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

Optimal. Leaf size=74 \[ \frac{16 \tanh (x)}{35 a^3 \sqrt{a \text{sech}^2(x)}}+\frac{8 \tanh (x)}{35 a^2 \left (a \text{sech}^2(x)\right )^{3/2}}+\frac{6 \tanh (x)}{35 a \left (a \text{sech}^2(x)\right )^{5/2}}+\frac{\tanh (x)}{7 \left (a \text{sech}^2(x)\right )^{7/2}} \]

[Out]

Tanh[x]/(7*(a*Sech[x]^2)^(7/2)) + (6*Tanh[x])/(35*a*(a*Sech[x]^2)^(5/2)) + (8*Tanh[x])/(35*a^2*(a*Sech[x]^2)^(
3/2)) + (16*Tanh[x])/(35*a^3*Sqrt[a*Sech[x]^2])

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Rubi [A]  time = 0.0403825, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4122, 192, 191} \[ \frac{16 \tanh (x)}{35 a^3 \sqrt{a \text{sech}^2(x)}}+\frac{8 \tanh (x)}{35 a^2 \left (a \text{sech}^2(x)\right )^{3/2}}+\frac{6 \tanh (x)}{35 a \left (a \text{sech}^2(x)\right )^{5/2}}+\frac{\tanh (x)}{7 \left (a \text{sech}^2(x)\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sech[x]^2)^(-7/2),x]

[Out]

Tanh[x]/(7*(a*Sech[x]^2)^(7/2)) + (6*Tanh[x])/(35*a*(a*Sech[x]^2)^(5/2)) + (8*Tanh[x])/(35*a^2*(a*Sech[x]^2)^(
3/2)) + (16*Tanh[x])/(35*a^3*Sqrt[a*Sech[x]^2])

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
]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0486604, size = 42, normalized size = 0.57 \[ \frac{(1225 \sinh (x)+245 \sinh (3 x)+49 \sinh (5 x)+5 \sinh (7 x)) \cosh (x) \sqrt{a \text{sech}^2(x)}}{2240 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sech[x]^2)^(-7/2),x]

[Out]

(Cosh[x]*Sqrt[a*Sech[x]^2]*(1225*Sinh[x] + 245*Sinh[3*x] + 49*Sinh[5*x] + 5*Sinh[7*x]))/(2240*a^4)

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Maple [B]  time = 0.057, size = 262, normalized size = 3.5 \begin{align*}{\frac{{{\rm e}^{8\,x}}}{896\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+{\frac{7\,{{\rm e}^{6\,x}}}{640\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+{\frac{7\,{{\rm e}^{4\,x}}}{128\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+{\frac{35\,{{\rm e}^{2\,x}}}{128\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-{\frac{35}{128\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-{\frac{7\,{{\rm e}^{-2\,x}}}{128\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-{\frac{7\,{{\rm e}^{-4\,x}}}{640\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-{\frac{{{\rm e}^{-6\,x}}}{896\,{a}^{3} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/896/a^3*exp(8*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)+7/640/a^3*exp(6*x)/(exp(2*x)+1)/(a*exp(2*x)/
(exp(2*x)+1)^2)^(1/2)+7/128/a^3*exp(4*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)+35/128/a^3*exp(2*x)/(e
xp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)-35/128/a^3/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)-7/128/a
^3*exp(-2*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)-7/640/a^3*exp(-4*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(
2*x)+1)^2)^(1/2)-1/896/a^3*exp(-6*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)

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Maxima [A]  time = 1.65516, size = 96, normalized size = 1.3 \begin{align*} \frac{e^{\left (7 \, x\right )}}{896 \, a^{\frac{7}{2}}} + \frac{7 \, e^{\left (5 \, x\right )}}{640 \, a^{\frac{7}{2}}} + \frac{7 \, e^{\left (3 \, x\right )}}{128 \, a^{\frac{7}{2}}} - \frac{35 \, e^{\left (-x\right )}}{128 \, a^{\frac{7}{2}}} - \frac{7 \, e^{\left (-3 \, x\right )}}{128 \, a^{\frac{7}{2}}} - \frac{7 \, e^{\left (-5 \, x\right )}}{640 \, a^{\frac{7}{2}}} - \frac{e^{\left (-7 \, x\right )}}{896 \, a^{\frac{7}{2}}} + \frac{35 \, e^{x}}{128 \, a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/896*e^(7*x)/a^(7/2) + 7/640*e^(5*x)/a^(7/2) + 7/128*e^(3*x)/a^(7/2) - 35/128*e^(-x)/a^(7/2) - 7/128*e^(-3*x)
/a^(7/2) - 7/640*e^(-5*x)/a^(7/2) - 1/896*e^(-7*x)/a^(7/2) + 35/128*e^x/a^(7/2)

