Integrand size = 22, antiderivative size = 139 \[ \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx=-\frac {320 (3 \cos (d+e x)-4 \sin (d+e x))}{3 e \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}}+\frac {16 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}}{3 e}-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}{5 e} \]
-2/5*(3*cos(e*x+d)-4*sin(e*x+d))*(-5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2)/e-32 0/3*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2)+16/ 3*(3*cos(e*x+d)-4*sin(e*x+d))*(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2)/e
Time = 5.84 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx=\frac {(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \left (11250 \cos \left (\frac {1}{2} (d+e x)\right )-1125 \cos \left (\frac {3}{2} (d+e x)\right )-9 \cos \left (\frac {5}{2} (d+e x)\right )+3750 \sin \left (\frac {1}{2} (d+e x)\right )-1625 \sin \left (\frac {3}{2} (d+e x)\right )+237 \sin \left (\frac {5}{2} (d+e x)\right )\right )}{30 e \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )^5} \]
((-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2)*(11250*Cos[(d + e*x)/2] - 11 25*Cos[(3*(d + e*x))/2] - 9*Cos[(5*(d + e*x))/2] + 3750*Sin[(d + e*x)/2] - 1625*Sin[(3*(d + e*x))/2] + 237*Sin[(5*(d + e*x))/2]))/(30*e*(Cos[(d + e* x)/2] - 3*Sin[(d + e*x)/2])^5)
Time = 0.41 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3592, 3042, 3592, 3042, 3591}
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.
\(\displaystyle \int (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}dx\) |
\(\Big \downarrow \) 3592 |
\(\displaystyle -8 \int (4 \cos (d+e x)+3 \sin (d+e x)-5)^{3/2}dx-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{5 e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -8 \int (4 \cos (d+e x)+3 \sin (d+e x)-5)^{3/2}dx-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{5 e}\) |
\(\Big \downarrow \) 3592 |
\(\displaystyle -8 \left (-\frac {20}{3} \int \sqrt {4 \cos (d+e x)+3 \sin (d+e x)-5}dx-\frac {2 \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5} (3 \cos (d+e x)-4 \sin (d+e x))}{3 e}\right )-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{5 e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -8 \left (-\frac {20}{3} \int \sqrt {4 \cos (d+e x)+3 \sin (d+e x)-5}dx-\frac {2 \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5} (3 \cos (d+e x)-4 \sin (d+e x))}{3 e}\right )-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{5 e}\) |
\(\Big \downarrow \) 3591 |
\(\displaystyle -\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{5 e}-8 \left (\frac {40 (3 \cos (d+e x)-4 \sin (d+e x))}{3 e \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}{3 e}\right )\) |
(-2*(3*Cos[d + e*x] - 4*Sin[d + e*x])*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x ])^(3/2))/(5*e) - 8*((40*(3*Cos[d + e*x] - 4*Sin[d + e*x]))/(3*e*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]]) - (2*(3*Cos[d + e*x] - 4*Sin[d + e*x])* Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])/(3*e))
3.5.24.3.1 Defintions of rubi rules used
Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_ )]], x_Symbol] :> Simp[-2*((c*Cos[d + e*x] - b*Sin[d + e*x])/(e*Sqrt[a + b* Cos[d + e*x] + c*Sin[d + e*x]])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^ 2 - b^2 - c^2, 0]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (n_), x_Symbol] :> Simp[(-(c*Cos[d + e*x] - b*Sin[d + e*x]))*((a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1)/(e*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e }, x] && EqQ[a^2 - b^2 - c^2, 0] && GtQ[n, 0]
Time = 0.84 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.53
method | result | size |
default | \(\frac {50 \left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \left (1+\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right ) \left (3 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )^{2}-14 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+43\right )}{3 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\) | \(74\) |
50/3*(sin(e*x+d+arctan(4/3))-1)*(1+sin(e*x+d+arctan(4/3)))*(3*sin(e*x+d+ar ctan(4/3))^2-14*sin(e*x+d+arctan(4/3))+43)/cos(e*x+d+arctan(4/3))/(-5+5*si n(e*x+d+arctan(4/3)))^(1/2)/e
Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.73 \[ \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx=-\frac {2 \, {\left (9 \, \cos \left (e x + d\right )^{3} + 567 \, \cos \left (e x + d\right )^{2} - {\left (237 \, \cos \left (e x + d\right )^{2} - 694 \, \cos \left (e x + d\right ) + 472\right )} \sin \left (e x + d\right ) - 2538 \, \cos \left (e x + d\right ) - 3096\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{15 \, {\left (e \cos \left (e x + d\right ) - 3 \, e \sin \left (e x + d\right ) + e\right )}} \]
-2/15*(9*cos(e*x + d)^3 + 567*cos(e*x + d)^2 - (237*cos(e*x + d)^2 - 694*c os(e*x + d) + 472)*sin(e*x + d) - 2538*cos(e*x + d) - 3096)*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5)/(e*cos(e*x + d) - 3*e*sin(e*x + d) + e)
Timed out. \[ \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx=\int { {\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {5}{2}} \,d x } \]
\[ \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx=\int { {\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx=\int {\left (4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5\right )}^{5/2} \,d x \]