Integrand size = 21, antiderivative size = 120 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx=\frac {130\ 13^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\arctan \left (\frac {3}{2}\right )\right ),2\right )}{21 d}-\frac {130 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}{21 d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}{7 d} \]
130/21*13^(3/4)*(cos(1/2*c+1/2*d*x-1/2*arctan(3/2))^2)^(1/2)/cos(1/2*c+1/2 *d*x-1/2*arctan(3/2))*EllipticF(sin(1/2*c+1/2*d*x-1/2*arctan(3/2)),2^(1/2) )/d-2/7*(3*cos(d*x+c)-2*sin(d*x+c))*(2*cos(d*x+c)+3*sin(d*x+c))^(5/2)/d-13 0/21*(3*cos(d*x+c)-2*sin(d*x+c))*(2*cos(d*x+c)+3*sin(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.46 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.28 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx=\frac {-\sqrt {2 \cos (c+d x)+3 \sin (c+d x)} (897 \cos (c+d x)+27 \cos (3 (c+d x))-598 \sin (c+d x)+138 \sin (3 (c+d x)))+260\ 13^{3/4} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\arctan \left (\frac {2}{3}\right )\right )\right ) \sec \left (c+d x+\arctan \left (\frac {2}{3}\right )\right ) \sqrt {-\left (\left (-1+\sin \left (c+d x+\arctan \left (\frac {2}{3}\right )\right )\right ) \sin \left (c+d x+\arctan \left (\frac {2}{3}\right )\right )\right )} \sqrt {1+\sin \left (c+d x+\arctan \left (\frac {2}{3}\right )\right )}}{42 d} \]
(-(Sqrt[2*Cos[c + d*x] + 3*Sin[c + d*x]]*(897*Cos[c + d*x] + 27*Cos[3*(c + d*x)] - 598*Sin[c + d*x] + 138*Sin[3*(c + d*x)])) + 260*13^(3/4)*Hypergeo metricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[2/3]]^2]*Sec[c + d*x + A rcTan[2/3]]*Sqrt[-((-1 + Sin[c + d*x + ArcTan[2/3]])*Sin[c + d*x + ArcTan[ 2/3]])]*Sqrt[1 + Sin[c + d*x + ArcTan[2/3]]])/(42*d)
Time = 0.49 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3552, 3042, 3552, 3042, 3556, 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.
\(\displaystyle \int (3 \sin (c+d x)+2 \cos (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (3 \sin (c+d x)+2 \cos (c+d x))^{7/2}dx\) |
\(\Big \downarrow \) 3552 |
\(\displaystyle \frac {65}{7} \int (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {65}{7} \int (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3552 |
\(\displaystyle \frac {65}{7} \left (\frac {13}{3} \int \frac {1}{\sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}{3 d}\right )-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {65}{7} \left (\frac {13}{3} \int \frac {1}{\sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}{3 d}\right )-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3556 |
\(\displaystyle \frac {65}{7} \left (\frac {1}{3} 13^{3/4} \int \frac {1}{\sqrt {\cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}{3 d}\right )-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {65}{7} \left (\frac {1}{3} 13^{3/4} \int \frac {1}{\sqrt {\sin \left (c+d x-\arctan \left (\frac {3}{2}\right )+\frac {\pi }{2}\right )}}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}{3 d}\right )-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {65}{7} \left (\frac {2\ 13^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\arctan \left (\frac {3}{2}\right )\right ),2\right )}{3 d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}{3 d}\right )-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}\) |
(-2*(3*Cos[c + d*x] - 2*Sin[c + d*x])*(2*Cos[c + d*x] + 3*Sin[c + d*x])^(5 /2))/(7*d) + (65*((2*13^(3/4)*EllipticF[(c + d*x - ArcTan[3/2])/2, 2])/(3* d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x])*Sqrt[2*Cos[c + d*x] + 3*Sin[c + d*x]])/(3*d)))/7
3.3.40.3.1 Defintions of rubi rules used
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[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(-(b*Cos[c + d*x] - a*Sin[c + d*x]))*((a*Cos[c + d*x] + b* Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[(n - 1)*((a^2 + b^2)/n) Int[(a*Co s[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && N eQ[a^2 + b^2, 0] && !IntegerQ[(n - 1)/2] && GtQ[n, 1]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(a^2 + b^2)^(n/2) Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && !(GeQ[n, 1] || LeQ[n, -1]) && GtQ[a^2 + b^2, 0]
Time = 1.13 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {\frac {338 \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )^{4} \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}{7}+\frac {845 \sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )}{21}-\frac {2704 \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )^{2} \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}{21}}{\cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \sqrt {\sqrt {13}\, \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, d}\) | \(128\) |
(338/7*cos(d*x+c+arctan(2/3))^4*sin(d*x+c+arctan(2/3))+845/21*(sin(d*x+c+a rctan(2/3))+1)^(1/2)*(-2*sin(d*x+c+arctan(2/3))+2)^(1/2)*(-sin(d*x+c+arcta n(2/3)))^(1/2)*EllipticF((sin(d*x+c+arctan(2/3))+1)^(1/2),1/2*2^(1/2))-270 4/21*cos(d*x+c+arctan(2/3))^2*sin(d*x+c+arctan(2/3)))/cos(d*x+c+arctan(2/3 ))/(13^(1/2)*sin(d*x+c+arctan(2/3)))^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx=\frac {\left (130 i + 195\right ) \, \sqrt {3 i + 2} \sqrt {2} {\rm weierstrassPInverse}\left (\frac {48}{13} i + \frac {20}{13}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - \left (130 i - 195\right ) \, \sqrt {2} \sqrt {-3 i + 2} {\rm weierstrassPInverse}\left (-\frac {48}{13} i + \frac {20}{13}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (27 \, \cos \left (d x + c\right )^{3} + 46 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 204 \, \cos \left (d x + c\right )\right )} \sqrt {2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )}}{21 \, d} \]
1/21*((130*I + 195)*sqrt(3*I + 2)*sqrt(2)*weierstrassPInverse(48/13*I + 20 /13, 0, cos(d*x + c) - I*sin(d*x + c)) - (130*I - 195)*sqrt(2)*sqrt(-3*I + 2)*weierstrassPInverse(-48/13*I + 20/13, 0, cos(d*x + c) + I*sin(d*x + c) ) - 2*(27*cos(d*x + c)^3 + 46*(3*cos(d*x + c)^2 - 4)*sin(d*x + c) + 204*co s(d*x + c))*sqrt(2*cos(d*x + c) + 3*sin(d*x + c)))/d
Timed out. \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx=\int { {\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {7}{2}} \,d x } \]
\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx=\int { {\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx=\int {\left (2\,\cos \left (c+d\,x\right )+3\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \]