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

Optimal. Leaf size=101 \[ \frac {16 \tanh (a+b x)}{35 b \sqrt {\text {sech}^2(a+b x)}}+\frac {8 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{3/2}}+\frac {6 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{5/2}}+\frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4122, 192, 191} \[ \frac {16 \tanh (a+b x)}{35 b \sqrt {\text {sech}^2(a+b x)}}+\frac {8 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{3/2}}+\frac {6 \tanh (a+b x)}{35 b \text {sech}^2(a+b x)^{5/2}}+\frac {\tanh (a+b x)}{7 b \text {sech}^2(a+b x)^{7/2}} \]

Antiderivative was successfully verified.

[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 191

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 192

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 4122

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*} \int \frac {1}{\text {sech}^2(a+b x)^{7/2}} \, dx &=\frac {\operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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*}

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Mathematica [A]  time = 0.14, size = 57, normalized size = 0.56 \[ \frac {\left (5 \sinh ^6(a+b x)+21 \sinh ^4(a+b x)+35 \sinh ^2(a+b x)+35\right ) \tanh (a+b x)}{35 b \sqrt {\text {sech}^2(a+b x)}} \]

Antiderivative was successfully verified.

[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]
)

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fricas [A]  time = 0.49, size = 108, normalized size = 1.07 \[ \frac {5 \, \sinh \left (b x + a\right )^{7} + 7 \, {\left (15 \, \cosh \left (b x + a\right )^{2} + 7\right )} \sinh \left (b x + a\right )^{5} + 35 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 14 \, \cosh \left (b x + a\right )^{2} + 7\right )} \sinh \left (b x + a\right )^{3} + 35 \, {\left (\cosh \left (b x + a\right )^{6} + 7 \, \cosh \left (b x + a\right )^{4} + 21 \, \cosh \left (b x + a\right )^{2} + 35\right )} \sinh \left (b x + a\right )}{2240 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.13, size = 92, normalized size = 0.91 \[ -\frac {{\left (1225 \, e^{\left (6 \, b x + 6 \, a\right )} + 245 \, e^{\left (4 \, b x + 4 \, a\right )} + 49 \, e^{\left (2 \, b x + 2 \, a\right )} + 5\right )} e^{\left (-7 \, b x - 7 \, a\right )} - 5 \, e^{\left (7 \, b x + 7 \, a\right )} - 49 \, e^{\left (5 \, b x + 5 \, a\right )} - 245 \, e^{\left (3 \, b x + 3 \, a\right )} - 1225 \, e^{\left (b x + a\right )}}{4480 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [B]  time = 0.42, size = 409, normalized size = 4.05 \[ \frac {{\mathrm e}^{8 b x +8 a}}{896 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {7 \,{\mathrm e}^{6 b x +6 a}}{640 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {7 \,{\mathrm e}^{4 b x +4 a}}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {35 \,{\mathrm e}^{2 b x +2 a}}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {35}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {7 \,{\mathrm e}^{-2 b x -2 a}}{128 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {7 \,{\mathrm e}^{-4 b x -4 a}}{640 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {{\mathrm e}^{-6 b x -6 a}}{896 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maxima [A]  time = 0.32, size = 100, normalized size = 0.99 \[ \frac {{\left (49 \, e^{\left (-2 \, b x - 2 \, a\right )} + 245 \, e^{\left (-4 \, b x - 4 \, a\right )} + 1225 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5\right )} e^{\left (7 \, b x + 7 \, a\right )}}{4480 \, b} - \frac {1225 \, e^{\left (-b x - a\right )} + 245 \, e^{\left (-3 \, b x - 3 \, a\right )} + 49 \, e^{\left (-5 \, b x - 5 \, a\right )} + 5 \, e^{\left (-7 \, b x - 7 \, a\right )}}{4480 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/cosh(a + b*x)^2)^(7/2),x)

[Out]

int(1/(1/cosh(a + b*x)^2)^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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