3.1540 \(\int \frac{d+e x}{9+12 x+4 x^2} \, dx\)

Optimal. Leaf size=30 \[ \frac{1}{4} e \log (2 x+3)-\frac{2 d-3 e}{4 (2 x+3)} \]

[Out]

-(2*d - 3*e)/(4*(3 + 2*x)) + (e*Log[3 + 2*x])/4

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Rubi [A]  time = 0.0168794, antiderivative size = 30, 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 = {27, 43} \[ \frac{1}{4} e \log (2 x+3)-\frac{2 d-3 e}{4 (2 x+3)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(9 + 12*x + 4*x^2),x]

[Out]

-(2*d - 3*e)/(4*(3 + 2*x)) + (e*Log[3 + 2*x])/4

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{d+e x}{9+12 x+4 x^2} \, dx &=\int \frac{d+e x}{(3+2 x)^2} \, dx\\ &=\int \left (\frac{2 d-3 e}{2 (3+2 x)^2}+\frac{e}{2 (3+2 x)}\right ) \, dx\\ &=-\frac{2 d-3 e}{4 (3+2 x)}+\frac{1}{4} e \log (3+2 x)\\ \end{align*}

Mathematica [A]  time = 0.0066698, size = 30, normalized size = 1. \[ \frac{3 e-2 d}{4 (2 x+3)}+\frac{1}{4} e \log (2 x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(9 + 12*x + 4*x^2),x]

[Out]

(-2*d + 3*e)/(4*(3 + 2*x)) + (e*Log[3 + 2*x])/4

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Maple [A]  time = 0.044, size = 31, normalized size = 1. \begin{align*}{\frac{e\ln \left ( 3+2\,x \right ) }{4}}-{\frac{d}{6+4\,x}}+{\frac{3\,e}{12+8\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)/(4*x^2+12*x+9),x)

[Out]

1/4*e*ln(3+2*x)-1/2/(3+2*x)*d+3/4*e/(3+2*x)

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Maxima [A]  time = 1.10739, size = 35, normalized size = 1.17 \begin{align*} \frac{1}{4} \, e \log \left (2 \, x + 3\right ) - \frac{2 \, d - 3 \, e}{4 \,{\left (2 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x^2+12*x+9),x, algorithm="maxima")

[Out]

1/4*e*log(2*x + 3) - 1/4*(2*d - 3*e)/(2*x + 3)

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Fricas [A]  time = 1.37845, size = 76, normalized size = 2.53 \begin{align*} \frac{{\left (2 \, e x + 3 \, e\right )} \log \left (2 \, x + 3\right ) - 2 \, d + 3 \, e}{4 \,{\left (2 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x^2+12*x+9),x, algorithm="fricas")

[Out]

1/4*((2*e*x + 3*e)*log(2*x + 3) - 2*d + 3*e)/(2*x + 3)

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Sympy [A]  time = 0.335154, size = 20, normalized size = 0.67 \begin{align*} \frac{e \log{\left (2 x + 3 \right )}}{4} - \frac{2 d - 3 e}{8 x + 12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x**2+12*x+9),x)

[Out]

e*log(2*x + 3)/4 - (2*d - 3*e)/(8*x + 12)

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Giac [A]  time = 1.13272, size = 39, normalized size = 1.3 \begin{align*} \frac{1}{4} \, e \log \left ({\left | 2 \, x + 3 \right |}\right ) - \frac{2 \, d - 3 \, e}{4 \,{\left (2 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x^2+12*x+9),x, algorithm="giac")

[Out]

1/4*e*log(abs(2*x + 3)) - 1/4*(2*d - 3*e)/(2*x + 3)