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Fricas [B]  time = 2.29884, size = 3340, normalized size = 45.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4480*(5*(e^(2*x) + 1)*sinh(x)^14 + 5*cosh(x)^14 + 70*(cosh(x)*e^(2*x) + cosh(x))*sinh(x)^13 + 7*(65*cosh(x)^
2 + (65*cosh(x)^2 + 7)*e^(2*x) + 7)*sinh(x)^12 + 49*cosh(x)^12 + 28*(65*cosh(x)^3 + (65*cosh(x)^3 + 21*cosh(x)
)*e^(2*x) + 21*cosh(x))*sinh(x)^11 + 7*(715*cosh(x)^4 + 462*cosh(x)^2 + (715*cosh(x)^4 + 462*cosh(x)^2 + 35)*e
^(2*x) + 35)*sinh(x)^10 + 245*cosh(x)^10 + 70*(143*cosh(x)^5 + 154*cosh(x)^3 + (143*cosh(x)^5 + 154*cosh(x)^3
+ 35*cosh(x))*e^(2*x) + 35*cosh(x))*sinh(x)^9 + 35*(429*cosh(x)^6 + 693*cosh(x)^4 + 315*cosh(x)^2 + (429*cosh(
x)^6 + 693*cosh(x)^4 + 315*cosh(x)^2 + 35)*e^(2*x) + 35)*sinh(x)^8 + 1225*cosh(x)^8 + 8*(2145*cosh(x)^7 + 4851
*cosh(x)^5 + 3675*cosh(x)^3 + (2145*cosh(x)^7 + 4851*cosh(x)^5 + 3675*cosh(x)^3 + 1225*cosh(x))*e^(2*x) + 1225
*cosh(x))*sinh(x)^7 + 7*(2145*cosh(x)^8 + 6468*cosh(x)^6 + 7350*cosh(x)^4 + 4900*cosh(x)^2 + (2145*cosh(x)^8 +
6468*cosh(x)^6 + 7350*cosh(x)^4 + 4900*cosh(x)^2 - 175)*e^(2*x) - 175)*sinh(x)^6 - 1225*cosh(x)^6 + 14*(715*c
osh(x)^9 + 2772*cosh(x)^7 + 4410*cosh(x)^5 + 4900*cosh(x)^3 + (715*cosh(x)^9 + 2772*cosh(x)^7 + 4410*cosh(x)^5
+ 4900*cosh(x)^3 - 525*cosh(x))*e^(2*x) - 525*cosh(x))*sinh(x)^5 + 35*(143*cosh(x)^10 + 693*cosh(x)^8 + 1470*
cosh(x)^6 + 2450*cosh(x)^4 - 525*cosh(x)^2 + (143*cosh(x)^10 + 693*cosh(x)^8 + 1470*cosh(x)^6 + 2450*cosh(x)^4
- 525*cosh(x)^2 - 7)*e^(2*x) - 7)*sinh(x)^4 - 245*cosh(x)^4 + 140*(13*cosh(x)^11 + 77*cosh(x)^9 + 210*cosh(x)
^7 + 490*cosh(x)^5 - 175*cosh(x)^3 + (13*cosh(x)^11 + 77*cosh(x)^9 + 210*cosh(x)^7 + 490*cosh(x)^5 - 175*cosh(
x)^3 - 7*cosh(x))*e^(2*x) - 7*cosh(x))*sinh(x)^3 + 7*(65*cosh(x)^12 + 462*cosh(x)^10 + 1575*cosh(x)^8 + 4900*c
osh(x)^6 - 2625*cosh(x)^4 - 210*cosh(x)^2 + (65*cosh(x)^12 + 462*cosh(x)^10 + 1575*cosh(x)^8 + 4900*cosh(x)^6
- 2625*cosh(x)^4 - 210*cosh(x)^2 - 7)*e^(2*x) - 7)*sinh(x)^2 - 49*cosh(x)^2 + (5*cosh(x)^14 + 49*cosh(x)^12 +
245*cosh(x)^10 + 1225*cosh(x)^8 - 1225*cosh(x)^6 - 245*cosh(x)^4 - 49*cosh(x)^2 - 5)*e^(2*x) + 14*(5*cosh(x)^1
3 + 42*cosh(x)^11 + 175*cosh(x)^9 + 700*cosh(x)^7 - 525*cosh(x)^5 - 70*cosh(x)^3 + (5*cosh(x)^13 + 42*cosh(x)^
11 + 175*cosh(x)^9 + 700*cosh(x)^7 - 525*cosh(x)^5 - 70*cosh(x)^3 - 7*cosh(x))*e^(2*x) - 7*cosh(x))*sinh(x) -
5)*sqrt(a/(e^(4*x) + 2*e^(2*x) + 1))*e^x/(a^4*cosh(x)^7*e^x + 7*a^4*cosh(x)^6*e^x*sinh(x) + 21*a^4*cosh(x)^5*e
^x*sinh(x)^2 + 35*a^4*cosh(x)^4*e^x*sinh(x)^3 + 35*a^4*cosh(x)^3*e^x*sinh(x)^4 + 21*a^4*cosh(x)^2*e^x*sinh(x)^
5 + 7*a^4*cosh(x)*e^x*sinh(x)^6 + a^4*e^x*sinh(x)^7)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.11404, size = 72, normalized size = 0.97 \begin{align*} -\frac{{\left (1225 \, e^{\left (6 \, x\right )} + 245 \, e^{\left (4 \, x\right )} + 49 \, e^{\left (2 \, x\right )} + 5\right )} e^{\left (-7 \, x\right )} - 5 \, e^{\left (7 \, x\right )} - 49 \, e^{\left (5 \, x\right )} - 245 \, e^{\left (3 \, x\right )} - 1225 \, e^{x}}{4480 \, a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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