\(\int \text {sech}^5(a+b x) \, dx\) [5]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 8, antiderivative size = 55 \[ \int \text {sech}^5(a+b x) \, dx=\frac {3 \arctan (\sinh (a+b x))}{8 b}+\frac {3 \text {sech}(a+b x) \tanh (a+b x)}{8 b}+\frac {\text {sech}^3(a+b x) \tanh (a+b x)}{4 b} \]

[Out]

3/8*arctan(sinh(b*x+a))/b+3/8*sech(b*x+a)*tanh(b*x+a)/b+1/4*sech(b*x+a)^3*tanh(b*x+a)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3853, 3855} \[ \int \text {sech}^5(a+b x) \, dx=\frac {3 \arctan (\sinh (a+b x))}{8 b}+\frac {\tanh (a+b x) \text {sech}^3(a+b x)}{4 b}+\frac {3 \tanh (a+b x) \text {sech}(a+b x)}{8 b} \]

[In]

Int[Sech[a + b*x]^5,x]

[Out]

(3*ArcTan[Sinh[a + b*x]])/(8*b) + (3*Sech[a + b*x]*Tanh[a + b*x])/(8*b) + (Sech[a + b*x]^3*Tanh[a + b*x])/(4*b
)

Rule 3853

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {sech}^3(a+b x) \tanh (a+b x)}{4 b}+\frac {3}{4} \int \text {sech}^3(a+b x) \, dx \\ & = \frac {3 \text {sech}(a+b x) \tanh (a+b x)}{8 b}+\frac {\text {sech}^3(a+b x) \tanh (a+b x)}{4 b}+\frac {3}{8} \int \text {sech}(a+b x) \, dx \\ & = \frac {3 \arctan (\sinh (a+b x))}{8 b}+\frac {3 \text {sech}(a+b x) \tanh (a+b x)}{8 b}+\frac {\text {sech}^3(a+b x) \tanh (a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \text {sech}^5(a+b x) \, dx=\frac {3 \arctan (\sinh (a+b x))}{8 b}+\frac {3 \text {sech}(a+b x) \tanh (a+b x)}{8 b}+\frac {\text {sech}^3(a+b x) \tanh (a+b x)}{4 b} \]

[In]

Integrate[Sech[a + b*x]^5,x]

[Out]

(3*ArcTan[Sinh[a + b*x]])/(8*b) + (3*Sech[a + b*x]*Tanh[a + b*x])/(8*b) + (Sech[a + b*x]^3*Tanh[a + b*x])/(4*b
)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.75

method result size
derivativedivides \(\frac {\left (\frac {\operatorname {sech}\left (b x +a \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (b x +a \right )}{8}\right ) \tanh \left (b x +a \right )+\frac {3 \arctan \left ({\mathrm e}^{b x +a}\right )}{4}}{b}\) \(41\)
default \(\frac {\left (\frac {\operatorname {sech}\left (b x +a \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (b x +a \right )}{8}\right ) \tanh \left (b x +a \right )+\frac {3 \arctan \left ({\mathrm e}^{b x +a}\right )}{4}}{b}\) \(41\)
risch \(\frac {{\mathrm e}^{b x +a} \left (3 \,{\mathrm e}^{6 b x +6 a}+11 \,{\mathrm e}^{4 b x +4 a}-11 \,{\mathrm e}^{2 b x +2 a}-3\right )}{4 b \left (1+{\mathrm e}^{2 b x +2 a}\right )^{4}}+\frac {3 i \ln \left ({\mathrm e}^{b x +a}+i\right )}{8 b}-\frac {3 i \ln \left ({\mathrm e}^{b x +a}-i\right )}{8 b}\) \(93\)
parallelrisch \(\frac {3 i \left (-3-\cosh \left (4 b x +4 a \right )-4 \cosh \left (2 b x +2 a \right )\right ) \ln \left (\tanh \left (\frac {b x}{2}+\frac {a}{2}\right )-i\right )+3 i \left (3+\cosh \left (4 b x +4 a \right )+4 \cosh \left (2 b x +2 a \right )\right ) \ln \left (\tanh \left (\frac {b x}{2}+\frac {a}{2}\right )+i\right )+22 \sinh \left (b x +a \right )+6 \sinh \left (3 b x +3 a \right )}{8 b \left (3+\cosh \left (4 b x +4 a \right )+4 \cosh \left (2 b x +2 a \right )\right )}\) \(128\)

[In]

int(sech(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/b*((1/4*sech(b*x+a)^3+3/8*sech(b*x+a))*tanh(b*x+a)+3/4*arctan(exp(b*x+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 812 vs. \(2 (49) = 98\).

