Integrand size = 22, antiderivative size = 139 \[ \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2} \, dx=\frac {16 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{3 e}+\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\arctan \left (\frac {5}{3}\right )\right ),\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{\sqrt {2+\sqrt {34}} e}-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}{3 e} \]
-2/3*(5*cos(e*x+d)-3*sin(e*x+d))*(2+3*cos(e*x+d)+5*sin(e*x+d))^(1/2)/e+20* (cos(1/2*d+1/2*e*x-1/2*arctan(5/3))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan( 5/3))*EllipticF(sin(1/2*d+1/2*e*x-1/2*arctan(5/3)),1/15*(510-30*34^(1/2))^ (1/2))/e/(2+34^(1/2))^(1/2)+16/3*(cos(1/2*d+1/2*e*x-1/2*arctan(5/3))^2)^(1 /2)/cos(1/2*d+1/2*e*x-1/2*arctan(5/3))*EllipticE(sin(1/2*d+1/2*e*x-1/2*arc tan(5/3)),1/15*(510-30*34^(1/2))^(1/2))*(2+34^(1/2))^(1/2)/e
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 2.86 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.51 \[ \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2} \, dx=\frac {2 \left (-60 \sqrt {30} \operatorname {AppellF1}\left (-\frac {1}{2},-\frac {1}{2},-\frac {1}{2},\frac {1}{2},\frac {\sqrt {34}+17 \cos \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )}{-17+\sqrt {34}},\frac {\sqrt {34}+17 \cos \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )}{17+\sqrt {34}}\right ) \sin \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )+\left (-15 (30 \cos (d+e x)+15 \cos (2 (d+e x))-18 \sin (d+e x)+8 \sin (2 (d+e x)))+23 \sqrt {30} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2},\frac {3}{2},\frac {\sqrt {34}+17 \sin \left (d+e x+\arctan \left (\frac {3}{5}\right )\right )}{-17+\sqrt {34}},\frac {\sqrt {34}+17 \sin \left (d+e x+\arctan \left (\frac {3}{5}\right )\right )}{17+\sqrt {34}}\right ) \sqrt {\cos ^2\left (d+e x+\arctan \left (\frac {3}{5}\right )\right )} \sqrt {2+\sqrt {34} \cos \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )} \sec \left (d+e x+\arctan \left (\frac {3}{5}\right )\right ) \sqrt {2+\sqrt {34} \sin \left (d+e x+\arctan \left (\frac {3}{5}\right )\right )}\right ) \sqrt {\sin ^2\left (d+e x-\arctan \left (\frac {5}{3}\right )\right )}\right )}{45 e \sqrt {2+\sqrt {34} \cos \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )} \sqrt {\sin ^2\left (d+e x-\arctan \left (\frac {5}{3}\right )\right )}} \]
(2*(-60*Sqrt[30]*AppellF1[-1/2, -1/2, -1/2, 1/2, (Sqrt[34] + 17*Cos[d + e* x - ArcTan[5/3]])/(-17 + Sqrt[34]), (Sqrt[34] + 17*Cos[d + e*x - ArcTan[5/ 3]])/(17 + Sqrt[34])]*Sin[d + e*x - ArcTan[5/3]] + (-15*(30*Cos[d + e*x] + 15*Cos[2*(d + e*x)] - 18*Sin[d + e*x] + 8*Sin[2*(d + e*x)]) + 23*Sqrt[30] *AppellF1[1/2, 1/2, 1/2, 3/2, (Sqrt[34] + 17*Sin[d + e*x + ArcTan[3/5]])/( -17 + Sqrt[34]), (Sqrt[34] + 17*Sin[d + e*x + ArcTan[3/5]])/(17 + Sqrt[34] )]*Sqrt[Cos[d + e*x + ArcTan[3/5]]^2]*Sqrt[2 + Sqrt[34]*Cos[d + e*x - ArcT an[5/3]]]*Sec[d + e*x + ArcTan[3/5]]*Sqrt[2 + Sqrt[34]*Sin[d + e*x + ArcTa n[3/5]]])*Sqrt[Sin[d + e*x - ArcTan[5/3]]^2]))/(45*e*Sqrt[2 + Sqrt[34]*Cos [d + e*x - ArcTan[5/3]]]*Sqrt[Sin[d + e*x - ArcTan[5/3]]^2])
Time = 0.70 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3599, 3042, 3628, 3042, 3597, 3042, 3132, 3605, 3042, 3140}
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.
