3.15 \(\int x \text{sech}^7(a+b x^2) \, dx\)
Optimal. Leaf size=90 \[ \frac{5 \tan ^{-1}\left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac{\tanh \left (a+b x^2\right ) \text{sech}^5\left (a+b x^2\right )}{12 b}+\frac{5 \tanh \left (a+b x^2\right ) \text{sech}^3\left (a+b x^2\right )}{48 b}+\frac{5 \tanh \left (a+b x^2\right ) \text{sech}\left (a+b x^2\right )}{32 b} \]
[Out]
(5*ArcTan[Sinh[a + b*x^2]])/(32*b) + (5*Sech[a + b*x^2]*Tanh[a + b*x^2])/(32*b) + (5*Sech[a + b*x^2]^3*Tanh[a
+ b*x^2])/(48*b) + (Sech[a + b*x^2]^5*Tanh[a + b*x^2])/(12*b)
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Rubi [A] time = 0.0795419, antiderivative size = 90, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{5436, 3768, 3770} \[ \frac{5 \tan ^{-1}\left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac{\tanh \left (a+b x^2\right ) \text{sech}^5\left (a+b x^2\right )}{12 b}+\frac{5 \tanh \left (a+b x^2\right ) \text{sech}^3\left (a+b x^2\right )}{48 b}+\frac{5 \tanh \left (a+b x^2\right ) \text{sech}\left (a+b x^2\right )}{32 b} \]
Antiderivative was successfully verified.
[In]
Int[x*Sech[a + b*x^2]^7,x]
[Out]
(5*ArcTan[Sinh[a + b*x^2]])/(32*b) + (5*Sech[a + b*x^2]*Tanh[a + b*x^2])/(32*b) + (5*Sech[a + b*x^2]^3*Tanh[a
+ b*x^2])/(48*b) + (Sech[a + b*x^2]^5*Tanh[a + b*x^2])/(12*b)
Rule 5436
Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sech[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]
Rule 3768
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int x \text{sech}^7\left (a+b x^2\right ) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \text{sech}^7(a+b x) \, dx,x,x^2\right )\\ &=\frac{\text{sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b}+\frac{5}{12} \operatorname{Subst}\left (\int \text{sech}^5(a+b x) \, dx,x,x^2\right )\\ &=\frac{5 \text{sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac{\text{sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b}+\frac{5}{16} \operatorname{Subst}\left (\int \text{sech}^3(a+b x) \, dx,x,x^2\right )\\ &=\frac{5 \text{sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac{5 \text{sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac{\text{sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b}+\frac{5}{32} \operatorname{Subst}\left (\int \text{sech}(a+b x) \, dx,x,x^2\right )\\ &=\frac{5 \tan ^{-1}\left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac{5 \text{sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac{5 \text{sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac{\text{sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b}\\ \end{align*}
Mathematica [A] time = 0.137687, size = 77, normalized size = 0.86 \[ \frac{15 \tan ^{-1}\left (\sinh \left (a+b x^2\right )\right )+8 \tanh \left (a+b x^2\right ) \text{sech}^5\left (a+b x^2\right )+10 \tanh \left (a+b x^2\right ) \text{sech}^3\left (a+b x^2\right )+15 \tanh \left (a+b x^2\right ) \text{sech}\left (a+b x^2\right )}{96 b} \]
Antiderivative was successfully verified.
