Integrand size = 8, antiderivative size = 223 \[ \int \frac {1}{1+\cos ^5(x)} \, dx=\frac {2 \arctan \left (\sqrt {\frac {1-(-1)^{2/5}}{1+(-1)^{2/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1-(-1)^{4/5}}}+\frac {2 \arctan \left (\sqrt {\frac {1-(-1)^{4/5}}{1+(-1)^{4/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1+(-1)^{3/5}}}-\frac {2 \text {arctanh}\left (\frac {\tan \left (\frac {x}{2}\right )}{\sqrt {-\frac {1-\sqrt [5]{-1}}{1+\sqrt [5]{-1}}}}\right )}{5 \sqrt {-1+(-1)^{2/5}}}-\frac {2 \sqrt {-\frac {1+(-1)^{3/5}}{1-(-1)^{3/5}}} \text {arctanh}\left (\sqrt {-\frac {1+(-1)^{3/5}}{1-(-1)^{3/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \left (1+(-1)^{3/5}\right )}+\frac {\sin (x)}{5 (1+\cos (x))} \]
1/5*sin(x)/(1+cos(x))-2/5*arctanh(tan(1/2*x)/((-1+(-1)^(1/5))/(1+(-1)^(1/5 )))^(1/2))/(-1+(-1)^(2/5))^(1/2)+2/5*arctan(((1-(-1)^(4/5))/(1+(-1)^(4/5)) )^(1/2)*tan(1/2*x))/(1+(-1)^(3/5))^(1/2)-2/5*arctanh(((-1-(-1)^(3/5))/(1-( -1)^(3/5)))^(1/2)*tan(1/2*x))*((-1-(-1)^(3/5))/(1-(-1)^(3/5)))^(1/2)/(1+(- 1)^(3/5))+2/5*arctan(((1-(-1)^(2/5))/(1+(-1)^(2/5)))^(1/2)*tan(1/2*x))/(1- (-1)^(4/5))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 5.07 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.70 \[ \int \frac {1}{1+\cos ^5(x)} \, dx=-\frac {1}{10} \text {RootSum}\left [1-2 \text {$\#$1}+8 \text {$\#$1}^2-14 \text {$\#$1}^3+30 \text {$\#$1}^4-14 \text {$\#$1}^5+8 \text {$\#$1}^6-2 \text {$\#$1}^7+\text {$\#$1}^8\&,\frac {2 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )-i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right )-8 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}+4 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}+30 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^2-15 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-80 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^3+40 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3+30 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^4-15 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4-8 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^5+4 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^5+2 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^6-i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-1+8 \text {$\#$1}-21 \text {$\#$1}^2+60 \text {$\#$1}^3-35 \text {$\#$1}^4+24 \text {$\#$1}^5-7 \text {$\#$1}^6+4 \text {$\#$1}^7}\&\right ]+\frac {1}{5} \tan \left (\frac {x}{2}\right ) \]
-1/10*RootSum[1 - 2*#1 + 8*#1^2 - 14*#1^3 + 30*#1^4 - 14*#1^5 + 8*#1^6 - 2 *#1^7 + #1^8 & , (2*ArcTan[Sin[x]/(Cos[x] - #1)] - I*Log[1 - 2*Cos[x]*#1 + #1^2] - 8*ArcTan[Sin[x]/(Cos[x] - #1)]*#1 + (4*I)*Log[1 - 2*Cos[x]*#1 + # 1^2]*#1 + 30*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^2 - (15*I)*Log[1 - 2*Cos[x]*# 1 + #1^2]*#1^2 - 80*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^3 + (40*I)*Log[1 - 2*C os[x]*#1 + #1^2]*#1^3 + 30*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^4 - (15*I)*Log[ 1 - 2*Cos[x]*#1 + #1^2]*#1^4 - 8*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^5 + (4*I) *Log[1 - 2*Cos[x]*#1 + #1^2]*#1^5 + 2*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^6 - I*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^6)/(-1 + 8*#1 - 21*#1^2 + 60*#1^3 - 35*#1 ^4 + 24*#1^5 - 7*#1^6 + 4*#1^7) & ] + Tan[x/2]/5
Time = 0.70 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3692, 2009}
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.
