Integrand size = 21, antiderivative size = 100 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {14 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b d \sqrt {\cos (c+d x)}}+\frac {14 (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 b^2 d}+\frac {2 (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^4 d} \]
14/45*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^2/d+2/9*(b*cos(d*x+c))^(7/2)*sin(d *x+c)/b^4/d+14/15*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic E(sin(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/b/d/cos(d*x+c)^(1/2)
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {168 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\cos (c+d x) (38 \sin (2 (c+d x))+5 \sin (4 (c+d x)))}{180 d \sqrt {b \cos (c+d x)}} \]
(168*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + Cos[c + d*x]*(38*Sin[2 *(c + d*x)] + 5*Sin[4*(c + d*x)]))/(180*d*Sqrt[b*Cos[c + d*x]])
Time = 0.44 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2030, 3042, 3115, 3042, 3115, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \cos (c+d x))^{9/2}dx}{b^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}dx}{b^5}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {7}{9} b^2 \int (b \cos (c+d x))^{5/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 d}}{b^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7}{9} b^2 \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 d}}{b^5}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {7}{9} b^2 \left (\frac {3}{5} b^2 \int \sqrt {b \cos (c+d x)}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 d}}{b^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7}{9} b^2 \left (\frac {3}{5} b^2 \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 d}}{b^5}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\frac {7}{9} b^2 \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 d}}{b^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7}{9} b^2 \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 d}}{b^5}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {7}{9} b^2 \left (\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 d}}{b^5}\) |
((2*b*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*d) + (7*b^2*((6*b^2*Sqrt[b*C os[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b*(b *Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)))/9)/b^5
3.2.4.3.1 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 = 3.90 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.20
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (160 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+616 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-432 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{45 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) | \(220\) |
-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)*(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))-2 4*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 = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} + 7 \, \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right ) + 21 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{45 \, b d} \]
1/45*(2*(5*cos(d*x + c)^3 + 7*cos(d*x + c))*sqrt(b*cos(d*x + c))*sin(d*x + c) + 21*I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*sqrt(b)*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b*d)
Timed out. \[ \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{5}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{5}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {\cos ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5}{\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]