\(\int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx\) [30]

Optimal result

Integrand size = 12, antiderivative size = 76 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {8 \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}} \]

[Out]

1/5*tanh(b*x+a)/b/(sech(b*x+a)^2)^(5/2)+4/15*tanh(b*x+a)/b/(sech(b*x+a)^2)^(3/2)+8/15*tanh(b*x+a)/b/(sech(b*x+
a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4207, 198, 197} \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {8 \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}} \]

[In]

Int[(Sech[a + b*x]^2)^(-5/2),x]

[Out]

Tanh[a + b*x]/(5*b*(Sech[a + b*x]^2)^(5/2)) + (4*Tanh[a + b*x])/(15*b*(Sech[a + b*x]^2)^(3/2)) + (8*Tanh[a + b
*x])/(15*b*Sqrt[Sech[a + b*x]^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 4207

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[b*(ff/
f), Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{7/2}} \, dx,x,\tanh (a+b x)\right )}{b} \\ & = \frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{5/2}} \, dx,x,\tanh (a+b x)\right )}{5 b} \\ & = \frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {8 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{3/2}} \, dx,x,\tanh (a+b x)\right )}{15 b} \\ & = \frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {8 \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {\left (15+10 \sinh ^2(a+b x)+3 \sinh ^4(a+b x)\right ) \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}} \]

[In]

Integrate[(Sech[a + b*x]^2)^(-5/2),x]

[Out]

((15 + 10*Sinh[a + b*x]^2 + 3*Sinh[a + b*x]^4)*Tanh[a + b*x])/(15*b*Sqrt[Sech[a + b*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(304\) vs. \(2(64)=128\).

Time = 0.47 (sec) , antiderivative size = 305, normalized size of antiderivative = 4.01

method	result	size
risch	\(\frac {{\mathrm e}^{6 b x +6 a}}{160 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{4 b x +4 a}}{96 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{2 b x +2 a}}{16 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {5}{16 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {5 \,{\mathrm e}^{-2 b x -2 a}}{96 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {{\mathrm e}^{-4 b x -4 a}}{160 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}\)	\(305\)

[In]

int(1/(sech(b*x+a)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/160/b/(1+exp(2*b*x+2*a))/(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*exp(6*b*x+6*a)+5/96/b/(1+exp(2*b*x+2*
a))/(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*exp(4*b*x+4*a)+5/16/b/(1+exp(2*b*x+2*a))/(1/(1+exp(2*b*x+2*a
))^2*exp(2*b*x+2*a))^(1/2)*exp(2*b*x+2*a)-5/16/b/(1+exp(2*b*x+2*a))/(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1
/2)-5/96/b/(1+exp(2*b*x+2*a))/(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*exp(-2*b*x-2*a)-1/160/b/(1+exp(2*b
*x+2*a))/(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*exp(-4*b*x-4*a)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {3 \, \sinh \left (b x + a\right )^{5} + 5 \, {\left (6 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 15 \, {\left (\cosh \left (b x + a\right )^{4} + 5 \, \cosh \left (b x + a\right )^{2} + 10\right )} \sinh \left (b x + a\right )}{240 \, b} \]

[In]

integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="fricas")

[Out]

1/240*(3*sinh(b*x + a)^5 + 5*(6*cosh(b*x + a)^2 + 5)*sinh(b*x + a)^3 + 15*(cosh(b*x + a)^4 + 5*cosh(b*x + a)^2
+ 10)*sinh(b*x + a))/b

Sympy [A] (verification not implemented)

Time = 1.79 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\begin {cases} \frac {8 \tanh ^{5}{\left (a + b x \right )}}{15 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} - \frac {4 \tanh ^{3}{\left (a + b x \right )}}{3 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} + \frac {\tanh {\left (a + b x \right )}}{b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\operatorname {sech}^{2}{\left (a \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(sech(b*x+a)**2)**(5/2),x)

[Out]

Piecewise((8*tanh(a + b*x)**5/(15*b*(sech(a + b*x)**2)**(5/2)) - 4*tanh(a + b*x)**3/(3*b*(sech(a + b*x)**2)**(
5/2)) + tanh(a + b*x)/(b*(sech(a + b*x)**2)**(5/2)), Ne(b, 0)), (x/(sech(a)**2)**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac {5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac {5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac {5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \]

[In]

integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="maxima")

[Out]

1/160*e^(5*b*x + 5*a)/b + 5/96*e^(3*b*x + 3*a)/b + 5/16*e^(b*x + a)/b - 5/16*e^(-b*x - a)/b - 5/96*e^(-3*b*x -
3*a)/b - 1/160*e^(-5*b*x - 5*a)/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=-\frac {{\left (150 \, e^{\left (4 \, b x + 4 \, a\right )} + 25 \, e^{\left (2 \, b x + 2 \, a\right )} + 3\right )} e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (5 \, b x + 5 \, a\right )} - 25 \, e^{\left (3 \, b x + 3 \, a\right )} - 150 \, e^{\left (b x + a\right )}}{480 \, b} \]

[In]

integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="giac")

[Out]

-1/480*((150*e^(4*b*x + 4*a) + 25*e^(2*b*x + 2*a) + 3)*e^(-5*b*x - 5*a) - 3*e^(5*b*x + 5*a) - 25*e^(3*b*x + 3*
a) - 150*e^(b*x + a))/b

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\int \frac {1}{{\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{5/2}} \,d x \]

[In]

int(1/(1/cosh(a + b*x)^2)^(5/2),x)

[Out]

int(1/(1/cosh(a + b*x)^2)^(5/2), x)