Integrand size = 12, antiderivative size = 76 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {\tanh (a+b x)}{5 b \text {sech}^2(a+b x)^{5/2}}+\frac {4 \tanh (a+b x)}{15 b \text {sech}^2(a+b x)^{3/2}}+\frac {8 \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}} \] Output:
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)
Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {\left (15+10 \sinh ^2(a+b x)+3 \sinh ^4(a+b x)\right ) \tanh (a+b x)}{15 b \sqrt {\text {sech}^2(a+b x)}} \] Input:
Integrate[(Sech[a + b*x]^2)^(-5/2),x]
Output:
((15 + 10*Sinh[a + b*x]^2 + 3*Sinh[a + b*x]^4)*Tanh[a + b*x])/(15*b*Sqrt[S ech[a + b*x]^2])
Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4610, 209, 209, 208}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\sec (i a+i b x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4610 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(a+b x)\right )^{7/2}}d\tanh (a+b x)}{b}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\frac {4}{5} \int \frac {1}{\left (1-\tanh ^2(a+b x)\right )^{5/2}}d\tanh (a+b x)+\frac {\tanh (a+b x)}{5 \left (1-\tanh ^2(a+b x)\right )^{5/2}}}{b}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\frac {4}{5} \left (\frac {2}{3} \int \frac {1}{\left (1-\tanh ^2(a+b x)\right )^{3/2}}d\tanh (a+b x)+\frac {\tanh (a+b x)}{3 \left (1-\tanh ^2(a+b x)\right )^{3/2}}\right )+\frac {\tanh (a+b x)}{5 \left (1-\tanh ^2(a+b x)\right )^{5/2}}}{b}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {\tanh (a+b x)}{5 \left (1-\tanh ^2(a+b x)\right )^{5/2}}+\frac {4}{5} \left (\frac {2 \tanh (a+b x)}{3 \sqrt {1-\tanh ^2(a+b x)}}+\frac {\tanh (a+b x)}{3 \left (1-\tanh ^2(a+b x)\right )^{3/2}}\right )}{b}\) |
Input:
Int[(Sech[a + b*x]^2)^(-5/2),x]
Output:
(Tanh[a + b*x]/(5*(1 - Tanh[a + b*x]^2)^(5/2)) + (4*(Tanh[a + b*x]/(3*(1 - Tanh[a + b*x]^2)^(3/2)) + (2*Tanh[a + b*x])/(3*Sqrt[1 - Tanh[a + b*x]^2]) ))/5)/b
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFac tors[Tan[e + f*x], x]}, Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(304\) vs. \(2(64)=128\).
Time = 0.26 (sec) , antiderivative size = 305, normalized size of antiderivative = 4.01
method | result | size |
risch | \(\frac {{\mathrm e}^{6 b x +6 a}}{160 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{4 b x +4 a}}{96 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{2 b x +2 a}}{16 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {5}{16 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {5 \,{\mathrm e}^{-2 b x -2 a}}{96 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {{\mathrm e}^{-4 b x -4 a}}{160 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}\) | \(305\) |
Input:
int(1/(sech(b*x+a)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
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)*e xp(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+ex p(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)
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {3 \, \sinh \left (b x + a\right )^{5} + 5 \, {\left (6 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 15 \, {\left (\cosh \left (b x + a\right )^{4} + 5 \, \cosh \left (b x + a\right )^{2} + 10\right )} \sinh \left (b x + a\right )}{240 \, b} \] Input:
integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="fricas")
Output:
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
Time = 1.96 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\begin {cases} \frac {8 \tanh ^{5}{\left (a + b x \right )}}{15 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} - \frac {4 \tanh ^{3}{\left (a + b x \right )}}{3 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} + \frac {\tanh {\left (a + b x \right )}}{b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\operatorname {sech}^{2}{\left (a \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(sech(b*x+a)**2)**(5/2),x)
Output:
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))
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac {5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac {5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac {5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \] Input:
integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="maxima")
Output:
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
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=-\frac {{\left (150 \, e^{\left (4 \, b x + 4 \, a\right )} + 25 \, e^{\left (2 \, b x + 2 \, a\right )} + 3\right )} e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (5 \, b x + 5 \, a\right )} - 25 \, e^{\left (3 \, b x + 3 \, a\right )} - 150 \, e^{\left (b x + a\right )}}{480 \, b} \] Input:
integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="giac")
Output:
-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
Timed out. \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\int \frac {1}{{\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{5/2}} \,d x \] Input:
int(1/(1/cosh(a + b*x)^2)^(5/2),x)
Output:
int(1/(1/cosh(a + b*x)^2)^(5/2), x)
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\text {sech}^2(a+b x)^{5/2}} \, dx=\frac {3 e^{10 b x +10 a}+25 e^{8 b x +8 a}+150 e^{6 b x +6 a}-150 e^{4 b x +4 a}-25 e^{2 b x +2 a}-3}{480 e^{5 b x +5 a} b} \] Input:
int(1/(sech(b*x+a)^2)^(5/2),x)
Output:
(3*e**(10*a + 10*b*x) + 25*e**(8*a + 8*b*x) + 150*e**(6*a + 6*b*x) - 150*e **(4*a + 4*b*x) - 25*e**(2*a + 2*b*x) - 3)/(480*e**(5*a + 5*b*x)*b)