Integrand size = 11, antiderivative size = 37 \[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx=-\frac {3 x}{2}+2 \log (\cosh (x))+\frac {3 \tanh (x)}{2}-\tanh ^2(x)+\frac {\tanh ^2(x)}{2 (1+\coth (x))} \] Output:
-3/2*x+2*ln(cosh(x))+3/2*tanh(x)-tanh(x)^2+tanh(x)^2/(2+2*coth(x))
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx=\frac {1}{2} \left (-3 \text {arctanh}(\tanh (x))+4 \log (\cosh (x))+3 \tanh (x)+\left (-2+\frac {1}{1+\coth (x)}\right ) \tanh ^2(x)\right ) \] Input:
Integrate[Tanh[x]^3/(1 + Coth[x]),x]
Output:
(-3*ArcTanh[Tanh[x]] + 4*Log[Cosh[x]] + 3*Tanh[x] + (-2 + (1 + Coth[x])^(- 1))*Tanh[x]^2)/2
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.62, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.636, Rules used = {3042, 26, 4035, 26, 3042, 26, 4012, 26, 3042, 25, 4012, 3042, 26, 4014, 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.
\(\displaystyle \int \frac {\tanh ^3(x)}{\coth (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i}{\left (1-i \tan \left (\frac {\pi }{2}+i x\right )\right ) \tan \left (\frac {\pi }{2}+i x\right )^3}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {1}{\left (1-i \tan \left (i x+\frac {\pi }{2}\right )\right ) \tan \left (i x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 4035 |
\(\displaystyle -i \left (\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}-\frac {1}{2} \int -i (4-3 \coth (x)) \tanh ^3(x)dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} i \int (4-3 \coth (x)) \tanh ^3(x)dx+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{2} i \int -\frac {i \left (3 i \tan \left (i x+\frac {\pi }{2}\right )+4\right )}{\tan \left (i x+\frac {\pi }{2}\right )^3}dx+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \int \frac {3 i \tan \left (i x+\frac {\pi }{2}\right )+4}{\tan \left (i x+\frac {\pi }{2}\right )^3}dx+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -i \left (\frac {1}{2} \left (\int -i (3-4 \coth (x)) \tanh ^2(x)dx-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \int (3-4 \coth (x)) \tanh ^2(x)dx-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \int -\frac {4 i \tan \left (i x+\frac {\pi }{2}\right )+3}{\tan \left (i x+\frac {\pi }{2}\right )^2}dx-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -i \left (\frac {1}{2} \left (i \int \frac {4 i \tan \left (i x+\frac {\pi }{2}\right )+3}{\tan \left (i x+\frac {\pi }{2}\right )^2}dx-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -i \left (\frac {1}{2} \left (i (\int (4-3 \coth (x)) \tanh (x)dx+3 \tanh (x))-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{2} \left (i \left (3 \tanh (x)+\int \frac {i \left (3 i \tan \left (i x+\frac {\pi }{2}\right )+4\right )}{\tan \left (i x+\frac {\pi }{2}\right )}dx\right )-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (i \left (3 \tanh (x)+i \int \frac {3 i \tan \left (i x+\frac {\pi }{2}\right )+4}{\tan \left (i x+\frac {\pi }{2}\right )}dx\right )-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle -i \left (\frac {1}{2} \left (i (3 \tanh (x)+i (4 \int -i \tanh (x)dx+3 i x))-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (i (3 \tanh (x)+i (3 i x-4 i \int \tanh (x)dx))-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{2} \left (i (3 \tanh (x)+i (3 i x-4 i \int -i \tan (i x)dx))-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (i (3 \tanh (x)+i (3 i x-4 \int \tan (i x)dx))-2 i \tanh ^2(x)\right )+\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -i \left (\frac {i \tanh ^2(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (i (3 \tanh (x)+i (3 i x-4 i \log (\cosh (x))))-2 i \tanh ^2(x)\right )\right )\) |
Input:
Int[Tanh[x]^3/(1 + Coth[x]),x]
Output:
