3.1.15 \(\int x \text {sech}^7(a+b x^2) \, dx\) [15]

Optimal. Leaf size=90 \[ \frac {5 \text {ArcTan}\left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac {5 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b} \]

[Out]

5/32*arctan(sinh(b*x^2+a))/b+5/32*sech(b*x^2+a)*tanh(b*x^2+a)/b+5/48*sech(b*x^2+a)^3*tanh(b*x^2+a)/b+1/12*sech
(b*x^2+a)^5*tanh(b*x^2+a)/b

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Rubi [A]
time = 0.06, antiderivative size = 90, 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 = {5544, 3853, 3855} \begin {gather*} \frac {5 \text {ArcTan}\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} \end {gather*}

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 3853

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 5544

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]

Rubi steps

\begin {align*} \int x \text {sech}^7\left (a+b x^2\right ) \, dx &=\frac {1}{2} \text {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} \text {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} \text {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} \text {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*}

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Mathematica [A]
time = 0.10, size = 77, normalized size = 0.86 \begin {gather*} \frac {15 \text {ArcTan}\left (\sinh \left (a+b x^2\right )\right )+15 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )+10 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )+8 \text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{96 b} \end {gather*}

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 [C] Result contains complex when optimal does not.
time = 1.23, size = 133, normalized size = 1.48

method	result	size
risch	\(\frac {{\mathrm e}^{x^{2} b +a} \left (15 \,{\mathrm e}^{10 x^{2} b +10 a}+85 \,{\mathrm e}^{8 x^{2} b +8 a}+198 \,{\mathrm e}^{6 x^{2} b +6 a}-198 \,{\mathrm e}^{4 x^{2} b +4 a}-85 \,{\mathrm e}^{2 x^{2} b +2 a}-15\right )}{48 b \left ({\mathrm e}^{2 x^{2} b +2 a}+1\right )^{6}}+\frac {5 i \ln \left ({\mathrm e}^{x^{2} b +a}+i\right )}{32 b}-\frac {5 i \ln \left ({\mathrm e}^{x^{2} b +a}-i\right )}{32 b}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*exp(b*x^2+a)*(15*exp(10*b*x^2+10*a)+85*exp(8*b*x^2+8*a)+198*exp(6*b*x^2+6*a)-198*exp(4*b*x^2+4*a)-85*exp(
2*b*x^2+2*a)-15)/b/(exp(2*b*x^2+2*a)+1)^6+5/32*I/b*ln(exp(b*x^2+a)+I)-5/32*I/b*ln(exp(b*x^2+a)-I)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (82) = 164\).
time = 0.49, size = 182, normalized size = 2.02 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 1918 vs. \(2 (82) = 164\).
time = 0.37, size = 1918, normalized size = 21.31 \begin {gather*} \text {Too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {sech}^{7}{\left (a + b x^{2} \right )}\, dx \end {gather*}

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 = 0.40, size = 146, normalized size = 1.62 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.10, size = 395, normalized size = 4.39 \begin {gather*} \frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^a\,{\mathrm {e}}^{b\,x^2}\,\sqrt {b^2}}{b}\right )}{16\,\sqrt {b^2}}-\frac {8\,{\mathrm {e}}^{3\,b\,x^2+3\,a}}{3\,b\,\left (5\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+10\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+10\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+5\,{\mathrm {e}}^{8\,b\,x^2+8\,a}+{\mathrm {e}}^{10\,b\,x^2+10\,a}+1\right )}-\frac {{\mathrm {e}}^{b\,x^2+a}}{b\,\left (4\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+6\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+4\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+{\mathrm {e}}^{8\,b\,x^2+8\,a}+1\right )}+\frac {5\,{\mathrm {e}}^{b\,x^2+a}}{24\,b\,\left (2\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+{\mathrm {e}}^{4\,b\,x^2+4\,a}+1\right )}-\frac {16\,{\mathrm {e}}^{5\,b\,x^2+5\,a}}{3\,b\,\left (6\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+15\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+20\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+15\,{\mathrm {e}}^{8\,b\,x^2+8\,a}+6\,{\mathrm {e}}^{10\,b\,x^2+10\,a}+{\mathrm {e}}^{12\,b\,x^2+12\,a}+1\right )}+\frac {{\mathrm {e}}^{b\,x^2+a}}{6\,b\,\left (3\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+3\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+{\mathrm {e}}^{6\,b\,x^2+6\,a}+1\right )}+\frac {5\,{\mathrm {e}}^{b\,x^2+a}}{16\,b\,\left ({\mathrm {e}}^{2\,b\,x^2+2\,a}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/cosh(a + b*x^2)^7,x)

[Out]

(5*atan((exp(a)*exp(b*x^2)*(b^2)^(1/2))/b))/(16*(b^2)^(1/2)) - (8*exp(3*a + 3*b*x^2))/(3*b*(5*exp(2*a + 2*b*x^
2) + 10*exp(4*a + 4*b*x^2) + 10*exp(6*a + 6*b*x^2) + 5*exp(8*a + 8*b*x^2) + exp(10*a + 10*b*x^2) + 1)) - exp(a
+ b*x^2)/(b*(4*exp(2*a + 2*b*x^2) + 6*exp(4*a + 4*b*x^2) + 4*exp(6*a + 6*b*x^2) + exp(8*a + 8*b*x^2) + 1)) +
(5*exp(a + b*x^2))/(24*b*(2*exp(2*a + 2*b*x^2) + exp(4*a + 4*b*x^2) + 1)) - (16*exp(5*a + 5*b*x^2))/(3*b*(6*ex
p(2*a + 2*b*x^2) + 15*exp(4*a + 4*b*x^2) + 20*exp(6*a + 6*b*x^2) + 15*exp(8*a + 8*b*x^2) + 6*exp(10*a + 10*b*x
^2) + exp(12*a + 12*b*x^2) + 1)) + exp(a + b*x^2)/(6*b*(3*exp(2*a + 2*b*x^2) + 3*exp(4*a + 4*b*x^2) + exp(6*a
+ 6*b*x^2) + 1)) + (5*exp(a + b*x^2))/(16*b*(exp(2*a + 2*b*x^2) + 1))

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