Integrand size = 13, antiderivative size = 99 \[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=-\frac {(a+2 b) \log (1-\tanh (x))}{4 (a+b)^2}+\frac {(a-2 b) \log (1+\tanh (x))}{4 (a-b)^2}+\frac {b^3 \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}+\frac {1}{4 (a+b) (1-\tanh (x))}-\frac {1}{4 (a-b) (1+\tanh (x))} \] Output:
-1/4*(a+2*b)*ln(1-tanh(x))/(a+b)^2+1/4*(a-2*b)*ln(1+tanh(x))/(a-b)^2+b^3*l n(a+b*tanh(x))/(a^2-b^2)^2+1/4/(a+b)/(1-tanh(x))-1/4/(a-b)/(1+tanh(x))
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.76 \[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=\frac {2 a^3 x-6 a b^2 x+\left (-a^2 b+b^3\right ) \cosh (2 x)+4 b^3 \log (a \cosh (x)+b \sinh (x))+a \left (a^2-b^2\right ) \sinh (2 x)}{4 (a-b)^2 (a+b)^2} \] Input:
Integrate[Cosh[x]^2/(a + b*Tanh[x]),x]
Output:
(2*a^3*x - 6*a*b^2*x + (-(a^2*b) + b^3)*Cosh[2*x] + 4*b^3*Log[a*Cosh[x] + b*Sinh[x]] + a*(a^2 - b^2)*Sinh[2*x])/(4*(a - b)^2*(a + b)^2)
Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3987, 27, 477, 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.
\(\displaystyle \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (i x)^2 (a-i b \tan (i x))}dx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {b^4}{(a+b \tanh (x)) \left (b^2-b^2 \tanh ^2(x)\right )^2}d(b \tanh (x))}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle b^3 \int \frac {1}{(a+b \tanh (x)) \left (b^2-b^2 \tanh ^2(x)\right )^2}d(b \tanh (x))\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\int \left (\frac {b^4}{\left (a^2-b^2\right )^2 (a+b \tanh (x))}+\frac {b^2}{4 (a+b) (b-b \tanh (x))^2}+\frac {b^2}{4 (a-b) (\tanh (x) b+b)^2}+\frac {(a+2 b) b}{4 (a+b)^2 (b-b \tanh (x))}+\frac {(a-2 b) b}{4 (a-b)^2 (\tanh (x) b+b)}\right )d(b \tanh (x))}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^4 \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}+\frac {b^2}{4 (a+b) (b-b \tanh (x))}-\frac {b^2}{4 (a-b) (b \tanh (x)+b)}-\frac {b (a+2 b) \log (b-b \tanh (x))}{4 (a+b)^2}+\frac {b (a-2 b) \log (b \tanh (x)+b)}{4 (a-b)^2}}{b}\) |
Input:
Int[Cosh[x]^2/(a + b*Tanh[x]),x]
Output:
(-1/4*(b*(a + 2*b)*Log[b - b*Tanh[x]])/(a + b)^2 + (b^4*Log[a + b*Tanh[x]] )/(a^2 - b^2)^2 + ((a - 2*b)*b*Log[b + b*Tanh[x]])/(4*(a - b)^2) + b^2/(4* (a + b)*(b - b*Tanh[x])) - b^2/(4*(a - b)*(b + b*Tanh[x])))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(b*f) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 1.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05
method | result | size |
risch | \(\frac {a x}{2 \left (a +b \right )^{2}}+\frac {x b}{\left (a +b \right )^{2}}+\frac {{\mathrm e}^{2 x}}{8 a +8 b}-\frac {{\mathrm e}^{-2 x}}{8 \left (a -b \right )}-\frac {2 b^{3} x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) | \(104\) |
default | \(\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a +2 b \tanh \left (\frac {x}{2}\right )+a \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {1}{\left (2 b +2 a \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {2}{\left (4 a +4 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-a -2 b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {1}{\left (2 a -2 b \right ) \left (1+\tanh \left (\frac {x}{2}\right )\right )^{2}}+\frac {2}{\left (4 a -4 b \right ) \left (1+\tanh \left (\frac {x}{2}\right )\right )}+\frac {\left (a -2 b \right ) \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 \left (a -b \right )^{2}}\) | \(153\) |
Input:
int(cosh(x)^2/(a+b*tanh(x)),x,method=_RETURNVERBOSE)
Output:
1/2*a*x/(a+b)^2+x/(a+b)^2*b+1/8/(a+b)*exp(2*x)-1/8/(a-b)*exp(-2*x)-2*b^3/( a^4-2*a^2*b^2+b^4)*x+b^3/(a^4-2*a^2*b^2+b^4)*ln(exp(2*x)+(a-b)/(a+b))
Leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (89) = 178\).
