3.199 \(\int \frac{1}{\sqrt{a+b \text{sech}^2(x)}} \, dx\)
Optimal. Leaf size=29 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a-b \tanh ^2(x)+b}}\right )}{\sqrt{a}} \]
[Out]
ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/Sqrt[a]
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Rubi [A] time = 0.0254476, antiderivative size = 29, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{4128, 377, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a-b \tanh ^2(x)+b}}\right )}{\sqrt{a}} \]
Antiderivative was successfully verified.
[In]
Int[1/Sqrt[a + b*Sech[x]^2],x]
[Out]
ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/Sqrt[a]
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 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}{\sqrt{a+b \text{sech}^2(x)}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b-b x^2}} \, dx,x,\tanh (x)\right )\\ &=\operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\tanh (x)}{\sqrt{a+b-b \tanh ^2(x)}}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b-b \tanh ^2(x)}}\right )}{\sqrt{a}}\\ \end{align*}
Mathematica [B] time = 0.0401439, size = 62, normalized size = 2.14 \[ \frac{\text{sech}(x) \sqrt{a \cosh (2 x)+a+2 b} \tanh ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{a \sinh ^2(x)+a+b}}\right )}{\sqrt{2} \sqrt{a} \sqrt{a+b \text{sech}^2(x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/Sqrt[a + b*Sech[x]^2],x]
[Out]
(ArcTanh[(Sqrt[a]*Sinh[x])/Sqrt[a + b + a*Sinh[x]^2]]*Sqrt[a + 2*b + a*Cosh[2*x]]*Sech[x])/(Sqrt[2]*Sqrt[a]*Sq
rt[a + b*Sech[x]^2])
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Maple [F] time = 0.149, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{a+b \left ({\rm sech} \left (x\right ) \right ) ^{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*sech(x)^2)^(1/2),x)
[Out]
int(1/(a+b*sech(x)^2)^(1/2),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \operatorname{sech}\left (x\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sech(x)^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(1/sqrt(b*sech(x)^2 + a), x)
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Fricas [B] time = 2.16516, size = 3023, normalized size = 104.24 \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)^(1/2),x, algorithm="fricas")
[Out]
[1/4*(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*si
nh(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*c
osh(x)^5 - 6*b^2*cosh(x)^3 - (a^2 + 4*a*b)*cosh(x))*sinh(x))*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*(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)) + sqrt(a)*log(
-(a*cosh(x)^4 + 4*a*cosh(x)*sinh(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*cosh(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)))/a, -1/2*(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*cosh(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))) + sqrt(-a)*arctan(sqrt(2)*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))/(a*cosh(x)^2 + 2*a*cosh
(x)*sinh(x) + a*sinh(x)^2 + a)))/a]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \operatorname{sech}^{2}{\left (x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sech(x)**2)**(1/2),x)
[Out]
Integral(1/sqrt(a + b*sech(x)**2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \operatorname{sech}\left (x\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sech(x)^2)^(1/2),x, algorithm="giac")
[Out]
integrate(1/sqrt(b*sech(x)^2 + a), x)