Integrand size = 22, antiderivative size = 49 \[ \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=-\frac {\sqrt {\frac {2}{5}} \arctan \left (\frac {\sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {-1+\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}}\right )}{e} \]
-1/5*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 = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.02 \[ \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\frac {\left (\frac {2}{5}+\frac {6 i}{5}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \text {arctanh}\left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \left (3+\tan \left (\frac {1}{4} (d+e x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )}{e \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \]
((2/5 + (6*I)/5)*Sqrt[-4/5 - (3*I)/5]*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]) )/(e*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {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.
\(\displaystyle \int \frac {1}{\sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}dx\) |
\(\Big \downarrow \) 3594 |
\(\displaystyle \int \frac {1}{\sqrt {5 \cos \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )-5}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {5 \sin \left (-\arctan \left (\frac {3}{4}\right )+d+e x+\frac {\pi }{2}\right )-5}}dx\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle -\frac {2 \int \frac {1}{-\frac {5 \sin ^2\left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-1}-10}d\left (-\frac {\sqrt {5} \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-1}}\right )}{e}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\sqrt {\frac {2}{5}} \arctan \left (\frac {\sin \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )}{\sqrt {2} \sqrt {\cos \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )-1}}\right )}{e}\) |
-((Sqrt[2/5]*ArcTan[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[-1 + Cos[d + e*x - ArcTan[3/4]]])])/e)
3.5.27.3.1 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]
Time = 2.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.57
method | result | size |
default | \(\frac {\left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right )}{5 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\) | \(77\) |
risch | \(\frac {2 i \left (5 \,{\mathrm e}^{i \left (e x +d \right )}-4-3 i\right ) \sqrt {2}\, \sqrt {\left (4-3 i\right ) \left (25 \,{\mathrm e}^{2 i \left (e x +d \right )}-30 i {\mathrm e}^{i \left (e x +d \right )}+7+24 i-40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, {\mathrm e}^{-i \left (e x +d \right )}}{e \sqrt {\left (100-75 i\right ) \left (25 \,{\mathrm e}^{2 i \left (e x +d \right )}-30 i {\mathrm e}^{i \left (e x +d \right )}+7+24 i-40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, \sqrt {-\left (3 i {\mathrm e}^{2 i \left (e x +d \right )}-4 \,{\mathrm e}^{2 i \left (e x +d \right )}-4-3 i+10 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{-i \left (e x +d \right )}}}-\frac {2 i \left (-5 \,{\mathrm e}^{i \left (e x +d \right )}+4+3 i\right ) \left (\sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (2500-1875 i\right ) {\mathrm e}^{i \left (e x +d \right )}}\, \sqrt {5}}{125}\right ) \sqrt {\left (2500-1875 i\right ) {\mathrm e}^{i \left (e x +d \right )}}-125\right ) \sqrt {2}\, \sqrt {\left (4-3 i\right ) \left (25 \,{\mathrm e}^{2 i \left (e x +d \right )}-30 i {\mathrm e}^{i \left (e x +d \right )}+7+24 i-40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, {\mathrm e}^{-i \left (e x +d \right )}}{125 e \sqrt {\left (-100+75 i\right ) \left (30 i {\mathrm e}^{i \left (e x +d \right )}-25 \,{\mathrm e}^{2 i \left (e x +d \right )}-7-24 i+40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, \sqrt {-\left (3 i {\mathrm e}^{2 i \left (e x +d \right )}-4 \,{\mathrm e}^{2 i \left (e x +d \right )}-4-3 i+10 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{-i \left (e x +d \right )}}}\) | \(448\) |
1/5*(sin(e*x+d+arctan(4/3))-1)*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2 )*arctan(1/10*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2))/cos(e*x+d+arct an(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. 88 vs. \(2 (38) = 76\).
Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.80 \[ \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\frac {\sqrt {5} \sqrt {2} \arctan \left (-\frac {{\left (3 \, \sqrt {5} \sqrt {2} \cos \left (e x + d\right ) + \sqrt {5} \sqrt {2} \sin \left (e x + d\right ) + 3 \, \sqrt {5} \sqrt {2}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{10 \, {\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}\right )}{5 \, e} \]
1/5*sqrt(5)*sqrt(2)*arctan(-1/10*(3*sqrt(5)*sqrt(2)*cos(e*x + d) + sqrt(5) *sqrt(2)*sin(e*x + d) + 3*sqrt(5)*sqrt(2))*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5)/(cos(e*x + d) - 3*sin(e*x + d) + 1))/e
\[ \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\int \frac {1}{\sqrt {3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} - 5}}\, dx \]
\[ \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\int { \frac {1}{\sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}} \,d x } \]
\[ \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\int { \frac {1}{\sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\int \frac {1}{\sqrt {4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5}} \,d x \]