Integrand size = 12, antiderivative size = 98 \[ \int (b \cos (c+d x))^{7/2} \, dx=\frac {10 b^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 b^3 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \]
2/7*b*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+10/21*b^4*(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)+10/21*b^3*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int (b \cos (c+d x))^{7/2} \, dx=\frac {b^3 \sqrt {b \cos (c+d x)} \left (20 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (23 \sin (c+d x)+3 \sin (3 (c+d x)))\right )}{42 d \sqrt {\cos (c+d x)}} \]
(b^3*Sqrt[b*Cos[c + d*x]]*(20*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x ]]*(23*Sin[c + d*x] + 3*Sin[3*(c + d*x)])))/(42*d*Sqrt[Cos[c + d*x]])
Time = 0.41 (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.
\(\displaystyle \int (b \cos (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{7} b^2 \int (b \cos (c+d x))^{3/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} b^2 \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {5}{7} b^2 \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} b^2 \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {5}{7} b^2 \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d}\) |
(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]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*b*Sqr t[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/7
3.2.2.3.1 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 = 3.62 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.14
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{4} \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) | \(210\) |
-2/21*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^4*(48*co s(1/2*d*x+1/2*c)^9-120*cos(1/2*d*x+1/2*c)^7+128*cos(1/2*d*x+1/2*c)^5-72*co s(1/2*d*x+1/2*c)^3+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))+16*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 = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.97 \[ \int (b \cos (c+d x))^{7/2} \, dx=\frac {-5 i \, \sqrt {2} b^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} b^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, b^{3} \cos \left (d x + c\right )^{2} + 5 \, b^{3}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{21 \, d} \]
1/21*(-5*I*sqrt(2)*b^(7/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin (d*x + c)) + 5*I*sqrt(2)*b^(7/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*b^3*cos(d*x + c)^2 + 5*b^3)*sqrt(b*cos(d*x + c))*s in(d*x + c))/d
Timed out. \[ \int (b \cos (c+d x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (b \cos (c+d x))^{7/2} \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
\[ \int (b \cos (c+d x))^{7/2} \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int (b \cos (c+d x))^{7/2} \, dx=\int {\left (b\,\cos \left (c+d\,x\right )\right )}^{7/2} \,d x \]