3.620 \(\int \frac {1}{(a \text {sech}(x)+b \tanh (x))^2} \, dx\)

Optimal. Leaf size=62 \[ \frac {2 a \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}-\frac {\cosh (x)}{b (a+b \sinh (x))}+\frac {x}{b^2} \]

[Out]

x/b^2-cosh(x)/b/(a+b*sinh(x))+2*a*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/b^2/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {4391, 2693, 2735, 2660, 618, 206} \[ \frac {2 a \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}-\frac {\cosh (x)}{b (a+b \sinh (x))}+\frac {x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sech[x] + b*Tanh[x])^(-2),x]

[Out]

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

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])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {1}{(a \text {sech}(x)+b \tanh (x))^2} \, dx &=\int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx\\ &=-\frac {\cosh (x)}{b (a+b \sinh (x))}+\frac {\int \frac {\sinh (x)}{a+b \sinh (x)} \, dx}{b}\\ &=\frac {x}{b^2}-\frac {\cosh (x)}{b (a+b \sinh (x))}-\frac {a \int \frac {1}{a+b \sinh (x)} \, dx}{b^2}\\ &=\frac {x}{b^2}-\frac {\cosh (x)}{b (a+b \sinh (x))}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^2}\\ &=\frac {x}{b^2}-\frac {\cosh (x)}{b (a+b \sinh (x))}+\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{b^2}\\ &=\frac {x}{b^2}+\frac {2 a \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}-\frac {\cosh (x)}{b (a+b \sinh (x))}\\ \end {align*}

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Mathematica [C]  time = 2.08, size = 502, normalized size = 8.10 \[ -\frac {\cosh (x) \left (\sqrt {a+i b} \sqrt {-\frac {b (\sinh (x)-i)}{a+i b}} \left (\sqrt {a-i b} \left (a^2+b^2\right ) \sqrt {1+i \sinh (x)} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}-2 \sqrt [4]{-1} b^{3/2} (b+i a) \sinh (x) \sin ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {b}}\right )-2 \sqrt [4]{-1} a \sqrt {b} (b+i a) \sin ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {b}}\right )\right )-2 a \sqrt {a-i b} \sqrt {a+i b} \sqrt {1+i \sinh (x)} (a+b \sinh (x)) \tanh ^{-1}\left (\frac {\sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {-\frac {b (\sinh (x)-i)}{a+i b}}}\right )+2 a (a-i b) \sqrt {1+i \sinh (x)} (a+b \sinh (x)) \tanh ^{-1}\left (\frac {\sqrt {a-i b} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {a+i b} \sqrt {-\frac {b (\sinh (x)-i)}{a+i b}}}\right )\right )}{b (a-i b)^{3/2} (a+i b)^{3/2} \sqrt {1+i \sinh (x)} \sqrt {-\frac {b (\sinh (x)-i)}{a+i b}} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}} (a+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sech[x] + b*Tanh[x])^(-2),x]

[Out]

-((Cosh[x]*(-2*a*Sqrt[a - I*b]*Sqrt[a + I*b]*ArcTanh[Sqrt[-((b*(I + Sinh[x]))/(a - I*b))]/Sqrt[-((b*(-I + Sinh
[x]))/(a + I*b))]]*Sqrt[1 + I*Sinh[x]]*(a + b*Sinh[x]) + 2*a*(a - I*b)*ArcTanh[(Sqrt[a - I*b]*Sqrt[-((b*(I + S
inh[x]))/(a - I*b))])/(Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))])]*Sqrt[1 + I*Sinh[x]]*(a + b*Sinh[x
]) + Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))]*(-2*(-1)^(1/4)*a*Sqrt[b]*(I*a + b)*ArcSin[((1/2 + I/2
)*Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]] - 2*(-1)^(1/4)*b^(3/2)*(I*a + b)*ArcSin[((1/2 +
I/2)*Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]]*Sinh[x] + Sqrt[a - I*b]*(a^2 + b^2)*Sqrt[1
+ I*Sinh[x]]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])))/((a - I*b)^(3/2)*(a + I*b)^(3/2)*b*Sqrt[1 + I*Sinh[x]]*Sq
rt[-((b*(-I + Sinh[x]))/(a + I*b))]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))]*(a + b*Sinh[x])))

