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

Optimal. Leaf size=90 \[ \frac {5 \sin ^{-1}(\tanh (a+b x))}{16 b}+\frac {\tanh (a+b x) \text {sech}^2(a+b x)^{5/2}}{6 b}+\frac {5 \tanh (a+b x) \text {sech}^2(a+b x)^{3/2}}{24 b}+\frac {5 \tanh (a+b x) \sqrt {\text {sech}^2(a+b x)}}{16 b} \]

[Out]

5/16*arcsin(tanh(b*x+a))/b+5/24*(sech(b*x+a)^2)^(3/2)*tanh(b*x+a)/b+1/6*(sech(b*x+a)^2)^(5/2)*tanh(b*x+a)/b+5/
16*(sech(b*x+a)^2)^(1/2)*tanh(b*x+a)/b

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Rubi [A]  time = 0.03, 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 = {4122, 195, 216} \[ \frac {5 \sin ^{-1}(\tanh (a+b x))}{16 b}+\frac {\tanh (a+b x) \text {sech}^2(a+b x)^{5/2}}{6 b}+\frac {5 \tanh (a+b x) \text {sech}^2(a+b x)^{3/2}}{24 b}+\frac {5 \tanh (a+b x) \sqrt {\text {sech}^2(a+b x)}}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[a + b*x]^2)^(7/2),x]

[Out]

(5*ArcSin[Tanh[a + b*x]])/(16*b) + (5*Sqrt[Sech[a + b*x]^2]*Tanh[a + b*x])/(16*b) + (5*(Sech[a + b*x]^2)^(3/2)
*Tanh[a + b*x])/(24*b) + ((Sech[a + b*x]^2)^(5/2)*Tanh[a + b*x])/(6*b)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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

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Mathematica [A]  time = 0.11, size = 81, normalized size = 0.90 \[ \frac {\cosh (a+b x) \sqrt {\text {sech}^2(a+b x)} \left (15 \tan ^{-1}(\sinh (a+b x))+8 \tanh (a+b x) \text {sech}^5(a+b x)+10 \tanh (a+b x) \text {sech}^3(a+b x)+15 \tanh (a+b x) \text {sech}(a+b x)\right )}{48 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[a + b*x]^2)^(7/2),x]

[Out]

(Cosh[a + b*x]*Sqrt[Sech[a + b*x]^2]*(15*ArcTan[Sinh[a + b*x]] + 15*Sech[a + b*x]*Tanh[a + b*x] + 10*Sech[a +
b*x]^3*Tanh[a + b*x] + 8*Sech[a + b*x]^5*Tanh[a + b*x]))/(48*b)

