3.3.22 \(\int \frac {1+2 x}{1+2 x+2 x^2} \, dx\) [222]

Optimal. Leaf size=15 \[ \frac {1}{2} \log \left (1+2 x+2 x^2\right ) \]

[Out]

1/2*ln(2*x^2+2*x+1)

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {642} \begin {gather*} \frac {1}{2} \log \left (2 x^2+2 x+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)/(1 + 2*x + 2*x^2),x]

[Out]

Log[1 + 2*x + 2*x^2]/2

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=\frac {1}{2} \log \left (1+2 x+2 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{2} \log \left (1+2 x+2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)/(1 + 2*x + 2*x^2),x]

[Out]

Log[1 + 2*x + 2*x^2]/2

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Maple [A]
time = 0.13, size = 14, normalized size = 0.93

method	result	size
parallelrisch	\(\frac {\ln \left (x^{2}+x +\frac {1}{2}\right )}{2}\)	\(10\)
default	\(\frac {\ln \left (2 x^{2}+2 x +1\right )}{2}\)	\(14\)
norman	\(\frac {\ln \left (2 x^{2}+2 x +1\right )}{2}\)	\(14\)
risch	\(\frac {\ln \left (2 x^{2}+2 x +1\right )}{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(2*x^2+2*x+1)

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Maxima [A]
time = 0.33, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(2*x^2 + 2*x + 1)

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Fricas [A]
time = 0.57, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(2*x^2 + 2*x + 1)

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Sympy [A]
time = 0.02, size = 12, normalized size = 0.80 \begin {gather*} \frac {\log {\left (2 x^{2} + 2 x + 1 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(2*x**2 + 2*x + 1)/2

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Giac [A]
time = 0.49, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(2*x^2 + 2*x + 1)

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Mupad [B]
time = 0.04, size = 9, normalized size = 0.60 \begin {gather*} \frac {\ln \left (x^2+x+\frac {1}{2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 1)/(2*x + 2*x^2 + 1),x)

[Out]

log(x + x^2 + 1/2)/2

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Chatgpt [F] Failed to verify
time = 1.00, size = 12, normalized size = 0.80 \begin {gather*} \frac {\ln \left (2 \left (x +\frac {1}{2}\right )^{2}+1\right )}{2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

int((1+2*x)/(2*x^2+2*x+1),x)

[Out]

1/2*ln(2*(x+1/2)^2+1)

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