Integrand size = 21, antiderivative size = 75 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=\frac {78 \sqrt [4]{13} E\left (\left .\frac {1}{2} \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right |2\right )}{5 d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}}{5 d} \] Output:
78/5*13^(1/4)*EllipticE(sin(1/2*c+1/2*d*x-1/2*arctan(3/2)),2^(1/2))/d-2/5* (3*cos(d*x+c)-2*sin(d*x+c))*(2*cos(d*x+c)+3*sin(d*x+c))^(3/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.77 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.65 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {2 \cos (c+d x)+3 \sin (c+d x)} (52-12 \cos (2 (c+d x))-5 \sin (2 (c+d x)))-\frac {13 \sqrt [4]{13} \left (4 \cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )-3 \sin \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right )}{\sqrt {\cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}}-\frac {39 \sqrt [4]{13} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right ) \sin \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}{\sqrt {-\left (\left (-1+\cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right ) \cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right )} \sqrt {1+\cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}}}{5 d} \] Input:
Integrate[(2*Cos[c + d*x] + 3*Sin[c + d*x])^(5/2),x]
Output:
(Sqrt[2*Cos[c + d*x] + 3*Sin[c + d*x]]*(52 - 12*Cos[2*(c + d*x)] - 5*Sin[2 *(c + d*x)]) - (13*13^(1/4)*(4*Cos[c + d*x - ArcTan[3/2]] - 3*Sin[c + d*x - ArcTan[3/2]]))/Sqrt[Cos[c + d*x - ArcTan[3/2]]] - (39*13^(1/4)*Hypergeom etricPFQ[{-1/2, -1/4}, {3/4}, Cos[c + d*x - ArcTan[3/2]]^2]*Sin[c + d*x - ArcTan[3/2]])/(Sqrt[-((-1 + Cos[c + d*x - ArcTan[3/2]])*Cos[c + d*x - ArcT an[3/2]])]*Sqrt[1 + Cos[c + d*x - ArcTan[3/2]]]))/(5*d)
Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3552, 3042, 3556, 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.
| \(\displaystyle \int (3 \sin (c+d x)+2 \cos (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 3552 |
| \(\displaystyle \frac {39}{5} \int \sqrt {2 \cos (c+d x)+3 \sin (c+d x)}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {39}{5} \int \sqrt {2 \cos (c+d x)+3 \sin (c+d x)}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3556 |
| \(\displaystyle \frac {39}{5} \sqrt [4]{13} \int \sqrt {\cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {39}{5} \sqrt [4]{13} \int \sqrt {\sin \left (c+d x-\arctan \left (\frac {3}{2}\right )+\frac {\pi }{2}\right )}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {78 \sqrt [4]{13} E\left (\left .\frac {1}{2} \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right |2\right )}{5 d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}{5 d}\) |
Input:
Int[(2*Cos[c + d*x] + 3*Sin[c + d*x])^(5/2),x]
Output:
(78*13^(1/4)*EllipticE[(c + d*x - ArcTan[3/2])/2, 2])/(5*d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x])*(2*Cos[c + d*x] + 3*Sin[c + d*x])^(3/2))/(5*d)
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[(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]
Leaf count of result is larger than twice the leaf count of optimal. \(173\) vs. \(2(67)=134\).
Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.32
| method | result | size |
| default | \(-\frac {13 \sqrt {13}\, \left (6 \sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )^{4}+2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )^{2}\right )}{5 \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \sqrt {\sqrt {13}\, \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, d}\) | \(174\) |
Input:
int((2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-13/5*13^(1/2)*(6*(sin(d*x+c+arctan(2/3))+1)^(1/2)*(-2*sin(d*x+c+arctan(2/ 3))+2)^(1/2)*(-sin(d*x+c+arctan(2/3)))^(1/2)*EllipticE((sin(d*x+c+arctan(2 /3))+1)^(1/2),1/2*2^(1/2))-3*(sin(d*x+c+arctan(2/3))+1)^(1/2)*(-2*sin(d*x+ c+arctan(2/3))+2)^(1/2)*(-sin(d*x+c+arctan(2/3)))^(1/2)*EllipticF((sin(d*x +c+arctan(2/3))+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c+arctan(2/3))^4+2*sin(d*x +c+arctan(2/3))^2)/cos(d*x+c+arctan(2/3))/(13^(1/2)*sin(d*x+c+arctan(2/3)) )^(1/2)/d
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.39 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left ({\left (12 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6\right )} \sqrt {2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )} + 39 i \, \sqrt {\frac {3}{2} i + 1} {\rm weierstrassZeta}\left (\frac {48}{13} i + \frac {20}{13}, 0, {\rm weierstrassPInverse}\left (\frac {48}{13} i + \frac {20}{13}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 39 i \, \sqrt {-\frac {3}{2} i + 1} {\rm weierstrassZeta}\left (-\frac {48}{13} i + \frac {20}{13}, 0, {\rm weierstrassPInverse}\left (-\frac {48}{13} i + \frac {20}{13}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right )\right )}}{5 \, d} \] Input:
integrate((2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/5*((12*cos(d*x + c)^2 + 5*cos(d*x + c)*sin(d*x + c) - 6)*sqrt(2*cos(d*x + c) + 3*sin(d*x + c)) + 39*I*sqrt(3/2*I + 1)*weierstrassZeta(48/13*I + 2 0/13, 0, weierstrassPInverse(48/13*I + 20/13, 0, cos(d*x + c) - I*sin(d*x + c))) - 39*I*sqrt(-3/2*I + 1)*weierstrassZeta(-48/13*I + 20/13, 0, weiers trassPInverse(-48/13*I + 20/13, 0, cos(d*x + c) + I*sin(d*x + c))))/d
Timed out. \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate((2*cos(d*x+c)+3*sin(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=\int { {\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(5/2), x)
\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=\int { {\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(5/2), x)
Timed out. \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=\int {\left (2\,\cos \left (c+d\,x\right )+3\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int((2*cos(c + d*x) + 3*sin(c + d*x))^(5/2),x)
Output:
int((2*cos(c + d*x) + 3*sin(c + d*x))^(5/2), x)
\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx=\frac {688 \sqrt {2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}+504 \sqrt {2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+2588 \sqrt {2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}+10985 \left (\int \frac {\sqrt {2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )}d x \right ) d}{255 d} \] Input:
int((2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x)
Output:
(688*sqrt(2*cos(c + d*x) + 3*sin(c + d*x))*cos(c + d*x)**2 + 504*sqrt(2*co s(c + d*x) + 3*sin(c + d*x))*cos(c + d*x)*sin(c + d*x) + 2588*sqrt(2*cos(c + d*x) + 3*sin(c + d*x))*sin(c + d*x)**2 + 10985*int((sqrt(2*cos(c + d*x) + 3*sin(c + d*x))*sin(c + d*x)**3)/(2*cos(c + d*x) + 3*sin(c + d*x)),x)*d )/(255*d)