\(\int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [103]

Optimal result

Integrand size = 21, antiderivative size = 125 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {30 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{77 d \sqrt {b \cos (c+d x)}}+\frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 b d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d} \]

[Out]

18/77*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^3/d+2/11*(b*cos(d*x+c))^(9/2)*sin(d*x+c)/b^5/d+30/77*(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)+3
0/77*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/b/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2715, 2721, 2720} \[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^5 d}+\frac {18 \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 b^3 d}+\frac {30 \sin (c+d x) \sqrt {b \cos (c+d x)}}{77 b d}+\frac {30 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{77 d \sqrt {b \cos (c+d x)}} \]

[In]

Int[Cos[c + d*x]^6/Sqrt[b*Cos[c + d*x]],x]

[Out]

(30*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*d*Sqrt[b*Cos[c + d*x]]) + (30*Sqrt[b*Cos[c + d*x]]*Sin[c
+ d*x])/(77*b*d) + (18*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*b^3*d) + (2*(b*Cos[c + d*x])^(9/2)*Sin[c + d*
x])/(11*b^5*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))^{11/2} \, dx}{b^6} \\ & = \frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac {9 \int (b \cos (c+d x))^{7/2} \, dx}{11 b^4} \\ & = \frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac {45 \int (b \cos (c+d x))^{3/2} \, dx}{77 b^2} \\ & = \frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 b d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac {15}{77} \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx \\ & = \frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 b d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d}+\frac {\left (15 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 \sqrt {b \cos (c+d x)}} \\ & = \frac {30 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{77 d \sqrt {b \cos (c+d x)}}+\frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 b d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^3 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.58 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {480 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+347 \sin (2 (c+d x))+64 \sin (4 (c+d x))+7 \sin (6 (c+d x))}{1232 d \sqrt {b \cos (c+d x)}} \]

[In]

Integrate[Cos[c + d*x]^6/Sqrt[b*Cos[c + d*x]],x]

[Out]

(480*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 347*Sin[2*(c + d*x)] + 64*Sin[4*(c + d*x)] + 7*Sin[6*(c +
d*x)])/(1232*d*Sqrt[b*Cos[c + d*x]])

Maple [A] (verified)

Time = 4.56 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.86

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (448 \left (\cos ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1568 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2384 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2040 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1084 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-370 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+62 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{77 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(233\)

[In]

int(cos(d*x+c)^6/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/77*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*(448*cos(1/2*d*x+1/2*c)^13-1568*cos(1/2*d*x+1/
2*c)^11+2384*cos(1/2*d*x+1/2*c)^9-2040*cos(1/2*d*x+1/2*c)^7+1084*cos(1/2*d*x+1/2*c)^5-370*cos(1/2*d*x+1/2*c)^3
+15*(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))+62*co
s(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 = 101, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \, {\left (7 \, \cos \left (d x + c\right )^{4} + 9 \, \cos \left (d x + c\right )^{2} + 15\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{77 \, b d} \]

[In]

integrate(cos(d*x+c)^6/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/77*(2*(7*cos(d*x + c)^4 + 9*cos(d*x + c)^2 + 15)*sqrt(b*cos(d*x + c))*sin(d*x + c) - 15*I*sqrt(2)*sqrt(b)*we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(
d*x + c) - I*sin(d*x + c)))/(b*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{6}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

integrate(cos(d*x+c)^6/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^6/sqrt(b*cos(d*x + c)), x)

Giac [F]

\[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{6}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]

[In]

integrate(cos(d*x+c)^6/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^6/sqrt(b*cos(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^6}{\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

int(cos(c + d*x)^6/(b*cos(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^6/(b*cos(c + d*x))^(1/2), x)