Integrand size = 22, antiderivative size = 48 \[ \int \frac {1}{\sqrt {5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\frac {\sqrt {\frac {2}{5}} \text {arctanh}\left (\frac {\sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {1+\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}}\right )}{e} \]
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.10 \[ \int \frac {1}{\sqrt {5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=-\frac {\left (\frac {2}{5}+\frac {6 i}{5}\right ) \sqrt {\frac {4}{5}+\frac {3 i}{5}} \arctan \left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {\frac {4}{5}+\frac {3 i}{5}} \left (-1+3 \tan \left (\frac {1}{4} (d+e x)\right )\right )\right ) \left (3 \cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )}{e \sqrt {5+4 \cos (d+e x)+3 \sin (d+e x)}} \]
((-2/5 - (6*I)/5)*Sqrt[4/5 + (3*I)/5]*ArcTan[(1/10 + (3*I)/10)*Sqrt[4/5 + (3*I)/5]*(-1 + 3*Tan[(d + e*x)/4])]*(3*Cos[(d + e*x)/2] + Sin[(d + e*x)/2] ))/(e*Sqrt[5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])
Time = 0.26 (sec) , antiderivative size = 48, 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, 219}
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 |
\(\displaystyle -\frac {2 \int \frac {1}{10-\frac {5 \sin ^2\left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )+1}}d\left (-\frac {\sqrt {5} \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )+1}}\right )}{e}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {\frac {2}{5}} \text {arctanh}\left (\frac {\sin \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )}{\sqrt {2} \sqrt {\cos \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )+1}}\right )}{e}\) |
(Sqrt[2/5]*ArcTanh[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[1 + Cos[d + e* x - ArcTan[3/4]]])])/e
3.5.20.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[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 = 2.90 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.60
method | result | size |
default | \(-\frac {\left (1+\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}\, \operatorname {arctanh}\left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right )}{5 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\) | \(77\) |
risch | \(\frac {2 i \left (5 \,{\mathrm e}^{i \left (e x +d \right )}+4+3 i\right ) \sqrt {2}\, \sqrt {\left (4-3 i\right ) \left (25 \,{\mathrm e}^{2 i \left (e x +d \right )}+30 i {\mathrm e}^{i \left (e x +d \right )}+7+24 i+40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, {\mathrm e}^{-i \left (e x +d \right )}}{e \sqrt {\left (100-75 i\right ) \left (25 \,{\mathrm e}^{2 i \left (e x +d \right )}+30 i {\mathrm e}^{i \left (e x +d \right )}+7+24 i+40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, \sqrt {-\left (3 i {\mathrm e}^{2 i \left (e x +d \right )}-4 \,{\mathrm e}^{2 i \left (e x +d \right )}-4-3 i-10 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{-i \left (e x +d \right )}}}-\frac {2 i \left (5 \,{\mathrm e}^{i \left (e x +d \right )}+4+3 i\right ) \left (\sqrt {5}\, \arctan \left (\frac {\sqrt {\left (2500-1875 i\right ) {\mathrm e}^{i \left (e x +d \right )}}\, \sqrt {5}}{125}\right ) \sqrt {\left (2500-1875 i\right ) {\mathrm e}^{i \left (e x +d \right )}}+125\right ) \sqrt {2}\, \sqrt {\left (4-3 i\right ) \left (25 \,{\mathrm e}^{2 i \left (e x +d \right )}+30 i {\mathrm e}^{i \left (e x +d \right )}+7+24 i+40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, {\mathrm e}^{-i \left (e x +d \right )}}{125 e \sqrt {\left (100-75 i\right ) \left (25 \,{\mathrm e}^{2 i \left (e x +d \right )}+30 i {\mathrm e}^{i \left (e x +d \right )}+7+24 i+40 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{i \left (e x +d \right )}}\, \sqrt {-\left (3 i {\mathrm e}^{2 i \left (e x +d \right )}-4 \,{\mathrm e}^{2 i \left (e x +d \right )}-4-3 i-10 \,{\mathrm e}^{i \left (e x +d \right )}\right ) {\mathrm e}^{-i \left (e x +d \right )}}}\) | \(448\) |
-1/5*(1+sin(e*x+d+arctan(4/3)))*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2)*10^(1/ 2)*arctanh(1/10*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2)*10^(1/2))/cos(e*x+d+ar ctan(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. 147 vs. \(2 (38) = 76\).
Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 3.06 \[ \int \frac {1}{\sqrt {5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\frac {\sqrt {5} \sqrt {2} \log \left (-\frac {9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) - 6\right )} \sin \left (e x + d\right ) + 2 \, {\left (\sqrt {5} \sqrt {2} \cos \left (e x + d\right ) - 3 \, \sqrt {5} \sqrt {2} \sin \left (e x + d\right ) + \sqrt {5} \sqrt {2}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5} - 33 \, \cos \left (e x + d\right ) - 42}{9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) + 14\right )} \sin \left (e x + d\right ) + 27 \, \cos \left (e x + d\right ) + 18}\right )}{10 \, e} \]
1/10*sqrt(5)*sqrt(2)*log(-(9*cos(e*x + d)^2 + (13*cos(e*x + d) - 6)*sin(e* x + d) + 2*(sqrt(5)*sqrt(2)*cos(e*x + d) - 3*sqrt(5)*sqrt(2)*sin(e*x + d) + sqrt(5)*sqrt(2))*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) + 5) - 33*cos(e*x + d) - 42)/(9*cos(e*x + d)^2 + (13*cos(e*x + d) + 14)*sin(e*x + d) + 27*co s(e*x + d) + 18))/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. \[ \int \frac {1}{\sqrt {5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx=\int \frac {1}{\sqrt {4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )+5}} \,d x \]