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

Optimal. Leaf size=63 \[ -\frac {a^2 \log (a \cosh (x)+b \sinh (x))}{b \left (a^2-b^2\right )}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}-\frac {a x}{b^2}+\frac {\log (\cosh (x))}{b} \]

[Out]

-a*x/b^2+a^3*x/b^2/(a^2-b^2)+ln(cosh(x))/b-a^2*ln(a*cosh(x)+b*sinh(x))/b/(a^2-b^2)

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Rubi [A]  time = 0.09, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3541, 3475, 3484, 3530} \[ \frac {a^3 x}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \log (a \cosh (x)+b \sinh (x))}{b \left (a^2-b^2\right )}-\frac {a x}{b^2}+\frac {\log (\cosh (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + b*Tanh[x]),x]

[Out]

-((a*x)/b^2) + (a^3*x)/(b^2*(a^2 - b^2)) + Log[Cosh[x]]/b - (a^2*Log[a*Cosh[x] + b*Sinh[x]])/(b*(a^2 - b^2))

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, 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 3541

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

Rubi steps

\begin {align*} \int \frac {\tanh ^2(x)}{a+b \tanh (x)} \, dx &=-\frac {a x}{b^2}+\frac {a^2 \int \frac {1}{a+b \tanh (x)} \, dx}{b^2}+\frac {\int \tanh (x) \, dx}{b}\\ &=-\frac {a x}{b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}+\frac {\log (\cosh (x))}{b}-\frac {\left (i a^2\right ) \int \frac {-i b-i a \tanh (x)}{a+b \tanh (x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {a x}{b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}+\frac {\log (\cosh (x))}{b}-\frac {a^2 \log (a \cosh (x)+b \sinh (x))}{b \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 49, normalized size = 0.78 \[ \frac {-a^2 \log (a \cosh (x)+b \sinh (x))+a^2 \log (\cosh (x))+a b x-b^2 \log (\cosh (x))}{a^2 b-b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*x + a^2*Log[Cosh[x]] - b^2*Log[Cosh[x]] - a^2*Log[a*Cosh[x] + b*Sinh[x]])/(a^2*b - b^3)

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fricas [A]  time = 0.60, size = 76, normalized size = 1.21 \[ -\frac {a^{2} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left (a b + b^{2}\right )} x - {\left (a^{2} - b^{2}\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} b - b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))) - (a*b + b^2)*x - (a^2 - b^2)*log(2*cosh(x)/(cosh(x)
- sinh(x))))/(a^2*b - b^3)

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giac [A]  time = 0.14, size = 58, normalized size = 0.92 \[ -\frac {a^{2} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} b - b^{3}} + \frac {x}{a - b} + \frac {\log \left (e^{\left (2 \, x\right )} + 1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^2*b - b^3) + x/(a - b) + log(e^(2*x) + 1)/b

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maple [A]  time = 0.05, size = 60, normalized size = 0.95 \[ -\frac {\ln \left (\tanh \relax (x )-1\right )}{2 b +2 a}+\frac {\ln \left (1+\tanh \relax (x )\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \tanh \relax (x )\right )}{\left (a +b \right ) \left (a -b \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 56, normalized size = 0.89 \[ -\frac {a^{2} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} b - b^{3}} + \frac {x}{a + b} + \frac {\log \left (e^{\left (-2 \, x\right )} + 1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*log(-(a - b)*e^(-2*x) - a - b)/(a^2*b - b^3) + x/(a + b) + log(e^(-2*x) + 1)/b

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mupad [B]  time = 0.12, size = 46, normalized size = 0.73 \[ -\frac {b^2\,\left (x-\ln \left (\mathrm {tanh}\relax (x)+1\right )\right )+a^2\,\ln \left (a+b\,\mathrm {tanh}\relax (x)\right )-a\,b\,x}{b\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a + b*tanh(x)),x)

[Out]

-(b^2*(x - log(tanh(x) + 1)) + a^2*log(a + b*tanh(x)) - a*b*x)/(b*(a^2 - b^2))

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sympy [A]  time = 0.64, size = 243, normalized size = 3.86 \[ \begin {cases} \tilde {\infty } \left (x - \log {\left (\tanh {\relax (x )} + 1 \right )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {3 x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} - \frac {3 x}{2 b \tanh {\relax (x )} - 2 b} - \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} + \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )}}{2 b \tanh {\relax (x )} - 2 b} + \frac {1}{2 b \tanh {\relax (x )} - 2 b} & \text {for}\: a = - b \\\frac {x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} + \frac {x}{2 b \tanh {\relax (x )} + 2 b} - \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} - \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )}}{2 b \tanh {\relax (x )} + 2 b} - \frac {1}{2 b \tanh {\relax (x )} + 2 b} & \text {for}\: a = b \\\frac {x - \tanh {\relax (x )}}{a} & \text {for}\: b = 0 \\- \frac {a^{2} \log {\left (\frac {a}{b} + \tanh {\relax (x )} \right )}}{a^{2} b - b^{3}} + \frac {a b x}{a^{2} b - b^{3}} - \frac {b^{2} x}{a^{2} b - b^{3}} + \frac {b^{2} \log {\left (\tanh {\relax (x )} + 1 \right )}}{a^{2} b - b^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(x - log(tanh(x) + 1)), Eq(a, 0) & Eq(b, 0)), (3*x*tanh(x)/(2*b*tanh(x) - 2*b) - 3*x/(2*b*tanh(
x) - 2*b) - 2*log(tanh(x) + 1)*tanh(x)/(2*b*tanh(x) - 2*b) + 2*log(tanh(x) + 1)/(2*b*tanh(x) - 2*b) + 1/(2*b*t
anh(x) - 2*b), Eq(a, -b)), (x*tanh(x)/(2*b*tanh(x) + 2*b) + x/(2*b*tanh(x) + 2*b) - 2*log(tanh(x) + 1)*tanh(x)
/(2*b*tanh(x) + 2*b) - 2*log(tanh(x) + 1)/(2*b*tanh(x) + 2*b) - 1/(2*b*tanh(x) + 2*b), Eq(a, b)), ((x - tanh(x
))/a, Eq(b, 0)), (-a**2*log(a/b + tanh(x))/(a**2*b - b**3) + a*b*x/(a**2*b - b**3) - b**2*x/(a**2*b - b**3) +
b**2*log(tanh(x) + 1)/(a**2*b - b**3), True))

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