Integrand size = 14, antiderivative size = 55 \[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx=\frac {a x}{b \left (a^2-b^2\right )}-\frac {\log (a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {x}{b (a+b \tanh (x))} \] Output:
a*x/b/(a^2-b^2)-ln(a*cosh(x)+b*sinh(x))/(a^2-b^2)-x/b/(a+b*tanh(x))
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx=\frac {b x-a \log (a \cosh (x)+b \sinh (x))}{a^3-a b^2}+\frac {x \sinh (x)}{a^2 \cosh (x)+a b \sinh (x)} \] Input:
Integrate[(x*Sech[x]^2)/(a + b*Tanh[x])^2,x]
Output:
(b*x - a*Log[a*Cosh[x] + b*Sinh[x]])/(a^3 - a*b^2) + (x*Sinh[x])/(a^2*Cosh [x] + a*b*Sinh[x])
Time = 0.41 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5989, 3042, 3965, 26, 3042, 4013}
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.
\(\displaystyle \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx\) |
\(\Big \downarrow \) 5989 |
\(\displaystyle \frac {\int \frac {1}{a+b \tanh (x)}dx}{b}-\frac {x}{b (a+b \tanh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {x}{b (a+b \tanh (x))}+\frac {\int \frac {1}{a-i b \tan (i x)}dx}{b}\) |
\(\Big \downarrow \) 3965 |
\(\displaystyle -\frac {x}{b (a+b \tanh (x))}+\frac {\frac {a x}{a^2-b^2}-\frac {i b \int -\frac {i (b+a \tanh (x))}{a+b \tanh (x)}dx}{a^2-b^2}}{b}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\frac {a x}{a^2-b^2}-\frac {b \int \frac {b+a \tanh (x)}{a+b \tanh (x)}dx}{a^2-b^2}}{b}-\frac {x}{b (a+b \tanh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {x}{b (a+b \tanh (x))}+\frac {\frac {a x}{a^2-b^2}-\frac {b \int \frac {b-i a \tan (i x)}{a-i b \tan (i x)}dx}{a^2-b^2}}{b}\) |
\(\Big \downarrow \) 4013 |
\(\displaystyle \frac {\frac {a x}{a^2-b^2}-\frac {b \log (a \cosh (x)+b \sinh (x))}{a^2-b^2}}{b}-\frac {x}{b (a+b \tanh (x))}\) |
Input:
Int[(x*Sech[x]^2)/(a + b*Tanh[x])^2,x]
Output:
((a*x)/(a^2 - b^2) - (b*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2))/b - x/(b* (a + b*Tanh[x]))
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[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[a*(x/(a^ 2 + b^2)), x] + Simp[b/(a^2 + b^2) Int[(b - a*Tan[c + d*x])/(a + b*Tan[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* (x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si n[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] && EqQ[a*c + b*d, 0]
Int[((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^2*((a_) + (b_.)*Tan h[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Simp[(e + f*x)^m*((a + b*Tanh[c + d*x])^(n + 1)/(b*d*(n + 1))), x] - Simp[f*(m/(b*d*(n + 1))) Int[(e + f* x)^(m - 1)*(a + b*Tanh[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f , n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Time = 8.96 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33
method | result | size |
risch | \(\frac {2 x}{a^{2}-b^{2}}-\frac {2 x}{\left ({\mathrm e}^{2 x} a +{\mathrm e}^{2 x} b +a -b \right ) \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{2}-b^{2}}\) | \(73\) |
Input:
int(x*sech(x)^2/(a+b*tanh(x))^2,x,method=_RETURNVERBOSE)
Output:
2/(a^2-b^2)*x-2*x/(exp(2*x)*a+exp(2*x)*b+a-b)/(a+b)-1/(a^2-b^2)*ln(exp(2*x )+(a-b)/(a+b))
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (55) = 110\).
Time = 0.10 (sec) , antiderivative size = 182, normalized size of antiderivative = 3.31 \[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx=\frac {2 \, {\left (a + b\right )} x \cosh \left (x\right )^{2} + 4 \, {\left (a + b\right )} x \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, {\left (a + b\right )} x \sinh \left (x\right )^{2} - {\left ({\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a + b\right )} \sinh \left (x\right )^{2} + a - b\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{3} - a^{2} b - a b^{2} + b^{3} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \sinh \left (x\right )^{2}} \] Input:
integrate(x*sech(x)^2/(a+b*tanh(x))^2,x, algorithm="fricas")
Output:
(2*(a + b)*x*cosh(x)^2 + 4*(a + b)*x*cosh(x)*sinh(x) + 2*(a + b)*x*sinh(x) ^2 - ((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))))/(a^3 - a^2*b - a*b^2 + b^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2 + 2*(a^3 + a^2*b - a*b ^2 - b^3)*cosh(x)*sinh(x) + (a^3 + a^2*b - a*b^2 - b^3)*sinh(x)^2)
\[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx=\int \frac {x \operatorname {sech}^{2}{\left (x \right )}}{\left (a + b \tanh {\left (x \right )}\right )^{2}}\, dx \] Input:
integrate(x*sech(x)**2/(a+b*tanh(x))**2,x)
Output:
Integral(x*sech(x)**2/(a + b*tanh(x))**2, x)
Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx=\frac {2 \, x e^{\left (2 \, x\right )}}{a^{2} - 2 \, a b + b^{2} + {\left (a^{2} - b^{2}\right )} e^{\left (2 \, x\right )}} - \frac {\log \left (\frac {{\left (a + b\right )} e^{\left (2 \, x\right )} + a - b}{a + b}\right )}{a^{2} - b^{2}} \] Input:
integrate(x*sech(x)^2/(a+b*tanh(x))^2,x, algorithm="maxima")
Output:
2*x*e^(2*x)/(a^2 - 2*a*b + b^2 + (a^2 - b^2)*e^(2*x)) - log(((a + b)*e^(2* x) + a - b)/(a + b))/(a^2 - b^2)
Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (55) = 110\).
