3.137 \(\int \frac {\tanh (x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=39 \[ \frac {a \log (a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {b x}{a^2-b^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3531, 3530} \[ \frac {a \log (a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {b x}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Tanh[x]),x]

[Out]

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

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 3531

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] + Dist[(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] && NeQ[a*c + b*d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 29, normalized size = 0.74 \[ \frac {a \log (a \cosh (x)+b \sinh (x))-b x}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(a + b*Tanh[x]),x]

[Out]

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

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fricas [A]  time = 0.56, size = 43, normalized size = 1.10 \[ -\frac {{\left (a + b\right )} x - a \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} - b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 43, normalized size = 1.10 \[ \frac {a \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} - b^{2}} - \frac {x}{a - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 55, normalized size = 1.41 \[ -\frac {\ln \left (\tanh \relax (x )-1\right )}{2 b +2 a}-\frac {\ln \left (1+\tanh \relax (x )\right )}{2 a -2 b}+\frac {a \ln \left (a +b \tanh \relax (x )\right )}{\left (a +b \right ) \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(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/(a+b)/(a-b)*ln(a+b*tanh(x))

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maxima [A]  time = 0.33, size = 40, normalized size = 1.03 \[ \frac {a \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} - b^{2}} + \frac {x}{a + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.06, size = 36, normalized size = 0.92 \[ -\frac {b\,x-a\,\left (x-\ln \left (\mathrm {tanh}\relax (x)+1\right )+\ln \left (a+b\,\mathrm {tanh}\relax (x)\right )\right )}{a^2-b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a + b*tanh(x)),x)

[Out]

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

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sympy [A]  time = 0.51, size = 141, normalized size = 3.62 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\\frac {x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} - \frac {x}{2 b \tanh {\relax (x )} - 2 b} + \frac {1}{2 b \tanh {\relax (x )} - 2 b} & \text {for}\: a = - b \\\frac {x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} + \frac {x}{2 b \tanh {\relax (x )} + 2 b} + \frac {1}{2 b \tanh {\relax (x )} + 2 b} & \text {for}\: a = b \\\frac {x - \log {\left (\tanh {\relax (x )} + 1 \right )}}{a} & \text {for}\: b = 0 \\\frac {a x}{a^{2} - b^{2}} + \frac {a \log {\left (\frac {a}{b} + \tanh {\relax (x )} \right )}}{a^{2} - b^{2}} - \frac {a \log {\left (\tanh {\relax (x )} + 1 \right )}}{a^{2} - b^{2}} - \frac {b x}{a^{2} - b^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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