Optimal. Leaf size=100 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^3 d}+\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 b^4 d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^6 d} \]

[Out]

10/21*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/b^4/d+2/7*(b*sec(d*x+c))^(7/2)*sin(d*x+c)/b^6/d+10/21*(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 = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, 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))^{7/2}}{7 b^6 d}+\frac {10 \sin (c+d x) (b \sec (c+d x))^{3/2}}{21 b^4 d}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(b*Sec[c + d*x])^(5/2),x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*b^3*d) + (10*(b*Sec[c + d*x])^(3/2)
*Sin[c + d*x])/(21*b^4*d) + (2*(b*Sec[c + d*x])^(7/2)*Sin[c + d*x])/(7*b^6*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 ^7(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {\int (b \sec (c+d x))^{9/2} \, dx}{b^7}\\ &=\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^6 d}+\frac {5 \int (b \sec (c+d x))^{5/2} \, dx}{7 b^5}\\ &=\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 b^4 d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^6 d}+\frac {5 \int \sqrt {b \sec (c+d x)} \, dx}{21 b^3}\\ &=\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 b^4 d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^6 d}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b^3}\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^3 d}+\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 b^4 d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^6 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.22, size = 64, normalized size = 0.64 \begin {gather*} \frac {(b \sec (c+d x))^{5/2} \left (10 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \sin (2 (c+d x))+6 \tan (c+d x)\right )}{21 b^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(b*Sec[c + d*x])^(5/2),x]

[Out]

((b*Sec[c + d*x])^(5/2)*(10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)] + 6*Tan[c + d*x]
))/(21*b^5*d)

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 52.51, size = 152, normalized size = 1.52

method	result	size
default	\(-\frac {2 \left (\cos \left (d x +c \right )+1\right )^{2} \left (\cos \left (d x +c \right )-1\right ) \left (5 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right )-5 \left (\cos ^{3}\left (d x +c \right )\right )+5 \left (\cos ^{2}\left (d x +c \right )\right )-3 \cos \left (d x +c \right )+3\right )}{21 d \sin \left (d x +c \right )^{3} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right )^{6}}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/21/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)*(5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipt
icF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^3*sin(d*x+c)-5*cos(d*x+c)^3+5*cos(d*x+c)^2-3*cos(d*x+c)+3)/sin(d
*x+c)^3/(b/cos(d*x+c))^(5/2)/cos(d*x+c)^6

________________________________________________________________________________________

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)^7/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^7/(b*sec(d*x + c))^(5/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.65, size = 117, normalized size = 1.17 \begin {gather*} \frac {-5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (5 \, \cos \left (d x + c\right )^{2} + 3\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, b^{3} d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/21*(-5*I*sqrt(2)*sqrt(b)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt
(2)*sqrt(b)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(5*cos(d*x + c)^2 + 3
)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^3)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{7}{\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)**7/(b*sec(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**7/(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)^7/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^7/(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 )}^7\,{\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)^7*(b/cos(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^7*(b/cos(c + d*x))^(5/2)), x)

________________________________________________________________________________________