3.1.25 \(\int \text {sech}^2(a+b x)^{5/2} \, dx\) [25]

Optimal. Leaf size=65 \[ \frac {3 \text {ArcSin}(\tanh (a+b x))}{8 b}+\frac {3 \sqrt {\text {sech}^2(a+b x)} \tanh (a+b x)}{8 b}+\frac {\text {sech}^2(a+b x)^{3/2} \tanh (a+b x)}{4 b} \]

[Out]

3/8*arcsin(tanh(b*x+a))/b+1/4*(sech(b*x+a)^2)^(3/2)*tanh(b*x+a)/b+3/8*(sech(b*x+a)^2)^(1/2)*tanh(b*x+a)/b

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Rubi [A]
time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4207, 201, 222} \begin {gather*} \frac {3 \text {ArcSin}(\tanh (a+b x))}{8 b}+\frac {\tanh (a+b x) \text {sech}^2(a+b x)^{3/2}}{4 b}+\frac {3 \tanh (a+b x) \sqrt {\text {sech}^2(a+b x)}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sech[a + b*x]^2)^(5/2),x]

[Out]

(3*ArcSin[Tanh[a + b*x]])/(8*b) + (3*Sqrt[Sech[a + b*x]^2]*Tanh[a + b*x])/(8*b) + ((Sech[a + b*x]^2)^(3/2)*Tan
h[a + b*x])/(4*b)

Rule 201

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 222

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 4207

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)^{5/2} \, dx &=\frac {\text {Subst}\left (\int \left (1-x^2\right )^{3/2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\text {sech}^2(a+b x)^{3/2} \tanh (a+b x)}{4 b}+\frac {3 \text {Subst}\left (\int \sqrt {1-x^2} \, dx,x,\tanh (a+b x)\right )}{4 b}\\ &=\frac {3 \sqrt {\text {sech}^2(a+b x)} \tanh (a+b x)}{8 b}+\frac {\text {sech}^2(a+b x)^{3/2} \tanh (a+b x)}{4 b}+\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,\tanh (a+b x)\right )}{8 b}\\ &=\frac {3 \sin ^{-1}(\tanh (a+b x))}{8 b}+\frac {3 \sqrt {\text {sech}^2(a+b x)} \tanh (a+b x)}{8 b}+\frac {\text {sech}^2(a+b x)^{3/2} \tanh (a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 55, normalized size = 0.85 \begin {gather*} \frac {\text {sech}^2(a+b x)^{3/2} \left (6 \text {ArcTan}(\sinh (a+b x)) \cosh ^3(a+b x)+3 \sinh (2 (a+b x))+4 \tanh (a+b x)\right )}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sech[a + b*x]^2)^(5/2),x]

[Out]

((Sech[a + b*x]^2)^(3/2)*(6*ArcTan[Sinh[a + b*x]]*Cosh[a + b*x]^3 + 3*Sinh[2*(a + b*x)] + 4*Tanh[a + b*x]))/(1
6*b)

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Maple [C] Result contains complex when optimal does not.
time = 2.35, size = 208, normalized size = 3.20

method	result	size
risch	\(\frac {\sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 b x +6 a}+11 \,{\mathrm e}^{4 b x +4 a}-11 \,{\mathrm e}^{2 b x +2 a}-3\right )}{4 \left ({\mathrm e}^{2 b x +2 a}+1\right )^{3} b}+\frac {3 i \ln \left ({\mathrm e}^{b x}+i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}\, \left ({\mathrm e}^{2 b x +2 a}+1\right ) {\mathrm e}^{-b x -a}}{8 b}-\frac {3 i \ln \left ({\mathrm e}^{b x}-i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}\, \left ({\mathrm e}^{2 b x +2 a}+1\right ) {\mathrm e}^{-b x -a}}{8 b}\)	\(208\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sech(b*x+a)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/4/(exp(2*b*x+2*a)+1)^3*(1/(exp(2*b*x+2*a)+1)^2*exp(2*b*x+2*a))^(1/2)*(3*exp(6*b*x+6*a)+11*exp(4*b*x+4*a)-11*
exp(2*b*x+2*a)-3)/b+3/8*I*ln(exp(b*x)+I*exp(-a))/b*(1/(exp(2*b*x+2*a)+1)^2*exp(2*b*x+2*a))^(1/2)*(exp(2*b*x+2*
a)+1)*exp(-b*x-a)-3/8*I*ln(exp(b*x)-I*exp(-a))/b*(1/(exp(2*b*x+2*a)+1)^2*exp(2*b*x+2*a))^(1/2)*(exp(2*b*x+2*a)
+1)*exp(-b*x-a)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (55) = 110\).
time = 0.46, size = 112, normalized size = 1.72 \begin {gather*} -\frac {3 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{4 \, b} + \frac {3 \, e^{\left (-b x - a\right )} + 11 \, e^{\left (-3 \, b x - 3 \, a\right )} - 11 \, e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(5/2),x, algorithm="maxima")

[Out]

