3.208 \(\int \frac{1}{(a+b \text{sech}^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=57 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a-b \tanh ^2(x)+b}}\right )}{a^{3/2}}-\frac{b \tanh (x)}{a (a+b) \sqrt{a-b \tanh ^2(x)+b}} \]

[Out]

ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/a^(3/2) - (b*Tanh[x])/(a*(a + b)*Sqrt[a + b - b*Tanh[x]^2
])

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Rubi [A]  time = 0.0451566, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4128, 382, 377, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a-b \tanh ^2(x)+b}}\right )}{a^{3/2}}-\frac{b \tanh (x)}{a (a+b) \sqrt{a-b \tanh ^2(x)+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[x]^2)^(-3/2),x]

[Out]

ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/a^(3/2) - (b*Tanh[x])/(a*(a + b)*Sqrt[a + b - b*Tanh[x]^2
])

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.810095, size = 107, normalized size = 1.88 \[ \frac{\text{sech}^3(x) (a \cosh (2 x)+a+2 b) \left ((a+b)^{3/2} \sinh ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{a+b}}\right ) \sqrt{\frac{a \cosh (2 x)+a+2 b}{a+b}}-\sqrt{2} \sqrt{a} b \sinh (x)\right )}{2 \sqrt{2} a^{3/2} (a+b) \left (a+b \text{sech}^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[x]^2)^(-3/2),x]

[Out]

((a + 2*b + a*Cosh[2*x])*Sech[x]^3*((a + b)^(3/2)*ArcSinh[(Sqrt[a]*Sinh[x])/Sqrt[a + b]]*Sqrt[(a + 2*b + a*Cos
h[2*x])/(a + b)] - Sqrt[2]*Sqrt[a]*b*Sinh[x]))/(2*Sqrt[2]*a^(3/2)*(a + b)*(a + b*Sech[x]^2)^(3/2))

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Maple [F]  time = 0.085, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ({\rm sech} \left (x\right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sech(x)^2)^(3/2),x)

[Out]

