### 3.225 $$\int \sqrt{1+\tanh ^2(x)} \, dx$$

Optimal. Leaf size=31 $\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{\tanh ^2(x)+1}}\right )-\sinh ^{-1}(\tanh (x))$

[Out]

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

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Rubi [A]  time = 0.0269331, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {3661, 402, 215, 377, 206} $\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{\tanh ^2(x)+1}}\right )-\sinh ^{-1}(\tanh (x))$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[1 + Tanh[x]^2],x]

[Out]

-ArcSinh[Tanh[x]] + Sqrt[2]*ArcTanh[(Sqrt[2]*Tanh[x])/Sqrt[1 + Tanh[x]^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 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 215

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

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

Mathematica [A]  time = 0.0580652, size = 51, normalized size = 1.65 $\frac{\cosh (x) \sqrt{\tanh ^2(x)+1} \left (\sqrt{2} \sinh ^{-1}\left (\sqrt{2} \sinh (x)\right )-\tanh ^{-1}\left (\frac{\sinh (x)}{\sqrt{\cosh (2 x)}}\right )\right )}{\sqrt{\cosh (2 x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[1 + Tanh[x]^2],x]

[Out]

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

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Maple [B]  time = 0.059, size = 97, normalized size = 3.1 \begin{align*}{\frac{1}{2}\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}-2\,\tanh \left ( x \right ) }}-{\it Arcsinh} \left ( \tanh \left ( x \right ) \right ) -{\frac{\sqrt{2}}{2}{\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{1}{2}\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}+2\,\tanh \left ( x \right ) }}+{\frac{\sqrt{2}}{2}{\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+tanh(x)^2)^(1/2),x)

[Out]

1/2*((1+tanh(x))^2-2*tanh(x))^(1/2)-arcsinh(tanh(x))-1/2*2^(1/2)*arctanh(1/4*(2-2*tanh(x))*2^(1/2)/((1+tanh(x)
)^2-2*tanh(x))^(1/2))-1/2*((tanh(x)-1)^2+2*tanh(x))^(1/2)+1/2*2^(1/2)*arctanh(1/4*(2*tanh(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 \sqrt{\tanh \left (x\right )^{2} + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(tanh(x)^2 + 1), x)

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Fricas [B]  time = 1.90972, size = 2319, normalized size = 74.81 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*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)) + 1/4*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)) - 1/2*log((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 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)) + 1/
2*log((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 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 \sqrt{\tanh ^{2}{\left (x \right )} + 1}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(tanh(x)**2 + 1), x)

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Giac [B]  time = 1.21573, size = 140, normalized size = 4.52 \begin{align*} -\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} \log \left (\frac{\sqrt{2} - \sqrt{e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1}{\sqrt{2} + \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 )} + 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+tanh(x)^2)^(1/2),x, algorithm="giac")

[Out]

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