Integrand size = 11, antiderivative size = 37 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=-\frac {3 x}{2}+2 \log (\cosh (x))+\frac {3 \tanh (x)}{2}-\tanh ^2(x)+\frac {\tanh ^3(x)}{2 (1+\tanh (x))} \] Output:
-3/2*x+2*ln(cosh(x))+3/2*tanh(x)-tanh(x)^2+tanh(x)^3/(2+2*tanh(x))
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {4 \log (\cosh (x))+(3+4 \log (\cosh (x))) \tanh (x)+\tanh ^2(x)-\tanh ^3(x)-3 \text {arctanh}(\tanh (x)) (1+\tanh (x))}{2 (1+\tanh (x))} \] Input:
Integrate[Tanh[x]^4/(1 + Tanh[x]),x]
Output:
(4*Log[Cosh[x]] + (3 + 4*Log[Cosh[x]])*Tanh[x] + Tanh[x]^2 - Tanh[x]^3 - 3 *ArcTanh[Tanh[x]]*(1 + Tanh[x]))/(2*(1 + Tanh[x]))
Result contains complex when optimal does not.
Time = 0.50 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.364, Rules used = {3042, 4033, 25, 3042, 25, 4011, 26, 26, 3042, 26, 4008, 26, 3042, 26, 3956}
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.
\(\displaystyle \int \frac {\tanh ^4(x)}{\tanh (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (i x)^4}{1-i \tan (i x)}dx\) |
\(\Big \downarrow \) 4033 |
\(\displaystyle \frac {1}{2} \int -\left ((3-4 \tanh (x)) \tanh ^2(x)\right )dx+\frac {\tanh ^3(x)}{2 (\tanh (x)+1)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}-\frac {1}{2} \int (3-4 \tanh (x)) \tanh ^2(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}-\frac {1}{2} \int -\left ((4 i \tan (i x)+3) \tan (i x)^2\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \int (4 i \tan (i x)+3) \tan (i x)^2dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)+\int i (3 i \tanh (x)-4 i) \tanh (x)dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)+i \int -i (4-3 \tanh (x)) \tanh (x)dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{2} \left (\int (4-3 \tanh (x)) \tanh (x)dx-2 \tanh ^2(x)\right )+\frac {\tanh ^3(x)}{2 (\tanh (x)+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)+\int -i (3 i \tan (i x)+4) \tan (i x)dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)-i \int (3 i \tan (i x)+4) \tan (i x)dx\right )\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)-i (4 \int i \tanh (x)dx-3 i x+3 i \tanh (x))\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)-i (4 i \int \tanh (x)dx-3 i x+3 i \tanh (x))\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)-i (4 i \int -i \tan (i x)dx-3 i x+3 i \tanh (x))\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)-i (4 \int \tan (i x)dx-3 i x+3 i \tanh (x))\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-2 \tanh ^2(x)-i (-3 i x+3 i \tanh (x)+4 i \log (\cosh (x)))\right )\) |
Input:
Int[Tanh[x]^4/(1 + Tanh[x]),x]
Output:
Tanh[x]^3/(2*(1 + Tanh[x])) + ((-I)*((-3*I)*x + (4*I)*Log[Cosh[x]] + (3*I) *Tanh[x]) - 2*Tanh[x]^2)/2
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), x] + Simp[(b*c + a*d) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 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] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/( 2*a*f*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a^2) Int[(c + d*Tan[e + f*x] )^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1]
Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {7 x}{2}-\frac {{\mathrm e}^{-2 x}}{4}-\frac {2}{\left ({\mathrm e}^{2 x}+1\right )^{2}}+2 \ln \left ({\mathrm e}^{2 x}+1\right )\) | \(30\) |
derivativedivides | \(-\frac {\tanh \left (x \right )^{2}}{2}+\tanh \left (x \right )-\frac {1}{2 \left (1+\tanh \left (x \right )\right )}-\frac {7 \ln \left (1+\tanh \left (x \right )\right )}{4}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}\) | \(32\) |
default | \(-\frac {\tanh \left (x \right )^{2}}{2}+\tanh \left (x \right )-\frac {1}{2 \left (1+\tanh \left (x \right )\right )}-\frac {7 \ln \left (1+\tanh \left (x \right )\right )}{4}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}\) | \(32\) |
parallelrisch | \(-\frac {3+\tanh \left (x \right )^{3}+4 \ln \left (1-\tanh \left (x \right )\right ) \tanh \left (x \right )+7 \tanh \left (x \right ) x -\tanh \left (x \right )^{2}+4 \ln \left (1-\tanh \left (x \right )\right )+7 x}{2 \left (1+\tanh \left (x \right )\right )}\) | \(49\) |
Input:
int(tanh(x)^4/(1+tanh(x)),x,method=_RETURNVERBOSE)
Output:
-7/2*x-1/4*exp(-2*x)-2/(exp(2*x)+1)^2+2*ln(exp(2*x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (31) = 62\).