| \(\displaystyle \int (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}dx\) |
| \(\Big \downarrow \) 3599 |
| \(\displaystyle \frac {2}{3} \int \frac {12 \cos (d+e x)+20 \sin (d+e x)+23}{\sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}}dx-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} \int \frac {12 \cos (d+e x)+20 \sin (d+e x)+23}{\sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}}dx-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3628 |
| \(\displaystyle \frac {2}{3} \left (15 \int \frac {1}{\sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}}dx+4 \int \sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}dx\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} \left (15 \int \frac {1}{\sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}}dx+4 \int \sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}dx\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3597 |
| \(\displaystyle \frac {2}{3} \left (4 \int \sqrt {\sqrt {34} \cos \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )+2}dx+15 \int \frac {1}{\sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}}dx\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} \left (4 \int \sqrt {\sqrt {34} \sin \left (d+e x-\arctan \left (\frac {5}{3}\right )+\frac {\pi }{2}\right )+2}dx+15 \int \frac {1}{\sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}}dx\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2}{3} \left (15 \int \frac {1}{\sqrt {3 \cos (d+e x)+5 \sin (d+e x)+2}}dx+\frac {8 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{e}\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3605 |
| \(\displaystyle \frac {2}{3} \left (15 \int \frac {1}{\sqrt {\sqrt {34} \cos \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )+2}}dx+\frac {8 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{e}\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} \left (15 \int \frac {1}{\sqrt {\sqrt {34} \sin \left (d+e x-\arctan \left (\frac {5}{3}\right )+\frac {\pi }{2}\right )+2}}dx+\frac {8 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{e}\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2}{3} \left (\frac {30 \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\arctan \left (\frac {5}{3}\right )\right ),\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{\sqrt {2+\sqrt {34}} e}+\frac {8 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\arctan \left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{e}\right )-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{3 e}\) |
(2*((8*Sqrt[2 + Sqrt[34]]*EllipticE[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sq rt[34]))/15])/e + (30*EllipticF[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[3 4]))/15])/(Sqrt[2 + Sqrt[34]]*e)))/3 - (2*(5*Cos[d + e*x] - 3*Sin[d + e*x] )*Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]])/(3*e)
3.5.4.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_ )]], x_Symbol] :> Int[Sqrt[a + Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]] , x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 + c^2, 0] && GtQ[a + Sqrt[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[1/n Int[Simp[n*a^2 + ( n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x ], x]*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]
Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*( x_)]], x_Symbol] :> Int[1/Sqrt[a + Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 + c^2, 0] && GtQ[a + Sqrt[ b^2 + c^2], 0]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]] , x_Symbol] :> Simp[B/b Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x] , x] + Simp[(A*b - a*B)/b Int[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]] , x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && EqQ[B*c - b*C, 0] && NeQ[ A*b - a*B, 0]
Result contains complex when optimal does not.
Time = 1.34 (sec) , antiderivative size = 806, normalized size of antiderivative = 5.80
