3.63.45 \(\int \frac {-3-3 x-19 x^2}{3 x+19 x^2} \, dx\)

Optimal. Leaf size=22 \[ 1-x-\log \left (\frac {5 e^2}{19+\frac {3}{x}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1593, 893} \begin {gather*} -x-\log (x)+\log (19 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 - 3*x - 19*x^2)/(3*x + 19*x^2),x]

[Out]

-x - Log[x] + Log[3 + 19*x]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3-3 x-19 x^2}{x (3+19 x)} \, dx\\ &=\int \left (-1-\frac {1}{x}+\frac {19}{3+19 x}\right ) \, dx\\ &=-x-\log (x)+\log (3+19 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.64 \begin {gather*} -x-\log (x)+\log (3+19 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 - 3*x - 19*x^2)/(3*x + 19*x^2),x]

[Out]

-x - Log[x] + Log[3 + 19*x]

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fricas [A]  time = 0.70, size = 14, normalized size = 0.64 \begin {gather*} -x + \log \left (19 \, x + 3\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-19*x^2-3*x-3)/(19*x^2+3*x),x, algorithm="fricas")

[Out]

-x + log(19*x + 3) - log(x)

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giac [A]  time = 0.18, size = 16, normalized size = 0.73 \begin {gather*} -x + \log \left ({\left | 19 \, x + 3 \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-19*x^2-3*x-3)/(19*x^2+3*x),x, algorithm="giac")

[Out]

-x + log(abs(19*x + 3)) - log(abs(x))

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maple [A]  time = 0.35, size = 15, normalized size = 0.68

method	result	size
default	\(-x -\ln \relax (x )+\ln \left (19 x +3\right )\)	\(15\)
norman	\(-x -\ln \relax (x )+\ln \left (19 x +3\right )\)	\(15\)
risch	\(-x -\ln \relax (x )+\ln \left (19 x +3\right )\)	\(15\)
meijerg	\(\ln \left (1+\frac {19 x}{3}\right )-\ln \relax (x )-\ln \left (19\right )+\ln \relax (3)-x\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-19*x^2-3*x-3)/(19*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

-x-ln(x)+ln(19*x+3)

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maxima [A]  time = 0.34, size = 14, normalized size = 0.64 \begin {gather*} -x + \log \left (19 \, x + 3\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-19*x^2-3*x-3)/(19*x^2+3*x),x, algorithm="maxima")

[Out]

-x + log(19*x + 3) - log(x)

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mupad [B]  time = 4.08, size = 12, normalized size = 0.55 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {38\,x}{3}+1\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 19*x^2 + 3)/(3*x + 19*x^2),x)

[Out]

2*atanh((38*x)/3 + 1) - x

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sympy [A]  time = 0.09, size = 10, normalized size = 0.45 \begin {gather*} - x - \log {\relax (x )} + \log {\left (x + \frac {3}{19} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-19*x**2-3*x-3)/(19*x**2+3*x),x)

[Out]

-x - log(x) + log(x + 3/19)

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