3.547 \(\int \frac{1}{d f+(e f+d g) x+e g x^2} \, dx\)

Optimal. Leaf size=36 \[ \frac{\log (d+e x)}{e f-d g}-\frac{\log (f+g x)}{e f-d g} \]

[Out]

Log[d + e*x]/(e*f - d*g) - Log[f + g*x]/(e*f - d*g)

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Rubi [A]  time = 0.0145575, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {616, 31} \[ \frac{\log (d+e x)}{e f-d g}-\frac{\log (f+g x)}{e f-d g} \]

Antiderivative was successfully verified.

[In]

Int[(d*f + (e*f + d*g)*x + e*g*x^2)^(-1),x]

[Out]

Log[d + e*x]/(e*f - d*g) - Log[f + g*x]/(e*f - d*g)

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{1}{d f+(e f+d g) x+e g x^2} \, dx &=-\frac{(e g) \int \frac{1}{e f+e g x} \, dx}{e f-d g}+\frac{(e g) \int \frac{1}{d g+e g x} \, dx}{e f-d g}\\ &=\frac{\log (d+e x)}{e f-d g}-\frac{\log (f+g x)}{e f-d g}\\ \end{align*}

Mathematica [A]  time = 0.0103459, size = 26, normalized size = 0.72 \[ \frac{\log (d+e x)-\log (f+g x)}{e f-d g} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*f + (e*f + d*g)*x + e*g*x^2)^(-1),x]

[Out]

(Log[d + e*x] - Log[f + g*x])/(e*f - d*g)

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Maple [A]  time = 0.007, size = 37, normalized size = 1. \begin{align*} -{\frac{\ln \left ( ex+d \right ) }{dg-fe}}+{\frac{\ln \left ( gx+f \right ) }{dg-fe}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*f+(d*g+e*f)*x+e*g*x^2),x)

[Out]

-1/(d*g-e*f)*ln(e*x+d)+1/(d*g-e*f)*ln(g*x+f)

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Maxima [A]  time = 0.974656, size = 49, normalized size = 1.36 \begin{align*} \frac{\log \left (e x + d\right )}{e f - d g} - \frac{\log \left (g x + f\right )}{e f - d g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="maxima")

[Out]

log(e*x + d)/(e*f - d*g) - log(g*x + f)/(e*f - d*g)

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Fricas [A]  time = 1.61521, size = 58, normalized size = 1.61 \begin{align*} \frac{\log \left (e x + d\right ) - \log \left (g x + f\right )}{e f - d g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="fricas")

[Out]

(log(e*x + d) - log(g*x + f))/(e*f - d*g)

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Sympy [B]  time = 0.307197, size = 128, normalized size = 3.56 \begin{align*} \frac{\log{\left (x + \frac{- \frac{d^{2} g^{2}}{d g - e f} + \frac{2 d e f g}{d g - e f} + d g - \frac{e^{2} f^{2}}{d g - e f} + e f}{2 e g} \right )}}{d g - e f} - \frac{\log{\left (x + \frac{\frac{d^{2} g^{2}}{d g - e f} - \frac{2 d e f g}{d g - e f} + d g + \frac{e^{2} f^{2}}{d g - e f} + e f}{2 e g} \right )}}{d g - e f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*f+(d*g+e*f)*x+e*g*x**2),x)

[Out]

log(x + (-d**2*g**2/(d*g - e*f) + 2*d*e*f*g/(d*g - e*f) + d*g - e**2*f**2/(d*g - e*f) + e*f)/(2*e*g))/(d*g - e
*f) - log(x + (d**2*g**2/(d*g - e*f) - 2*d*e*f*g/(d*g - e*f) + d*g + e**2*f**2/(d*g - e*f) + e*f)/(2*e*g))/(d*
g - e*f)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError