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

3.2.3.1 Optimal result
3.2.3.2 Mathematica [A] (verified)
3.2.3.3 Rubi [A] (verified)
3.2.3.4 Maple [A] (verified)
3.2.3.5 Fricas [C] (verification not implemented)
3.2.3.6 Sympy [F(-1)]
3.2.3.7 Maxima [F]
3.2.3.8 Giac [F]
3.2.3.9 Mupad [F(-1)]

3.2.3.1 Optimal result

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

output
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)*Ellipti 
cF(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)+30/ 
77*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/b/d
 
3.2.3.2 Mathematica [A] (verified)

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

input
Integrate[Cos[c + d*x]^6/Sqrt[b*Cos[c + d*x]],x]
 
output
(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]])
 
3.2.3.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {2030, 3042, 3115, 3042, 3115, 3042, 3115, 3042, 3121, 3042, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{11/2}dx}{b^6}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{11/2}dx}{b^6}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {9}{11} b^2 \int (b \cos (c+d x))^{7/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {9}{11} b^2 \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \int (b \cos (c+d x))^{3/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^6}\)

input
Int[Cos[c + d*x]^6/Sqrt[b*Cos[c + d*x]],x]
 
output
((2*b*(b*Cos[c + d*x])^(9/2)*Sin[c + d*x])/(11*d) + (9*b^2*((2*b*(b*Cos[c 
+ d*x])^(5/2)*Sin[c + d*x])/(7*d) + (5*b^2*((2*b^2*Sqrt[Cos[c + d*x]]*Elli 
pticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*b*Sqrt[b*Cos[c + d* 
x]]*Sin[c + d*x])/(3*d)))/7))/11)/b^6
 

3.2.3.3.1 Defintions of rubi rules used

rule 2030
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115
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] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3120
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 3121
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 
3.2.3.4 Maple [A] (verified)

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

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

input
int(cos(d*x+c)^6/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)
 
output
-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+1 
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))+62*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
 
3.2.3.5 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 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \, {\left (7 \, \cos \left (d x + c\right )^{4} + 9 \, \cos \left (d x + c\right )^{2} + 15\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{77 \, b d} \]

input
integrate(cos(d*x+c)^6/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")
 
output
1/77*(2*(7*cos(d*x + c)^4 + 9*cos(d*x + c)^2 + 15)*sqrt(b*cos(d*x + c))*si 
n(d*x + c) - 15*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-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)
 
3.2.3.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**6/(b*cos(d*x+c))**(1/2),x)
 
output
Timed out
 
3.2.3.7 Maxima [F]

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

input
integrate(cos(d*x+c)^6/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")
 
output
integrate(cos(d*x + c)^6/sqrt(b*cos(d*x + c)), x)
 
3.2.3.8 Giac [F]

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

input
integrate(cos(d*x+c)^6/(b*cos(d*x+c))^(1/2),x, algorithm="giac")
 
output
integrate(cos(d*x + c)^6/sqrt(b*cos(d*x + c)), x)
 
3.2.3.9 Mupad [F(-1)]

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

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