3.375 \(\int \frac{\text{sech}(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=46 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \sinh (e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}}\right )}{f \sqrt{a-b}} \]

[Out]

ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]]/(Sqrt[a - b]*f)

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Rubi [A]  time = 0.0688678, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3190, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \sinh (e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}}\right )}{f \sqrt{a-b}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[e + f*x]/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]]/(Sqrt[a - b]*f)

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0307613, size = 46, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \sinh (e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}}\right )}{f \sqrt{a-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[e + f*x]/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]]/(Sqrt[a - b]*f)

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Maple [C]  time = 0.079, size = 35, normalized size = 0.8 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{1}{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2),x)

[Out]

`int/indef0`(1/cosh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sech(f*x + e)/sqrt(b*sinh(f*x + e)^2 + a), x)

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Fricas [B]  time = 1.87393, size = 1615, normalized size = 35.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-a + b)*log(((a - 2*b)*cosh(f*x + e)^4 + 4*(a - 2*b)*cosh(f*x + e)*sinh(f*x + e)^3 + (a - 2*b)*sinh
(f*x + e)^4 - 2*(3*a - 2*b)*cosh(f*x + e)^2 + 2*(3*(a - 2*b)*cosh(f*x + e)^2 - 3*a + 2*b)*sinh(f*x + e)^2 - 2*
sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2 - 1)*sqrt(-a + b)*sqrt((b*cosh(f*x
+ e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)) + 4
*((a - 2*b)*cosh(f*x + e)^3 - (3*a - 2*b)*cosh(f*x + e))*sinh(f*x + e) + a - 2*b)/(cosh(f*x + e)^4 + 4*cosh(f*
x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 + 1)*sinh(f*x + e)^2 + 2*cosh(f*x + e)^2 + 4*(
cosh(f*x + e)^3 + cosh(f*x + e))*sinh(f*x + e) + 1))/((a - b)*f), arctan(sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x
+ e)*sinh(f*x + e) + sinh(f*x + e)^2 - 1)*sqrt(a - b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/
(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/(b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*si
nh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(2*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x +
e)^2 + 4*(b*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + b))/(sqrt(a - b)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(sech(e + f*x)/sqrt(a + b*sinh(e + f*x)**2), x)

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Giac [B]  time = 1.43875, size = 113, normalized size = 2.46 \begin{align*} \frac{2 \, \arctan \left (-\frac{\sqrt{b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt{b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b} + \sqrt{b}}{2 \, \sqrt{a - b}}\right )}{\sqrt{a - b} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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