\(\int (b \cos (c+d x))^{5/2} \sec (c+d x) \, dx\) [94]

Optimal result

Integrand size = 19, antiderivative size = 72 \[ \int (b \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\frac {2 b^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {16, 2715, 2721, 2720} \[ \int (b \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\frac {2 b^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b^2 \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d} \]

[In]

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

[Out]

(2*b^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*b^2*Sqrt[b*Cos[c + d*x]]*
Sin[c + d*x])/(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 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 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 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 (b \cos (c+d x))^{3/2} \, dx \\ & = \frac {2 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} b^3 \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx \\ & = \frac {2 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {\left (b^3 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 \sqrt {b \cos (c+d x)}} \\ & = \frac {2 b^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82 \[ \int (b \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\frac {2 b (b \cos (c+d x))^{3/2} \left (\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} \sin (c+d x)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(189\) vs. \(2(88)=176\).

Time = 2.77 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.64

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(190\)

[In]

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

[Out]

-2/3*((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*(4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-
2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(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)*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 = 79, normalized size of antiderivative = 1.10 \[ \int (b \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\frac {-i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} b^{2} \sin \left (d x + c\right )}{3 \, d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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