3.282 \(\int \frac{1}{(\sec (x)+\tan (x))^5} \, dx\)
Optimal. Leaf size=22 \[ \frac{4}{\sin (x)+1}-\frac{2}{(\sin (x)+1)^2}+\log (\sin (x)+1) \]
[Out]
Log[1 + Sin[x]] - 2/(1 + Sin[x])^2 + 4/(1 + Sin[x])
________________________________________________________________________________________
Rubi [A] time = 0.0479451, antiderivative size = 22, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used =
{4391, 2667, 43} \[ \frac{4}{\sin (x)+1}-\frac{2}{(\sin (x)+1)^2}+\log (\sin (x)+1) \]
Antiderivative was successfully verified.
[In]
Int[(Sec[x] + Tan[x])^(-5),x]
[Out]
Log[1 + Sin[x]] - 2/(1 + Sin[x])^2 + 4/(1 + Sin[x])
Rule 4391
Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[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]
Rule 2667
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[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]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/2])
Rule 43
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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{1}{(\sec (x)+\tan (x))^5} \, dx &=\int \frac{\cos ^5(x)}{(1+\sin (x))^5} \, dx\\ &=\operatorname{Subst}\left (\int \frac{(1-x)^2}{(1+x)^3} \, dx,x,\sin (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{4}{(1+x)^3}-\frac{4}{(1+x)^2}+\frac{1}{1+x}\right ) \, dx,x,\sin (x)\right )\\ &=\log (1+\sin (x))-\frac{2}{(1+\sin (x))^2}+\frac{4}{1+\sin (x)}\\ \end{align*}
Mathematica [A] time = 0.0465485, size = 39, normalized size = 1.77 \[ \frac{4 \sin (x)+2}{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^4}+2 \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(Sec[x] + Tan[x])^(-5),x]
[Out]
2*Log[Cos[x/2] + Sin[x/2]] + (2 + 4*Sin[x])/(Cos[x/2] + Sin[x/2])^4
________________________________________________________________________________________
Maple [A] time = 0.109, size = 23, normalized size = 1.1 \begin{align*} \ln \left ( 1+\sin \left ( x \right ) \right ) -2\, \left ( 1+\sin \left ( x \right ) \right ) ^{-2}+4\, \left ( 1+\sin \left ( x \right ) \right ) ^{-1} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(sec(x)+tan(x))^5,x)
[Out]
ln(1+sin(x))-2/(1+sin(x))^2+4/(1+sin(x))
________________________________________________________________________________________
Maxima [B] time = 1.5119, size = 124, normalized size = 5.64 \begin{align*} -\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(sec(x)+tan(x))^5,x, algorithm="maxima")
[Out]
-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(sin(x)/(cos(x) + 1) + 1) - log(sin(x)^2/(cos(x) + 1)^2 + 1)
________________________________________________________________________________________
Fricas [A] time = 2.09642, size = 116, normalized size = 5.27 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(sec(x)+tan(x))^5,x, algorithm="fricas")
[Out]
((cos(x)^2 - 2*sin(x) - 2)*log(sin(x) + 1) - 4*sin(x) - 2)/(cos(x)^2 - 2*sin(x) - 2)
________________________________________________________________________________________
Sympy [B] time = 5.48686, size = 1064, normalized size = 48.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(sec(x)+tan(x))**5,x)
[Out]
12*log(tan(x) + sec(x))*tan(x)**4/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec
(x)**3 + 12*sec(x)**4) + 48*log(tan(x) + sec(x))*tan(x)**3*sec(x)/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan
(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) + 72*log(tan(x) + sec(x))*tan(x)**2*sec(x)**2/(12*tan(x
)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) + 48*log(tan(x) + se
c(x))*tan(x)*sec(x)**3/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12
*sec(x)**4) + 12*log(tan(x) + sec(x))*sec(x)**4/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 +
48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 6*log(tan(x)**2 + 1)*tan(x)**4/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72
*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 24*log(tan(x)**2 + 1)*tan(x)**3*sec(x)/(12*tan(x)
**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 36*log(tan(x)**2 +
1)*tan(x)**2*sec(x)**2/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12
*sec(x)**4) - 24*log(tan(x)**2 + 1)*tan(x)*sec(x)**3/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)
**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 6*log(tan(x)**2 + 1)*sec(x)**4/(12*tan(x)**4 + 48*tan(x)**3*sec(x)
+ 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 16*tan(x)**4/(12*tan(x)**4 + 48*tan(x)**3*se
c(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 44*tan(x)**3*sec(x)/(12*tan(x)**4 + 48*t
an(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 40*tan(x)**2*sec(x)**2/(12*ta
n(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) + 6*tan(x)**2/(12
*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) - 12*tan(x)*se
c(x)**3/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4) + 8
*tan(x)*sec(x)/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)*
*4) + 2*sec(x)**2/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(
x)**4) - 3/(12*tan(x)**4 + 48*tan(x)**3*sec(x) + 72*tan(x)**2*sec(x)**2 + 48*tan(x)*sec(x)**3 + 12*sec(x)**4)
________________________________________________________________________________________
Giac [B] time = 1.12812, size = 86, normalized size = 3.91 \begin{align*} -\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(sec(x)+tan(x))^5,x, algorithm="giac")
[Out]
-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(ta
n(1/2*x)^2 + 1) + 2*log(abs(tan(1/2*x) + 1))