[In]
Integrate[x*Sech[a + b*x^2]^7,x]
[Out]
(15*ArcTan[Sinh[a + b*x^2]] + 15*Sech[a + b*x^2]*Tanh[a + b*x^2] + 10*Sech[a + b*x^2]^3*Tanh[a + b*x^2] + 8*Se
ch[a + b*x^2]^5*Tanh[a + b*x^2])/(96*b)
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Maple [A] time = 0.053, size = 83, normalized size = 0.9 \begin{align*}{\frac{ \left ({\rm sech} \left (b{x}^{2}+a\right ) \right ) ^{5}\tanh \left ( b{x}^{2}+a \right ) }{12\,b}}+{\frac{5\, \left ({\rm sech} \left (b{x}^{2}+a\right ) \right ) ^{3}\tanh \left ( b{x}^{2}+a \right ) }{48\,b}}+{\frac{5\,{\rm sech} \left (b{x}^{2}+a\right )\tanh \left ( b{x}^{2}+a \right ) }{32\,b}}+{\frac{5\,\arctan \left ({{\rm e}^{b{x}^{2}+a}} \right ) }{16\,b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*sech(b*x^2+a)^7,x)
[Out]
1/12*sech(b*x^2+a)^5*tanh(b*x^2+a)/b+5/48*sech(b*x^2+a)^3*tanh(b*x^2+a)/b+5/32*sech(b*x^2+a)*tanh(b*x^2+a)/b+5
/16/b*arctan(exp(b*x^2+a))
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Maxima [B] time = 1.65228, size = 246, normalized size = 2.73 \begin{align*} -\frac{5 \, \arctan \left (e^{\left (-b x^{2} - a\right )}\right )}{16 \, b} + \frac{15 \, e^{\left (-b x^{2} - a\right )} + 85 \, e^{\left (-3 \, b x^{2} - 3 \, a\right )} + 198 \, e^{\left (-5 \, b x^{2} - 5 \, a\right )} - 198 \, e^{\left (-7 \, b x^{2} - 7 \, a\right )} - 85 \, e^{\left (-9 \, b x^{2} - 9 \, a\right )} - 15 \, e^{\left (-11 \, b x^{2} - 11 \, a\right )}}{48 \, b{\left (6 \, e^{\left (-2 \, b x^{2} - 2 \, a\right )} + 15 \, e^{\left (-4 \, b x^{2} - 4 \, a\right )} + 20 \, e^{\left (-6 \, b x^{2} - 6 \, a\right )} + 15 \, e^{\left (-8 \, b x^{2} - 8 \, a\right )} + 6 \, e^{\left (-10 \, b x^{2} - 10 \, a\right )} + e^{\left (-12 \, b x^{2} - 12 \, a\right )} + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sech(b*x^2+a)^7,x, algorithm="maxima")
[Out]
-5/16*arctan(e^(-b*x^2 - a))/b + 1/48*(15*e^(-b*x^2 - a) + 85*e^(-3*b*x^2 - 3*a) + 198*e^(-5*b*x^2 - 5*a) - 19
8*e^(-7*b*x^2 - 7*a) - 85*e^(-9*b*x^2 - 9*a) - 15*e^(-11*b*x^2 - 11*a))/(b*(6*e^(-2*b*x^2 - 2*a) + 15*e^(-4*b*
x^2 - 4*a) + 20*e^(-6*b*x^2 - 6*a) + 15*e^(-8*b*x^2 - 8*a) + 6*e^(-10*b*x^2 - 10*a) + e^(-12*b*x^2 - 12*a) + 1
))
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Fricas [B] time = 2.27907, size = 5025, normalized size = 55.83 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sech(b*x^2+a)^7,x, algorithm="fricas")
[Out]