| \(\displaystyle \int \frac {1}{\cos ^5(x)+1} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right )^5+1}dx\) |
| \(\Big \downarrow \) 3692 |
| \(\displaystyle \int \left (-\frac {1}{5 \left (\sqrt [5]{-1} \cos (x)-1\right )}-\frac {1}{5 \left (-(-1)^{2/5} \cos (x)-1\right )}-\frac {1}{5 \left ((-1)^{3/5} \cos (x)-1\right )}-\frac {1}{5 \left (-(-1)^{4/5} \cos (x)-1\right )}-\frac {1}{5 (-\cos (x)-1)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \arctan \left (\sqrt {\frac {1-(-1)^{2/5}}{1+(-1)^{2/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1-(-1)^{4/5}}}+\frac {2 \arctan \left (\sqrt {\frac {1-(-1)^{4/5}}{1+(-1)^{4/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \sqrt {1+(-1)^{3/5}}}-\frac {2 \text {arctanh}\left (\frac {\tan \left (\frac {x}{2}\right )}{\sqrt {-\frac {1-\sqrt [5]{-1}}{1+\sqrt [5]{-1}}}}\right )}{5 \sqrt {(-1)^{2/5}-1}}-\frac {2 \sqrt {-\frac {1+(-1)^{3/5}}{1-(-1)^{3/5}}} \text {arctanh}\left (\sqrt {-\frac {1+(-1)^{3/5}}{1-(-1)^{3/5}}} \tan \left (\frac {x}{2}\right )\right )}{5 \left (1+(-1)^{3/5}\right )}+\frac {\sin (x)}{5 (\cos (x)+1)}\) |
(2*ArcTan[Sqrt[(1 - (-1)^(2/5))/(1 + (-1)^(2/5))]*Tan[x/2]])/(5*Sqrt[1 - ( -1)^(4/5)]) + (2*ArcTan[Sqrt[(1 - (-1)^(4/5))/(1 + (-1)^(4/5))]*Tan[x/2]]) /(5*Sqrt[1 + (-1)^(3/5)]) - (2*ArcTanh[Tan[x/2]/Sqrt[-((1 - (-1)^(1/5))/(1 + (-1)^(1/5)))]])/(5*Sqrt[-1 + (-1)^(2/5)]) - (2*Sqrt[-((1 + (-1)^(3/5))/ (1 - (-1)^(3/5)))]*ArcTanh[Sqrt[-((1 + (-1)^(3/5))/(1 - (-1)^(3/5)))]*Tan[ x/2]])/(5*(1 + (-1)^(3/5))) + Sin[x]/(5*(1 + Cos[x]))
3.1.80.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f , n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.35 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.28
| method | result | size |
| default | \(\frac {\tan \left (\frac {x}{2}\right )}{5}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (5 \textit {\_Z}^{8}+10 \textit {\_Z}^{4}+1\right )}{\sum }\frac {\left (5 \textit {\_R}^{6}+5 \textit {\_R}^{4}+5 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7}+\textit {\_R}^{3}}\right )}{50}\) | \(62\) |
| risch | \(\frac {2 i}{5 \left ({\mathrm e}^{i x}+1\right )}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1953125 \textit {\_Z}^{8}+156250 \textit {\_Z}^{6}+6250 \textit {\_Z}^{4}+125 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i x}-2343750 i \textit {\_R}^{7}+234375 \textit {\_R}^{6}-140625 i \textit {\_R}^{5}+15625 \textit {\_R}^{4}-4375 i \textit {\_R}^{3}+500 \textit {\_R}^{2}-50 i \textit {\_R} +6\right )\right )\) | \(87\) |
1/5*tan(1/2*x)+1/50*sum((5*_R^6+5*_R^4+5*_R^2+1)/(_R^7+_R^3)*ln(tan(1/2*x) -_R),_R=RootOf(5*_Z^8+10*_Z^4+1))
Leaf count of result is larger than twice the leaf count of optimal. 789 vs. \(2 (150) = 300\).