int(1/(a+b*sech(x)^2)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{sech}\left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(x)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.81514, size = 5858, normalized size = 102.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(x)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh(x)*sinh(x)^3 + (a^2 + a*b)*sinh(x)^4 + 2*(a^2 + 3*a*b + 2*b^
2)*cosh(x)^2 + 2*(3*(a^2 + a*b)*cosh(x)^2 + a^2 + 3*a*b + 2*b^2)*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x
)^3 + (a^2 + 3*a*b + 2*b^2)*cosh(x))*sinh(x))*sqrt(a)*log((a*b^2*cosh(x)^8 + 8*a*b^2*cosh(x)*sinh(x)^7 + a*b^2
*sinh(x)^8 - 2*(a*b^2 - b^3)*cosh(x)^6 + 2*(14*a*b^2*cosh(x)^2 - a*b^2 + b^3)*sinh(x)^6 + 4*(14*a*b^2*cosh(x)^
3 - 3*(a*b^2 - b^3)*cosh(x))*sinh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^4 + (70*a*b^2*cosh(x)^4 + a^3 + 4*a
^2*b + 9*a*b^2 - 30*(a*b^2 - b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*a*b^2*cosh(x)^5 - 10*(a*b^2 - b^3)*cosh(x)^3 +
(a^3 + 4*a^2*b + 9*a*b^2)*cosh(x))*sinh(x)^3 + a^3 + 2*(a^3 + 3*a^2*b)*cosh(x)^2 + 2*(14*a*b^2*cosh(x)^6 - 15*
(a*b^2 - b^3)*cosh(x)^4 + a^3 + 3*a^2*b + 3*(a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^2)*sinh(x)^2 + sqrt(2)*(b^2*cosh
(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 - 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4 + 4*(5
*b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 + 4*a*b)*cosh(x)^2 + (15*b^2*cosh(x)^4 - 18*b^2*cosh(x)^2 - a
^2 - 4*a*b)*sinh(x)^2 - a^2 + 2*(3*b^2*cosh(x)^5 - 6*b^2*cosh(x)^3 - (a^2 + 4*a*b)*cosh(x))*sinh(x))*sqrt(a)*s
qrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*a*b^2*cosh(x)^7
- 3*(a*b^2 - b^3)*cosh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^3 + (a^3 + 3*a^2*b)*cosh(x))*sinh(x))/(cosh(x)
^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x
)*sinh(x)^5 + sinh(x)^6)) + ((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh(x)*sinh(x)^3 + (a^2 + a*b)*sinh(x)^4 +
2*(a^2 + 3*a*b + 2*b^2)*cosh(x)^2 + 2*(3*(a^2 + a*b)*cosh(x)^2 + a^2 + 3*a*b + 2*b^2)*sinh(x)^2 + a^2 + a*b +
4*((a^2 + a*b)*cosh(x)^3 + (a^2 + 3*a*b + 2*b^2)*cosh(x))*sinh(x))*sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)*si
nh(x)^3 + a*sinh(x)^4 + 2*(a + b)*cosh(x)^2 + 2*(3*a*cosh(x)^2 + a + b)*sinh(x)^2 + sqrt(2)*(cosh(x)^2 + 2*cos
h(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x
) + sinh(x)^2)) + 4*(a*cosh(x)^3 + (a + b)*cosh(x))*sinh(x) + a)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2))
- 4*sqrt(2)*(a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 - a*b)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a
+ 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^4 + a^3*b)*cosh(x)^4 + 4*(a^4 + a^3*b)*cosh(x)*sinh(x
)^3 + (a^4 + a^3*b)*sinh(x)^4 + a^4 + a^3*b + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*cosh(x)^2 + 2*(a^4 + 3*a^3*b + 2*a
^2*b^2 + 3*(a^4 + a^3*b)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + a^3*b)*cosh(x)^3 + (a^4 + 3*a^3*b + 2*a^2*b^2)*cosh(
x))*sinh(x)), -1/2*(((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh(x)*sinh(x)^3 + (a^2 + a*b)*sinh(x)^4 + 2*(a^2
+ 3*a*b + 2*b^2)*cosh(x)^2 + 2*(3*(a^2 + a*b)*cosh(x)^2 + a^2 + 3*a*b + 2*b^2)*sinh(x)^2 + a^2 + a*b + 4*((a^2
+ a*b)*cosh(x)^3 + (a^2 + 3*a*b + 2*b^2)*cosh(x))*sinh(x))*sqrt(-a)*arctan(sqrt(2)*(b*cosh(x)^2 + 2*b*cosh(x)
*sinh(x) + b*sinh(x)^2 + a)*sqrt(-a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x)
+ sinh(x)^2))/(a*b*cosh(x)^4 + 4*a*b*cosh(x)*sinh(x)^3 + a*b*sinh(x)^4 - (a^2 + 3*a*b)*cosh(x)^2 + (6*a*b*cos
h(x)^2 - a^2 - 3*a*b)*sinh(x)^2 - a^2 + 2*(2*a*b*cosh(x)^3 - (a^2 + 3*a*b)*cosh(x))*sinh(x))) + ((a^2 + a*b)*c
osh(x)^4 + 4*(a^2 + a*b)*cosh(x)*sinh(x)^3 + (a^2 + a*b)*sinh(x)^4 + 2*(a^2 + 3*a*b + 2*b^2)*cosh(x)^2 + 2*(3*
(a^2 + a*b)*cosh(x)^2 + a^2 + 3*a*b + 2*b^2)*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x)^3 + (a^2 + 3*a*b +
2*b^2)*cosh(x))*sinh(x))*sqrt(-a)*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(-a)*sqr
t((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(a*cosh(x)^4 + 4*a*cosh(x
)*sinh(x)^3 + a*sinh(x)^4 + 2*(a + 2*b)*cosh(x)^2 + 2*(3*a*cosh(x)^2 + a + 2*b)*sinh(x)^2 + 4*(a*cosh(x)^3 + (
a + 2*b)*cosh(x))*sinh(x) + a)) + 2*sqrt(2)*(a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 - a*b)*sqrt
((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^4 + a^3*b)*cosh(x)^4
+ 4*(a^4 + a^3*b)*cosh(x)*sinh(x)^3 + (a^4 + a^3*b)*sinh(x)^4 + a^4 + a^3*b + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*co
sh(x)^2 + 2*(a^4 + 3*a^3*b + 2*a^2*b^2 + 3*(a^4 + a^3*b)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + a^3*b)*cosh(x)^3 + (
a^4 + 3*a^3*b + 2*a^2*b^2)*cosh(x))*sinh(x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{sech}^{2}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(x)**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{sech}\left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(x)^2)^(3/2),x, algorithm="giac")

[Out]

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