\(\int \cos ^{\frac {5}{2}}(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1175]

Optimal result

Integrand size = 30, antiderivative size = 61 \[ \int \cos ^{\frac {5}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

2*C*(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))/d+2/3*B*(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))/d+2/3*B*sin(d*x+c)*cos(d*x+c)^(1/2)/
d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4149, 2827, 2719, 2715, 2720} \[ \int \cos ^{\frac {5}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d} \]

[In]

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

[Out]

(2*C*EllipticE[(c + d*x)/2, 2])/d + (2*B*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*B*Sqrt[Cos[c + d*x]]*Sin[c + d*
x])/(3*d)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 4149

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {\cos (c+d x)} (C+B \cos (c+d x)) \, dx \\ & = B \int \cos ^{\frac {3}{2}}(c+d x) \, dx+C \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} B \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \cos ^{\frac {5}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (3 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+B \left (\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} \sin (c+d x)\right )\right )}{3 d} \]

[In]

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

[Out]

(2*(3*C*EllipticE[(c + d*x)/2, 2] + B*(EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*Sin[c + d*x])))/(3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(227\) vs. \(2(107)=214\).

Time = 11.12 (sec) , antiderivative size = 228, normalized size of antiderivative = 3.74

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 B \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(228\)

[In]

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

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*B*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-3*C*(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(co
s(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*
d*x+1/2*c)^2-1)^(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 = 125, normalized size of antiderivative = 2.05 \[ \int \cos ^{\frac {5}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, B \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} C {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} C {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, d} \]

[In]

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

[Out]

1/3*(2*B*sqrt(cos(d*x + c))*sin(d*x + c) - I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
)) + I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*C*weierstrassZeta(-4,
0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*C*weierstrassZeta(-4, 0, weierstr
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \cos ^{\frac {5}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,B\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d} \]

[In]

int(cos(c + d*x)^(5/2)*(B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*B*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*ellipticE(c/2 + (d*x)/2, 2))/d + (2*B*cos(c + d*x)^(1/2)*sin(c
+ d*x))/(3*d)