3.143 \(\int \frac{1}{a+b \text{sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=46 \[ \frac{x}{a}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a d \sqrt{a+b}} \]

[Out]

x/a - (Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]*d)

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Rubi [A]  time = 0.0463049, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4127, 3181, 208} \[ \frac{x}{a}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)^(-1),x]

[Out]

x/a - (Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]*d)

Rule 4127

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> Simp[x/a, x] - Dist[b/a, Int[1/(b + a*Cos[e +
f*x]^2), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0]

Rule 3181

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

Rule 208

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

Rubi steps

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

Mathematica [B]  time = 0.246705, size = 172, normalized size = 3.74 \[ \frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (d x \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}+b (\sinh (2 c)-\cosh (2 c)) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )\right )}{2 a d \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4} \left (a+b \text{sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[c + d*x]^2)^(-1),x]

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(Sqrt[a + b]*d*x*Sqrt[b*(Cosh[c] - Sinh[c])^4] + b*ArcTanh[(S
ech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - S
inh[c])^4])]*(-Cosh[2*c] + Sinh[2*c])))/(2*a*Sqrt[a + b]*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^
4])

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Maple [B]  time = 0.052, size = 152, normalized size = 3.3 \begin{align*}{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{2\,da}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{1}{2\,da}\sqrt{b}\ln \left ( -\sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}-\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}-{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)-1/2/d*b^(1/2)/a/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*
x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/2/d/a*b^(1/2)/(a+b)^(1/2)*ln(-(a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x
+1/2*c)*b^(1/2)-(a+b)^(1/2))-1/d/a*ln(tanh(1/2*d*x+1/2*c)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.30798, size = 1122, normalized size = 24.39 \begin{align*} \left [\frac{2 \, d x + \sqrt{\frac{b}{a + b}} \log \left (\frac{a^{2} \cosh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} \sinh \left (d x + c\right )^{4} + 2 \,{\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + a^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2} + 4 \,{\left (a^{2} \cosh \left (d x + c\right )^{3} +{\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 4 \,{\left ({\left (a^{2} + a b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{2} + a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{2} + a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt{\frac{b}{a + b}}}{a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} + 2 \,{\left (a + 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a \cosh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (a \cosh \left (d x + c\right )^{3} +{\left (a + 2 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a}\right )}{2 \, a d}, \frac{d x - \sqrt{-\frac{b}{a + b}} \arctan \left (\frac{{\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sqrt{-\frac{b}{a + b}}}{2 \, b}\right )}{a d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*d*x + sqrt(b/(a + b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x +
c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b
+ 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^
2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/(a +
b)))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2
+ 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*
x + c) + a)))/(a*d), (d*x - sqrt(-b/(a + b))*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) +
a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b))/(a*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a + b*sech(c + d*x)**2), x)

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Giac [A]  time = 1.17124, size = 88, normalized size = 1.91 \begin{align*} -\frac{b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} a d} + \frac{d x + c}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a*d) + (d*x + c)/(a*d)