Integrand size = 12, antiderivative size = 98 \[ \int (c \cos (a+b x))^{7/2} \, dx=\frac {10 c^4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{21 b \sqrt {c \cos (a+b x)}}+\frac {10 c^3 \sqrt {c \cos (a+b x)} \sin (a+b x)}{21 b}+\frac {2 c (c \cos (a+b x))^{5/2} \sin (a+b x)}{7 b} \]
2/7*c*(c*cos(b*x+a))^(5/2)*sin(b*x+a)/b+10/21*c^4*(cos(1/2*a+1/2*b*x)^2)^( 1/2)/cos(1/2*a+1/2*b*x)*EllipticF(sin(1/2*a+1/2*b*x),2^(1/2))*cos(b*x+a)^( 1/2)/b/(c*cos(b*x+a))^(1/2)+10/21*c^3*sin(b*x+a)*(c*cos(b*x+a))^(1/2)/b
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int (c \cos (a+b x))^{7/2} \, dx=\frac {c^3 \sqrt {c \cos (a+b x)} \left (20 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )+\sqrt {\cos (a+b x)} (23 \sin (a+b x)+3 \sin (3 (a+b x)))\right )}{42 b \sqrt {\cos (a+b x)}} \]
(c^3*Sqrt[c*Cos[a + b*x]]*(20*EllipticF[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x ]]*(23*Sin[a + b*x] + 3*Sin[3*(a + b*x)])))/(42*b*Sqrt[Cos[a + b*x]])
Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {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.
\(\displaystyle \int (c \cos (a+b x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (c \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{7/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{7} c^2 \int (c \cos (a+b x))^{3/2}dx+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{5/2}}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} c^2 \int \left (c \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{5/2}}{7 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{7} c^2 \left (\frac {1}{3} c^2 \int \frac {1}{\sqrt {c \cos (a+b x)}}dx+\frac {2 c \sin (a+b x) \sqrt {c \cos (a+b x)}}{3 b}\right )+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{5/2}}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} c^2 \left (\frac {1}{3} c^2 \int \frac {1}{\sqrt {c \sin \left (a+b x+\frac {\pi }{2}\right )}}dx+\frac {2 c \sin (a+b x) \sqrt {c \cos (a+b x)}}{3 b}\right )+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{5/2}}{7 b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {5}{7} c^2 \left (\frac {c^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 \sqrt {c \cos (a+b x)}}+\frac {2 c \sin (a+b x) \sqrt {c \cos (a+b x)}}{3 b}\right )+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{5/2}}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} c^2 \left (\frac {c^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {c \cos (a+b x)}}+\frac {2 c \sin (a+b x) \sqrt {c \cos (a+b x)}}{3 b}\right )+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{5/2}}{7 b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {5}{7} c^2 \left (\frac {2 c^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {c \cos (a+b x)}}+\frac {2 c \sin (a+b x) \sqrt {c \cos (a+b x)}}{3 b}\right )+\frac {2 c \sin (a+b x) (c \cos (a+b x))^{5/2}}{7 b}\) |
(2*c*(c*Cos[a + b*x])^(5/2)*Sin[a + b*x])/(7*b) + (5*c^2*((2*c^2*Sqrt[Cos[ a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*Sqrt[c*Cos[a + b*x]]) + (2*c*Sqr t[c*Cos[a + b*x]]*Sin[a + b*x])/(3*b)))/7
3.1.17.3.1 Defintions of rubi rules used
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 = 4.01 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.14
method | result | size |
default | \(-\frac {2 \sqrt {c \left (-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, c^{4} \left (48 \left (\cos ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{21 \sqrt {-c \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {c \left (-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, b}\) | \(210\) |
-2/21*(c*(-1+2*cos(1/2*b*x+1/2*a)^2)*sin(1/2*b*x+1/2*a)^2)^(1/2)*c^4*(48*c os(1/2*b*x+1/2*a)^9-120*cos(1/2*b*x+1/2*a)^7+128*cos(1/2*b*x+1/2*a)^5-72*c os(1/2*b*x+1/2*a)^3+5*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(-2*cos(1/2*b*x+1/2*a)^ 2+1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))+16*cos(1/2*b*x+1/2*a))/(- c*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/ (c*(-1+2*cos(1/2*b*x+1/2*a)^2))^(1/2)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.97 \[ \int (c \cos (a+b x))^{7/2} \, dx=\frac {-5 i \, \sqrt {2} c^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 5 i \, \sqrt {2} c^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, {\left (3 \, c^{3} \cos \left (b x + a\right )^{2} + 5 \, c^{3}\right )} \sqrt {c \cos \left (b x + a\right )} \sin \left (b x + a\right )}{21 \, b} \]
1/21*(-5*I*sqrt(2)*c^(7/2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin (b*x + a)) + 5*I*sqrt(2)*c^(7/2)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + 2*(3*c^3*cos(b*x + a)^2 + 5*c^3)*sqrt(c*cos(b*x + a))*s in(b*x + a))/b
Timed out. \[ \int (c \cos (a+b x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (c \cos (a+b x))^{7/2} \, dx=\int { \left (c \cos \left (b x + a\right )\right )^{\frac {7}{2}} \,d x } \]
\[ \int (c \cos (a+b x))^{7/2} \, dx=\int { \left (c \cos \left (b x + a\right )\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int (c \cos (a+b x))^{7/2} \, dx=\int {\left (c\,\cos \left (a+b\,x\right )\right )}^{7/2} \,d x \]