Time = 0.11 (sec) , antiderivative size = 331, normalized size of antiderivative = 3.34 \[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=\frac {{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{4} + 4 \, {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x \cosh \left (x\right )^{2} - a^{3} - a^{2} b + a b^{2} + b^{3} + 2 \, {\left (3 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x\right )} \sinh \left (x\right )^{2} + 8 \, {\left (b^{3} \cosh \left (x\right )^{2} + 2 \, b^{3} \cosh \left (x\right ) \sinh \left (x\right ) + b^{3} \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{3} + 2 \, {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right )}{8 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}\right )}} \] Input:
integrate(cosh(x)^2/(a+b*tanh(x)),x, algorithm="fricas")
Output:
1/8*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 - a^2*b - a*b^2 + b^3) *cosh(x)*sinh(x)^3 + (a^3 - a^2*b - a*b^2 + b^3)*sinh(x)^4 + 4*(a^3 - 3*a* b^2 - 2*b^3)*x*cosh(x)^2 - a^3 - a^2*b + a*b^2 + b^3 + 2*(3*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^2 + 2*(a^3 - 3*a*b^2 - 2*b^3)*x)*sinh(x)^2 + 8*(b^3* cosh(x)^2 + 2*b^3*cosh(x)*sinh(x) + b^3*sinh(x)^2)*log(2*(a*cosh(x) + b*si nh(x))/(cosh(x) - sinh(x))) + 4*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^3 + 2 *(a^3 - 3*a*b^2 - 2*b^3)*x*cosh(x))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4)*cosh (x)^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) + (a^4 - 2*a^2*b^2 + b^4 )*sinh(x)^2)
\[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=\int \frac {\cosh ^{2}{\left (x \right )}}{a + b \tanh {\left (x \right )}}\, dx \] Input:
integrate(cosh(x)**2/(a+b*tanh(x)),x)
Output:
Integral(cosh(x)**2/(a + b*tanh(x)), x)
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.87 \[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=\frac {b^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a + 2 \, b\right )} x}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} - \frac {e^{\left (-2 \, x\right )}}{8 \, {\left (a - b\right )}} \] Input:
integrate(cosh(x)^2/(a+b*tanh(x)),x, algorithm="maxima")
Output:
b^3*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - 2*a^2*b^2 + b^4) + 1/2*(a + 2*b) *x/(a^2 + 2*a*b + b^2) + 1/8*e^(2*x)/(a + b) - 1/8*e^(-2*x)/(a - b)
Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.12 \[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=\frac {b^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a - 2 \, b\right )} x}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (2 \, a e^{\left (2 \, x\right )} - 4 \, b e^{\left (2 \, x\right )} + a - b\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} \] Input:
integrate(cosh(x)^2/(a+b*tanh(x)),x, algorithm="giac")
Output:
b^3*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - 2*a^2*b^2 + b^4) + 1/2* (a - 2*b)*x/(a^2 - 2*a*b + b^2) - 1/8*(2*a*e^(2*x) - 4*b*e^(2*x) + a - b)* e^(-2*x)/(a^2 - 2*a*b + b^2) + 1/8*e^(2*x)/(a + b)
Time = 2.51 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,a+8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a-8\,b}+\frac {b^3\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {x\,\left (a-2\,b\right )}{2\,{\left (a-b\right )}^2} \] Input:
int(cosh(x)^2/(a + b*tanh(x)),x)
Output:
exp(2*x)/(8*a + 8*b) - exp(-2*x)/(8*a - 8*b) + (b^3*log(a - b + a*exp(2*x) + b*exp(2*x)))/(a^4 + b^4 - 2*a^2*b^2) + (x*(a - 2*b))/(2*(a - b)^2)
Time = 0.24 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.52 \[ \int \frac {\cosh ^2(x)}{a+b \tanh (x)} \, dx=\frac {e^{4 x} a^{3}-e^{4 x} a^{2} b -e^{4 x} a \,b^{2}+e^{4 x} b^{3}+8 e^{2 x} \mathrm {log}\left (e^{2 x} a +e^{2 x} b +a -b \right ) b^{3}+4 e^{2 x} a^{3} x -12 e^{2 x} a \,b^{2} x -8 e^{2 x} b^{3} x -a^{3}-a^{2} b +a \,b^{2}+b^{3}}{8 e^{2 x} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(cosh(x)^2/(a+b*tanh(x)),x)
Output:
(e**(4*x)*a**3 - e**(4*x)*a**2*b - e**(4*x)*a*b**2 + e**(4*x)*b**3 + 8*e** (2*x)*log(e**(2*x)*a + e**(2*x)*b + a - b)*b**3 + 4*e**(2*x)*a**3*x - 12*e **(2*x)*a*b**2*x - 8*e**(2*x)*b**3*x - a**3 - a**2*b + a*b**2 + b**3)/(8*e **(2*x)*(a**4 - 2*a**2*b**2 + b**4))