Integrand size = 7, antiderivative size = 22 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=\log (1+\sin (x))-\frac {2}{(1+\sin (x))^2}+\frac {4}{1+\sin (x)} \] Output:
ln(1+sin(x))-2/(1+sin(x))^2+4/(1+sin(x))
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=2 \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {2+4 \sin (x)}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^4} \] Input:
Integrate[(Sec[x] + Tan[x])^(-5),x]
Output:
2*Log[Cos[x/2] + Sin[x/2]] + (2 + 4*Sin[x])/(Cos[x/2] + Sin[x/2])^4
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 4891, 3042, 3146, 49, 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.
\(\displaystyle \int \frac {1}{(\tan (x)+\sec (x))^5} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(\tan (x)+\sec (x))^5}dx\) |
\(\Big \downarrow \) 4891 |
\(\displaystyle \int \frac {\cos ^5(x)}{(\sin (x)+1)^5}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (x)^5}{(\sin (x)+1)^5}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \int \frac {(1-\sin (x))^2}{(\sin (x)+1)^3}d\sin (x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (\frac {1}{\sin (x)+1}-\frac {4}{(\sin (x)+1)^2}+\frac {4}{(\sin (x)+1)^3}\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{\sin (x)+1}-\frac {2}{(\sin (x)+1)^2}+\log (\sin (x)+1)\) |
Input:
Int[(Sec[x] + Tan[x])^(-5),x]
Output:
Log[1 + Sin[x]] - 2/(1 + Sin[x])^2 + 4/(1 + Sin[x])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x _)]^(n_.))^(p_), x_Symbol] :> Int[ActivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a *Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]
Time = 7.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05
method | result | size |
default | \(\ln \left (1+\sin \left (x \right )\right )-\frac {2}{\left (1+\sin \left (x \right )\right )^{2}}+\frac {4}{1+\sin \left (x \right )}\) | \(23\) |
risch | \(-i x +\frac {8 i \left (i {\mathrm e}^{2 i x}+{\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{\left ({\mathrm e}^{i x}+i\right )^{4}}+2 \ln \left ({\mathrm e}^{i x}+i\right )\) | \(51\) |
Input:
int(1/(sec(x)+tan(x))^5,x,method=_RETURNVERBOSE)
Output:
ln(1+sin(x))-2/(1+sin(x))^2+4/(1+sin(x))
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=\frac {{\left (\cos \left (x\right )^{2} - 2 \, \sin \left (x\right ) - 2\right )} \log \left (\sin \left (x\right ) + 1\right ) - 4 \, \sin \left (x\right ) - 2}{\cos \left (x\right )^{2} - 2 \, \sin \left (x\right ) - 2} \] Input:
integrate(1/(sec(x)+tan(x))^5,x, algorithm="fricas")
Output:
((cos(x)^2 - 2*sin(x) - 2)*log(sin(x) + 1) - 4*sin(x) - 2)/(cos(x)^2 - 2*s in(x) - 2)
Leaf count of result is larger than twice the leaf count of optimal. 1059 vs. \(2 (20) = 40\).
Time = 1.16 (sec) , antiderivative size = 1059, normalized size of antiderivative = 48.14 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=\text {Too large to display} \] Input:
integrate(1/(sec(x)+tan(x))**5,x)
Output:
36*log(tan(x) + sec(x))*tan(x)**4/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 2 16*tan(x)**2*sec(x)**2 + 144*tan(x)*sec(x)**3 + 36*sec(x)**4) + 144*log(ta n(x) + sec(x))*tan(x)**3*sec(x)/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216 *tan(x)**2*sec(x)**2 + 144*tan(x)*sec(x)**3 + 36*sec(x)**4) + 216*log(tan( x) + sec(x))*tan(x)**2*sec(x)**2/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 21 6*tan(x)**2*sec(x)**2 + 144*tan(x)*sec(x)**3 + 36*sec(x)**4) + 144*log(tan (x) + sec(x))*tan(x)*sec(x)**3/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216* tan(x)**2*sec(x)**2 + 144*tan(x)*sec(x)**3 + 36*sec(x)**4) + 36*log(tan(x) + sec(x))*sec(x)**4/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216*tan(x)**2* sec(x)**2 + 144*tan(x)*sec(x)**3 + 36*sec(x)**4) - 18*log(tan(x)**2 + 1)*t an(x)**4/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216*tan(x)**2*sec(x)**2 + 144*tan(x)*sec(x)**3 + 36*sec(x)**4) - 72*log(tan(x)**2 + 1)*tan(x)**3*sec (x)/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216*tan(x)**2*sec(x)**2 + 144*t an(x)*sec(x)**3 + 36*sec(x)**4) - 108*log(tan(x)**2 + 1)*tan(x)**2*sec(x)* *2/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216*tan(x)**2*sec(x)**2 + 144*ta n(x)*sec(x)**3 + 36*sec(x)**4) - 72*log(tan(x)**2 + 1)*tan(x)*sec(x)**3/(3 6*tan(x)**4 + 144*tan(x)**3*sec(x) + 216*tan(x)**2*sec(x)**2 + 144*tan(x)* sec(x)**3 + 36*sec(x)**4) - 18*log(tan(x)**2 + 1)*sec(x)**4/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216*tan(x)**2*sec(x)**2 + 144*tan(x)*sec(x)**3 + 36*sec(x)**4) - 28*tan(x)**4/(36*tan(x)**4 + 144*tan(x)**3*sec(x) + 216...