Time = 0.24 (sec) , antiderivative size = 812, normalized size of antiderivative = 14.76 \[ \int \text {sech}^5(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate(sech(b*x+a)^5,x, algorithm="fricas")

[Out]

1/4*(3*cosh(b*x + a)^7 + 21*cosh(b*x + a)*sinh(b*x + a)^6 + 3*sinh(b*x + a)^7 + (63*cosh(b*x + a)^2 + 11)*sinh
(b*x + a)^5 + 11*cosh(b*x + a)^5 + 5*(21*cosh(b*x + a)^3 + 11*cosh(b*x + a))*sinh(b*x + a)^4 + (105*cosh(b*x +
 a)^4 + 110*cosh(b*x + a)^2 - 11)*sinh(b*x + a)^3 - 11*cosh(b*x + a)^3 + (63*cosh(b*x + a)^5 + 110*cosh(b*x +
a)^3 - 33*cosh(b*x + a))*sinh(b*x + a)^2 + 3*(cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a
)^8 + 4*(7*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 + 3*cosh(b*x + a))*
sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 + 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*c
osh(b*x + a)^5 + 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 + 15*cosh(b*x +
a)^4 + 9*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 + 3*cosh(b*x + a)^5 + 3
*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (21*cosh(b*x + a)
^6 + 55*cosh(b*x + a)^4 - 33*cosh(b*x + a)^2 - 3)*sinh(b*x + a) - 3*cosh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*co
sh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 + 4*b*cosh(b*x + a)^6 + 4*(7*b*cosh(b*x + a)^2 + b)*sinh(b*x +
 a)^6 + 8*(7*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x +
 a)^4 + 30*b*cosh(b*x + a)^2 + 3*b)*sinh(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 + 10*b*cosh(b*x + a)^3 + 3*b*cosh
(b*x + a))*sinh(b*x + a)^3 + 4*b*cosh(b*x + a)^2 + 4*(7*b*cosh(b*x + a)^6 + 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*
x + a)^2 + b)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a)^7 + 3*b*cosh(b*x + a)^5 + 3*b*cosh(b*x + a)^3 + b*cosh(b*x
+ a))*sinh(b*x + a) + b)

Sympy [F]

\[ \int \text {sech}^5(a+b x) \, dx=\int \operatorname {sech}^{5}{\left (a + b x \right )}\, dx \]

[In]

integrate(sech(b*x+a)**5,x)

[Out]

Integral(sech(a + b*x)**5, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (49) = 98\).

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.04 \[ \int \text {sech}^5(a+b x) \, dx=-\frac {3 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{4 \, b} + \frac {3 \, e^{\left (-b x - a\right )} + 11 \, e^{\left (-3 \, b x - 3 \, a\right )} - 11 \, e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1\right )}} \]

[In]

integrate(sech(b*x+a)^5,x, algorithm="maxima")

[Out]

-3/4*arctan(e^(-b*x - a))/b + 1/4*(3*e^(-b*x - a) + 11*e^(-3*b*x - 3*a) - 11*e^(-5*b*x - 5*a) - 3*e^(-7*b*x -
7*a))/(b*(4*e^(-2*b*x - 2*a) + 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) + e^(-8*b*x - 8*a) + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (49) = 98\).

Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.85 \[ \int \text {sech}^5(a+b x) \, dx=\frac {3 \, \pi + \frac {4 \, {\left (3 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} + 20 \, e^{\left (b x + a\right )} - 20 \, e^{\left (-b x - a\right )}\right )}}{{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )}^{2}} + 6 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{16 \, b} \]

[In]

integrate(sech(b*x+a)^5,x, algorithm="giac")

[Out]

1/16*(3*pi + 4*(3*(e^(b*x + a) - e^(-b*x - a))^3 + 20*e^(b*x + a) - 20*e^(-b*x - a))/((e^(b*x + a) - e^(-b*x -
 a))^2 + 4)^2 + 6*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b

Mupad [B] (verification not implemented)

Time = 2.02 (sec) , antiderivative size = 189, normalized size of antiderivative = 3.44 \[ \int \text {sech}^5(a+b x) \, dx=\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{4\,\sqrt {b^2}}+\frac {{\mathrm {e}}^{a+b\,x}}{2\,b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1\right )}-\frac {4\,{\mathrm {e}}^{3\,a+3\,b\,x}}{b\,\left (4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1\right )}+\frac {3\,{\mathrm {e}}^{a+b\,x}}{4\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

[In]

int(1/cosh(a + b*x)^5,x)

[Out]

(3*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(4*(b^2)^(1/2)) + exp(a + b*x)/(2*b*(2*exp(2*a + 2*b*x) + exp(4*a +
4*b*x) + 1)) - (2*exp(a + b*x))/(b*(3*exp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) + 1)) - (4*exp(
3*a + 3*b*x))/(b*(4*exp(2*a + 2*b*x) + 6*exp(4*a + 4*b*x) + 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1)) + (3*e
xp(a + b*x))/(4*b*(exp(2*a + 2*b*x) + 1))