3.1476 \(\int \frac{(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx\)

Optimal. Leaf size=33 \[ -\frac{125 x^2}{12}-\frac{1225 x}{36}-\frac{1331}{56} \log (1-2 x)-\frac{1}{189} \log (3 x+2) \]

[Out]

(-1225*x)/36 - (125*x^2)/12 - (1331*Log[1 - 2*x])/56 - Log[2 + 3*x]/189

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Rubi [A]  time = 0.0132522, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {72} \[ -\frac{125 x^2}{12}-\frac{1225 x}{36}-\frac{1331}{56} \log (1-2 x)-\frac{1}{189} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^3/((1 - 2*x)*(2 + 3*x)),x]

[Out]

(-1225*x)/36 - (125*x^2)/12 - (1331*Log[1 - 2*x])/56 - Log[2 + 3*x]/189

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx &=\int \left (-\frac{1225}{36}-\frac{125 x}{6}-\frac{1331}{28 (-1+2 x)}-\frac{1}{63 (2+3 x)}\right ) \, dx\\ &=-\frac{1225 x}{36}-\frac{125 x^2}{12}-\frac{1331}{56} \log (1-2 x)-\frac{1}{189} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.014913, size = 35, normalized size = 1.06 \[ \frac{-1050 \left (15 x^2+49 x+24\right )-35937 \log (5-10 x)-8 \log (5 (3 x+2))}{1512} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^3/((1 - 2*x)*(2 + 3*x)),x]

[Out]

(-1050*(24 + 49*x + 15*x^2) - 35937*Log[5 - 10*x] - 8*Log[5*(2 + 3*x)])/1512

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Maple [A]  time = 0.006, size = 26, normalized size = 0.8 \begin{align*} -{\frac{125\,{x}^{2}}{12}}-{\frac{1225\,x}{36}}-{\frac{1331\,\ln \left ( 2\,x-1 \right ) }{56}}-{\frac{\ln \left ( 2+3\,x \right ) }{189}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3+5*x)^3/(1-2*x)/(2+3*x),x)

[Out]

-125/12*x^2-1225/36*x-1331/56*ln(2*x-1)-1/189*ln(2+3*x)

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Maxima [A]  time = 1.10596, size = 34, normalized size = 1.03 \begin{align*} -\frac{125}{12} \, x^{2} - \frac{1225}{36} \, x - \frac{1}{189} \, \log \left (3 \, x + 2\right ) - \frac{1331}{56} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/12*x^2 - 1225/36*x - 1/189*log(3*x + 2) - 1331/56*log(2*x - 1)

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Fricas [A]  time = 1.26176, size = 93, normalized size = 2.82 \begin{align*} -\frac{125}{12} \, x^{2} - \frac{1225}{36} \, x - \frac{1}{189} \, \log \left (3 \, x + 2\right ) - \frac{1331}{56} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/12*x^2 - 1225/36*x - 1/189*log(3*x + 2) - 1331/56*log(2*x - 1)

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Sympy [A]  time = 0.121128, size = 31, normalized size = 0.94 \begin{align*} - \frac{125 x^{2}}{12} - \frac{1225 x}{36} - \frac{1331 \log{\left (x - \frac{1}{2} \right )}}{56} - \frac{\log{\left (x + \frac{2}{3} \right )}}{189} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125*x**2/12 - 1225*x/36 - 1331*log(x - 1/2)/56 - log(x + 2/3)/189

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Giac [A]  time = 1.48798, size = 36, normalized size = 1.09 \begin{align*} -\frac{125}{12} \, x^{2} - \frac{1225}{36} \, x - \frac{1}{189} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac{1331}{56} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/12*x^2 - 1225/36*x - 1/189*log(abs(3*x + 2)) - 1331/56*log(abs(2*x - 1))