### 3.255 $$\int \frac{1}{\sqrt{1+\tanh ^2(x)}} \, dx$$

Optimal. Leaf size=25 $\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{\tanh ^2(x)+1}}\right )}{\sqrt{2}}$

[Out]

ArcTanh[(Sqrt[2]*Tanh[x])/Sqrt[1 + Tanh[x]^2]]/Sqrt[2]

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Rubi [A]  time = 0.0185679, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {3661, 377, 206} $\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{\tanh ^2(x)+1}}\right )}{\sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/Sqrt[1 + Tanh[x]^2],x]

[Out]

ArcTanh[(Sqrt[2]*Tanh[x])/Sqrt[1 + Tanh[x]^2]]/Sqrt[2]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

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{1+\tanh ^2(x)}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{1+x^2}} \, dx,x,\tanh (x)\right )\\ &=\operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\frac{\tanh (x)}{\sqrt{1+\tanh ^2(x)}}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{1+\tanh ^2(x)}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0249142, size = 35, normalized size = 1.4 $\frac{\sinh ^{-1}\left (\sqrt{2} \sinh (x)\right ) \sqrt{\cosh (2 x)} \text{sech}(x)}{\sqrt{2} \sqrt{\tanh ^2(x)+1}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/Sqrt[1 + Tanh[x]^2],x]

[Out]

(ArcSinh[Sqrt[2]*Sinh[x]]*Sqrt[Cosh[2*x]]*Sech[x])/(Sqrt[2]*Sqrt[1 + Tanh[x]^2])

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Maple [B]  time = 0.047, size = 62, normalized size = 2.5 \begin{align*} -{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{ \left ( 2-2\,\tanh \left ( x \right ) \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}-2\,\tanh \left ( x \right ) }}}} \right ) }+{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{ \left ( 2\,\tanh \left ( x \right ) +2 \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}+2\,\tanh \left ( x \right ) }}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+tanh(x)^2)^(1/2),x)

[Out]

-1/4*2^(1/2)*arctanh(1/4*(2-2*tanh(x))*2^(1/2)/((1+tanh(x))^2-2*tanh(x))^(1/2))+1/4*2^(1/2)*arctanh(1/4*(2*tan
h(x)+2)*2^(1/2)/((tanh(x)-1)^2+2*tanh(x))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\tanh \left (x\right )^{2} + 1}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(tanh(x)^2 + 1), x)

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Fricas [B]  time = 2.54609, size = 1831, normalized size = 73.24 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/8*sqrt(2)*log(-2*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2 - 3)*sinh(x)^6 - 3*cosh(x)^6 +
2*(28*cosh(x)^3 - 9*cosh(x))*sinh(x)^5 + 5*(14*cosh(x)^4 - 9*cosh(x)^2 + 1)*sinh(x)^4 + 5*cosh(x)^4 + 4*(14*c
osh(x)^5 - 15*cosh(x)^3 + 5*cosh(x))*sinh(x)^3 + (28*cosh(x)^6 - 45*cosh(x)^4 + 30*cosh(x)^2 - 4)*sinh(x)^2 -
4*cosh(x)^2 + 2*(4*cosh(x)^7 - 9*cosh(x)^5 + 10*cosh(x)^3 - 4*cosh(x))*sinh(x) + (sqrt(2)*cosh(x)^6 + 6*sqrt(2
)*cosh(x)*sinh(x)^5 + sqrt(2)*sinh(x)^6 + 3*(5*sqrt(2)*cosh(x)^2 - sqrt(2))*sinh(x)^4 - 3*sqrt(2)*cosh(x)^4 +
4*(5*sqrt(2)*cosh(x)^3 - 3*sqrt(2)*cosh(x))*sinh(x)^3 + (15*sqrt(2)*cosh(x)^4 - 18*sqrt(2)*cosh(x)^2 + 4*sqrt(
2))*sinh(x)^2 + 4*sqrt(2)*cosh(x)^2 + 2*(3*sqrt(2)*cosh(x)^5 - 6*sqrt(2)*cosh(x)^3 + 4*sqrt(2)*cosh(x))*sinh(x
) - 4*sqrt(2))*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4)/(cosh(x)^6 + 6*c
osh(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)) + 1/8*sqrt(2)*log(2*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sinh(x)
^2 + cosh(x)^2 + 2*(2*cosh(x)^3 + cosh(x))*sinh(x) + (sqrt(2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*
sinh(x)^2 + sqrt(2))*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1)/(cosh(x)^2
+ 2*cosh(x)*sinh(x) + sinh(x)^2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\tanh ^{2}{\left (x \right )} + 1}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x)**2)**(1/2),x)

[Out]

Integral(1/sqrt(tanh(x)**2 + 1), x)

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Giac [B]  time = 1.30805, size = 78, normalized size = 3.12 \begin{align*} -\frac{1}{4} \, \sqrt{2}{\left (\log \left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) + \log \left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) - \log \left (-\sqrt{e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right )\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x)^2)^(1/2),x, algorithm="giac")

[Out]

-1/4*sqrt(2)*(log(sqrt(e^(4*x) + 1) - e^(2*x) + 1) + log(sqrt(e^(4*x) + 1) - e^(2*x)) - log(-sqrt(e^(4*x) + 1)
+ e^(2*x) + 1))