1/48*(15*cosh(b*x^2 + a)^11 + 165*cosh(b*x^2 + a)*sinh(b*x^2 + a)^10 + 15*sinh(b*x^2 + a)^11 + 5*(165*cosh(b*x
^2 + a)^2 + 17)*sinh(b*x^2 + a)^9 + 85*cosh(b*x^2 + a)^9 + 45*(55*cosh(b*x^2 + a)^3 + 17*cosh(b*x^2 + a))*sinh
(b*x^2 + a)^8 + 18*(275*cosh(b*x^2 + a)^4 + 170*cosh(b*x^2 + a)^2 + 11)*sinh(b*x^2 + a)^7 + 198*cosh(b*x^2 + a
)^7 + 42*(165*cosh(b*x^2 + a)^5 + 170*cosh(b*x^2 + a)^3 + 33*cosh(b*x^2 + a))*sinh(b*x^2 + a)^6 + 18*(385*cosh
(b*x^2 + a)^6 + 595*cosh(b*x^2 + a)^4 + 231*cosh(b*x^2 + a)^2 - 11)*sinh(b*x^2 + a)^5 - 198*cosh(b*x^2 + a)^5
+ 90*(55*cosh(b*x^2 + a)^7 + 119*cosh(b*x^2 + a)^5 + 77*cosh(b*x^2 + a)^3 - 11*cosh(b*x^2 + a))*sinh(b*x^2 + a
)^4 + 5*(495*cosh(b*x^2 + a)^8 + 1428*cosh(b*x^2 + a)^6 + 1386*cosh(b*x^2 + a)^4 - 396*cosh(b*x^2 + a)^2 - 17)
*sinh(b*x^2 + a)^3 - 85*cosh(b*x^2 + a)^3 + 3*(275*cosh(b*x^2 + a)^9 + 1020*cosh(b*x^2 + a)^7 + 1386*cosh(b*x^
2 + a)^5 - 660*cosh(b*x^2 + a)^3 - 85*cosh(b*x^2 + a))*sinh(b*x^2 + a)^2 + 15*(cosh(b*x^2 + a)^12 + 12*cosh(b*
x^2 + a)*sinh(b*x^2 + a)^11 + sinh(b*x^2 + a)^12 + 6*(11*cosh(b*x^2 + a)^2 + 1)*sinh(b*x^2 + a)^10 + 6*cosh(b*
x^2 + a)^10 + 20*(11*cosh(b*x^2 + a)^3 + 3*cosh(b*x^2 + a))*sinh(b*x^2 + a)^9 + 15*(33*cosh(b*x^2 + a)^4 + 18*
cosh(b*x^2 + a)^2 + 1)*sinh(b*x^2 + a)^8 + 15*cosh(b*x^2 + a)^8 + 24*(33*cosh(b*x^2 + a)^5 + 30*cosh(b*x^2 + a
)^3 + 5*cosh(b*x^2 + a))*sinh(b*x^2 + a)^7 + 4*(231*cosh(b*x^2 + a)^6 + 315*cosh(b*x^2 + a)^4 + 105*cosh(b*x^2
+ a)^2 + 5)*sinh(b*x^2 + a)^6 + 20*cosh(b*x^2 + a)^6 + 24*(33*cosh(b*x^2 + a)^7 + 63*cosh(b*x^2 + a)^5 + 35*c
osh(b*x^2 + a)^3 + 5*cosh(b*x^2 + a))*sinh(b*x^2 + a)^5 + 15*(33*cosh(b*x^2 + a)^8 + 84*cosh(b*x^2 + a)^6 + 70
*cosh(b*x^2 + a)^4 + 20*cosh(b*x^2 + a)^2 + 1)*sinh(b*x^2 + a)^4 + 15*cosh(b*x^2 + a)^4 + 20*(11*cosh(b*x^2 +
a)^9 + 36*cosh(b*x^2 + a)^7 + 42*cosh(b*x^2 + a)^5 + 20*cosh(b*x^2 + a)^3 + 3*cosh(b*x^2 + a))*sinh(b*x^2 + a)
^3 + 6*(11*cosh(b*x^2 + a)^10 + 45*cosh(b*x^2 + a)^8 + 70*cosh(b*x^2 + a)^6 + 50*cosh(b*x^2 + a)^4 + 15*cosh(b
*x^2 + a)^2 + 1)*sinh(b*x^2 + a)^2 + 6*cosh(b*x^2 + a)^2 + 12*(cosh(b*x^2 + a)^11 + 5*cosh(b*x^2 + a)^9 + 10*c
osh(b*x^2 + a)^7 + 10*cosh(b*x^2 + a)^5 + 5*cosh(b*x^2 + a)^3 + cosh(b*x^2 + a))*sinh(b*x^2 + a) + 1)*arctan(c
osh(b*x^2 + a) + sinh(b*x^2 + a)) + 3*(55*cosh(b*x^2 + a)^10 + 255*cosh(b*x^2 + a)^8 + 462*cosh(b*x^2 + a)^6 -
330*cosh(b*x^2 + a)^4 - 85*cosh(b*x^2 + a)^2 - 5)*sinh(b*x^2 + a) - 15*cosh(b*x^2 + a))/(b*cosh(b*x^2 + a)^12