Time = 0.10 (sec) , antiderivative size = 354, normalized size of antiderivative = 9.57 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx =\text {Too large to display} \] Input:
integrate(tanh(x)^4/(1+tanh(x)),x, algorithm="fricas")
Output:
-1/4*(14*x*cosh(x)^6 + 84*x*cosh(x)*sinh(x)^5 + 14*x*sinh(x)^6 + (28*x + 1 )*cosh(x)^4 + (210*x*cosh(x)^2 + 28*x + 1)*sinh(x)^4 + 4*(70*x*cosh(x)^3 + (28*x + 1)*cosh(x))*sinh(x)^3 + 2*(7*x + 5)*cosh(x)^2 + 2*(105*x*cosh(x)^ 4 + 3*(28*x + 1)*cosh(x)^2 + 7*x + 5)*sinh(x)^2 - 8*(cosh(x)^6 + 6*cosh(x) *sinh(x)^5 + sinh(x)^6 + (15*cosh(x)^2 + 2)*sinh(x)^4 + 2*cosh(x)^4 + 4*(5 *cosh(x)^3 + 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4 + 12*cosh(x)^2 + 1)*sinh (x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 + 4*cosh(x)^3 + cosh(x))*sinh(x))*log(2 *cosh(x)/(cosh(x) - sinh(x))) + 4*(21*x*cosh(x)^5 + (28*x + 1)*cosh(x)^3 + (7*x + 5)*cosh(x))*sinh(x) + 1)/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x )^6 + (15*cosh(x)^2 + 2)*sinh(x)^4 + 2*cosh(x)^4 + 4*(5*cosh(x)^3 + 2*cosh (x))*sinh(x)^3 + (15*cosh(x)^4 + 12*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 + 4*cosh(x)^3 + cosh(x))*sinh(x))
Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (34) = 68\).
Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.30 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {x \tanh {\left (x \right )}}{2 \tanh {\left (x \right )} + 2} + \frac {x}{2 \tanh {\left (x \right )} + 2} - \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{2 \tanh {\left (x \right )} + 2} - \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{2 \tanh {\left (x \right )} + 2} - \frac {\tanh ^{3}{\left (x \right )}}{2 \tanh {\left (x \right )} + 2} + \frac {\tanh ^{2}{\left (x \right )}}{2 \tanh {\left (x \right )} + 2} - \frac {3}{2 \tanh {\left (x \right )} + 2} \] Input:
integrate(tanh(x)**4/(1+tanh(x)),x)
Output:
x*tanh(x)/(2*tanh(x) + 2) + x/(2*tanh(x) + 2) - 4*log(tanh(x) + 1)*tanh(x) /(2*tanh(x) + 2) - 4*log(tanh(x) + 1)/(2*tanh(x) + 2) - tanh(x)**3/(2*tanh (x) + 2) + tanh(x)**2/(2*tanh(x) + 2) - 3/(2*tanh(x) + 2)
Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x + \frac {2 \, {\left (2 \, e^{\left (-2 \, x\right )} + 1\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} - \frac {1}{4} \, e^{\left (-2 \, x\right )} + 2 \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \] Input:
integrate(tanh(x)^4/(1+tanh(x)),x, algorithm="maxima")
Output:
1/2*x + 2*(2*e^(-2*x) + 1)/(2*e^(-2*x) + e^(-4*x) + 1) - 1/4*e^(-2*x) + 2* log(e^(-2*x) + 1)
Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=-\frac {7}{2} \, x - \frac {{\left (e^{\left (4 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} + 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \] Input:
integrate(tanh(x)^4/(1+tanh(x)),x, algorithm="giac")
Output:
-7/2*x - 1/4*(e^(4*x) + 10*e^(2*x) + 1)*e^(-2*x)/(e^(2*x) + 1)^2 + 2*log(e ^(2*x) + 1)
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {x}{2}-2\,\ln \left (\mathrm {tanh}\left (x\right )+1\right )+\mathrm {tanh}\left (x\right )-\frac {{\mathrm {tanh}\left (x\right )}^2}{2}-\frac {1}{2\,\left (\mathrm {tanh}\left (x\right )+1\right )} \] Input:
int(tanh(x)^4/(tanh(x) + 1),x)
Output:
x/2 - 2*log(tanh(x) + 1) + tanh(x) - tanh(x)^2/2 - 1/(2*(tanh(x) + 1))
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.38 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {-8 e^{2 x} \mathrm {log}\left (\tanh \left (x \right )+1\right )-2 e^{2 x} \tanh \left (x \right )^{2}+4 e^{2 x} \tanh \left (x \right )+2 e^{2 x} x -1}{4 e^{2 x}} \] Input:
int(tanh(x)^4/(1+tanh(x)),x)
Output:
( - 8*e**(2*x)*log(tanh(x) + 1) - 2*e**(2*x)*tanh(x)**2 + 4*e**(2*x)*tanh( x) + 2*e**(2*x)*x - 1)/(4*e**(2*x))