Integrand size = 21, antiderivative size = 123 \[ \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx=\frac {30 b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{77 d \sqrt {b \cos (c+d x)}}+\frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^2 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^4 d} \]
18/77*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^2/d+2/11*(b*cos(d*x+c))^(9/2)*sin( d*x+c)/b^4/d+30/77*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip ticF(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)/d
Time = 0.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx=\frac {\sqrt {b \cos (c+d x)} \left (240 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (290 \sin (c+d x)+57 \sin (3 (c+d x))+7 \sin (5 (c+d x)))\right )}{616 d \sqrt {\cos (c+d x)}} \]
(Sqrt[b*Cos[c + d*x]]*(240*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]* (290*Sin[c + d*x] + 57*Sin[3*(c + d*x)] + 7*Sin[5*(c + d*x)])))/(616*d*Sqr t[Cos[c + d*x]])
Time = 0.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15, 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.
| \(\displaystyle \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle \frac {\int (b \cos (c+d x))^{11/2}dx}{b^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{11/2}dx}{b^5}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \int (b \cos (c+d x))^{7/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \int (b \cos (c+d x))^{3/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {9}{11} b^2 \left (\frac {5}{7} b^2 \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 d}}{b^5}\) |
((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^5
3.1.66.3.1 Defintions of rubi rules used
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]
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]
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]
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]
Time = 6.76 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.90
| method | result | size |
| default | \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \left (448 \left (\cos ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1568 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2384 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2040 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1084 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-370 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+62 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{77 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) | \(234\) |
-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)*b*(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)^9-20 40*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)*Ellipti cF(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
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx=\frac {2 \, {\left (7 \, \cos \left (d x + c\right )^{4} + 9 \, \cos \left (d x + c\right )^{2} + 15\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{77 \, d} \]
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)))/d
Timed out. \[ \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx=\text {Timed out} \]
\[ \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{5} \,d x } \]
\[ \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(c+d x) \sqrt {b \cos (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^5\,\sqrt {b\,\cos \left (c+d\,x\right )} \,d x \]