-3/4*arctan(e^(-b*x - a))/b + 1/4*(3*e^(-b*x - a) + 11*e^(-3*b*x - 3*a) - 11*e^(-5*b*x - 5*a) - 3*e^(-7*b*x -
7*a))/(b*(4*e^(-2*b*x - 2*a) + 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) + e^(-8*b*x - 8*a) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 812 vs. \(2 (55) = 110\).
time = 0.35, size = 812, normalized size = 12.49 \begin {gather*} \frac {3 \, \cosh \left (b x + a\right )^{7} + 21 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + 3 \, \sinh \left (b x + a\right )^{7} + {\left (63 \, \cosh \left (b x + a\right )^{2} + 11\right )} \sinh \left (b x + a\right )^{5} + 11 \, \cosh \left (b x + a\right )^{5} + 5 \, {\left (21 \, \cosh \left (b x + a\right )^{3} + 11 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + {\left (105 \, \cosh \left (b x + a\right )^{4} + 110 \, \cosh \left (b x + a\right )^{2} - 11\right )} \sinh \left (b x + a\right )^{3} - 11 \, \cosh \left (b x + a\right )^{3} + {\left (63 \, \cosh \left (b x + a\right )^{5} + 110 \, \cosh \left (b x + a\right )^{3} - 33 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 3 \, {\left (\cosh \left (b x + a\right )^{8} + 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{7} + \sinh \left (b x + a\right )^{8} + 4 \, {\left (7 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{6} + 4 \, \cosh \left (b x + a\right )^{6} + 8 \, {\left (7 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{5} + 2 \, {\left (35 \, \cosh \left (b x + a\right )^{4} + 30 \, \cosh \left (b x + a\right )^{2} + 3\right )} \sinh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{4} + 8 \, {\left (7 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 4 \, {\left (7 \, \cosh \left (b x + a\right )^{6} + 15 \, \cosh \left (b x + a\right )^{4} + 9 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 4 \, \cosh \left (b x + a\right )^{2} + 8 \, {\left (\cosh \left (b x + a\right )^{7} + 3 \, \cosh \left (b x + a\right )^{5} + 3 \, \cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + {\left (21 \, \cosh \left (b x + a\right )^{6} + 55 \, \cosh \left (b x + a\right )^{4} - 33 \, \cosh \left (b x + a\right )^{2} - 3\right )} \sinh \left (b x + a\right ) - 3 \, \cosh \left (b x + a\right )}{4 \, {\left (b \cosh \left (b x + a\right )^{8} + 8 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{7} + b \sinh \left (b x + a\right )^{8} + 4 \, b \cosh \left (b x + a\right )^{6} + 4 \, {\left (7 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{6} + 8 \, {\left (7 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{5} + 6 \, b \cosh \left (b x + a\right )^{4} + 2 \, {\left (35 \, b \cosh \left (b x + a\right )^{4} + 30 \, b \cosh \left (b x + a\right )^{2} + 3 \, b\right )} \sinh \left (b x + a\right )^{4} + 8 \, {\left (7 \, b \cosh \left (b x + a\right )^{5} + 10 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 4 \, b \cosh \left (b x + a\right )^{2} + 4 \, {\left (7 \, b \cosh \left (b x + a\right )^{6} + 15 \, b \cosh \left (b x + a\right )^{4} + 9 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 8 \, {\left (b \cosh \left (b x + a\right )^{7} + 3 \, b \cosh \left (b x + a\right )^{5} + 3 \, b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(5/2),x, algorithm="fricas")

[Out]

1/4*(3*cosh(b*x + a)^7 + 21*cosh(b*x + a)*sinh(b*x + a)^6 + 3*sinh(b*x + a)^7 + (63*cosh(b*x + a)^2 + 11)*sinh
(b*x + a)^5 + 11*cosh(b*x + a)^5 + 5*(21*cosh(b*x + a)^3 + 11*cosh(b*x + a))*sinh(b*x + a)^4 + (105*cosh(b*x +
a)^4 + 110*cosh(b*x + a)^2 - 11)*sinh(b*x + a)^3 - 11*cosh(b*x + a)^3 + (63*cosh(b*x + a)^5 + 110*cosh(b*x +
a)^3 - 33*cosh(b*x + a))*sinh(b*x + a)^2 + 3*(cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a
)^8 + 4*(7*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 + 3*cosh(b*x + a))*
sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 + 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*c
osh(b*x + a)^5 + 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 + 15*cosh(b*x +
a)^4 + 9*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 + 3*cosh(b*x + a)^5 + 3
*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (21*cosh(b*x + a)
^6 + 55*cosh(b*x + a)^4 - 33*cosh(b*x + a)^2 - 3)*sinh(b*x + a) - 3*cosh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*co
sh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 + 4*b*cosh(b*x + a)^6 + 4*(7*b*cosh(b*x + a)^2 + b)*sinh(b*x +
a)^6 + 8*(7*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x +
a)^4 + 30*b*cosh(b*x + a)^2 + 3*b)*sinh(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 + 10*b*cosh(b*x + a)^3 + 3*b*cosh
(b*x + a))*sinh(b*x + a)^3 + 4*b*cosh(b*x + a)^2 + 4*(7*b*cosh(b*x + a)^6 + 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*
x + a)^2 + b)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a)^7 + 3*b*cosh(b*x + a)^5 + 3*b*cosh(b*x + a)^3 + b*cosh(b*x
+ a))*sinh(b*x + a) + b)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)**2)**(5/2),x)

[Out]

Integral((sech(a + b*x)**2)**(5/2), x)

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Giac [A]
time = 0.39, size = 102, normalized size = 1.57 \begin {gather*} \frac {3 \, \pi + \frac {4 \, {\left (3 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} + 20 \, e^{\left (b x + a\right )} - 20 \, 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 )}^{2}} + 6 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(5/2),x, algorithm="giac")

[Out]

1/16*(3*pi + 4*(3*(e^(b*x + a) - e^(-b*x - a))^3 + 20*e^(b*x + a) - 20*e^(-b*x - a))/((e^(b*x + a) - e^(-b*x -
a))^2 + 4)^2 + 6*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cosh(a + b*x)^2)^(5/2),x)

[Out]

int((1/cosh(a + b*x)^2)^(5/2), x)

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