3.3.61 \(\int \frac {1}{x (4+6 x)^2} \, dx\) [261]

Optimal. Leaf size=28 \[ \frac {1}{8 (2+3 x)}+\frac {\log (x)}{16}-\frac {1}{16} \log (2+3 x) \]

[Out]

1/8/(2+3*x)+1/16*ln(x)-1/16*ln(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \begin {gather*} \frac {1}{8 (3 x+2)}+\frac {\log (x)}{16}-\frac {1}{16} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(4 + 6*x)^2),x]

[Out]

1/(8*(2 + 3*x)) + Log[x]/16 - Log[2 + 3*x]/16

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {1}{x (4+6 x)^2} \, dx &=\int \left (\frac {1}{16 x}-\frac {3}{8 (2+3 x)^2}-\frac {3}{16 (2+3 x)}\right ) \, dx\\ &=\frac {1}{8 (2+3 x)}+\frac {\log (x)}{16}-\frac {1}{16} \log (2+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 26, normalized size = 0.93 \begin {gather*} \frac {1}{16} \left (\frac {2}{2+3 x}+\log (-6 x)-\log (4+6 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(4 + 6*x)^2),x]

[Out]

(2/(2 + 3*x) + Log[-6*x] - Log[4 + 6*x])/16

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Mathics [A]
time = 1.67, size = 26, normalized size = 0.93 \begin {gather*} \frac {2+\left (2+3 x\right ) \left (\text {Log}\left [x\right ]-\text {Log}\left [\frac {2}{3}+x\right ]\right )}{32+48 x} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(x^1*(4 + 6*x)^2),x]')

[Out]

(2 + (2 + 3 x) (Log[x] - Log[2 / 3 + x])) / (16 (2 + 3 x))

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Maple [A]
time = 0.09, size = 23, normalized size = 0.82

method	result	size
risch	\(\frac {1}{16+24 x}+\frac {\ln \left (x \right )}{16}-\frac {\ln \left (2+3 x \right )}{16}\)	\(21\)
default	\(\frac {1}{16+24 x}+\frac {\ln \left (x \right )}{16}-\frac {\ln \left (2+3 x \right )}{16}\)	\(23\)
norman	\(-\frac {3 x}{16 \left (2+3 x \right )}+\frac {\ln \left (x \right )}{16}-\frac {\ln \left (2+3 x \right )}{16}\)	\(24\)
meijerg	\(\frac {1}{16}+\frac {\ln \left (x \right )}{16}-\frac {\ln \left (2\right )}{16}+\frac {\ln \left (3\right )}{16}-\frac {3 x}{16 \left (2+3 x \right )}-\frac {\ln \left (1+\frac {3 x}{2}\right )}{16}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(4+6*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/8/(2+3*x)+1/16*ln(x)-1/16*ln(2+3*x)

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Maxima [A]
time = 0.24, size = 22, normalized size = 0.79 \begin {gather*} \frac {1}{8 \, {\left (3 \, x + 2\right )}} - \frac {1}{16} \, \log \left (3 \, x + 2\right ) + \frac {1}{16} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^2,x, algorithm="maxima")

[Out]

1/8/(3*x + 2) - 1/16*log(3*x + 2) + 1/16*log(x)

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Fricas [A]
time = 0.31, size = 32, normalized size = 1.14 \begin {gather*} -\frac {{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - {\left (3 \, x + 2\right )} \log \left (x\right ) - 2}{16 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^2,x, algorithm="fricas")

[Out]

-1/16*((3*x + 2)*log(3*x + 2) - (3*x + 2)*log(x) - 2)/(3*x + 2)

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Sympy [A]
time = 0.06, size = 19, normalized size = 0.68 \begin {gather*} \frac {\log {\left (x \right )}}{16} - \frac {\log {\left (x + \frac {2}{3} \right )}}{16} + \frac {1}{24 x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)**2,x)

[Out]

log(x)/16 - log(x + 2/3)/16 + 1/(24*x + 16)

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Giac [A]
time = 0.00, size = 29, normalized size = 1.04 \begin {gather*} \frac {\ln \left |x\right |}{16}-\frac {\ln \left |3 x+2\right |}{16}+\frac {\frac {1}{16}\cdot 2}{3 x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^2,x)

[Out]

1/8/(3*x + 2) - 1/16*log(abs(3*x + 2)) + 1/16*log(abs(x))

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Mupad [B]
time = 0.06, size = 20, normalized size = 0.71 \begin {gather*} \frac {1}{8\,\left (3\,x+2\right )}-\frac {\ln \left (\frac {6\,x+4}{x}\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(6*x + 4)^2),x)

[Out]

1/(8*(3*x + 2)) - log((6*x + 4)/x)/16

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