### 3.215 $$\int \frac{1+2 x}{-7+12 x+4 x^2} \, dx$$

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

[Out]

Log[1 - 2*x]/8 + (3*Log[7 + 2*x])/8

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Rubi [A]  time = 0.0054708, antiderivative size = 21, normalized size of antiderivative = 1., 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 = {632, 31} $\frac{1}{8} \log (1-2 x)+\frac{3}{8} \log (2 x+7)$

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 632

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]

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{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*}

Mathematica [A]  time = 0.0040953, size = 21, normalized size = 1. $\frac{1}{8} \log (1-2 x)+\frac{3}{8} \log (2 x+7)$

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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Maple [A]  time = 0.005, size = 18, normalized size = 0.9 \begin{align*}{\frac{3\,\ln \left ( 7+2\,x \right ) }{8}}+{\frac{\ln \left ( 2\,x-1 \right ) }{8}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.943763, size = 23, normalized size = 1.1 \begin{align*} \frac{3}{8} \, \log \left (2 \, x + 7\right ) + \frac{1}{8} \, \log \left (2 \, x - 1\right ) \end{align*}

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 = 1.86837, size = 50, normalized size = 2.38 \begin{align*} \frac{3}{8} \, \log \left (2 \, x + 7\right ) + \frac{1}{8} \, \log \left (2 \, x - 1\right ) \end{align*}

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 = 0.122183, size = 17, normalized size = 0.81 \begin{align*} \frac{\log{\left (x - \frac{1}{2} \right )}}{8} + \frac{3 \log{\left (x + \frac{7}{2} \right )}}{8} \end{align*}

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 = 1.05618, size = 26, normalized size = 1.24 \begin{align*} \frac{3}{8} \, \log \left ({\left | 2 \, x + 7 \right |}\right ) + \frac{1}{8} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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