Time = 0.13 (sec) , antiderivative size = 174, normalized size of antiderivative = 3.16 \[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx=\frac {2 \, a x e^{\left (2 \, x\right )} + 2 \, b x e^{\left (2 \, x\right )} - a e^{\left (2 \, x\right )} \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right ) - b e^{\left (2 \, x\right )} \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right ) - a \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right ) + b \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right )}{a^{3} e^{\left (2 \, x\right )} + a^{2} b e^{\left (2 \, x\right )} - a b^{2} e^{\left (2 \, x\right )} - b^{3} e^{\left (2 \, x\right )} + a^{3} - a^{2} b - a b^{2} + b^{3}} \] Input:
integrate(x*sech(x)^2/(a+b*tanh(x))^2,x, algorithm="giac")
Output:
(2*a*x*e^(2*x) + 2*b*x*e^(2*x) - a*e^(2*x)*log(-a*e^(2*x) - b*e^(2*x) - a + b) - b*e^(2*x)*log(-a*e^(2*x) - b*e^(2*x) - a + b) - a*log(-a*e^(2*x) - b*e^(2*x) - a + b) + b*log(-a*e^(2*x) - b*e^(2*x) - a + b))/(a^3*e^(2*x) + a^2*b*e^(2*x) - a*b^2*e^(2*x) - b^3*e^(2*x) + a^3 - a^2*b - a*b^2 + b^3)
Time = 2.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25 \[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx=\frac {2\,x}{a^2-b^2}-\frac {\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2}-\frac {2\,x}{\left (a+b\right )\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )} \] Input:
int(x/(cosh(x)^2*(a + b*tanh(x))^2),x)
Output:
(2*x)/(a^2 - b^2) - log(a - b + a*exp(2*x) + b*exp(2*x))/(a^2 - b^2) - (2* x)/((a + b)*(a - b + exp(2*x)*(a + b)))
Time = 0.27 (sec) , antiderivative size = 2513, normalized size of antiderivative = 45.69 \[ \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx =\text {Too large to display} \] Input:
int(x*sech(x)^2/(a+b*tanh(x))^2,x)
Output:
( - 18*e**(6*x)*log(e**(2*x)*a + e**(2*x)*b + a - b)*tanh(x)*a**4*b**2 - 2 4*e**(6*x)*log(e**(2*x)*a + e**(2*x)*b + a - b)*tanh(x)*a**3*b**3 - 6*e**( 6*x)*log(e**(2*x)*a + e**(2*x)*b + a - b)*tanh(x)*a**2*b**4 - 18*e**(6*x)* log(e**(2*x)*a + e**(2*x)*b + a - b)*a**5*b - 24*e**(6*x)*log(e**(2*x)*a + e**(2*x)*b + a - b)*a**4*b**2 - 6*e**(6*x)*log(e**(2*x)*a + e**(2*x)*b + a - b)*a**3*b**3 + 3*e**(6*x)*sech(x)**2*tanh(x)*a**4*b**2 + 4*e**(6*x)*se ch(x)**2*tanh(x)*a**3*b**3 - 2*e**(6*x)*sech(x)**2*tanh(x)*a**2*b**4 - 4*e **(6*x)*sech(x)**2*tanh(x)*a*b**5 - e**(6*x)*sech(x)**2*tanh(x)*b**6 - 18* e**(6*x)*sech(x)**2*a**6*x - 24*e**(6*x)*sech(x)**2*a**5*b*x - 9*e**(6*x)* sech(x)**2*a**5*b + 12*e**(6*x)*sech(x)**2*a**4*b**2*x - 12*e**(6*x)*sech( x)**2*a**4*b**2 + 24*e**(6*x)*sech(x)**2*a**3*b**3*x + 6*e**(6*x)*sech(x)* *2*a**3*b**3 + 6*e**(6*x)*sech(x)**2*a**2*b**4*x + 12*e**(6*x)*sech(x)**2* a**2*b**4 + 3*e**(6*x)*sech(x)**2*a*b**5 + 36*e**(6*x)*tanh(x)*a**4*b**2*x - 12*e**(6*x)*tanh(x)*a**4*b**2 + 48*e**(6*x)*tanh(x)*a**3*b**3*x - 8*e** (6*x)*tanh(x)*a**3*b**3 + 12*e**(6*x)*tanh(x)*a**2*b**4*x + 16*e**(6*x)*ta nh(x)*a**2*b**4 + 8*e**(6*x)*tanh(x)*a*b**5 - 4*e**(6*x)*tanh(x)*b**6 + 36 *e**(6*x)*a**5*b*x - 12*e**(6*x)*a**5*b + 48*e**(6*x)*a**4*b**2*x - 8*e**( 6*x)*a**4*b**2 + 12*e**(6*x)*a**3*b**3*x + 16*e**(6*x)*a**3*b**3 + 8*e**(6 *x)*a**2*b**4 - 4*e**(6*x)*a*b**5 - 54*e**(4*x)*log(e**(2*x)*a + e**(2*x)* b + a - b)*tanh(x)*a**4*b**2 - 36*e**(4*x)*log(e**(2*x)*a + e**(2*x)*b ...