Integrand size = 12, antiderivative size = 65 \[ \int \text {sech}^2(a+b x)^{5/2} \, dx=\frac {3 \arcsin (\tanh (a+b x))}{8 b}+\frac {3 \sqrt {\text {sech}^2(a+b x)} \tanh (a+b x)}{8 b}+\frac {\text {sech}^2(a+b x)^{3/2} \tanh (a+b x)}{4 b} \] Output:
3/8*arcsin(tanh(b*x+a))/b+3/8*(sech(b*x+a)^2)^(1/2)*tanh(b*x+a)/b+1/4*(sec h(b*x+a)^2)^(3/2)*tanh(b*x+a)/b
Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \text {sech}^2(a+b x)^{5/2} \, dx=\frac {\text {sech}(a+b x) \left (3 \arctan (\sinh (a+b x))+3 \text {sech}(a+b x) \tanh (a+b x)+2 \text {sech}^3(a+b x) \tanh (a+b x)\right )}{8 b \sqrt {\text {sech}^2(a+b x)}} \] Input:
Integrate[(Sech[a + b*x]^2)^(5/2),x]
Output:
(Sech[a + b*x]*(3*ArcTan[Sinh[a + b*x]] + 3*Sech[a + b*x]*Tanh[a + b*x] + 2*Sech[a + b*x]^3*Tanh[a + b*x]))/(8*b*Sqrt[Sech[a + b*x]^2])
Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4610, 211, 211, 223}
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.
\(\displaystyle \int \text {sech}^2(a+b x)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (\sec (i a+i b x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4610 |
\(\displaystyle \frac {\int \left (1-\tanh ^2(a+b x)\right )^{3/2}d\tanh (a+b x)}{b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {3}{4} \int \sqrt {1-\tanh ^2(a+b x)}d\tanh (a+b x)+\frac {1}{4} \tanh (a+b x) \left (1-\tanh ^2(a+b x)\right )^{3/2}}{b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-\tanh ^2(a+b x)}}d\tanh (a+b x)+\frac {1}{2} \sqrt {1-\tanh ^2(a+b x)} \tanh (a+b x)\right )+\frac {1}{4} \tanh (a+b x) \left (1-\tanh ^2(a+b x)\right )^{3/2}}{b}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \arcsin (\tanh (a+b x))+\frac {1}{2} \tanh (a+b x) \sqrt {1-\tanh ^2(a+b x)}\right )+\frac {1}{4} \tanh (a+b x) \left (1-\tanh ^2(a+b x)\right )^{3/2}}{b}\) |
Input:
Int[(Sech[a + b*x]^2)^(5/2),x]
Output:
((Tanh[a + b*x]*(1 - Tanh[a + b*x]^2)^(3/2))/4 + (3*(ArcSin[Tanh[a + b*x]] /2 + (Tanh[a + b*x]*Sqrt[1 - Tanh[a + b*x]^2])/2))/4)/b
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
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]
Result contains complex when optimal does not.
Time = 6.65 (sec) , antiderivative size = 208, normalized size of antiderivative = 3.20
method | result | size |
risch | \(\frac {\sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 b x +6 a}+11 \,{\mathrm e}^{4 b x +4 a}-11 \,{\mathrm e}^{2 b x +2 a}-3\right )}{4 \left (1+{\mathrm e}^{2 b x +2 a}\right )^{3} b}+\frac {3 i \ln \left ({\mathrm e}^{b x}+i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}\, \left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{-b x -a}}{8 b}-\frac {3 i \ln \left ({\mathrm e}^{b x}-i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}\, \left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{-b x -a}}{8 b}\) | \(208\) |
Input:
int((sech(b*x+a)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/4/(1+exp(2*b*x+2*a))^3*(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*(3* exp(6*b*x+6*a)+11*exp(4*b*x+4*a)-11*exp(2*b*x+2*a)-3)/b+3/8*I*ln(exp(b*x)+ I*exp(-a))/b*(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*(1+exp(2*b*x+2* a))*exp(-b*x-a)-3/8*I*ln(exp(b*x)-I*exp(-a))/b*(1/(1+exp(2*b*x+2*a))^2*exp (2*b*x+2*a))^(1/2)*(1+exp(2*b*x+2*a))*exp(-b*x-a)
Leaf count of result is larger than twice the leaf count of optimal. 812 vs. \(2 (55) = 110\).
