3.3.57 \(\int \frac {1}{x^2 (4+6 x)} \, dx\) [257]

Optimal. Leaf size=24 \[ -\frac {1}{4 x}-\frac {3 \log (x)}{8}+\frac {3}{8} \log (2+3 x) \]

[Out]

-1/4/x-3/8*ln(x)+3/8*ln(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 24, 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}{4 x}-\frac {3 \log (x)}{8}+\frac {3}{8} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(4 + 6*x)),x]

[Out]

-1/4*1/x - (3*Log[x])/8 + (3*Log[2 + 3*x])/8

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^2 (4+6 x)} \, dx &=\int \left (\frac {1}{4 x^2}-\frac {3}{8 x}+\frac {9}{8 (2+3 x)}\right ) \, dx\\ &=-\frac {1}{4 x}-\frac {3 \log (x)}{8}+\frac {3}{8} \log (2+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 24, normalized size = 1.00 \begin {gather*} -\frac {1}{4 x}-\frac {3 \log (x)}{8}+\frac {3}{8} \log (2+3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(4 + 6*x)),x]

[Out]

-1/4*1/x - (3*Log[x])/8 + (3*Log[2 + 3*x])/8

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Maple [A]
time = 0.09, size = 19, normalized size = 0.79

method	result	size
default	\(-\frac {1}{4 x}-\frac {3 \ln \left (x \right )}{8}+\frac {3 \ln \left (2+3 x \right )}{8}\)	\(19\)
norman	\(-\frac {1}{4 x}-\frac {3 \ln \left (x \right )}{8}+\frac {3 \ln \left (2+3 x \right )}{8}\)	\(19\)
risch	\(-\frac {1}{4 x}-\frac {3 \ln \left (x \right )}{8}+\frac {3 \ln \left (2+3 x \right )}{8}\)	\(19\)
meijerg	\(-\frac {1}{4 x}-\frac {3 \ln \left (x \right )}{8}+\frac {3 \ln \left (2\right )}{8}-\frac {3 \ln \left (3\right )}{8}+\frac {3 \ln \left (1+\frac {3 x}{2}\right )}{8}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/x-3/8*ln(x)+3/8*ln(2+3*x)

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Maxima [A]
time = 0.27, size = 18, normalized size = 0.75 \begin {gather*} -\frac {1}{4 \, x} + \frac {3}{8} \, \log \left (3 \, x + 2\right ) - \frac {3}{8} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/x + 3/8*log(3*x + 2) - 3/8*log(x)

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Fricas [A]
time = 0.94, size = 21, normalized size = 0.88 \begin {gather*} \frac {3 \, x \log \left (3 \, x + 2\right ) - 3 \, x \log \left (x\right ) - 2}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.03, size = 20, normalized size = 0.83 \begin {gather*} - \frac {3 \log {\left (x \right )}}{8} + \frac {3 \log {\left (x + \frac {2}{3} \right )}}{8} - \frac {1}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x)/8 + 3*log(x + 2/3)/8 - 1/(4*x)

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Giac [A]
time = 1.47, size = 20, normalized size = 0.83 \begin {gather*} -\frac {1}{4 \, x} + \frac {3}{8} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {3}{8} \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/x + 3/8*log(abs(3*x + 2)) - 3/8*log(abs(x))

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Mupad [B]
time = 0.05, size = 18, normalized size = 0.75 \begin {gather*} -\frac {3\,\ln \left (\frac {x}{6\,x+4}\right )}{8}-\frac {1}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(6*x + 4)),x)

[Out]

- (3*log(x/(6*x + 4)))/8 - 1/(4*x)

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