3.3.58 \(\int \frac {1}{x^3 (4+6 x)} \, dx\) [258]

Optimal. Leaf size=31 \[ -\frac {1}{8 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (2+3 x) \]

[Out]

-1/8/x^2+3/8/x+9/16*ln(x)-9/16*ln(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 31, 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 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(4 + 6*x)),x]

[Out]

-1/8*1/x^2 + 3/(8*x) + (9*Log[x])/16 - (9*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^3 (4+6 x)} \, dx &=\int \left (\frac {1}{4 x^3}-\frac {3}{8 x^2}+\frac {9}{16 x}-\frac {27}{16 (2+3 x)}\right ) \, dx\\ &=-\frac {1}{8 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (2+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 31, normalized size = 1.00 \begin {gather*} -\frac {1}{8 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (2+3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(4 + 6*x)),x]

[Out]

-1/8*1/x^2 + 3/(8*x) + (9*Log[x])/16 - (9*Log[2 + 3*x])/16

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Maple [A]
time = 0.09, size = 24, normalized size = 0.77

method	result	size
norman	\(\frac {-\frac {1}{8}+\frac {3 x}{8}}{x^{2}}+\frac {9 \ln \left (x \right )}{16}-\frac {9 \ln \left (2+3 x \right )}{16}\)	\(23\)
risch	\(\frac {-\frac {1}{8}+\frac {3 x}{8}}{x^{2}}+\frac {9 \ln \left (x \right )}{16}-\frac {9 \ln \left (2+3 x \right )}{16}\)	\(23\)
default	\(-\frac {1}{8 x^{2}}+\frac {3}{8 x}+\frac {9 \ln \left (x \right )}{16}-\frac {9 \ln \left (2+3 x \right )}{16}\)	\(24\)
meijerg	\(-\frac {1}{8 x^{2}}+\frac {3}{8 x}+\frac {9 \ln \left (x \right )}{16}-\frac {9 \ln \left (2\right )}{16}+\frac {9 \ln \left (3\right )}{16}-\frac {9 \ln \left (1+\frac {3 x}{2}\right )}{16}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(4+6*x),x,method=_RETURNVERBOSE)

[Out]

-1/8/x^2+3/8/x+9/16*ln(x)-9/16*ln(2+3*x)

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Maxima [A]
time = 0.29, size = 23, normalized size = 0.74 \begin {gather*} \frac {3 \, x - 1}{8 \, x^{2}} - \frac {9}{16} \, \log \left (3 \, x + 2\right ) + \frac {9}{16} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(4+6*x),x, algorithm="maxima")

[Out]

1/8*(3*x - 1)/x^2 - 9/16*log(3*x + 2) + 9/16*log(x)

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Fricas [A]
time = 0.78, size = 28, normalized size = 0.90 \begin {gather*} -\frac {9 \, x^{2} \log \left (3 \, x + 2\right ) - 9 \, x^{2} \log \left (x\right ) - 6 \, x + 2}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(4+6*x),x, algorithm="fricas")

[Out]

-1/16*(9*x^2*log(3*x + 2) - 9*x^2*log(x) - 6*x + 2)/x^2

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Sympy [A]
time = 0.04, size = 26, normalized size = 0.84 \begin {gather*} \frac {9 \log {\left (x \right )}}{16} - \frac {9 \log {\left (x + \frac {2}{3} \right )}}{16} + \frac {3 x - 1}{8 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(4+6*x),x)

[Out]

9*log(x)/16 - 9*log(x + 2/3)/16 + (3*x - 1)/(8*x**2)

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Giac [A]
time = 1.27, size = 25, normalized size = 0.81 \begin {gather*} \frac {3 \, x - 1}{8 \, x^{2}} - \frac {9}{16} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {9}{16} \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(4+6*x),x, algorithm="giac")

[Out]

1/8*(3*x - 1)/x^2 - 9/16*log(abs(3*x + 2)) + 9/16*log(abs(x))

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Mupad [B]
time = 0.04, size = 18, normalized size = 0.58 \begin {gather*} \frac {\frac {3\,x}{8}-\frac {1}{8}}{x^2}-\frac {9\,\mathrm {atanh}\left (3\,x+1\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(6*x + 4)),x)

[Out]

((3*x)/8 - 1/8)/x^2 - (9*atanh(3*x + 1))/8

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