3.2.63 \(\int \frac {7}{-12+5 x+2 x^2} \, dx\) [163]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, 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, 630, 31} \begin {gather*} \frac {7}{11} \log (3-2 x)-\frac {7}{11} \log (x+4) \end {gather*}

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 31

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

Rule 630

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]

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*}

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Mathematica [A]
time = 0.00, size = 21, normalized size = 1.11 \begin {gather*} 7 \left (\frac {1}{11} \log (3-2 x)-\frac {1}{11} \log (4+x)\right ) \end {gather*}

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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Mathics [A]
time = 1.75, size = 13, normalized size = 0.68 \begin {gather*} \frac {-7 \text {Log}\left [4+x\right ]}{11}+\frac {7 \text {Log}\left [-\frac {3}{2}+x\right ]}{11} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[7/(-12 + 5*x + 2*x^2),x]')

[Out]

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

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Maple [A]
time = 0.06, size = 16, normalized size = 0.84

method	result	size
default	\(\frac {7 \ln \left (2 x -3\right )}{11}-\frac {7 \ln \left (4+x \right )}{11}\)	\(16\)
norman	\(\frac {7 \ln \left (2 x -3\right )}{11}-\frac {7 \ln \left (4+x \right )}{11}\)	\(16\)
risch	\(\frac {7 \ln \left (2 x -3\right )}{11}-\frac {7 \ln \left (4+x \right )}{11}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 15, normalized size = 0.79 \begin {gather*} \frac {7}{11} \, \log \left (2 \, x - 3\right ) - \frac {7}{11} \, \log \left (x + 4\right ) \end {gather*}

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 = 0.32, size = 15, normalized size = 0.79 \begin {gather*} \frac {7}{11} \, \log \left (2 \, x - 3\right ) - \frac {7}{11} \, \log \left (x + 4\right ) \end {gather*}

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 = 0.05, size = 17, normalized size = 0.89 \begin {gather*} \frac {7 \log {\left (x - \frac {3}{2} \right )}}{11} - \frac {7 \log {\left (x + 4 \right )}}{11} \end {gather*}

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 = 0.00, size = 22, normalized size = 1.16 \begin {gather*} 7 \left (-\frac {\ln \left |x+4\right |}{11}+\frac {\ln \left |2 x-3\right |}{11}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.12, size = 8, normalized size = 0.42 \begin {gather*} -\frac {14\,\mathrm {atanh}\left (\frac {4\,x}{11}+\frac {5}{11}\right )}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(7/(5*x + 2*x^2 - 12),x)

[Out]

-(14*atanh((4*x)/11 + 5/11))/11

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