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (22) = 44\).
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 4.18 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=-\frac {8 \, \sin \left (x\right )^{2}}{{\left (\frac {4 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {6 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {4 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {\sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + 1\right )} {\left (\cos \left (x\right ) + 1\right )}^{2}} + 2 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right ) - \log \left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right ) \] Input:
integrate(1/(sec(x)+tan(x))^5,x, algorithm="maxima")
Output:
-8*sin(x)^2/((4*sin(x)/(cos(x) + 1) + 6*sin(x)^2/(cos(x) + 1)^2 + 4*sin(x) ^3/(cos(x) + 1)^3 + sin(x)^4/(cos(x) + 1)^4 + 1)*(cos(x) + 1)^2) + 2*log(s in(x)/(cos(x) + 1) + 1) - log(sin(x)^2/(cos(x) + 1)^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (22) = 44\).
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.91 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=-\frac {25 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 100 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 198 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 100 \, \tan \left (\frac {1}{2} \, x\right ) + 25}{6 \, {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{4}} - \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) + 2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) \] Input:
integrate(1/(sec(x)+tan(x))^5,x, algorithm="giac")
Output:
-1/6*(25*tan(1/2*x)^4 + 100*tan(1/2*x)^3 + 198*tan(1/2*x)^2 + 100*tan(1/2* x) + 25)/(tan(1/2*x) + 1)^4 - log(tan(1/2*x)^2 + 1) + 2*log(abs(tan(1/2*x) + 1))
Time = 15.38 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.77 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=2\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )-\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )-\frac {8\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+4\,\mathrm {tan}\left (\frac {x}{2}\right )+1} \] Input:
int(1/(tan(x) + 1/cos(x))^5,x)
Output:
2*log(tan(x/2) + 1) - log(tan(x/2)^2 + 1) - (8*tan(x/2)^2)/(4*tan(x/2) + 6 *tan(x/2)^2 + 4*tan(x/2)^3 + tan(x/2)^4 + 1)
Time = 0.18 (sec) , antiderivative size = 233, normalized size of antiderivative = 10.59 \[ \int \frac {1}{(\sec (x)+\tan (x))^5} \, dx=\frac {-6 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right )^{4}-24 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right )^{3} \tan \left (x \right )-36 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right )^{2} \tan \left (x \right )^{2}-24 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right ) \tan \left (x \right )^{3}-6 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \tan \left (x \right )^{4}+12 \,\mathrm {log}\left (\sec \left (x \right )+\tan \left (x \right )\right ) \sec \left (x \right )^{4}+48 \,\mathrm {log}\left (\sec \left (x \right )+\tan \left (x \right )\right ) \sec \left (x \right )^{3} \tan \left (x \right )+72 \,\mathrm {log}\left (\sec \left (x \right )+\tan \left (x \right )\right ) \sec \left (x \right )^{2} \tan \left (x \right )^{2}+48 \,\mathrm {log}\left (\sec \left (x \right )+\tan \left (x \right )\right ) \sec \left (x \right ) \tan \left (x \right )^{3}+12 \,\mathrm {log}\left (\sec \left (x \right )+\tan \left (x \right )\right ) \tan \left (x \right )^{4}+3 \sec \left (x \right )^{4}-22 \sec \left (x \right )^{2} \tan \left (x \right )^{2}+2 \sec \left (x \right )^{2}-32 \sec \left (x \right ) \tan \left (x \right )^{3}+8 \sec \left (x \right ) \tan \left (x \right )-13 \tan \left (x \right )^{4}+6 \tan \left (x \right )^{2}-3}{12 \sec \left (x \right )^{4}+48 \sec \left (x \right )^{3} \tan \left (x \right )+72 \sec \left (x \right )^{2} \tan \left (x \right )^{2}+48 \sec \left (x \right ) \tan \left (x \right )^{3}+12 \tan \left (x \right )^{4}} \] Input:
int(1/(sec(x)+tan(x))^5,x)
Output:
( - 6*log(tan(x)**2 + 1)*sec(x)**4 - 24*log(tan(x)**2 + 1)*sec(x)**3*tan(x ) - 36*log(tan(x)**2 + 1)*sec(x)**2*tan(x)**2 - 24*log(tan(x)**2 + 1)*sec( x)*tan(x)**3 - 6*log(tan(x)**2 + 1)*tan(x)**4 + 12*log(sec(x) + tan(x))*se c(x)**4 + 48*log(sec(x) + tan(x))*sec(x)**3*tan(x) + 72*log(sec(x) + tan(x ))*sec(x)**2*tan(x)**2 + 48*log(sec(x) + tan(x))*sec(x)*tan(x)**3 + 12*log (sec(x) + tan(x))*tan(x)**4 + 3*sec(x)**4 - 22*sec(x)**2*tan(x)**2 + 2*sec (x)**2 - 32*sec(x)*tan(x)**3 + 8*sec(x)*tan(x) - 13*tan(x)**4 + 6*tan(x)** 2 - 3)/(12*(sec(x)**4 + 4*sec(x)**3*tan(x) + 6*sec(x)**2*tan(x)**2 + 4*sec (x)*tan(x)**3 + tan(x)**4))