Time = 0.09 (sec) , antiderivative size = 812, normalized size of antiderivative = 12.49 \[ \int \text {sech}^2(a+b x)^{5/2} \, dx=\text {Too large to display} \] Input:
integrate((sech(b*x+a)^2)^(5/2),x, algorithm="fricas")
Output:
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*cos h(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*cosh(b* x + a)^5 + 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*co sh(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*cosh(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*co sh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3 + 4*b*cosh(b*x + a)^...
\[ \int \text {sech}^2(a+b x)^{5/2} \, dx=\int \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((sech(b*x+a)**2)**(5/2),x)
Output:
Integral((sech(a + b*x)**2)**(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (55) = 110\).
Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.72 \[ \int \text {sech}^2(a+b x)^{5/2} \, dx=-\frac {3 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{4 \, b} + \frac {3 \, e^{\left (-b x - a\right )} + 11 \, e^{\left (-3 \, b x - 3 \, a\right )} - 11 \, e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1\right )}} \] Input:
integrate((sech(b*x+a)^2)^(5/2),x, algorithm="maxima")
Output:
-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))
Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.57 \[ \int \text {sech}^2(a+b x)^{5/2} \, dx=\frac {3 \, \pi + \frac {4 \, {\left (3 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} + 20 \, e^{\left (b x + a\right )} - 20 \, e^{\left (-b x - a\right )}\right )}}{{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )}^{2}} + 6 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{16 \, b} \] Input:
integrate((sech(b*x+a)^2)^(5/2),x, algorithm="giac")
Output:
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
Timed out. \[ \int \text {sech}^2(a+b x)^{5/2} \, dx=\int {\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{5/2} \,d x \] Input:
int((1/cosh(a + b*x)^2)^(5/2),x)
Output:
int((1/cosh(a + b*x)^2)^(5/2), x)
Time = 0.23 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.94 \[ \int \text {sech}^2(a+b x)^{5/2} \, dx=\frac {3 e^{8 b x +8 a} \mathit {atan} \left (e^{b x +a}\right )+12 e^{6 b x +6 a} \mathit {atan} \left (e^{b x +a}\right )+18 e^{4 b x +4 a} \mathit {atan} \left (e^{b x +a}\right )+12 e^{2 b x +2 a} \mathit {atan} \left (e^{b x +a}\right )+3 \mathit {atan} \left (e^{b x +a}\right )+3 e^{7 b x +7 a}+11 e^{5 b x +5 a}-11 e^{3 b x +3 a}-3 e^{b x +a}}{4 b \left (e^{8 b x +8 a}+4 e^{6 b x +6 a}+6 e^{4 b x +4 a}+4 e^{2 b x +2 a}+1\right )} \] Input:
int((sech(b*x+a)^2)^(5/2),x)
Output:
(3*e**(8*a + 8*b*x)*atan(e**(a + b*x)) + 12*e**(6*a + 6*b*x)*atan(e**(a + b*x)) + 18*e**(4*a + 4*b*x)*atan(e**(a + b*x)) + 12*e**(2*a + 2*b*x)*atan( e**(a + b*x)) + 3*atan(e**(a + b*x)) + 3*e**(7*a + 7*b*x) + 11*e**(5*a + 5 *b*x) - 11*e**(3*a + 3*b*x) - 3*e**(a + b*x))/(4*b*(e**(8*a + 8*b*x) + 4*e **(6*a + 6*b*x) + 6*e**(4*a + 4*b*x) + 4*e**(2*a + 2*b*x) + 1))