+ 12*b*cosh(b*x^2 + a)*sinh(b*x^2 + a)^11 + b*sinh(b*x^2 + a)^12 + 6*b*cosh(b*x^2 + a)^10 + 6*(11*b*cosh(b*x^
2 + a)^2 + b)*sinh(b*x^2 + a)^10 + 20*(11*b*cosh(b*x^2 + a)^3 + 3*b*cosh(b*x^2 + a))*sinh(b*x^2 + a)^9 + 15*b*
cosh(b*x^2 + a)^8 + 15*(33*b*cosh(b*x^2 + a)^4 + 18*b*cosh(b*x^2 + a)^2 + b)*sinh(b*x^2 + a)^8 + 24*(33*b*cosh
(b*x^2 + a)^5 + 30*b*cosh(b*x^2 + a)^3 + 5*b*cosh(b*x^2 + a))*sinh(b*x^2 + a)^7 + 20*b*cosh(b*x^2 + a)^6 + 4*(
231*b*cosh(b*x^2 + a)^6 + 315*b*cosh(b*x^2 + a)^4 + 105*b*cosh(b*x^2 + a)^2 + 5*b)*sinh(b*x^2 + a)^6 + 24*(33*
b*cosh(b*x^2 + a)^7 + 63*b*cosh(b*x^2 + a)^5 + 35*b*cosh(b*x^2 + a)^3 + 5*b*cosh(b*x^2 + a))*sinh(b*x^2 + a)^5
+ 15*b*cosh(b*x^2 + a)^4 + 15*(33*b*cosh(b*x^2 + a)^8 + 84*b*cosh(b*x^2 + a)^6 + 70*b*cosh(b*x^2 + a)^4 + 20*
b*cosh(b*x^2 + a)^2 + b)*sinh(b*x^2 + a)^4 + 20*(11*b*cosh(b*x^2 + a)^9 + 36*b*cosh(b*x^2 + a)^7 + 42*b*cosh(b
*x^2 + a)^5 + 20*b*cosh(b*x^2 + a)^3 + 3*b*cosh(b*x^2 + a))*sinh(b*x^2 + a)^3 + 6*b*cosh(b*x^2 + a)^2 + 6*(11*
b*cosh(b*x^2 + a)^10 + 45*b*cosh(b*x^2 + a)^8 + 70*b*cosh(b*x^2 + a)^6 + 50*b*cosh(b*x^2 + a)^4 + 15*b*cosh(b*
x^2 + a)^2 + b)*sinh(b*x^2 + a)^2 + 12*(b*cosh(b*x^2 + a)^11 + 5*b*cosh(b*x^2 + a)^9 + 10*b*cosh(b*x^2 + a)^7
+ 10*b*cosh(b*x^2 + a)^5 + 5*b*cosh(b*x^2 + a)^3 + b*cosh(b*x^2 + a))*sinh(b*x^2 + a) + b)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{sech}^{7}{\left (a + b x^{2} \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sech(b*x**2+a)**7,x)
[Out]
Integral(x*sech(a + b*x**2)**7, x)
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Giac [A] time = 1.18181, size = 197, normalized size = 2.19 \begin{align*} \frac{5 \,{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, b x^{2} + 2 \, a\right )} - 1\right )} e^{\left (-b x^{2} - a\right )}\right )\right )}}{64 \, b} + \frac{15 \,{\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{5} + 160 \,{\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{3} + 528 \, e^{\left (b x^{2} + a\right )} - 528 \, e^{\left (-b x^{2} - a\right )}}{48 \,{\left ({\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{2} + 4\right )}^{3} b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*sech(b*x^2+a)^7,x, algorithm="giac")
[Out]
5/64*(pi + 2*arctan(1/2*(e^(2*b*x^2 + 2*a) - 1)*e^(-b*x^2 - a)))/b + 1/48*(15*(e^(b*x^2 + a) - e^(-b*x^2 - a))
^5 + 160*(e^(b*x^2 + a) - e^(-b*x^2 - a))^3 + 528*e^(b*x^2 + a) - 528*e^(-b*x^2 - a))/(((e^(b*x^2 + a) - e^(-b
*x^2 - a))^2 + 4)^3*b)