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

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Rubi [A]  time = 0.0143809, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 4, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, 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 = 0.113231, 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

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Maple [A]  time = 0.032, size = 25, normalized size = 1. \begin{align*} - \left ( -{\frac{ \left ( \sec \left ( x \right ) \right ) ^{3}}{4}}-{\frac{3\,\sec \left ( x \right ) }{8}} \right ) \tan \left ( x \right ) +{\frac{3\,\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) }{8}} \end{align*}

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))

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Maxima [B]  time = 0.946055, size = 57, normalized size = 2.19 \begin{align*} -\frac{3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )}{8 \,{\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac{3}{16} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{3}{16} \, \log \left (\sin \left (x\right ) - 1\right ) \end{align*}

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)

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Fricas [B]  time = 2.16144, size = 138, normalized size = 5.31 \begin{align*} \frac{3 \, \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - 3 \, \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (3 \, \cos \left (x\right )^{2} + 2\right )} \sin \left (x\right )}{16 \, \cos \left (x\right )^{4}} \end{align*}

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

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Sympy [A]  time = 0.135195, size = 46, normalized size = 1.77 \begin{align*} - \frac{3 \sin ^{3}{\left (x \right )} - 5 \sin{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} - \frac{3 \log{\left (\sin{\left (x \right )} - 1 \right )}}{16} + \frac{3 \log{\left (\sin{\left (x \right )} + 1 \right )}}{16} \end{align*}

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

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Giac [A]  time = 1.08199, size = 51, normalized size = 1.96 \begin{align*} -\frac{3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )}{8 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{2}} + \frac{3}{16} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{3}{16} \, \log \left (-\sin \left (x\right ) + 1\right ) \end{align*}

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)