3.355 \(\int \sec ^5(x) \, dx\)

Optimal. Leaf size=26 \[ \frac {3}{8} \tanh ^{-1}(\sin (x))+\frac {1}{4} \tan (x) \sec ^3(x)+\frac {3}{8} \tan (x) \sec (x) \]

[Out]

3/8*arctanh(sin(x))+3/8*sec(x)*tan(x)+1/4*sec(x)^3*tan(x)

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3768, 3770} \[ \frac {3}{8} \tanh ^{-1}(\sin (x))+\frac {1}{4} \tan (x) \sec ^3(x)+\frac {3}{8} \tan (x) \sec (x) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^5,x]

[Out]

(3*ArcTanh[Sin[x]])/8 + (3*Sec[x]*Tan[x])/8 + (Sec[x]^3*Tan[x])/4

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \sec ^5(x) \, dx &=\frac {1}{4} \sec ^3(x) \tan (x)+\frac {3}{4} \int \sec ^3(x) \, dx\\ &=\frac {3}{8} \sec (x) \tan (x)+\frac {1}{4} \sec ^3(x) \tan (x)+\frac {3}{8} \int \sec (x) \, dx\\ &=\frac {3}{8} \tanh ^{-1}(\sin (x))+\frac {3}{8} \sec (x) \tan (x)+\frac {1}{4} \sec ^3(x) \tan (x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.12, size = 58, normalized size = 2.23 \[ \frac {1}{16} \left (\frac {1}{2} (11 \sin (x)+3 \sin (3 x)) \sec ^4(x)-6 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+6 \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^5,x]

[Out]

(-6*Log[Cos[x/2] - Sin[x/2]] + 6*Log[Cos[x/2] + Sin[x/2]] + (Sec[x]^4*(11*Sin[x] + 3*Sin[3*x]))/2)/16

________________________________________________________________________________________

fricas [B]  time = 0.46, size = 43, normalized size = 1.65 \[ \frac {3 \, \cos \relax (x)^{4} \log \left (\sin \relax (x) + 1\right ) - 3 \, \cos \relax (x)^{4} \log \left (-\sin \relax (x) + 1\right ) + 2 \, {\left (3 \, \cos \relax (x)^{2} + 2\right )} \sin \relax (x)}{16 \, \cos \relax (x)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^5,x, algorithm="fricas")

[Out]

1/16*(3*cos(x)^4*log(sin(x) + 1) - 3*cos(x)^4*log(-sin(x) + 1) + 2*(3*cos(x)^2 + 2)*sin(x))/cos(x)^4

________________________________________________________________________________________

giac [A]  time = 0.90, size = 38, normalized size = 1.46 \[ -\frac {3 \, \sin \relax (x)^{3} - 5 \, \sin \relax (x)}{8 \, {\left (\sin \relax (x)^{2} - 1\right )}^{2}} + \frac {3}{16} \, \log \left (\sin \relax (x) + 1\right ) - \frac {3}{16} \, \log \left (-\sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^5,x, algorithm="giac")

[Out]

-1/8*(3*sin(x)^3 - 5*sin(x))/(sin(x)^2 - 1)^2 + 3/16*log(sin(x) + 1) - 3/16*log(-sin(x) + 1)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 25, normalized size = 0.96 \[ \frac {3 \ln \left (\sec \relax (x )+\tan \relax (x )\right )}{8}-\left (-\frac {\left (\sec ^{3}\relax (x )\right )}{4}-\frac {3 \sec \relax (x )}{8}\right ) \tan \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^5,x)

[Out]

-(-1/4*sec(x)^3-3/8*sec(x))*tan(x)+3/8*ln(sec(x)+tan(x))

________________________________________________________________________________________

maxima [B]  time = 0.65, size = 42, normalized size = 1.62 \[ -\frac {3 \, \sin \relax (x)^{3} - 5 \, \sin \relax (x)}{8 \, {\left (\sin \relax (x)^{4} - 2 \, \sin \relax (x)^{2} + 1\right )}} + \frac {3}{16} \, \log \left (\sin \relax (x) + 1\right ) - \frac {3}{16} \, \log \left (\sin \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^5,x, algorithm="maxima")

[Out]

-1/8*(3*sin(x)^3 - 5*sin(x))/(sin(x)^4 - 2*sin(x)^2 + 1) + 3/16*log(sin(x) + 1) - 3/16*log(sin(x) - 1)

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 29, normalized size = 1.12 \[ \frac {3\,\ln \left (\frac {\sin \relax (x)+1}{\cos \relax (x)}\right )}{8}+\sin \relax (x)\,\left (\frac {3}{8\,{\cos \relax (x)}^2}+\frac {1}{4\,{\cos \relax (x)}^4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(x)^5,x)

[Out]

(3*log((sin(x) + 1)/cos(x)))/8 + sin(x)*(3/(8*cos(x)^2) + 1/(4*cos(x)^4))

________________________________________________________________________________________

sympy [A]  time = 0.16, size = 46, normalized size = 1.77 \[ - \frac {3 \sin ^{3}{\relax (x )} - 5 \sin {\relax (x )}}{8 \sin ^{4}{\relax (x )} - 16 \sin ^{2}{\relax (x )} + 8} - \frac {3 \log {\left (\sin {\relax (x )} - 1 \right )}}{16} + \frac {3 \log {\left (\sin {\relax (x )} + 1 \right )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**5,x)

[Out]

-(3*sin(x)**3 - 5*sin(x))/(8*sin(x)**4 - 16*sin(x)**2 + 8) - 3*log(sin(x) - 1)/16 + 3*log(sin(x) + 1)/16

________________________________________________________________________________________