Integrand size = 22, antiderivative size = 142 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=-\frac {3 \arctan \left (\frac {\sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {-1+\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}}\right )}{400 \sqrt {10} e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \]
1/20*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(5/2)-3/ 400*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2)-3/4 000*arctan(1/2*sin(d+e*x-arctan(3/4))*2^(1/2)/(-1+cos(d+e*x-arctan(3/4)))^ (1/2))*10^(1/2)/e
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\frac {\left (\frac {1}{10000}+\frac {i}{20000}\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right ) \left ((6+6 i) \sqrt {-20-15 i} \text {arctanh}\left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \left (3+\tan \left (\frac {1}{4} (d+e x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )^4+(10-5 i) \left (165 \cos \left (\frac {1}{2} (d+e x)\right )-27 \cos \left (\frac {3}{2} (d+e x)\right )+55 \sin \left (\frac {1}{2} (d+e x)\right )-39 \sin \left (\frac {3}{2} (d+e x)\right )\right )\right )}{e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \]
((1/10000 + I/20000)*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])*((6 + 6*I)*Sq rt[-20 - 15*I]*ArcTanh[(1/10 + (3*I)/10)*Sqrt[-4/5 - (3*I)/5]*(3 + Tan[(d + e*x)/4])]*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])^4 + (10 - 5*I)*(165*Co s[(d + e*x)/2] - 27*Cos[(3*(d + e*x))/2] + 55*Sin[(d + e*x)/2] - 39*Sin[(3 *(d + e*x))/2])))/(e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2))
Time = 0.50 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {3042, 3595, 3042, 3595, 3042, 3594, 3042, 3128, 217}
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.
| \(\displaystyle \int \frac {1}{(3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}dx\) |
| \(\Big \downarrow \) 3595 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \int \frac {1}{(4 \cos (d+e x)+3 \sin (d+e x)-5)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \int \frac {1}{(4 \cos (d+e x)+3 \sin (d+e x)-5)^{3/2}}dx\) |
| \(\Big \downarrow \) 3595 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {4 \cos (d+e x)+3 \sin (d+e x)-5}}dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {4 \cos (d+e x)+3 \sin (d+e x)-5}}dx\right )\) |
| \(\Big \downarrow \) 3594 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {5 \cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-5}}dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {5 \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )+\frac {\pi }{2}\right )-5}}dx\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {\int \frac {1}{-\frac {5 \sin ^2\left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-1}-10}d\left (-\frac {\sqrt {5} \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-1}}\right )}{10 e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {\arctan \left (\frac {\sin \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )}{\sqrt {2} \sqrt {\cos \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )-1}}\right )}{10 \sqrt {10} e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}\right )\) |
(3*Cos[d + e*x] - 4*Sin[d + e*x])/(20*e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e *x])^(5/2)) - (3*(ArcTan[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[-1 + Cos [d + e*x - ArcTan[3/4]]])]/(10*Sqrt[10]*e) + (3*Cos[d + e*x] - 4*Sin[d + e *x])/(10*e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))))/40
3.5.29.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
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] && 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/(a*e*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) 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] && LtQ[n, -1]
Time = 0.88 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.34
| method | result | size |
| default | \(\frac {\left (3 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )^{2}-6 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+3 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right )-6 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+14 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}}{4000 \left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\) | \(190\) |
1/4000*(3*10^(1/2)*arctan(1/10*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2 ))*sin(e*x+d+arctan(4/3))^2-6*10^(1/2)*arctan(1/10*(-5*sin(e*x+d+arctan(4/ 3))-5)^(1/2)*10^(1/2))*sin(e*x+d+arctan(4/3))+3*10^(1/2)*arctan(1/10*(-5*s in(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2))-6*(-5*sin(e*x+d+arctan(4/3))-5)^( 1/2)*sin(e*x+d+arctan(4/3))+14*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2))*(-5*si n(e*x+d+arctan(4/3))-5)^(1/2)/(sin(e*x+d+arctan(4/3))-1)/cos(e*x+d+arctan( 4/3))/(-5+5*sin(e*x+d+arctan(4/3)))^(1/2)/e
Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (123) = 246\).
Time = 0.25 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.97 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\frac {3 \, {\left (79 \, \sqrt {10} \cos \left (e x + d\right )^{3} - 123 \, \sqrt {10} \cos \left (e x + d\right )^{2} + 3 \, {\left (\sqrt {10} \cos \left (e x + d\right )^{2} + 38 \, \sqrt {10} \cos \left (e x + d\right ) - 44 \, \sqrt {10}\right )} \sin \left (e x + d\right ) - 78 \, \sqrt {10} \cos \left (e x + d\right ) + 124 \, \sqrt {10}\right )} \arctan \left (-\frac {{\left (3 \, \sqrt {10} \cos \left (e x + d\right ) + \sqrt {10} \sin \left (e x + d\right ) + 3 \, \sqrt {10}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{10 \, {\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}\right ) + 10 \, {\left (27 \, \cos \left (e x + d\right )^{2} + {\left (39 \, \cos \left (e x + d\right ) - 8\right )} \sin \left (e x + d\right ) - 69 \, \cos \left (e x + d\right ) - 96\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{4000 \, {\left (79 \, e \cos \left (e x + d\right )^{3} - 123 \, e \cos \left (e x + d\right )^{2} - 78 \, e \cos \left (e x + d\right ) + 3 \, {\left (e \cos \left (e x + d\right )^{2} + 38 \, e \cos \left (e x + d\right ) - 44 \, e\right )} \sin \left (e x + d\right ) + 124 \, e\right )}} \]
1/4000*(3*(79*sqrt(10)*cos(e*x + d)^3 - 123*sqrt(10)*cos(e*x + d)^2 + 3*(s qrt(10)*cos(e*x + d)^2 + 38*sqrt(10)*cos(e*x + d) - 44*sqrt(10))*sin(e*x + d) - 78*sqrt(10)*cos(e*x + d) + 124*sqrt(10))*arctan(-1/10*(3*sqrt(10)*co s(e*x + d) + sqrt(10)*sin(e*x + d) + 3*sqrt(10))*sqrt(4*cos(e*x + d) + 3*s in(e*x + d) - 5)/(cos(e*x + d) - 3*sin(e*x + d) + 1)) + 10*(27*cos(e*x + d )^2 + (39*cos(e*x + d) - 8)*sin(e*x + d) - 69*cos(e*x + d) - 96)*sqrt(4*co s(e*x + d) + 3*sin(e*x + d) - 5))/(79*e*cos(e*x + d)^3 - 123*e*cos(e*x + d )^2 - 78*e*cos(e*x + d) + 3*(e*cos(e*x + d)^2 + 38*e*cos(e*x + d) - 44*e)* sin(e*x + d) + 124*e)
\[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int \frac {1}{\left (3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} - 5\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int { \frac {1}{{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int \frac {1}{{\left (4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5\right )}^{5/2}} \,d x \]