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fricas [B]  time = 0.42, size = 362, normalized size = 5.84 \[ -\frac {{\left (a^{2} b + b^{3}\right )} x \cosh \relax (x)^{2} + {\left (a^{2} b + b^{3}\right )} x \sinh \relax (x)^{2} - 2 \, a^{2} b - 2 \, b^{3} + {\left (a b \cosh \relax (x)^{2} + a b \sinh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) - a b + 2 \, {\left (a b \cosh \relax (x) + a^{2}\right )} \sinh \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) - b}\right ) - {\left (a^{2} b + b^{3}\right )} x + 2 \, {\left (a^{3} + a b^{2} + {\left (a^{3} + a b^{2}\right )} x\right )} \cosh \relax (x) + 2 \, {\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} x \cosh \relax (x) + {\left (a^{3} + a b^{2}\right )} x\right )} \sinh \relax (x)}{a^{2} b^{3} + b^{5} - {\left (a^{2} b^{3} + b^{5}\right )} \cosh \relax (x)^{2} - {\left (a^{2} b^{3} + b^{5}\right )} \sinh \relax (x)^{2} - 2 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x) - 2 \, {\left (a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \cosh \relax (x)\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2*b + b^3)*x*cosh(x)^2 + (a^2*b + b^3)*x*sinh(x)^2 - 2*a^2*b - 2*b^3 + (a*b*cosh(x)^2 + a*b*sinh(x)^2 + 2
*a^2*cosh(x) - a*b + 2*(a*b*cosh(x) + a^2)*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b
*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cos
h(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) - (a^2*b + b^3)*x + 2*(a^3 + a*b^2 + (a^3
+ a*b^2)*x)*cosh(x) + 2*(a^3 + a*b^2 + (a^2*b + b^3)*x*cosh(x) + (a^3 + a*b^2)*x)*sinh(x))/(a^2*b^3 + b^5 - (
a^2*b^3 + b^5)*cosh(x)^2 - (a^2*b^3 + b^5)*sinh(x)^2 - 2*(a^3*b^2 + a*b^4)*cosh(x) - 2*(a^3*b^2 + a*b^4 + (a^2
*b^3 + b^5)*cosh(x))*sinh(x))

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giac [A]  time = 0.14, size = 97, normalized size = 1.56 \[ -\frac {a \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} + \frac {x}{b^{2}} + \frac {2 \, {\left (a e^{x} - b\right )}}{{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^2) +
x/b^2 + 2*(a*e^x - b)/((b*e^(2*x) + 2*a*e^x - b)*b^2)

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maple [B]  time = 0.28, size = 119, normalized size = 1.92 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}}+\frac {2 \tanh \left (\frac {x}{2}\right )}{\left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right ) a}+\frac {2}{b \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}-\frac {2 a \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{2} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 100, normalized size = 1.61 \[ -\frac {2 \, {\left (a e^{\left (-x\right )} + b\right )}}{2 \, a b^{2} e^{\left (-x\right )} - b^{3} e^{\left (-2 \, x\right )} + b^{3}} - \frac {a \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} + \frac {x}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a*e^(-x) + b)/(2*a*b^2*e^(-x) - b^3*e^(-2*x) + b^3) - a*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a
+ sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^2) + x/b^2

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mupad [B]  time = 1.74, size = 132, normalized size = 2.13 \[ \frac {x}{b^2}-\frac {\frac {2}{b}-\frac {2\,a\,{\mathrm {e}}^x}{b^2}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,\sqrt {a^2+b^2}}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}+\frac {2\,a\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tanh(x) + a/cosh(x))^2,x)

[Out]

x/b^2 - (2/b - (2*a*exp(x))/b^2)/(2*a*exp(x) - b + b*exp(2*x)) - (a*log((2*a*exp(x))/b^3 - (2*a*(b - a*exp(x))
)/(b^3*(a^2 + b^2)^(1/2))))/(b^2*(a^2 + b^2)^(1/2)) + (a*log((2*a*exp(x))/b^3 + (2*a*(b - a*exp(x)))/(b^3*(a^2
+ b^2)^(1/2))))/(b^2*(a^2 + b^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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