3.273 \(\int (\sec (x)+\tan (x))^5 \, dx\)

Optimal. Leaf size=30 \[ -\frac{4}{1-\sin (x)}+\frac{2}{(1-\sin (x))^2}-\log (1-\sin (x)) \]

[Out]

-Log[1 - Sin[x]] + 2/(1 - Sin[x])^2 - 4/(1 - Sin[x])

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Rubi [A]  time = 0.0499651, antiderivative size = 30, 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} \[ -\frac{4}{1-\sin (x)}+\frac{2}{(1-\sin (x))^2}-\log (1-\sin (x)) \]

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 (\sec (x)+\tan (x))^5 \, dx &=\int \sec ^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{1}{1-x}-\frac{4}{(-1+x)^3}-\frac{4}{(-1+x)^2}\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 = 0.111897, size = 54, normalized size = 1.8 \[ \frac{11 \tan ^4(x)}{4}-\frac{\tan ^2(x)}{2}+\frac{5 \sec ^4(x)}{4}+\tanh ^{-1}(\sin (x))-\log (\cos (x))-\tan (x) \sec ^3(x)+5 \tan ^3(x) \sec (x)+\tan (x) \sec (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x] + Tan[x])^5,x]

[Out]

ArcTanh[Sin[x]] - Log[Cos[x]] + (5*Sec[x]^4)/4 + Sec[x]*Tan[x] - Sec[x]^3*Tan[x] - Tan[x]^2/2 + 5*Sec[x]*Tan[x
]^3 + (11*Tan[x]^4)/4

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Maple [B]  time = 0.064, size = 106, normalized size = 3.5 \begin{align*} - \left ( -{\frac{ \left ( \sec \left ( x \right ) \right ) ^{3}}{4}}-{\frac{3\,\sec \left ( x \right ) }{8}} \right ) \tan \left ( x \right ) +\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) +{\frac{5}{4\, \left ( \cos \left ( x \right ) \right ) ^{4}}}+{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{3}}{2\, \left ( \cos \left ( x \right ) \right ) ^{4}}}+{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{3}}{4\, \left ( \cos \left ( x \right ) \right ) ^{2}}}-{\frac{5\,\sin \left ( x \right ) }{8}}+{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{4}}{2\, \left ( \cos \left ( x \right ) \right ) ^{4}}}+{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{5}}{4\, \left ( \cos \left ( x \right ) \right ) ^{4}}}-{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{5}}{8\, \left ( \cos \left ( x \right ) \right ) ^{2}}}-{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{3}}{8}}+{\frac{ \left ( \tan \left ( x \right ) \right ) ^{4}}{4}}-{\frac{ \left ( \tan \left ( x \right ) \right ) ^{2}}{2}}-\ln \left ( \cos \left ( x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sec(x)+tan(x))^5,x)

[Out]

-(-1/4*sec(x)^3-3/8*sec(x))*tan(x)+ln(sec(x)+tan(x))+5/4/cos(x)^4+5/2*sin(x)^3/cos(x)^4+5/4*sin(x)^3/cos(x)^2-
5/8*sin(x)+5/2*sin(x)^4/cos(x)^4+5/4*sin(x)^5/cos(x)^4-5/8*sin(x)^5/cos(x)^2-5/8*sin(x)^3+1/4*tan(x)^4-1/2*tan
(x)^2-ln(cos(x))

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Maxima [B]  time = 0.995535, size = 190, normalized size = 6.33 \begin{align*} \frac{5}{2} \, \tan \left (x\right )^{4} + \frac{5 \,{\left (5 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )\right )}}{8 \,{\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} - \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{5 \,{\left (\sin \left (x\right )^{3} + \sin \left (x\right )\right )}}{4 \,{\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac{4 \, \sin \left (x\right )^{2} - 3}{4 \,{\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac{5}{4 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{2}} - \frac{1}{2} \, \log \left (\sin \left (x\right )^{2} - 1\right ) + \frac{1}{2} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{1}{2} \, \log \left (\sin \left (x\right ) - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.077, size = 119, normalized size = 3.97 \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((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)

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Sympy [B]  time = 8.26483, size = 68, normalized size = 2.27 \begin{align*} - \frac{\log{\left (\sin{\left (x \right )} - 1 \right )}}{2} + \frac{\log{\left (\sin{\left (x \right )} + 1 \right )}}{2} + \frac{\log{\left (\sec ^{2}{\left (x \right )} \right )}}{2} + \frac{5 \tan ^{4}{\left (x \right )}}{2} + \frac{3 \sec ^{4}{\left (x \right )}}{2} - \sec ^{2}{\left (x \right )} + \frac{32 \sin ^{3}{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sin(x) - 1)/2 + log(sin(x) + 1)/2 + log(sec(x)**2)/2 + 5*tan(x)**4/2 + 3*sec(x)**4/2 - sec(x)**2 + 32*sin
(x)**3/(8*sin(x)**4 - 16*sin(x)**2 + 8)

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Giac [B]  time = 1.12623, size = 84, normalized size = 2.8 \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((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(tan
(1/2*x)^2 + 1) - 2*log(abs(tan(1/2*x) - 1))