3.3.15 \(\int \frac {1+2 x}{-7+12 x+4 x^2} \, dx\) [215]

Optimal. Leaf size=21 \[ \frac {1}{8} \log (1-2 x)+\frac {3}{8} \log (7+2 x) \]

[Out]

1/8*ln(1-2*x)+3/8*ln(7+2*x)

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Rubi [A]
time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {646, 31} \begin {gather*} \frac {1}{8} \log (1-2 x)+\frac {3}{8} \log (2 x+7) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)/(-7 + 12*x + 4*x^2),x]

[Out]

Log[1 - 2*x]/8 + (3*Log[7 + 2*x])/8

Rule 31

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

Rule 646

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

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 21, normalized size = 1.00 \begin {gather*} \frac {1}{8} \log (1-2 x)+\frac {3}{8} \log (7+2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)/(-7 + 12*x + 4*x^2),x]

[Out]

Log[1 - 2*x]/8 + (3*Log[7 + 2*x])/8

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Mathics [A]
time = 1.85, size = 13, normalized size = 0.62 \begin {gather*} \frac {\text {Log}\left [-\frac {1}{2}+x\right ]}{8}+\frac {3 \text {Log}\left [\frac {7}{2}+x\right ]}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(2*x + 1)/(4*x^2 + 12*x - 7),x]')

[Out]

Log[-1 / 2 + x] / 8 + 3 Log[7 / 2 + x] / 8

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Maple [A]
time = 0.06, size = 18, normalized size = 0.86

method	result	size
default	\(\frac {\ln \left (2 x -1\right )}{8}+\frac {3 \ln \left (7+2 x \right )}{8}\)	\(18\)
norman	\(\frac {\ln \left (2 x -1\right )}{8}+\frac {3 \ln \left (7+2 x \right )}{8}\)	\(18\)
risch	\(\frac {\ln \left (2 x -1\right )}{8}+\frac {3 \ln \left (7+2 x \right )}{8}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x)/(4*x^2+12*x-7),x,method=_RETURNVERBOSE)

[Out]

1/8*ln(2*x-1)+3/8*ln(7+2*x)

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Maxima [A]
time = 0.26, size = 17, normalized size = 0.81 \begin {gather*} \frac {3}{8} \, \log \left (2 \, x + 7\right ) + \frac {1}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)/(4*x^2+12*x-7),x, algorithm="maxima")

[Out]

3/8*log(2*x + 7) + 1/8*log(2*x - 1)

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Fricas [A]
time = 0.33, size = 17, normalized size = 0.81 \begin {gather*} \frac {3}{8} \, \log \left (2 \, x + 7\right ) + \frac {1}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)/(4*x^2+12*x-7),x, algorithm="fricas")

[Out]

3/8*log(2*x + 7) + 1/8*log(2*x - 1)

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Sympy [A]
time = 0.05, size = 17, normalized size = 0.81 \begin {gather*} \frac {\log {\left (x - \frac {1}{2} \right )}}{8} + \frac {3 \log {\left (x + \frac {7}{2} \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)/(4*x**2+12*x-7),x)

[Out]

log(x - 1/2)/8 + 3*log(x + 7/2)/8

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Giac [A]
time = 0.00, size = 22, normalized size = 1.05 \begin {gather*} \frac {\ln \left |2 x-1\right |}{8}+\frac {3}{8} \ln \left |2 x+7\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)/(4*x^2+12*x-7),x)

[Out]

3/8*log(abs(2*x + 7)) + 1/8*log(abs(2*x - 1))

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Mupad [B]
time = 0.20, size = 13, normalized size = 0.62 \begin {gather*} \frac {\ln \left (x-\frac {1}{2}\right )}{8}+\frac {3\,\ln \left (x+\frac {7}{2}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 1)/(12*x + 4*x^2 - 7),x)

[Out]

log(x - 1/2)/8 + (3*log(x + 7/2))/8

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