\(\int (\sec (x)+\tan (x))^5 \, dx\) [273]

Optimal result

Integrand size = 7, antiderivative size = 30 \[ \int (\sec (x)+\tan (x))^5 \, dx=-\log (1-\sin (x))+\frac {2}{(1-\sin (x))^2}-\frac {4}{1-\sin (x)} \]

[Out]

-ln(1-sin(x))+2/(1-sin(x))^2-4/(1-sin(x))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4476, 2746, 45} \[ \int (\sec (x)+\tan (x))^5 \, dx=-\frac {4}{1-\sin (x)}+\frac {2}{(1-\sin (x))^2}-\log (1-\sin (x)) \]

[In]

Int[(Sec[x] + Tan[x])^5,x]

[Out]

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

Rule 45

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])

Rule 2746

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 4476

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]

Rubi steps \begin{align*} \text {integral}& = \int \sec ^5(x) (1+\sin (x))^5 \, dx \\ & = \text {Subst}\left (\int \frac {(1+x)^2}{(1-x)^3} \, dx,x,\sin (x)\right ) \\ & = \text {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] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int (\sec (x)+\tan (x))^5 \, dx=\text {arctanh}(\sin (x))-\log (\cos (x))+\frac {5 \sec ^4(x)}{4}+\sec (x) \tan (x)-\sec ^3(x) \tan (x)-\frac {\tan ^2(x)}{2}+5 \sec (x) \tan ^3(x)+\frac {11 \tan ^4(x)}{4} \]

[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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 8.77 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70

method	result	size
risch	\(i x +\frac {8 i \left (-i {\mathrm e}^{2 i x}+{\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{\left ({\mathrm e}^{i x}-i\right )^{4}}-2 \ln \left ({\mathrm e}^{i x}-i\right )\)	\(51\)
parts	\(-\left (-\frac {\sec \left (x \right )^{3}}{4}-\frac {3 \sec \left (x \right )}{8}\right ) \tan \left (x \right )+\ln \left (\sec \left (x \right )+\tan \left (x \right )\right )+\frac {11 \tan \left (x \right )^{4}}{4}-\frac {\tan \left (x \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+\frac {5 \sin \left (x \right )^{3}}{2 \cos \left (x \right )^{4}}+\frac {5 \sin \left (x \right )^{3}}{4 \cos \left (x \right )^{2}}-\frac {5 \sin \left (x \right )}{8}+\frac {5 \sec \left (x \right )^{4}}{4}+\frac {5 \sin \left (x \right )^{5}}{4 \cos \left (x \right )^{4}}-\frac {5 \sin \left (x \right )^{5}}{8 \cos \left (x \right )^{2}}-\frac {5 \sin \left (x \right )^{3}}{8}\)	\(100\)
default	\(-\left (-\frac {\sec \left (x \right )^{3}}{4}-\frac {3 \sec \left (x \right )}{8}\right ) \tan \left (x \right )+\ln \left (\sec \left (x \right )+\tan \left (x \right )\right )+\frac {5}{4 \cos \left (x \right )^{4}}+\frac {5 \sin \left (x \right )^{3}}{2 \cos \left (x \right )^{4}}+\frac {5 \sin \left (x \right )^{3}}{4 \cos \left (x \right )^{2}}-\frac {5 \sin \left (x \right )}{8}+\frac {5 \sin \left (x \right )^{4}}{2 \cos \left (x \right )^{4}}+\frac {5 \sin \left (x \right )^{5}}{4 \cos \left (x \right )^{4}}-\frac {5 \sin \left (x \right )^{5}}{8 \cos \left (x \right )^{2}}-\frac {5 \sin \left (x \right )^{3}}{8}+\frac {\tan \left (x \right )^{4}}{4}-\frac {\tan \left (x \right )^{2}}{2}-\ln \left (\cos \left (x \right )\right )\)	\(106\)

[In]

int((sec(x)+tan(x))^5,x,method=_RETURNVERBOSE)

[Out]

I*x+8*I/(exp(I*x)-I)^4*(-I*exp(2*I*x)+exp(3*I*x)-exp(I*x))-2*ln(exp(I*x)-I)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int (\sec (x)+\tan (x))^5 \, dx=-\frac {{\left (\cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 2\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 4 \, \sin \left (x\right ) - 2}{\cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 2} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (20) = 40\).

Time = 1.98 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int (\sec (x)+\tan (x))^5 \, dx=- \frac {\log {\left (\sin {\left (x \right )} - 1 \right )}}{2} + \frac {\log {\left (\sin {\left (x \right )} + 1 \right )}}{2} + \frac {\log {\left (\sec ^{2}{\left (x \right )} \right )}}{2} + \frac {5 \tan ^{4}{\left (x \right )}}{2} + \frac {3 \sec ^{4}{\left (x \right )}}{2} - \sec ^{2}{\left (x \right )} + \frac {32 \sin ^{3}{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (26) = 52\).

Time = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 4.70 \[ \int (\sec (x)+\tan (x))^5 \, dx=\frac {5}{2} \, \tan \left (x\right )^{4} + \frac {5 \, {\left (5 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )\right )}}{8 \, {\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} - \frac {3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )}{8 \, {\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac {5 \, {\left (\sin \left (x\right )^{3} + \sin \left (x\right )\right )}}{4 \, {\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac {4 \, \sin \left (x\right )^{2} - 3}{4 \, {\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac {5}{4 \, {\left (\sin \left (x\right )^{2} - 1\right )}^{2}} - \frac {1}{2} \, \log \left (\sin \left (x\right )^{2} - 1\right ) + \frac {1}{2} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac {1}{2} \, \log \left (\sin \left (x\right ) - 1\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (26) = 52\).

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.07 \[ \int (\sec (x)+\tan (x))^5 \, dx=\frac {25 \, \tan \left (\frac {1}{2} \, x\right )^{4} - 100 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 198 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 100 \, \tan \left (\frac {1}{2} \, x\right ) + 25}{6 \, {\left (\tan \left (\frac {1}{2} \, x\right ) - 1\right )}^{4}} + \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) - 2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 29.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.97 \[ \int (\sec (x)+\tan (x))^5 \, dx=\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )-2\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )+\frac {8\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {x}{2}\right )+1} \]

[In]

int((tan(x) + 1/cos(x))^5,x)

[Out]

log(tan(x/2)^2 + 1) - 2*log(tan(x/2) - 1) + (8*tan(x/2)^2)/(6*tan(x/2)^2 - 4*tan(x/2) - 4*tan(x/2)^3 + tan(x/2
)^4 + 1)