(76/17*((17*sin(e*x+d+arctan(3/5))+34^(1/2))/(34^(1/2)+17))^(1/2)*17^(1/2) *((sin(e*x+d+arctan(3/5))+1)/(-34^(1/2)+17))^(1/2)*(-17*(sin(e*x+d+arctan( 3/5))-1)/(34^(1/2)+17))^(1/2)*EllipticF(((17*sin(e*x+d+arctan(3/5))+34^(1/ 2))/(34^(1/2)+17))^(1/2),I*(1/(-34^(1/2)+17)*(34^(1/2)+17))^(1/2))*34^(1/2 )+36*((17*sin(e*x+d+arctan(3/5))+34^(1/2))/(34^(1/2)+17))^(1/2)*17^(1/2)*( (sin(e*x+d+arctan(3/5))+1)/(-34^(1/2)+17))^(1/2)*(-17*(sin(e*x+d+arctan(3/ 5))-1)/(34^(1/2)+17))^(1/2)*EllipticF(((17*sin(e*x+d+arctan(3/5))+34^(1/2) )/(34^(1/2)+17))^(1/2),I*(1/(-34^(1/2)+17)*(34^(1/2)+17))^(1/2))+68/3*sin( e*x+d+arctan(3/5))^3-68/3*sin(e*x+d+arctan(3/5))+4/3*34^(1/2)*sin(e*x+d+ar ctan(3/5))^2-4/3*34^(1/2)-40/17*34^(1/2)*((17*sin(e*x+d+arctan(3/5))+34^(1 /2))/(34^(1/2)+17))^(1/2)*17^(1/2)*((sin(e*x+d+arctan(3/5))+1)/(-34^(1/2)+ 17))^(1/2)*(-17*(sin(e*x+d+arctan(3/5))-1)/(34^(1/2)+17))^(1/2)*EllipticE( ((17*sin(e*x+d+arctan(3/5))+34^(1/2))/(34^(1/2)+17))^(1/2),I*(1/(-34^(1/2) +17)*(34^(1/2)+17))^(1/2))+120/17*34^(1/2)*17^(1/2)*((sin(e*x+d+arctan(3/5 ))+1)/(-34^(1/2)+17))^(1/2)*(-17*(sin(e*x+d+arctan(3/5))-1)/(34^(1/2)+17)) ^(1/2)*(-(17*sin(e*x+d+arctan(3/5))+34^(1/2))/(-34^(1/2)+17))^(1/2)*Ellipt icE((-(17*sin(e*x+d+arctan(3/5))+34^(1/2))/(-34^(1/2)+17))^(1/2),I*((-34^( 1/2)+17)/(34^(1/2)+17))^(1/2))+16*(-(17*sin(e*x+d+arctan(3/5))+34^(1/2))/( -34^(1/2)+17))^(1/2)*(-17*(sin(e*x+d+arctan(3/5))-1)/(34^(1/2)+17))^(1/2)* 17^(1/2)*((sin(e*x+d+arctan(3/5))+1)/(-34^(1/2)+17))^(1/2)*EllipticF((-...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14 \[ \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2} \, dx=\frac {\left (159 i + 265\right ) \, \sqrt {5 i + 3} \sqrt {2} {\rm weierstrassPInverse}\left (\frac {860}{289} i + \frac {1376}{867}, -\frac {5480}{132651} i - \frac {12056}{14739}, \cos \left (e x + d\right ) - i \, \sin \left (e x + d\right ) - \frac {10}{51} i + \frac {2}{17}\right ) - \left (159 i - 265\right ) \, \sqrt {-5 i + 3} \sqrt {2} {\rm weierstrassPInverse}\left (-\frac {860}{289} i + \frac {1376}{867}, \frac {5480}{132651} i - \frac {12056}{14739}, \cos \left (e x + d\right ) + i \, \sin \left (e x + d\right ) + \frac {10}{51} i + \frac {2}{17}\right ) - 408 i \, \sqrt {5 i + 3} \sqrt {2} {\rm weierstrassZeta}\left (\frac {860}{289} i + \frac {1376}{867}, -\frac {5480}{132651} i - \frac {12056}{14739}, {\rm weierstrassPInverse}\left (\frac {860}{289} i + \frac {1376}{867}, -\frac {5480}{132651} i - \frac {12056}{14739}, \cos \left (e x + d\right ) - i \, \sin \left (e x + d\right ) - \frac {10}{51} i + \frac {2}{17}\right )\right ) + 408 i \, \sqrt {-5 i + 3} \sqrt {2} {\rm weierstrassZeta}\left (-\frac {860}{289} i + \frac {1376}{867}, \frac {5480}{132651} i - \frac {12056}{14739}, {\rm weierstrassPInverse}\left (-\frac {860}{289} i + \frac {1376}{867}, \frac {5480}{132651} i - \frac {12056}{14739}, \cos \left (e x + d\right ) + i \, \sin \left (e x + d\right ) + \frac {10}{51} i + \frac {2}{17}\right )\right ) - 102 \, {\left (5 \, \cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right )\right )} \sqrt {3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2}}{153 \, e} \]
1/153*((159*I + 265)*sqrt(5*I + 3)*sqrt(2)*weierstrassPInverse(860/289*I + 1376/867, -5480/132651*I - 12056/14739, cos(e*x + d) - I*sin(e*x + d) - 1 0/51*I + 2/17) - (159*I - 265)*sqrt(-5*I + 3)*sqrt(2)*weierstrassPInverse( -860/289*I + 1376/867, 5480/132651*I - 12056/14739, cos(e*x + d) + I*sin(e *x + d) + 10/51*I + 2/17) - 408*I*sqrt(5*I + 3)*sqrt(2)*weierstrassZeta(86 0/289*I + 1376/867, -5480/132651*I - 12056/14739, weierstrassPInverse(860/ 289*I + 1376/867, -5480/132651*I - 12056/14739, cos(e*x + d) - I*sin(e*x + d) - 10/51*I + 2/17)) + 408*I*sqrt(-5*I + 3)*sqrt(2)*weierstrassZeta(-860 /289*I + 1376/867, 5480/132651*I - 12056/14739, weierstrassPInverse(-860/2 89*I + 1376/867, 5480/132651*I - 12056/14739, cos(e*x + d) + I*sin(e*x + d ) + 10/51*I + 2/17)) - 102*(5*cos(e*x + d) - 3*sin(e*x + d))*sqrt(3*cos(e* x + d) + 5*sin(e*x + d) + 2))/e
\[ \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2} \, dx=\int \left (5 \sin {\left (d + e x \right )} + 3 \cos {\left (d + e x \right )} + 2\right )^{\frac {3}{2}}\, dx \]
\[ \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2} \, dx=\int { {\left (3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2} \, dx=\int { {\left (3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2} \, dx=\int {\left (3\,\cos \left (d+e\,x\right )+5\,\sin \left (d+e\,x\right )+2\right )}^{3/2} \,d x \]