3.1.66 \(\int \frac {1}{4-6 \tanh (c+d x)} \, dx\) [66]

Optimal. Leaf size=31 \[ -\frac {x}{5}-\frac {3 \log (2 \cosh (c+d x)-3 \sinh (c+d x))}{10 d} \]

[Out]

-1/5*x-3/10*ln(2*cosh(d*x+c)-3*sinh(d*x+c))/d

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Rubi [A]
time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3565, 3611} \begin {gather*} -\frac {3 \log (2 \cosh (c+d x)-3 \sinh (c+d x))}{10 d}-\frac {x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 6*Tanh[c + d*x])^(-1),x]

[Out]

-1/5*x - (3*Log[2*Cosh[c + d*x] - 3*Sinh[c + d*x]])/(10*d)

Rule 3565

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 3611

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*Sin[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]

Rubi steps

\begin {align*} \int \frac {1}{4-6 \tanh (c+d x)} \, dx &=-\frac {x}{5}-\frac {3}{10} i \int \frac {6 i-4 i \tanh (c+d x)}{4-6 \tanh (c+d x)} \, dx\\ &=-\frac {x}{5}-\frac {3 \log (2 \cosh (c+d x)-3 \sinh (c+d x))}{10 d}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 55, normalized size = 1.77 \begin {gather*} -\frac {3 \log (4-6 \tanh (c+d x))}{10 d}+\frac {\log (6-6 \tanh (c+d x))}{4 d}+\frac {\log (6+6 \tanh (c+d x))}{20 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 6*Tanh[c + d*x])^(-1),x]

[Out]

(-3*Log[4 - 6*Tanh[c + d*x]])/(10*d) + Log[6 - 6*Tanh[c + d*x]]/(4*d) + Log[6 + 6*Tanh[c + d*x]]/(20*d)

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Maple [A]
time = 0.64, size = 42, normalized size = 1.35

method	result	size
risch	\(\frac {x}{10}+\frac {3 c}{5 d}-\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}-5\right )}{10 d}\)	\(28\)
derivativedivides	\(\frac {\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{10}-\frac {3 \ln \left (-2+3 \tanh \left (d x +c \right )\right )}{5}+\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}}{2 d}\)	\(42\)
default	\(\frac {\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{10}-\frac {3 \ln \left (-2+3 \tanh \left (d x +c \right )\right )}{5}+\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}}{2 d}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4-6*tanh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/2/d*(1/10*ln(tanh(d*x+c)+1)-3/5*ln(-2+3*tanh(d*x+c))+1/2*ln(tanh(d*x+c)-1))

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Maxima [A]
time = 0.28, size = 29, normalized size = 0.94 \begin {gather*} -\frac {1}{2} \, x - \frac {c}{2 \, d} - \frac {3 \, \log \left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-6*tanh(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*x - 1/2*c/d - 3/10*log(5*e^(-2*d*x - 2*c) - 1)/d

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Fricas [A]
time = 0.35, size = 48, normalized size = 1.55 \begin {gather*} \frac {d x - 3 \, \log \left (-\frac {2 \, {\left (2 \, \cosh \left (d x + c\right ) - 3 \, \sinh \left (d x + c\right )\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-6*tanh(d*x+c)),x, algorithm="fricas")

[Out]

1/10*(d*x - 3*log(-2*(2*cosh(d*x + c) - 3*sinh(d*x + c))/(cosh(d*x + c) - sinh(d*x + c))))/d

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Sympy [A]
time = 0.23, size = 42, normalized size = 1.35 \begin {gather*} \begin {cases} - \frac {x}{2} + \frac {3 \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{10 d} - \frac {3 \log {\left (3 \tanh {\left (c + d x \right )} - 2 \right )}}{10 d} & \text {for}\: d \neq 0 \\\frac {x}{4 - 6 \tanh {\left (c \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-6*tanh(d*x+c)),x)

[Out]

Piecewise((-x/2 + 3*log(tanh(c + d*x) + 1)/(10*d) - 3*log(3*tanh(c + d*x) - 2)/(10*d), Ne(d, 0)), (x/(4 - 6*ta
nh(c)), True))

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Giac [A]
time = 0.41, size = 25, normalized size = 0.81 \begin {gather*} \frac {d x + c - 3 \, \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 5 \right |}\right )}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-6*tanh(d*x+c)),x, algorithm="giac")

[Out]

1/10*(d*x + c - 3*log(abs(e^(2*d*x + 2*c) - 5)))/d

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Mupad [B]
time = 0.12, size = 33, normalized size = 1.06 \begin {gather*} \frac {\frac {3\,\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )}{10}-\frac {3\,\ln \left (3\,\mathrm {tanh}\left (c+d\,x\right )-2\right )}{10}}{d}-\frac {x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(6*tanh(c + d*x) - 4),x)

[Out]

((3*log(tanh(c + d*x) + 1))/10 - (3*log(3*tanh(c + d*x) - 2))/10)/d - x/2

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