(-I)*(((I/2)*Tanh[x]^2)/(1 + Coth[x]) + ((-2*I)*Tanh[x]^2 + I*(I*((3*I)*x - (4*I)*Log[Cosh[x]]) + 3*Tanh[x]))/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_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*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] && N eQ[a*c + b*d, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-a)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b* c - a*d)*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d)) Int[(c + d *Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !GtQ[n, 0]
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {7 x}{2}-\frac {{\mathrm e}^{-2 x}}{4}-\frac {2}{\left ({\mathrm e}^{2 x}+1\right )^{2}}+2 \ln \left ({\mathrm e}^{2 x}+1\right )\) | \(30\) |
parallelrisch | \(\frac {\left (-4 \tanh \left (x \right )-4\right ) \ln \left (1-\tanh \left (x \right )\right )-\tanh \left (x \right )^{3}-7 \tanh \left (x \right ) x +\tanh \left (x \right )^{2}-7 x -3}{2 \tanh \left (x \right )+2}\) | \(44\) |
default | \(-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-\frac {7 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}+\frac {2 \tanh \left (\frac {x}{2}\right )^{3}-2 \tanh \left (\frac {x}{2}\right )^{2}+2 \tanh \left (\frac {x}{2}\right )}{\left (\tanh \left (\frac {x}{2}\right )^{2}+1\right )^{2}}+2 \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+1\right )-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}\) | \(80\) |
Input:
int(tanh(x)^3/(1+coth(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 = 0.09 (sec) , antiderivative size = 354, normalized size of antiderivative = 9.57 \[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx =\text {Too large to display} \] Input:
integrate(tanh(x)^3/(1+coth(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))
\[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\coth {\left (x \right )} + 1}\, dx \] Input:
integrate(tanh(x)**3/(1+coth(x)),x)
Output:
Integral(tanh(x)**3/(coth(x) + 1), x)
Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16 \[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx=\frac {1}{2} \, x + \frac {2 \, {\left (2 \, e^{\left (-2 \, x\right )} + 1\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} - \frac {1}{4} \, e^{\left (-2 \, x\right )} + 2 \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \] Input:
integrate(tanh(x)^3/(1+coth(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 = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx=-\frac {7}{2} \, x - \frac {{\left (e^{\left (4 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} + 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \] Input:
integrate(tanh(x)^3/(1+coth(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 = 2.46 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx=2\,\ln \left ({\mathrm {e}}^{2\,x}+1\right )-\frac {7\,x}{2}-\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {2}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1} \] Input:
int(tanh(x)^3/(coth(x) + 1),x)
Output:
2*log(exp(2*x) + 1) - (7*x)/2 - exp(-2*x)/4 - 2/(2*exp(2*x) + exp(4*x) + 1 )
Time = 0.24 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.92 \[ \int \frac {\tanh ^3(x)}{1+\coth (x)} \, dx=\frac {16 e^{6 x} \mathrm {log}\left (e^{2 x}+1\right )-28 e^{6 x} x +e^{6 x}+32 e^{4 x} \mathrm {log}\left (e^{2 x}+1\right )-56 e^{4 x} x +16 e^{2 x} \mathrm {log}\left (e^{2 x}+1\right )-28 e^{2 x} x -19 e^{2 x}-2}{8 e^{2 x} \left (e^{4 x}+2 e^{2 x}+1\right )} \] Input:
int(tanh(x)^3/(1+coth(x)),x)
Output:
(16*e**(6*x)*log(e**(2*x) + 1) - 28*e**(6*x)*x + e**(6*x) + 32*e**(4*x)*lo g(e**(2*x) + 1) - 56*e**(4*x)*x + 16*e**(2*x)*log(e**(2*x) + 1) - 28*e**(2 *x)*x - 19*e**(2*x) - 2)/(8*e**(2*x)*(e**(4*x) + 2*e**(2*x) + 1))