3.2.0 ∫ tanh(x)∘ _______ 1 + tanh(x) dx [126]

Integrand size = 11, antiderivative size = 32 \[ \int \tanh (x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-2 \sqrt {1+\tanh (x)} \]

Rubi [A] (veriﬁed)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3608, 3561, 212} \[ \int \tanh (x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-2 \sqrt {\tanh (x)+1} \]

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]

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq

rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*(

(a + b*Tan[e + f*x])^m/(f*m)), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,

d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && !LtQ[m, 0]

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

Mathematica [A] (veriﬁed)

Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \tanh (x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-2 \sqrt {1+\tanh (x)} \]

Maple [A] (veriﬁed)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81


method	result	size

derivativedivides	\(\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}-2 \sqrt {1+\tanh \left (x \right )}\)	\(26\)
default	\(\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}-2 \sqrt {1+\tanh \left (x \right )}\)	\(26\)

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 4.03 \[ \int \tanh (x) \sqrt {1+\tanh (x)} \, dx=-\frac {4 \, \sqrt {2} {\left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right )\right )} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - {\left (\sqrt {2} \cosh \left (x\right )^{2} + 2 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt {2} \sinh \left (x\right )^{2} + \sqrt {2}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right )}{2 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )}} \]

-1/2*(4*sqrt(2)*(sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(cosh(x)/(cosh(x) - sinh(x))) - (sqrt(2)*cosh(x)^2 + 2

*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 + sqrt(2))*log(-2*sqrt(2)*sqrt(cosh(x)/(cosh(x) - sinh(x)))*(cosh

(x) + sinh(x)) - 2*cosh(x)^2 - 4*cosh(x)*sinh(x) - 2*sinh(x)^2 - 1))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^

Sympy [A] (veriﬁcation not implemented)

Time = 0.73 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \tanh (x) \sqrt {1+\tanh (x)} \, dx=- \frac {\sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right )}{2} - 2 \sqrt {\tanh {\left (x \right )} + 1} \]

-sqrt(2)*(log(sqrt(tanh(x) + 1) - sqrt(2)) - log(sqrt(tanh(x) + 1) + sqrt(2)))/2 - 2*sqrt(tanh(x) + 1)

Maxima [F]

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

-sqrt(2)/sqrt(e^(-2*x) + 1) + integrate(sqrt(2)*e^(-x)/((e^(-x) + e^(-3*x))*sqrt(e^(-2*x) + 1)), x)

Giac [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.66 \[ \int \tanh (x) \sqrt {1+\tanh (x)} \, dx=\frac {1}{2} \, \sqrt {2} {\left (\frac {4}{\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1} - \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

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

Mupad [B] (veriﬁcation not implemented)

Time = 1.70 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \tanh (x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )-2\,\sqrt {\mathrm {tanh}\left (x\right )+1} \]

2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2) - 2*(tanh(x) + 1)^(1/2)

\(\int \tanh (x) \sqrt {1+\tanh (x)} \, dx\) [126]

Optimal result