3.1.26 \(\int \text {sech}^2(a+b x)^{3/2} \, dx\) [26]

Optimal. Leaf size=40 \[ \frac {\text {ArcSin}(\tanh (a+b x))}{2 b}+\frac {\sqrt {\text {sech}^2(a+b x)} \tanh (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {\text {ArcSin}(\tanh (a+b x))}{2 b}+\frac {\tanh (a+b x) \sqrt {\text {sech}^2(a+b x)}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sech[a + b*x]^2)^(3/2),x]

[Out]

ArcSin[Tanh[a + b*x]]/(2*b) + (Sqrt[Sech[a + b*x]^2]*Tanh[a + b*x])/(2*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)^{3/2} \, dx &=\frac {\text {Subst}\left (\int \sqrt {1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\sqrt {\text {sech}^2(a+b x)} \tanh (a+b x)}{2 b}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=\frac {\sin ^{-1}(\tanh (a+b x))}{2 b}+\frac {\sqrt {\text {sech}^2(a+b x)} \tanh (a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 46, normalized size = 1.15 \begin {gather*} \frac {\text {sech}(a+b x) (\text {ArcTan}(\sinh (a+b x))+\text {sech}(a+b x) \tanh (a+b x))}{2 b \sqrt {\text {sech}^2(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sech[a + b*x]^2)^(3/2),x]

[Out]

(Sech[a + b*x]*(ArcTan[Sinh[a + b*x]] + Sech[a + b*x]*Tanh[a + b*x]))/(2*b*Sqrt[Sech[a + b*x]^2])

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Maple [C] Result contains complex when optimal does not.
time = 2.42, size = 183, normalized size = 4.58

method	result	size
risch	\(\frac {\sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}\, \left ({\mathrm e}^{2 b x +2 a}-1\right )}{\left ({\mathrm e}^{2 b x +2 a}+1\right ) b}+\frac {i \ln \left ({\mathrm e}^{b x}+i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}\, \left ({\mathrm e}^{2 b x +2 a}+1\right ) {\mathrm e}^{-b x -a}}{2 b}-\frac {i \ln \left ({\mathrm e}^{b x}-i {\mathrm e}^{-a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}}\, \left ({\mathrm e}^{2 b x +2 a}+1\right ) {\mathrm e}^{-b x -a}}{2 b}\)	\(183\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sech(b*x+a)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/(exp(2*b*x+2*a)+1)*(1/(exp(2*b*x+2*a)+1)^2*exp(2*b*x+2*a))^(1/2)*(exp(2*b*x+2*a)-1)/b+1/2*I*ln(exp(b*x)+I*ex
p(-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)-1/2*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 [A]
time = 0.48, size = 65, normalized size = 1.62 \begin {gather*} -\frac {\arctan \left (e^{\left (-b x - a\right )}\right )}{b} + \frac {e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (34) = 68\).
time = 0.37, size = 267, normalized size = 6.68 \begin {gather*} \frac {\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} + {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\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 (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(3/2),x, algorithm="fricas")

[Out]

(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh
(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x +
a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (3*cosh(b*x + a)^2 - 1)*sinh(
b*x + a) - cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 + 2*b*cos
h(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 4*(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 {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)**2)**(3/2),x)

[Out]

Integral((sech(a + b*x)**2)**(3/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (34) = 68\).
time = 0.40, size = 76, normalized size = 1.90 \begin {gather*} \frac {\pi + \frac {4 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}}{{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4} + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(3/2),x, algorithm="giac")

[Out]

1/4*(pi + 4*(e^(b*x + a) - e^(-b*x - a))/((e^(b*x + a) - e^(-b*x - a))^2 + 4) + 2*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 )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cosh(a + b*x)^2)^(3/2),x)

[Out]

int((1/cosh(a + b*x)^2)^(3/2), x)

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