Time = 0.40 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.54 \[ \int \frac {1}{1+\cos ^5(x)} \, dx=\text {Too large to display} \]
1/100*((sqrt(5)*cos(x) + sqrt(5))*sqrt(2*sqrt(5)*sqrt(2*sqrt(5) - 5) - 10) *log(sqrt(2*sqrt(5)*sqrt(2*sqrt(5) - 5) - 10)*(3*sqrt(5) + 5)*sqrt(2*sqrt( 5) - 5)*sin(x) - 5*sqrt(2*sqrt(5) - 5)*(sqrt(5) + 3)*cos(x) - 5*(sqrt(5) - 1)*cos(x) - 20) - (sqrt(5)*cos(x) + sqrt(5))*sqrt(2*sqrt(5)*sqrt(2*sqrt(5 ) - 5) - 10)*log(-sqrt(2*sqrt(5)*sqrt(2*sqrt(5) - 5) - 10)*(3*sqrt(5) + 5) *sqrt(2*sqrt(5) - 5)*sin(x) - 5*sqrt(2*sqrt(5) - 5)*(sqrt(5) + 3)*cos(x) - 5*(sqrt(5) - 1)*cos(x) - 20) + (sqrt(5)*cos(x) + sqrt(5))*sqrt(-2*sqrt(5) *sqrt(2*sqrt(5) - 5) - 10)*log(sqrt(-2*sqrt(5)*sqrt(2*sqrt(5) - 5) - 10)*( 3*sqrt(5) + 5)*sqrt(2*sqrt(5) - 5)*sin(x) - 5*sqrt(2*sqrt(5) - 5)*(sqrt(5) + 3)*cos(x) + 5*(sqrt(5) - 1)*cos(x) + 20) - (sqrt(5)*cos(x) + sqrt(5))*s qrt(-2*sqrt(5)*sqrt(2*sqrt(5) - 5) - 10)*log(-sqrt(-2*sqrt(5)*sqrt(2*sqrt( 5) - 5) - 10)*(3*sqrt(5) + 5)*sqrt(2*sqrt(5) - 5)*sin(x) - 5*sqrt(2*sqrt(5 ) - 5)*(sqrt(5) + 3)*cos(x) + 5*(sqrt(5) - 1)*cos(x) + 20) - (sqrt(5)*cos( x) + sqrt(5))*sqrt(2*sqrt(5)*sqrt(-2*sqrt(5) - 5) - 10)*log(sqrt(2*sqrt(5) *sqrt(-2*sqrt(5) - 5) - 10)*(3*sqrt(5) - 5)*sqrt(-2*sqrt(5) - 5)*sin(x) - 5*(sqrt(5) - 3)*sqrt(-2*sqrt(5) - 5)*cos(x) + 5*(sqrt(5) + 1)*cos(x) - 20) + (sqrt(5)*cos(x) + sqrt(5))*sqrt(2*sqrt(5)*sqrt(-2*sqrt(5) - 5) - 10)*lo g(-sqrt(2*sqrt(5)*sqrt(-2*sqrt(5) - 5) - 10)*(3*sqrt(5) - 5)*sqrt(-2*sqrt( 5) - 5)*sin(x) - 5*(sqrt(5) - 3)*sqrt(-2*sqrt(5) - 5)*cos(x) + 5*(sqrt(5) + 1)*cos(x) - 20) - (sqrt(5)*cos(x) + sqrt(5))*sqrt(-2*sqrt(5)*sqrt(-2*...
Timed out. \[ \int \frac {1}{1+\cos ^5(x)} \, dx=\text {Timed out} \]
\[ \int \frac {1}{1+\cos ^5(x)} \, dx=\int { \frac {1}{\cos \left (x\right )^{5} + 1} \,d x } \]
-1/5*(5*(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1)*integrate(-2/5*((cos(7*x) - 4 *cos(6*x) + 15*cos(5*x) - 40*cos(4*x) + 15*cos(3*x) - 4*cos(2*x) + cos(x)) *cos(8*x) + (16*cos(6*x) - 44*cos(5*x) + 110*cos(4*x) - 44*cos(3*x) + 16*c os(2*x) - 4*cos(x) + 1)*cos(7*x) - 2*cos(7*x)^2 + 4*(44*cos(5*x) - 110*cos (4*x) + 44*cos(3*x) - 16*cos(2*x) + 4*cos(x) - 1)*cos(6*x) - 32*cos(6*x)^2 + (1010*cos(4*x) - 420*cos(3*x) + 176*cos(2*x) - 44*cos(x) + 15)*cos(5*x) - 210*cos(5*x)^2 + 10*(101*cos(3*x) - 44*cos(2*x) + 11*cos(x) - 4)*cos(4* x) - 1200*cos(4*x)^2 + (176*cos(2*x) - 44*cos(x) + 15)*cos(3*x) - 210*cos( 3*x)^2 + 4*(4*cos(x) - 1)*cos(2*x) - 32*cos(2*x)^2 - 2*cos(x)^2 + (sin(7*x ) - 4*sin(6*x) + 15*sin(5*x) - 40*sin(4*x) + 