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

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Rubi [A]  time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [B]  time = 0.25, 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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fricas [B]  time = 0.44, size = 436, normalized size = 9.48 \[ \left [\frac {2 \, d x + \sqrt {\frac {b}{a + b}} \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + a^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 4 \, {\left ({\left (a^{2} + a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} + a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} + a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}}}{a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} + 2 \, {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a \cosh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a \cosh \left (d x + c\right )^{3} + {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a}\right )}{2 \, a d}, \frac {d x - \sqrt {-\frac {b}{a + b}} \arctan \left (\frac {{\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sqrt {-\frac {b}{a + b}}}{2 \, b}\right )}{a d}\right ] \]

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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giac [A]  time = 0.39, size = 64, normalized size = 1.39 \[ -\frac {\frac {b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a} - \frac {d x + c}{a}}{d} \]

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*x + c)/a)/d

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maple [B]  time = 0.28, size = 149, normalized size = 3.24 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}+\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \sqrt {a +b}}-\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \sqrt {a +b}} \]

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

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maxima [B]  time = 0.47, size = 83, normalized size = 1.80 \[ \frac {b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} a d} + \frac {d x + c}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.16, size = 470, normalized size = 10.22 \[ \frac {x}{a}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\left (a^5\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}+a^4\,b\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}\right )\,\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {2\,\left (a^2+8\,a\,b+8\,b^2\right )\,\left (8\,b^{5/2}\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}+8\,a\,b^{3/2}\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}+a^2\,\sqrt {b}\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}\right )}{a^8\,d\,{\left (a+b\right )}^2\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}}+\frac {4\,\sqrt {b}\,\left (2\,a+4\,b\right )\,\left (4\,d\,a^3\,b+12\,d\,a^2\,b^2+8\,d\,a\,b^3\right )}{a^7\,\left (a+b\right )\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}\,\sqrt {-a^2\,d^2\,\left (a+b\right )}}\right )+\frac {2\,\left (2\,a\,b^{3/2}\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}+a^2\,\sqrt {b}\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}\right )\,\left (a^2+8\,a\,b+8\,b^2\right )}{a^8\,d\,{\left (a+b\right )}^2\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}}+\frac {4\,\sqrt {b}\,\left (2\,d\,a^3\,b+2\,d\,a^2\,b^2\right )\,\left (2\,a+4\,b\right )}{a^7\,\left (a+b\right )\,\sqrt {-a^3\,d^2-b\,a^2\,d^2}\,\sqrt {-a^2\,d^2\,\left (a+b\right )}}\right )}{4\,b}\right )}{\sqrt {-a^3\,d^2-b\,a^2\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cosh(c + d*x)^2),x)

[Out]

x/a + (b^(1/2)*atan(((a^5*(- a^3*d^2 - a^2*b*d^2)^(1/2) + a^4*b*(- a^3*d^2 - a^2*b*d^2)^(1/2))*(exp(2*c)*exp(2
*d*x)*((2*(8*a*b + a^2 + 8*b^2)*(8*b^(5/2)*(- a^3*d^2 - a^2*b*d^2)^(1/2) + 8*a*b^(3/2)*(- a^3*d^2 - a^2*b*d^2)
^(1/2) + a^2*b^(1/2)*(- a^3*d^2 - a^2*b*d^2)^(1/2)))/(a^8*d*(a + b)^2*(- a^3*d^2 - a^2*b*d^2)^(1/2)) + (4*b^(1
/2)*(2*a + 4*b)*(12*a^2*b^2*d + 8*a*b^3*d + 4*a^3*b*d))/(a^7*(a + b)*(- a^3*d^2 - a^2*b*d^2)^(1/2)*(-a^2*d^2*(
a + b))^(1/2))) + (2*(2*a*b^(3/2)*(- a^3*d^2 - a^2*b*d^2)^(1/2) + a^2*b^(1/2)*(- a^3*d^2 - a^2*b*d^2)^(1/2))*(
8*a*b + a^2 + 8*b^2))/(a^8*d*(a + b)^2*(- a^3*d^2 - a^2*b*d^2)^(1/2)) + (4*b^(1/2)*(2*a^2*b^2*d + 2*a^3*b*d)*(
2*a + 4*b))/(a^7*(a + b)*(- a^3*d^2 - a^2*b*d^2)^(1/2)*(-a^2*d^2*(a + b))^(1/2))))/(4*b)))/(- a^3*d^2 - a^2*b*
d^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

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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