\(\int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx\) [68]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\frac {10 b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^2 d} \]

[Out]

2/7*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^2/d+10/21*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+10/21*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 95, 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, 2720} \[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\frac {2 \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^2 d}+\frac {10 \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 d}+\frac {10 b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}} \]

[In]

Int[Cos[c + d*x]^3*Sqrt[b*Cos[c + d*x]],x]

[Out]

(10*b*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (10*Sqrt[b*Cos[c + d*x]]*Sin
[c + d*x])/(21*d) + (2*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b^2*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}& = \frac {\int (b \cos (c+d x))^{7/2} \, dx}{b^3} \\ & = \frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^2 d}+\frac {5 \int (b \cos (c+d x))^{3/2} \, dx}{7 b} \\ & = \frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^2 d}+\frac {1}{21} (5 b) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx \\ & = \frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^2 d}+\frac {\left (5 b \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}} \\ & = \frac {10 b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\frac {\sqrt {b \cos (c+d x)} \left (20 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (23 \sin (c+d x)+3 \sin (3 (c+d x)))\right )}{42 d \sqrt {\cos (c+d x)}} \]

[In]

Integrate[Cos[c + d*x]^3*Sqrt[b*Cos[c + d*x]],x]

[Out]

(Sqrt[b*Cos[c + d*x]]*(20*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(23*Sin[c + d*x] + 3*Sin[3*(c + d*x)]
)))/(42*d*Sqrt[Cos[c + d*x]])

Maple [A] (verified)

Time = 3.46 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.19

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(208\)

[In]

int(cos(d*x+c)^3*(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/21*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(48*cos(1/2*d*x+1/2*c)^9-120*cos(1/2*d*x+1/2
*c)^7+128*cos(1/2*d*x+1/2*c)^5-72*cos(1/2*d*x+1/2*c)^3+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2
+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+16*cos(1/2*d*x+1/2*c))/(-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 = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\frac {2 \, \sqrt {b \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) - 5 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{21 \, d} \]

[In]

integrate(cos(d*x+c)^3*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cos(d*x+c)^3*(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*cos(d*x + c))*cos(d*x + c)^3, x)

Giac [F]

\[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cos(d*x+c)^3*(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*cos(d*x + c))*cos(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) \sqrt {b \cos (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^3\,\sqrt {b\,\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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