15*sin(3*x) - 4*sin(2*x) + si n(x))*sin(8*x) + 2*(8*sin(6*x) - 22*sin(5*x) + 55*sin(4*x) - 22*sin(3*x) + 8*sin(2*x) - 2*sin(x))*sin(7*x) - 2*sin(7*x)^2 + 8*(22*sin(5*x) - 55*sin( 4*x) + 22*sin(3*x) - 8*sin(2*x) + 2*sin(x))*sin(6*x) - 32*sin(6*x)^2 + 2*( 505*sin(4*x) - 210*sin(3*x) + 88*sin(2*x) - 22*sin(x))*sin(5*x) - 210*sin( 5*x)^2 + 10*(101*sin(3*x) - 44*sin(2*x) + 11*sin(x))*sin(4*x) - 1200*sin(4 *x)^2 + 44*(4*sin(2*x) - sin(x))*sin(3*x) - 210*sin(3*x)^2 - 32*sin(2*x)^2 + 16*sin(2*x)*sin(x) - 2*sin(x)^2 + cos(x))/(2*(2*cos(7*x) - 8*cos(6*x) + 14*cos(5*x) - 30*cos(4*x) + 14*cos(3*x) - 8*cos(2*x) + 2*cos(x) - 1)*cos( 8*x) - cos(8*x)^2 + 4*(8*cos(6*x) - 14*cos(5*x) + 30*cos(4*x) - 14*cos(3*x ) + 8*cos(2*x) - 2*cos(x) + 1)*cos(7*x) - 4*cos(7*x)^2 + 16*(14*cos(5*x...
\[ \int \frac {1}{1+\cos ^5(x)} \, dx=\int { \frac {1}{\cos \left (x\right )^{5} + 1} \,d x } \]
Time = 3.28 (sec) , antiderivative size = 535, normalized size of antiderivative = 2.40 \[ \int \frac {1}{1+\cos ^5(x)} \, dx=\text {Too large to display} \]
tan(x/2)/5 + 2*atanh((603979776*tan(x/2)*(- (- (2*5^(1/2))/5 - 1)^(1/2)/50 - 1/50)^(1/2))/(244140625*((33554432*5^(1/2)*(- (2*5^(1/2))/5 - 1)^(1/2)) /1220703125 - (134217728*5^(1/2))/1220703125 + (67108864*(- (2*5^(1/2))/5 - 1)^(1/2))/1220703125 - 301989888/1220703125)) + (268435456*5^(1/2)*tan(x /2)*(- (- (2*5^(1/2))/5 - 1)^(1/2)/50 - 1/50)^(1/2))/(244140625*((33554432 *5^(1/2)*(- (2*5^(1/2))/5 - 1)^(1/2))/1220703125 - (134217728*5^(1/2))/122 0703125 + (67108864*(- (2*5^(1/2))/5 - 1)^(1/2))/1220703125 - 301989888/12 20703125)))*(- (- (2*5^(1/2))/5 - 1)^(1/2)/50 - 1/50)^(1/2) - 2*atanh((603 979776*tan(x/2)*((- (2*5^(1/2))/5 - 1)^(1/2)/50 - 1/50)^(1/2))/(244140625* ((33554432*5^(1/2)*(- (2*5^(1/2))/5 - 1)^(1/2))/1220703125 + (134217728*5^ (1/2))/1220703125 + (67108864*(- (2*5^(1/2))/5 - 1)^(1/2))/1220703125 + 30 1989888/1220703125)) + (268435456*5^(1/2)*tan(x/2)*((- (2*5^(1/2))/5 - 1)^ (1/2)/50 - 1/50)^(1/2))/(244140625*((33554432*5^(1/2)*(- (2*5^(1/2))/5 - 1 )^(1/2))/1220703125 + (134217728*5^(1/2))/1220703125 + (67108864*(- (2*5^( 1/2))/5 - 1)^(1/2))/1220703125 + 301989888/1220703125)))*((- (2*5^(1/2))/5 - 1)^(1/2)/50 - 1/50)^(1/2) - 2*atanh((603979776*tan(x/2)*(- ((2*5^(1/2)) /5 - 1)^(1/2)/50 - 1/50)^(1/2))/(244140625*((33554432*5^(1/2)*((2*5^(1/2)) /5 - 1)^(1/2))/1220703125 - (134217728*5^(1/2))/1220703125 - (67108864*((2 *5^(1/2))/5 - 1)^(1/2))/1220703125 + 301989888/1220703125)) - (268435456*5 ^(1/2)*tan(x/2)*(- ((2*5^(1/2))/5 - 1)^(1/2)/50 - 1/50)^(1/2))/(2441406...