\(\int (c \cos (a+b x))^{5/2} \, dx\) [18]

Optimal result

Integrand size = 12, antiderivative size = 70 \[ \int (c \cos (a+b x))^{5/2} \, dx=\frac {6 c^2 \sqrt {c \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)}}+\frac {2 c (c \cos (a+b x))^{3/2} \sin (a+b x)}{5 b} \]

[Out]

2/5*c*(c*cos(b*x+a))^(3/2)*sin(b*x+a)/b+6/5*c^2*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(
1/2*a+1/2*b*x),2^(1/2))*(c*cos(b*x+a))^(1/2)/b/cos(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 2721, 2719} \[ \int (c \cos (a+b x))^{5/2} \, dx=\frac {6 c^2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {c \cos (a+b x)}}{5 b \sqrt {\cos (a+b x)}}+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{3/2}}{5 b} \]

[In]

Int[(c*Cos[a + b*x])^(5/2),x]

[Out]

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

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 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}& = \frac {2 c (c \cos (a+b x))^{3/2} \sin (a+b x)}{5 b}+\frac {1}{5} \left (3 c^2\right ) \int \sqrt {c \cos (a+b x)} \, dx \\ & = \frac {2 c (c \cos (a+b x))^{3/2} \sin (a+b x)}{5 b}+\frac {\left (3 c^2 \sqrt {c \cos (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \, dx}{5 \sqrt {\cos (a+b x)}} \\ & = \frac {6 c^2 \sqrt {c \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)}}+\frac {2 c (c \cos (a+b x))^{3/2} \sin (a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89 \[ \int (c \cos (a+b x))^{5/2} \, dx=\frac {(c \cos (a+b x))^{5/2} \left (6 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\sqrt {\cos (a+b x)} \sin (2 (a+b x))\right )}{5 b \cos ^{\frac {5}{2}}(a+b x)} \]

[In]

Integrate[(c*Cos[a + b*x])^(5/2),x]

[Out]

((c*Cos[a + b*x])^(5/2)*(6*EllipticE[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x]]*Sin[2*(a + b*x)]))/(5*b*Cos[a + b*x]
^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(212\) vs. \(2(86)=172\).

Time = 2.78 (sec) , antiderivative size = 213, normalized size of antiderivative = 3.04

method	result	size
default	\(-\frac {2 \sqrt {c \left (-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, c^{3} \left (-8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-c \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {c \left (-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, b}\)	\(213\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.30 \[ \int (c \cos (a+b x))^{5/2} \, dx=\frac {2 \, \sqrt {c \cos \left (b x + a\right )} c^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 i \, \sqrt {2} c^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) - 3 i \, \sqrt {2} c^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}{5 \, b} \]

[In]

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

[Out]

1/5*(2*sqrt(c*cos(b*x + a))*c^2*cos(b*x + a)*sin(b*x + a) + 3*I*sqrt(2)*c^(5/2)*weierstrassZeta(-4, 0, weierst
rassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) - 3*I*sqrt(2)*c^(5/2)*weierstrassZeta(-4, 0, weierstrassPI
nverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))))/b

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate((c*cos(b*x + a))^(5/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*cos(b*x + a))^(5/2), x)

Mupad [F(-1)]

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

[In]

int((c*cos(a + b*x))^(5/2),x)

[Out]

int((c*cos(a + b*x))^(5/2), x)