Integrand size = 15, antiderivative size = 69 \[ \int \frac {1+\sinh ^2(x)}{1+\cosh (x)+\sinh (x)} \, dx=\frac {1}{4} \log \left (1-\tanh \left (\frac {x}{2}\right )\right )+\frac {3}{4} \log \left (1+\tanh \left (\frac {x}{2}\right )\right )+\frac {1}{2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}-\frac {1}{2 \left (1+\tanh \left (\frac {x}{2}\right )\right )^2}+\frac {1}{1+\tanh \left (\frac {x}{2}\right )} \]
1/4*ln(1-tanh(1/2*x))+3/4*ln(1+tanh(1/2*x))+1/2/(1-tanh(1/2*x))-1/2/(1+tan h(1/2*x))^2+1/(1+tanh(1/2*x))
Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.54 \[ \int \frac {1+\sinh ^2(x)}{1+\cosh (x)+\sinh (x)} \, dx=\frac {x}{4}+\frac {\cosh (x)}{2}-\frac {1}{8} \cosh (2 x)-\log \left (\cosh \left (\frac {x}{2}\right )\right )+\frac {1}{8} \sinh (2 x) \]
Time = 0.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 4897, 3042, 4902, 27, 652, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sinh ^2(x)+1}{\sinh (x)+\cosh (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1-\sin (i x)^2}{-i \sin (i x)+\cos (i x)+1}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\cosh ^2(x)}{\sinh (x)+\cosh (x)+1}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i x)^2}{-i \sin (i x)+\cos (i x)+1}dx\) |
\(\Big \downarrow \) 4902 |
\(\displaystyle 2 \int \frac {\left (\tanh ^2\left (\frac {x}{2}\right )+1\right )^2}{2 \left (1-\tanh \left (\frac {x}{2}\right )\right )^2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^3}d\tanh \left (\frac {x}{2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\left (\tanh ^2\left (\frac {x}{2}\right )+1\right )^2}{\left (1-\tanh \left (\frac {x}{2}\right )\right )^2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^3}d\tanh \left (\frac {x}{2}\right )\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {3}{4 \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^2}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^3}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^2}\right )d\tanh \left (\frac {x}{2}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}+\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^2}+\frac {1}{4} \log \left (1-\tanh \left (\frac {x}{2}\right )\right )+\frac {3}{4} \log \left (\tanh \left (\frac {x}{2}\right )+1\right )\) |
Log[1 - Tanh[x/2]]/4 + (3*Log[1 + Tanh[x/2]])/4 + 1/(2*(1 - Tanh[x/2])) - 1/(2*(1 + Tanh[x/2])^2) + (1 + Tanh[x/2])^(-1)
3.11.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
Int[u_, x_Symbol] :> With[{w = Block[{$ShowSteps = False, $StepCounter = Nu ll}, Int[SubstFor[1/(1 + FreeFactors[Tan[FunctionOfTrig[u, x]/2], x]^2*x^2) , Tan[FunctionOfTrig[u, x]/2]/FreeFactors[Tan[FunctionOfTrig[u, x]/2], x], u, x], x]]}, Module[{v = FunctionOfTrig[u, x], d}, Simp[d = FreeFactors[Tan [v/2], x]; 2*(d/Coefficient[v, x, 1]) Subst[Int[SubstFor[1/(1 + d^2*x^2), Tan[v/2]/d, u, x], x], x, Tan[v/2]/d], x]] /; CalculusFreeQ[w, x]] /; Inve rseFunctionFreeQ[u, x] && !FalseQ[FunctionOfTrig[u, x]]
Time = 0.41 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.41
method | result | size |
risch | \(\frac {3 x}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{-x}}{4}-\frac {{\mathrm e}^{-2 x}}{8}-\ln \left (1+{\mathrm e}^{x}\right )\) | \(28\) |
default | \(-\frac {1}{2 \left (1+\tanh \left (\frac {x}{2}\right )\right )^{2}}+\frac {1}{1+\tanh \left (\frac {x}{2}\right )}+\frac {3 \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{4}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{4}\) | \(48\) |
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {1+\sinh ^2(x)}{1+\cosh (x)+\sinh (x)} \, dx=\frac {6 \, x \cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{3} + 6 \, {\left (x + \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 2 \, \sinh \left (x\right )^{3} - 8 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + 2 \, {\left (6 \, x \cosh \left (x\right ) + 3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right ) + 2 \, \cosh \left (x\right ) - 1}{8 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}} \]
1/8*(6*x*cosh(x)^2 + 2*cosh(x)^3 + 6*(x + cosh(x))*sinh(x)^2 + 2*sinh(x)^3 - 8*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)*log(cosh(x) + sinh(x) + 1 ) + 2*(6*x*cosh(x) + 3*cosh(x)^2 + 1)*sinh(x) + 2*cosh(x) - 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (51) = 102\).
Time = 0.66 (sec) , antiderivative size = 381, normalized size of antiderivative = 5.52 \[ \int \frac {1+\sinh ^2(x)}{1+\cosh (x)+\sinh (x)} \, dx=- \frac {x \tanh ^{3}{\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} - \frac {x \tanh ^{2}{\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} + \frac {x \tanh {\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} + \frac {x}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} + \frac {4 \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} \tanh ^{3}{\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} + \frac {4 \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} \tanh ^{2}{\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} - \frac {4 \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} \tanh {\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} - \frac {4 \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} + \frac {2 \tanh ^{2}{\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} - \frac {6 \tanh {\left (\frac {x}{2} \right )}}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} - \frac {4}{4 \tanh ^{3}{\left (\frac {x}{2} \right )} + 4 \tanh ^{2}{\left (\frac {x}{2} \right )} - 4 \tanh {\left (\frac {x}{2} \right )} - 4} \]
-x*tanh(x/2)**3/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) - x*ta nh(x/2)**2/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) + x*tanh(x/ 2)/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) + x/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) + 4*log(tanh(x/2) + 1)*tanh(x/2)**3/( 4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) + 4*log(tanh(x/2) + 1)* tanh(x/2)**2/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) - 4*log(t anh(x/2) + 1)*tanh(x/2)/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4 ) - 4*log(tanh(x/2) + 1)/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) + 2*tanh(x/2)**2/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) - 6*tanh(x/2)/(4*tanh(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4) - 4/(4*tan h(x/2)**3 + 4*tanh(x/2)**2 - 4*tanh(x/2) - 4)
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.42 \[ \int \frac {1+\sinh ^2(x)}{1+\cosh (x)+\sinh (x)} \, dx=-\frac {1}{4} \, x + \frac {1}{4} \, e^{\left (-x\right )} - \frac {1}{8} \, e^{\left (-2 \, x\right )} + \frac {1}{4} \, e^{x} - \log \left (e^{\left (-x\right )} + 1\right ) \]
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.39 \[ \int \frac {1+\sinh ^2(x)}{1+\cosh (x)+\sinh (x)} \, dx=\frac {1}{8} \, {\left (2 \, e^{x} - 1\right )} e^{\left (-2 \, x\right )} + \frac {3}{4} \, x + \frac {1}{4} \, e^{x} - \log \left (e^{x} + 1\right ) \]
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.39 \[ \int \frac {1+\sinh ^2(x)}{1+\cosh (x)+\sinh (x)} \, dx=\frac {3\,x}{4}+\frac {{\mathrm {e}}^{-x}}{4}-\frac {{\mathrm {e}}^{-2\,x}}{8}-\ln \left ({\mathrm {e}}^x+1\right )+\frac {{\mathrm {e}}^x}{4} \]