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fricas [B]  time = 0.42, size = 1604, normalized size = 17.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*cosh(b*x + a)^11 + 165*cosh(b*x + a)*sinh(b*x + a)^10 + 15*sinh(b*x + a)^11 + 5*(165*cosh(b*x + a)^2
+ 17)*sinh(b*x + a)^9 + 85*cosh(b*x + a)^9 + 45*(55*cosh(b*x + a)^3 + 17*cosh(b*x + a))*sinh(b*x + a)^8 + 18*(
275*cosh(b*x + a)^4 + 170*cosh(b*x + a)^2 + 11)*sinh(b*x + a)^7 + 198*cosh(b*x + a)^7 + 42*(165*cosh(b*x + a)^
5 + 170*cosh(b*x + a)^3 + 33*cosh(b*x + a))*sinh(b*x + a)^6 + 18*(385*cosh(b*x + a)^6 + 595*cosh(b*x + a)^4 +
231*cosh(b*x + a)^2 - 11)*sinh(b*x + a)^5 - 198*cosh(b*x + a)^5 + 90*(55*cosh(b*x + a)^7 + 119*cosh(b*x + a)^5
+ 77*cosh(b*x + a)^3 - 11*cosh(b*x + a))*sinh(b*x + a)^4 + 5*(495*cosh(b*x + a)^8 + 1428*cosh(b*x + a)^6 + 13
86*cosh(b*x + a)^4 - 396*cosh(b*x + a)^2 - 17)*sinh(b*x + a)^3 - 85*cosh(b*x + a)^3 + 3*(275*cosh(b*x + a)^9 +
1020*cosh(b*x + a)^7 + 1386*cosh(b*x + a)^5 - 660*cosh(b*x + a)^3 - 85*cosh(b*x + a))*sinh(b*x + a)^2 + 15*(c
osh(b*x + a)^12 + 12*cosh(b*x + a)*sinh(b*x + a)^11 + sinh(b*x + a)^12 + 6*(11*cosh(b*x + a)^2 + 1)*sinh(b*x +
a)^10 + 6*cosh(b*x + a)^10 + 20*(11*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^9 + 15*(33*cosh(b*x + a)
^4 + 18*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^8 + 15*cosh(b*x + a)^8 + 24*(33*cosh(b*x + a)^5 + 30*cosh(b*x + a)^
3 + 5*cosh(b*x + a))*sinh(b*x + a)^7 + 4*(231*cosh(b*x + a)^6 + 315*cosh(b*x + a)^4 + 105*cosh(b*x + a)^2 + 5)
*sinh(b*x + a)^6 + 20*cosh(b*x + a)^6 + 24*(33*cosh(b*x + a)^7 + 63*cosh(b*x + a)^5 + 35*cosh(b*x + a)^3 + 5*c
osh(b*x + a))*sinh(b*x + a)^5 + 15*(33*cosh(b*x + a)^8 + 84*cosh(b*x + a)^6 + 70*cosh(b*x + a)^4 + 20*cosh(b*x
+ a)^2 + 1)*sinh(b*x + a)^4 + 15*cosh(b*x + a)^4 + 20*(11*cosh(b*x + a)^9 + 36*cosh(b*x + a)^7 + 42*cosh(b*x
+ a)^5 + 20*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 6*(11*cosh(b*x + a)^10 + 45*cosh(b*x + a)^8 +
70*cosh(b*x + a)^6 + 50*cosh(b*x + a)^4 + 15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 6*cosh(b*x + a)^2 + 12*(c
osh(b*x + a)^11 + 5*cosh(b*x + a)^9 + 10*cosh(b*x + a)^7 + 10*cosh(b*x + a)^5 + 5*cosh(b*x + a)^3 + cosh(b*x +
a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + 3*(55*cosh(b*x + a)^10 + 255*cosh(b*x + a)^8 +
462*cosh(b*x + a)^6 - 330*cosh(b*x + a)^4 - 85*cosh(b*x + a)^2 - 5)*sinh(b*x + a) - 15*cosh(b*x + a))/(b*cosh
(b*x + a)^12 + 12*b*cosh(b*x + a)*sinh(b*x + a)^11 + b*sinh(b*x + a)^12 + 6*b*cosh(b*x + a)^10 + 6*(11*b*cosh(
b*x + a)^2 + b)*sinh(b*x + a)^10 + 20*(11*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^9 + 15*b*cosh(b
*x + a)^8 + 15*(33*b*cosh(b*x + a)^4 + 18*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^8 + 24*(33*b*cosh(b*x + a)^5 +
30*b*cosh(b*x + a)^3 + 5*b*cosh(b*x + a))*sinh(b*x + a)^7 + 20*b*cosh(b*x + a)^6 + 4*(231*b*cosh(b*x + a)^6 +
315*b*cosh(b*x + a)^4 + 105*b*cosh(b*x + a)^2 + 5*b)*sinh(b*x + a)^6 + 24*(33*b*cosh(b*x + a)^7 + 63*b*cosh(b*
x + a)^5 + 35*b*cosh(b*x + a)^3 + 5*b*cosh(b*x + a))*sinh(b*x + a)^5 + 15*b*cosh(b*x + a)^4 + 15*(33*b*cosh(b*
x + a)^8 + 84*b*cosh(b*x + a)^6 + 70*b*cosh(b*x + a)^4 + 20*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^4 + 20*(11*b*
cosh(b*x + a)^9 + 36*b*cosh(b*x + a)^7 + 42*b*cosh(b*x + a)^5 + 20*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh
(b*x + a)^3 + 6*b*cosh(b*x + a)^2 + 6*(11*b*cosh(b*x + a)^10 + 45*b*cosh(b*x + a)^8 + 70*b*cosh(b*x + a)^6 + 5
0*b*cosh(b*x + a)^4 + 15*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 12*(b*cosh(b*x + a)^11 + 5*b*cosh(b*x + a)^9
+ 10*b*cosh(b*x + a)^7 + 10*b*cosh(b*x + a)^5 + 5*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a) + b)

