Integrand size = 14, antiderivative size = 119 \[ \int (a+a \cos (c+d x))^{7/2} \, dx=\frac {256 a^4 \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {64 a^3 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{35 d}+\frac {24 a^2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 a (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \]
24/35*a^2*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*a*(a+a*cos(d*x+c))^(5/2) *sin(d*x+c)/d+256/35*a^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+64/35*a^3*sin (d*x+c)*(a+a*cos(d*x+c))^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70 \[ \int (a+a \cos (c+d x))^{7/2} \, dx=\frac {a^3 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (1225 \sin \left (\frac {1}{2} (c+d x)\right )+245 \sin \left (\frac {3}{2} (c+d x)\right )+49 \sin \left (\frac {5}{2} (c+d x)\right )+5 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{140 d} \]
(a^3*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(1225*Sin[(c + d*x)/2] + 245*Sin[(3*(c + d*x))/2] + 49*Sin[(5*(c + d*x))/2] + 5*Sin[(7*(c + d*x))/2 ]))/(140*d)
Time = 0.44 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3126, 3042, 3126, 3042, 3126, 3042, 3125}
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.
\(\displaystyle \int (a \cos (c+d x)+a)^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{7/2}dx\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {12}{7} a \int (\cos (c+d x) a+a)^{5/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {12}{7} a \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \int (\cos (c+d x) a+a)^{3/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \left (\frac {8 a^2 \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d}\) |
(2*a*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d) + (12*a*((2*a*(a + a*C os[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (8*a*((8*a^2*Sin[c + d*x])/(3*d*S qrt[a + a*Cos[c + d*x]]) + (2*a*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3* d)))/5))/7
3.1.1.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Time = 0.78 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {16 a^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (5 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16\right ) \sqrt {2}}{35 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(86\) |
16/35*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)*(5*cos(1/2*d*x+1/2*c)^6+6* cos(1/2*d*x+1/2*c)^4+8*cos(1/2*d*x+1/2*c)^2+16)*2^(1/2)/(a*cos(1/2*d*x+1/2 *c)^2)^(1/2)/d
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.63 \[ \int (a+a \cos (c+d x))^{7/2} \, dx=\frac {2 \, {\left (5 \, a^{3} \cos \left (d x + c\right )^{3} + 27 \, a^{3} \cos \left (d x + c\right )^{2} + 71 \, a^{3} \cos \left (d x + c\right ) + 177 \, a^{3}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{35 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/35*(5*a^3*cos(d*x + c)^3 + 27*a^3*cos(d*x + c)^2 + 71*a^3*cos(d*x + c) + 177*a^3)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int (a+a \cos (c+d x))^{7/2} \, dx=\text {Timed out} \]
Time = 0.47 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.65 \[ \int (a+a \cos (c+d x))^{7/2} \, dx=\frac {{\left (5 \, \sqrt {2} a^{3} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 49 \, \sqrt {2} a^{3} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 245 \, \sqrt {2} a^{3} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1225 \, \sqrt {2} a^{3} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{140 \, d} \]
1/140*(5*sqrt(2)*a^3*sin(7/2*d*x + 7/2*c) + 49*sqrt(2)*a^3*sin(5/2*d*x + 5 /2*c) + 245*sqrt(2)*a^3*sin(3/2*d*x + 3/2*c) + 1225*sqrt(2)*a^3*sin(1/2*d* x + 1/2*c))*sqrt(a)/d
Time = 0.42 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.91 \[ \int (a+a \cos (c+d x))^{7/2} \, dx=\frac {\sqrt {2} {\left (5 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 49 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 245 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1225 \, a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{140 \, d} \]
1/140*sqrt(2)*(5*a^3*sgn(cos(1/2*d*x + 1/2*c))*sin(7/2*d*x + 7/2*c) + 49*a ^3*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 245*a^3*sgn(cos(1/2*d* x + 1/2*c))*sin(3/2*d*x + 3/2*c) + 1225*a^3*sgn(cos(1/2*d*x + 1/2*c))*sin( 1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{7/2} \, dx=\int {\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2} \,d x \]