\(\int \cos ^2(c+d x) (b \cos (c+d x))^{5/2} \, dx\) [91]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \cos ^2(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {14 b^2 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {14 b (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d} \]

[Out]

14/45*b*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/9*(b*cos(d*x+c))^(7/2)*sin(d*x+c)/b/d+14/15*b^2*(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)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2715, 2721, 2719} \[ \int \cos ^2(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {14 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{15 d \sqrt {\cos (c+d x)}}+\frac {2 \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}+\frac {14 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{45 d} \]

[In]

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

[Out]

(14*b^2*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]) + (14*b*(b*Cos[c + d*x])^(3/
2)*Sin[c + d*x])/(45*d) + (2*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*b*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 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 {\int (b \cos (c+d x))^{9/2} \, dx}{b^2} \\ & = \frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac {7}{9} \int (b \cos (c+d x))^{5/2} \, dx \\ & = \frac {14 b (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac {1}{15} \left (7 b^2\right ) \int \sqrt {b \cos (c+d x)} \, dx \\ & = \frac {14 b (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac {\left (7 b^2 \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 \sqrt {\cos (c+d x)}} \\ & = \frac {14 b^2 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {14 b (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {(b \cos (c+d x))^{5/2} \left (168 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} (38 \sin (2 (c+d x))+5 \sin (4 (c+d x)))\right )}{180 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(168*EllipticE[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(38*Sin[2*(c + d*x)] + 5*Sin[4*(c
+ d*x)])))/(180*d*Cos[c + d*x]^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(222\) vs. \(2(110)=220\).

Time = 5.66 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.28

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (160 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+616 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-432 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{45 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(223\)

[In]

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

[Out]

-2/45*((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*(160*cos(1/2*d*x+1/2*c)^11-480*cos(1/2*d*x
+1/2*c)^9+616*cos(1/2*d*x+1/2*c)^7-432*cos(1/2*d*x+1/2*c)^5+160*cos(1/2*d*x+1/2*c)^3-21*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-24*cos(1/2*d*x+1/2*c))/(-b*(2*s
in(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 = 107, normalized size of antiderivative = 1.09 \[ \int \cos ^2(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {21 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (5 \, b^{2} \cos \left (d x + c\right )^{3} + 7 \, b^{2} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, d} \]

[In]

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

[Out]

1/45*(21*I*sqrt(2)*b^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) -
21*I*sqrt(2)*b^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(5
*b^2*cos(d*x + c)^3 + 7*b^2*cos(d*x + c))*sqrt(b*cos(d*x + c))*sin(d*x + c))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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