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giac [A]  time = 0.12, size = 124, normalized size = 1.38 \[ \frac {15 \, \pi + \frac {4 \, {\left (15 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{5} + 160 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} + 528 \, e^{\left (b x + a\right )} - 528 \, e^{\left (-b x - a\right )}\right )}}{{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )}^{3}} + 30 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{96 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(15*pi + 4*(15*(e^(b*x + a) - e^(-b*x - a))^5 + 160*(e^(b*x + a) - e^(-b*x - a))^3 + 528*e^(b*x + a) - 52
8*e^(-b*x - a))/((e^(b*x + a) - e^(-b*x - a))^2 + 4)^3 + 30*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b

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maple [C]  time = 0.49, size = 230, normalized size = 2.56 \[ \frac {\sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}\, \left (15 \,{\mathrm e}^{10 b x +10 a}+85 \,{\mathrm e}^{8 b x +8 a}+198 \,{\mathrm e}^{6 b x +6 a}-198 \,{\mathrm e}^{4 b x +4 a}-85 \,{\mathrm e}^{2 b x +2 a}-15\right )}{24 \left (1+{\mathrm e}^{2 b x +2 a}\right )^{5} b}+\frac {5 i \ln \left ({\mathrm e}^{b x}+i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}\, \left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{-b x -a}}{16 b}-\frac {5 i \ln \left ({\mathrm e}^{b x}-i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}\, \left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{-b x -a}}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/(1+exp(2*b*x+2*a))^5*(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*(15*exp(10*b*x+10*a)+85*exp(8*b*x+8*a)
+198*exp(6*b*x+6*a)-198*exp(4*b*x+4*a)-85*exp(2*b*x+2*a)-15)/b+5/16*I*ln(exp(b*x)+I*exp(-a))/b*(1/(1+exp(2*b*x
+2*a))^2*exp(2*b*x+2*a))^(1/2)*(1+exp(2*b*x+2*a))*exp(-b*x-a)-5/16*I*ln(exp(b*x)-I*exp(-a))/b*(1/(1+exp(2*b*x+
2*a))^2*exp(2*b*x+2*a))^(1/2)*(1+exp(2*b*x+2*a))*exp(-b*x-a)

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maxima [B]  time = 0.48, size = 156, normalized size = 1.73 \[ -\frac {5 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{8 \, b} + \frac {15 \, e^{\left (-b x - a\right )} + 85 \, e^{\left (-3 \, b x - 3 \, a\right )} + 198 \, e^{\left (-5 \, b x - 5 \, a\right )} - 198 \, e^{\left (-7 \, b x - 7 \, a\right )} - 85 \, e^{\left (-9 \, b x - 9 \, a\right )} - 15 \, e^{\left (-11 \, b x - 11 \, a\right )}}{24 \, b {\left (6 \, e^{\left (-2 \, b x - 2 \, a\right )} + 15 \, e^{\left (-4 \, b x - 4 \, a\right )} + 20 \, e^{\left (-6 \, b x - 6 \, a\right )} + 15 \, e^{\left (-8 \, b x - 8 \, a\right )} + 6 \, e^{\left (-10 \, b x - 10 \, a\right )} + e^{\left (-12 \, b x - 12 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*arctan(e^(-b*x - a))/b + 1/24*(15*e^(-b*x - a) + 85*e^(-3*b*x - 3*a) + 198*e^(-5*b*x - 5*a) - 198*e^(-7*b
*x - 7*a) - 85*e^(-9*b*x - 9*a) - 15*e^(-11*b*x - 11*a))/(b*(6*e^(-2*b*x - 2*a) + 15*e^(-4*b*x - 4*a) + 20*e^(
-6*b*x - 6*a) + 15*e^(-8*b*x - 8*a) + 6*e^(-10*b*x - 10*a) + e^(-12*b*x - 12*a) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cosh(a + b*x)^2)^(7/2),x)

[Out]

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

[Out]

Timed out

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