\(\int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\) [136]

Optimal result

Integrand size = 19, antiderivative size = 97 \[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=-\frac {6 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt {\cos (c+d x)}}+\frac {2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {6 \sin (c+d x)}{5 b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {16, 2716, 2721, 2719} \[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 b^3 d \sqrt {\cos (c+d x)}}+\frac {6 \sin (c+d x)}{5 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}} \]

[In]

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

[Out]

(-6*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b^3*d*Sqrt[Cos[c + d*x]]) + (2*Sin[c + d*x])/(5*d*(b*Co
s[c + d*x])^(5/2)) + (6*Sin[c + d*x])/(5*b^2*d*Sqrt[b*Cos[c + d*x]])

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 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = b \int \frac {1}{(b \cos (c+d x))^{7/2}} \, dx \\ & = \frac {2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx}{5 b} \\ & = \frac {2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {6 \sin (c+d x)}{5 b^2 d \sqrt {b \cos (c+d x)}}-\frac {3 \int \sqrt {b \cos (c+d x)} \, dx}{5 b^3} \\ & = \frac {2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {6 \sin (c+d x)}{5 b^2 d \sqrt {b \cos (c+d x)}}-\frac {\left (3 \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^3 \sqrt {\cos (c+d x)}} \\ & = -\frac {6 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt {\cos (c+d x)}}+\frac {2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {6 \sin (c+d x)}{5 b^2 d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.70 \[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {-6 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 \sin (c+d x)+2 \sec (c+d x) \tan (c+d x)}{5 b^2 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(-6*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 6*Sin[c + d*x] + 2*Sec[c + d*x]*Tan[c + d*x])/(5*b^2*d*Sqrt
[b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(366\) vs. \(2(109)=218\).

Time = 2.78 (sec) , antiderivative size = 367, normalized size of antiderivative = 3.78

method	result	size
default	\(-\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{5 b^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(367\)

[In]

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

[Out]

-2/5*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)/b^3/sin(1/2*d*x+1/2*c)^3/(8*sin(1/2*d*x+1/2*c
)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/
2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^
4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+
1/2*c)^4*b+b*sin(1/2*d*x+1/2*c)^2)^(1/2)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.25 \[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {-3 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right )}{5 \, b^{3} d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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