3.163 \(\int \frac{7}{-12+5 x+2 x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac{7}{11} \log (3-2 x)-\frac{7}{11} \log (x+4) \]

[Out]

(7*Log[3 - 2*x])/11 - (7*Log[4 + x])/11

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Rubi [A]  time = 0.0066137, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {12, 616, 31} \[ \frac{7}{11} \log (3-2 x)-\frac{7}{11} \log (x+4) \]

Antiderivative was successfully verified.

[In]

Int[7/(-12 + 5*x + 2*x^2),x]

[Out]

(7*Log[3 - 2*x])/11 - (7*Log[4 + x])/11

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{7}{-12+5 x+2 x^2} \, dx &=7 \int \frac{1}{-12+5 x+2 x^2} \, dx\\ &=\frac{14}{11} \int \frac{1}{-3+2 x} \, dx-\frac{14}{11} \int \frac{1}{8+2 x} \, dx\\ &=\frac{7}{11} \log (3-2 x)-\frac{7}{11} \log (4+x)\\ \end{align*}

Mathematica [A]  time = 0.0033927, size = 21, normalized size = 1.11 \[ 7 \left (\frac{1}{11} \log (3-2 x)-\frac{1}{11} \log (x+4)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[7/(-12 + 5*x + 2*x^2),x]

[Out]

7*(Log[3 - 2*x]/11 - Log[4 + x]/11)

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Maple [A]  time = 0.006, size = 16, normalized size = 0.8 \begin{align*}{\frac{7\,\ln \left ( -3+2\,x \right ) }{11}}-{\frac{7\,\ln \left ( 4+x \right ) }{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(7/(2*x^2+5*x-12),x)

[Out]

7/11*ln(-3+2*x)-7/11*ln(4+x)

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Maxima [A]  time = 0.919794, size = 20, normalized size = 1.05 \begin{align*} \frac{7}{11} \, \log \left (2 \, x - 3\right ) - \frac{7}{11} \, \log \left (x + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(7/(2*x^2+5*x-12),x, algorithm="maxima")

[Out]

7/11*log(2*x - 3) - 7/11*log(x + 4)

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Fricas [A]  time = 1.88224, size = 50, normalized size = 2.63 \begin{align*} \frac{7}{11} \, \log \left (2 \, x - 3\right ) - \frac{7}{11} \, \log \left (x + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(7/(2*x^2+5*x-12),x, algorithm="fricas")

[Out]

7/11*log(2*x - 3) - 7/11*log(x + 4)

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Sympy [A]  time = 0.090972, size = 17, normalized size = 0.89 \begin{align*} \frac{7 \log{\left (x - \frac{3}{2} \right )}}{11} - \frac{7 \log{\left (x + 4 \right )}}{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(7/(2*x**2+5*x-12),x)

[Out]

7*log(x - 3/2)/11 - 7*log(x + 4)/11

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Giac [A]  time = 1.05879, size = 23, normalized size = 1.21 \begin{align*} \frac{7}{11} \, \log \left ({\left | 2 \, x - 3 \right |}\right ) - \frac{7}{11} \, \log \left ({\left | x + 4 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(7/(2*x^2+5*x-12),x, algorithm="giac")

[Out]

7/11*log(abs(2*x - 3)) - 7/11*log(abs(x + 4))