3.143 \(\int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx\)

Optimal. Leaf size=55 \[ \frac {a x}{b \left (a^2-b^2\right )}-\frac {\log (a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {x}{b (a+b \tanh (x))} \]

[Out]

a*x/b/(a^2-b^2)-ln(a*cosh(x)+b*sinh(x))/(a^2-b^2)-x/b/(a+b*tanh(x))

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Rubi [A]  time = 0.09, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5466, 3484, 3530} \[ \frac {a x}{b \left (a^2-b^2\right )}-\frac {\log (a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {x}{b (a+b \tanh (x))} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sech[x]^2)/(a + b*Tanh[x])^2,x]

[Out]

(a*x)/(b*(a^2 - b^2)) - Log[a*Cosh[x] + b*Sinh[x]]/(a^2 - b^2) - x/(b*(a + b*Tanh[x]))

Rule 3484

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[(a*x)/(a^2 + b^2), x] + Dist[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]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), 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]

Rule 5466

Int[((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^2*((a_) + (b_.)*Tanh[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Simp[((e + f*x)^m*(a + b*Tanh[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(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]

Rubi steps

\begin {align*} \int \frac {x \text {sech}^2(x)}{(a+b \tanh (x))^2} \, dx &=-\frac {x}{b (a+b \tanh (x))}+\frac {\int \frac {1}{a+b \tanh (x)} \, dx}{b}\\ &=\frac {a x}{b \left (a^2-b^2\right )}-\frac {x}{b (a+b \tanh (x))}-\frac {i \int \frac {-i b-i a \tanh (x)}{a+b \tanh (x)} \, dx}{a^2-b^2}\\ &=\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))}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 49, normalized size = 0.89 \[ \frac {b x-a \log (a \cosh (x)+b \sinh (x))}{a^3-a b^2}+\frac {x \sinh (x)}{a^2 \cosh (x)+a b \sinh (x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sech[x]^2)/(a + b*Tanh[x])^2,x]

[Out]

(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])

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fricas [B]  time = 0.96, size = 182, normalized size = 3.31 \[ \frac {2 \, {\left (a + b\right )} x \cosh \relax (x)^{2} + 4 \, {\left (a + b\right )} x \cosh \relax (x) \sinh \relax (x) + 2 \, {\left (a + b\right )} x \sinh \relax (x)^{2} - {\left ({\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} + a - b\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{3} - a^{2} b - a b^{2} + b^{3} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \sinh \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(x)^2/(a+b*tanh(x))^2,x, algorithm="fricas")

[Out]

(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)

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giac [B]  time = 0.13, size = 174, normalized size = 3.16 \[ \frac {2 \, a x e^{\left (2 \, x\right )} + 2 \, b x e^{\left (2 \, x\right )} - a e^{\left (2 \, x\right )} \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right ) - b e^{\left (2 \, x\right )} \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right ) - a \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right ) + b \log \left (-a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} - a + b\right )}{a^{3} e^{\left (2 \, x\right )} + a^{2} b e^{\left (2 \, x\right )} - a b^{2} e^{\left (2 \, x\right )} - b^{3} e^{\left (2 \, x\right )} + a^{3} - a^{2} b - a b^{2} + b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(x)^2/(a+b*tanh(x))^2,x, algorithm="giac")

[Out]

(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)

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maple [A]  time = 0.28, size = 73, normalized size = 1.33 \[ \frac {2 x}{a^{2}-b^{2}}-\frac {2 x}{\left (a \,{\mathrm e}^{2 x}+b \,{\mathrm e}^{2 x}+a -b \right ) \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{2}-b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sech(x)^2/(a+b*tanh(x))^2,x)

[Out]

2/(a^2-b^2)*x-2*x/(a*exp(2*x)+b*exp(2*x)+a-b)/(a+b)-1/(a^2-b^2)*ln(exp(2*x)+(a-b)/(a+b))

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maxima [A]  time = 0.59, size = 68, normalized size = 1.24 \[ \frac {2 \, x e^{\left (2 \, x\right )}}{a^{2} - 2 \, a b + b^{2} + {\left (a^{2} - b^{2}\right )} e^{\left (2 \, x\right )}} - \frac {\log \left (\frac {{\left (a + b\right )} e^{\left (2 \, x\right )} + a - b}{a + b}\right )}{a^{2} - b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(x)^2/(a+b*tanh(x))^2,x, algorithm="maxima")

[Out]

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)

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mupad [B]  time = 1.15, size = 69, normalized size = 1.25 \[ \frac {2\,x}{a^2-b^2}-\frac {\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2}-\frac {2\,x}{\left (a+b\right )\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(cosh(x)^2*(a + b*tanh(x))^2),x)

[Out]

(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
)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \operatorname {sech}^{2}{\relax (x )}}{\left (a + b \tanh {\relax (x )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(x)**2/(a+b*tanh(x))**2,x)

[Out]

Integral(x*sech(x)**2/(a + b*tanh(x))**2, x)

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