\(\int \sqrt {1+\tanh ^2(x)} \, dx\) [225]

Optimal result

Integrand size = 10, antiderivative size = 31 \[ \int \sqrt {1+\tanh ^2(x)} \, dx=-\text {arcsinh}(\tanh (x))+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \tanh (x)}{\sqrt {1+\tanh ^2(x)}}\right ) \]

[Out]

-arcsinh(tanh(x))+arctanh(2^(1/2)*tanh(x)/(1+tanh(x)^2)^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3742, 399, 221, 385, 212} \[ \int \sqrt {1+\tanh ^2(x)} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \tanh (x)}{\sqrt {\tanh ^2(x)+1}}\right )-\text {arcsinh}(\tanh (x)) \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 221

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 385

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 399

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

Rule 3742

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])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\sqrt {1+x^2}}{1-x^2} \, dx,x,\tanh (x)\right ) \\ & = 2 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {1+x^2}} \, dx,x,\tanh (x)\right )-\text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,\tanh (x)\right ) \\ & = -\text {arcsinh}(\tanh (x))+2 \text {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {1+\tanh ^2(x)}}\right ) \\ & = -\text {arcsinh}(\tanh (x))+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \tanh (x)}{\sqrt {1+\tanh ^2(x)}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \sqrt {1+\tanh ^2(x)} \, dx=\frac {\left (\sqrt {2} \text {arcsinh}\left (\sqrt {2} \sinh (x)\right )-\text {arctanh}\left (\frac {\sinh (x)}{\sqrt {\cosh (2 x)}}\right )\right ) \cosh (x) \sqrt {1+\tanh ^2(x)}}{\sqrt {\cosh (2 x)}} \]

[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]]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(25)=50\).

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.13

method	result	size
derivativedivides	\(-\frac {\sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}{2}-\operatorname {arcsinh}\left (\tanh \left (x \right )\right )+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2+2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}\right )}{2}+\frac {\sqrt {\left (1+\tanh \left (x \right )\right )^{2}-2 \tanh \left (x \right )}}{2}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2-2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (1+\tanh \left (x \right )\right )^{2}-2 \tanh \left (x \right )}}\right )}{2}\)	\(97\)
default	\(-\frac {\sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}{2}-\operatorname {arcsinh}\left (\tanh \left (x \right )\right )+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2+2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}\right )}{2}+\frac {\sqrt {\left (1+\tanh \left (x \right )\right )^{2}-2 \tanh \left (x \right )}}{2}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2-2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (1+\tanh \left (x \right )\right )^{2}-2 \tanh \left (x \right )}}\right )}{2}\)	\(97\)

[In]

int((1+tanh(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*((tanh(x)-1)^2+2*tanh(x))^(1/2)-arcsinh(tanh(x))+1/2*2^(1/2)*arctanh(1/4*(2+2*tanh(x))*2^(1/2)/((tanh(x)-
1)^2+2*tanh(x))^(1/2))+1/2*((1+tanh(x))^2-2*tanh(x))^(1/2)-1/2*2^(1/2)*arctanh(1/4*(2-2*tanh(x))*2^(1/2)/((1+t
anh(x))^2-2*tanh(x))^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 679 vs. \(2 (25) = 50\).

Time = 0.27 (sec) , antiderivative size = 679, normalized size of antiderivative = 21.90 \[ \int \sqrt {1+\tanh ^2(x)} \, dx=\text {Too large to display} \]

[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))

Sympy [F]

\[ \int \sqrt {1+\tanh ^2(x)} \, dx=\int \sqrt {\tanh ^{2}{\left (x \right )} + 1}\, dx \]

[In]

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

[Out]

Integral(sqrt(tanh(x)**2 + 1), x)

Maxima [F]

\[ \int \sqrt {1+\tanh ^2(x)} \, dx=\int { \sqrt {\tanh \left (x\right )^{2} + 1} \,d x } \]

[In]

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

[Out]

integrate(sqrt(tanh(x)^2 + 1), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (25) = 50\).

Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.35 \[ \int \sqrt {1+\tanh ^2(x)} \, dx=-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1}{\sqrt {2} + \sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} - 1}\right ) + \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) + \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) - \log \left (-\sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right )\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.19 \[ \int \sqrt {1+\tanh ^2(x)} \, dx=\frac {\sqrt {2}\,\left (\ln \left (\mathrm {tanh}\left (x\right )+1\right )-\ln \left (\sqrt {2}\,\sqrt {{\mathrm {tanh}\left (x\right )}^2+1}-\mathrm {tanh}\left (x\right )+1\right )\right )}{2}-\mathrm {asinh}\left (\mathrm {tanh}\left (x\right )\right )+\frac {\sqrt {2}\,\left (\ln \left (\mathrm {tanh}\left (x\right )+\sqrt {2}\,\sqrt {{\mathrm {tanh}\left (x\right )}^2+1}+1\right )-\ln \left (\mathrm {tanh}\left (x\right )-1\right )\right )}{2} \]

[In]

int((tanh(x)^2 + 1)^(1/2),x)

[Out]

(2^(1/2)*(log(tanh(x) + 1) - log(2^(1/2)*(tanh(x)^2 + 1)^(1/2) - tanh(x) + 1)))/2 - asinh(tanh(x)) + (2^(1/2)*
(log(tanh(x) + 2^(1/2)*(tanh(x)^2 + 1)^(1/2) + 1) - log(tanh(x) - 1)))/2