3.1.81 \(\int \frac {1}{2+5 x+3 x^2} \, dx\) [81]

Optimal. Leaf size=13 \[ -\log (1+x)+\log (2+3 x) \]

[Out]

-ln(1+x)+ln(2+3*x)

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Rubi [A]
time = 0.00, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {630, 31} \begin {gather*} \log (3 x+2)-\log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 5*x + 3*x^2)^(-1),x]

[Out]

-Log[1 + x] + Log[2 + 3*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 {1}{2+5 x+3 x^2} \, dx &=3 \int \frac {1}{2+3 x} \, dx-3 \int \frac {1}{3+3 x} \, dx\\ &=-\log (1+x)+\log (2+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} -\log (1+x)+\log (2+3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 5*x + 3*x^2)^(-1),x]

[Out]

-Log[1 + x] + Log[2 + 3*x]

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Maple [A]
time = 0.45, size = 14, normalized size = 1.08

method	result	size
default	\(-\ln \left (x +1\right )+\ln \left (2+3 x \right )\)	\(14\)
norman	\(-\ln \left (x +1\right )+\ln \left (2+3 x \right )\)	\(14\)
risch	\(-\ln \left (x +1\right )+\ln \left (2+3 x \right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2+5*x+2),x,method=_RETURNVERBOSE)

[Out]

-ln(x+1)+ln(2+3*x)

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Maxima [A]
time = 0.28, size = 13, normalized size = 1.00 \begin {gather*} \log \left (3 \, x + 2\right ) - \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+5*x+2),x, algorithm="maxima")

[Out]

log(3*x + 2) - log(x + 1)

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Fricas [A]
time = 1.31, size = 13, normalized size = 1.00 \begin {gather*} \log \left (3 \, x + 2\right ) - \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+5*x+2),x, algorithm="fricas")

[Out]

log(3*x + 2) - log(x + 1)

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Sympy [A]
time = 0.03, size = 10, normalized size = 0.77 \begin {gather*} \log {\left (x + \frac {2}{3} \right )} - \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x**2+5*x+2),x)

[Out]

log(x + 2/3) - log(x + 1)

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Giac [A]
time = 0.63, size = 15, normalized size = 1.15 \begin {gather*} \log \left ({\left | 3 \, x + 2 \right |}\right ) - \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+5*x+2),x, algorithm="giac")

[Out]

log(abs(3*x + 2)) - log(abs(x + 1))

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Mupad [B]
time = 0.08, size = 8, normalized size = 0.62 \begin {gather*} -2\,\mathrm {atanh}\left (6\,x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x + 3*x^2 + 2),x)

[Out]

-2*atanh(6*x + 5)

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