\(\int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx\) [31]

Optimal result

Integrand size = 12, antiderivative size = 101 \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=\frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}}+\frac {6 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{5/2}}+\frac {8 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{3/2}}+\frac {16 \tanh (a+b x)}{35 b \sqrt {\text {sech}^2(a+b x)}} \]

[Out]

1/7*tanh(b*x+a)/b/(sech(b*x+a)^2)^(7/2)+6/35*tanh(b*x+a)/b/(sech(b*x+a)^2)^(5/2)+8/35*tanh(b*x+a)/b/(sech(b*x+
a)^2)^(3/2)+16/35*tanh(b*x+a)/b/(sech(b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4207, 198, 197} \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=\frac {16 \tanh (a+b x)}{35 b \sqrt {\text {sech}^2(a+b x)}}+\frac {8 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{3/2}}+\frac {6 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{5/2}}+\frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}} \]

[In]

Int[(Sech[a + b*x]^2)^(-7/2),x]

[Out]

Tanh[a + b*x]/(7*b*(Sech[a + b*x]^2)^(7/2)) + (6*Tanh[a + b*x])/(35*b*(Sech[a + b*x]^2)^(5/2)) + (8*Tanh[a + b
*x])/(35*b*(Sech[a + b*x]^2)^(3/2)) + (16*Tanh[a + b*x])/(35*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 )^{9/2}} \, dx,x,\tanh (a+b x)\right )}{b} \\ & = \frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}}+\frac {6 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{7/2}} \, dx,x,\tanh (a+b x)\right )}{7 b} \\ & = \frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}}+\frac {6 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{5/2}}+\frac {24 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{5/2}} \, dx,x,\tanh (a+b x)\right )}{35 b} \\ & = \frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}}+\frac {6 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{5/2}}+\frac {8 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{3/2}}+\frac {16 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{3/2}} \, dx,x,\tanh (a+b x)\right )}{35 b} \\ & = \frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}}+\frac {6 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{5/2}}+\frac {8 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{3/2}}+\frac {16 \tanh (a+b x)}{35 b \sqrt {\text {sech}^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=\frac {\left (35+35 \sinh ^2(a+b x)+21 \sinh ^4(a+b x)+5 \sinh ^6(a+b x)\right ) \tanh (a+b x)}{35 b \sqrt {\text {sech}^2(a+b x)}} \]

[In]

Integrate[(Sech[a + b*x]^2)^(-7/2),x]

[Out]

((35 + 35*Sinh[a + b*x]^2 + 21*Sinh[a + b*x]^4 + 5*Sinh[a + b*x]^6)*Tanh[a + b*x])/(35*b*Sqrt[Sech[a + b*x]^2]
)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(408\) vs. \(2(85)=170\).

Time = 0.48 (sec) , antiderivative size = 409, normalized size of antiderivative = 4.05

method	result	size
risch	\(\frac {{\mathrm e}^{8 b x +8 a}}{896 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {7 \,{\mathrm e}^{6 b x +6 a}}{640 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {7 \,{\mathrm e}^{4 b x +4 a}}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {35 \,{\mathrm e}^{2 b x +2 a}}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {35}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {7 \,{\mathrm e}^{-2 b x -2 a}}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {7 \,{\mathrm e}^{-4 b x -4 a}}{640 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {{\mathrm e}^{-6 b x -6 a}}{896 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}\)	\(409\)

[In]

int(1/(sech(b*x+a)^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/896/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(8*b*x+8*a)+7/640/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)+7/128/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)+35/128/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)-35/128/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)-7/128/b/(1+ex
p(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)-7/640/b/(1+exp(2*b*x+2*a))/(1/(1+e
xp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*exp(-4*b*x-4*a)-1/896/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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=\frac {5 \, \sinh \left (b x + a\right )^{7} + 7 \, {\left (15 \, \cosh \left (b x + a\right )^{2} + 7\right )} \sinh \left (b x + a\right )^{5} + 35 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 14 \, \cosh \left (b x + a\right )^{2} + 7\right )} \sinh \left (b x + a\right )^{3} + 35 \, {\left (\cosh \left (b x + a\right )^{6} + 7 \, \cosh \left (b x + a\right )^{4} + 21 \, \cosh \left (b x + a\right )^{2} + 35\right )} \sinh \left (b x + a\right )}{2240 \, b} \]

[In]

integrate(1/(sech(b*x+a)^2)^(7/2),x, algorithm="fricas")

[Out]

1/2240*(5*sinh(b*x + a)^7 + 7*(15*cosh(b*x + a)^2 + 7)*sinh(b*x + a)^5 + 35*(5*cosh(b*x + a)^4 + 14*cosh(b*x +
a)^2 + 7)*sinh(b*x + a)^3 + 35*(cosh(b*x + a)^6 + 7*cosh(b*x + a)^4 + 21*cosh(b*x + a)^2 + 35)*sinh(b*x + a))
/b

Sympy [A] (verification not implemented)

Time = 17.86 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=\begin {cases} - \frac {16 \tanh ^{7}{\left (a + b x \right )}}{35 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {7}{2}}} + \frac {8 \tanh ^{5}{\left (a + b x \right )}}{5 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {7}{2}}} - \frac {2 \tanh ^{3}{\left (a + b x \right )}}{b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {7}{2}}} + \frac {\tanh {\left (a + b x \right )}}{b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {7}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\operatorname {sech}^{2}{\left (a \right )}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(sech(b*x+a)**2)**(7/2),x)

[Out]

Piecewise((-16*tanh(a + b*x)**7/(35*b*(sech(a + b*x)**2)**(7/2)) + 8*tanh(a + b*x)**5/(5*b*(sech(a + b*x)**2)*
*(7/2)) - 2*tanh(a + b*x)**3/(b*(sech(a + b*x)**2)**(7/2)) + tanh(a + b*x)/(b*(sech(a + b*x)**2)**(7/2)), Ne(b
, 0)), (x/(sech(a)**2)**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=\frac {{\left (49 \, e^{\left (-2 \, b x - 2 \, a\right )} + 245 \, e^{\left (-4 \, b x - 4 \, a\right )} + 1225 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5\right )} e^{\left (7 \, b x + 7 \, a\right )}}{4480 \, b} - \frac {1225 \, e^{\left (-b x - a\right )} + 245 \, e^{\left (-3 \, b x - 3 \, a\right )} + 49 \, e^{\left (-5 \, b x - 5 \, a\right )} + 5 \, e^{\left (-7 \, b x - 7 \, a\right )}}{4480 \, b} \]

[In]

integrate(1/(sech(b*x+a)^2)^(7/2),x, algorithm="maxima")

[Out]

1/4480*(49*e^(-2*b*x - 2*a) + 245*e^(-4*b*x - 4*a) + 1225*e^(-6*b*x - 6*a) + 5)*e^(7*b*x + 7*a)/b - 1/4480*(12
25*e^(-b*x - a) + 245*e^(-3*b*x - 3*a) + 49*e^(-5*b*x - 5*a) + 5*e^(-7*b*x - 7*a))/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=-\frac {{\left (1225 \, e^{\left (6 \, b x + 6 \, a\right )} + 245 \, e^{\left (4 \, b x + 4 \, a\right )} + 49 \, e^{\left (2 \, b x + 2 \, a\right )} + 5\right )} e^{\left (-7 \, b x - 7 \, a\right )} - 5 \, e^{\left (7 \, b x + 7 \, a\right )} - 49 \, e^{\left (5 \, b x + 5 \, a\right )} - 245 \, e^{\left (3 \, b x + 3 \, a\right )} - 1225 \, e^{\left (b x + a\right )}}{4480 \, b} \]

[In]

integrate(1/(sech(b*x+a)^2)^(7/2),x, algorithm="giac")

[Out]

-1/4480*((1225*e^(6*b*x + 6*a) + 245*e^(4*b*x + 4*a) + 49*e^(2*b*x + 2*a) + 5)*e^(-7*b*x - 7*a) - 5*e^(7*b*x +
7*a) - 49*e^(5*b*x + 5*a) - 245*e^(3*b*x + 3*a) - 1225*e^(b*x + a))/b

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx=\int \frac {1}{{\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{7/2}} \,d x \]

[In]

int(1/(1/cosh(a + b*x)^2)^(7/2),x)

[Out]

int(1/(1/cosh(a + b*x)^2)^(7/2), x)