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*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)
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Rubi [A] time = 0.0297275, 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 =
{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 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 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]
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*}
Mathematica [A] time = 0.100585, size = 81, normalized size = 0.9 \[ \frac{\cosh (a+b x) \sqrt{\text{sech}^2(a+b x)} \left (15 \tan ^{-1}(\sinh (a+b x))+8 \tanh (a+b x) \text{sech}^5(a+b x)+10 \tanh (a+b x) \text{sech}^3(a+b x)+15 \tanh (a+b x) \text{sech}(a+b x)\right )}{48 b} \]
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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Maple [C] time = 0.177, size = 230, normalized size = 2.6 \begin{align*}{\frac{15\,{{\rm e}^{10\,bx+10\,a}}+85\,{{\rm e}^{8\,bx+8\,a}}+198\,{{\rm e}^{6\,bx+6\,a}}-198\,{{\rm e}^{4\,bx+4\,a}}-85\,{{\rm e}^{2\,bx+2\,a}}-15}{24\, \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{5}b}\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}}+{\frac{{\frac{5\,i}{16}} \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) \ln \left ({{\rm e}^{bx}}+i{{\rm e}^{-a}} \right ){{\rm e}^{-bx-a}}}{b}\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}}-{\frac{{\frac{5\,i}{16}} \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) \ln \left ({{\rm e}^{bx}}-i{{\rm e}^{-a}} \right ){{\rm e}^{-bx-a}}}{b}\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}} \end{align*}
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*(1+exp(2*b*x+2*a))*(1/(1+exp(2*b*x+2*a))
^2*exp(2*b*x+2*a))^(1/2)/b*ln(exp(b*x)+I*exp(-a))*exp(-b*x-a)-5/16*I*(1+exp(2*b*x+2*a))*(1/(1+exp(2*b*x+2*a))^
2*exp(2*b*x+2*a))^(1/2)/b*ln(exp(b*x)-I*exp(-a))*exp(-b*x-a)
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Maxima [B] time = 1.52539, size = 211, normalized size = 2.34 \begin{align*} -\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 )}} \end{align*}
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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Fricas [B] time = 2.3338, size = 4601, normalized size = 51.12 \begin{align*} \text{result too large to display} \end{align*}
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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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((sech(b*x+a)**2)**(7/2),x)
[Out]
Timed out
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Giac [A] time = 1.106, size = 170, normalized size = 1.89 \begin{align*} \frac{5 \,{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )\right )}}{32 \, b} + \frac{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 )}}{24 \,{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )}^{3} b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((sech(b*x+a)^2)^(7/2),x, algorithm="giac")
[Out]
5/32*(pi + 2*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b + 1/24*(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) - 528*e^(-b*x - a))/(((e^(b*x + a) - e^(-b*x - a))^2 + 4)^3*
b)