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

Optimal. Leaf size=37 \[ -\frac {8 x}{75}-\frac {1331}{125 (5 x+3)}+\frac {343}{9} \log (3 x+2)-\frac {4719}{125} \log (5 x+3) \]

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 = {88} \begin {gather*} -\frac {8 x}{75}-\frac {1331}{125 (5 x+3)}+\frac {343}{9} \log (3 x+2)-\frac {4719}{125} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*x)/75 - 1331/(125*(3 + 5*x)) + (343*Log[2 + 3*x])/9 - (4719*Log[3 + 5*x])/125

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx &=\int \left (-\frac {8}{75}+\frac {343}{3 (2+3 x)}+\frac {1331}{25 (3+5 x)^2}-\frac {4719}{25 (3+5 x)}\right ) \, dx\\ &=-\frac {8 x}{75}-\frac {1331}{125 (3+5 x)}+\frac {343}{9} \log (2+3 x)-\frac {4719}{125} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.97 \begin {gather*} \frac {-120 x-\frac {11979}{5 x+3}+42875 \log (3 x+2)-42471 \log (-3 (5 x+3))-80}{1125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-80 - 120*x - 11979/(3 + 5*x) + 42875*Log[2 + 3*x] - 42471*Log[-3*(3 + 5*x)])/1125

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.29, size = 45, normalized size = 1.22 \begin {gather*} -\frac {600 \, x^{2} + 42471 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 42875 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 360 \, x + 11979}{1125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1125*(600*x^2 + 42471*(5*x + 3)*log(5*x + 3) - 42875*(5*x + 3)*log(3*x + 2) + 360*x + 11979)/(5*x + 3)

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giac [A]  time = 1.07, size = 47, normalized size = 1.27 \begin {gather*} -\frac {8}{75} \, x - \frac {1331}{125 \, {\left (5 \, x + 3\right )}} - \frac {404}{1125} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) + \frac {343}{9} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) - \frac {8}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/75*x - 1331/125/(5*x + 3) - 404/1125*log(1/5*abs(5*x + 3)/(5*x + 3)^2) + 343/9*log(abs(-1/(5*x + 3) - 3)) -
8/125

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maple [A]  time = 0.01, size = 30, normalized size = 0.81 \begin {gather*} -\frac {8 x}{75}+\frac {343 \ln \left (3 x +2\right )}{9}-\frac {4719 \ln \left (5 x +3\right )}{125}-\frac {1331}{125 \left (5 x +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/75*x-1331/125/(5*x+3)+343/9*ln(3*x+2)-4719/125*ln(5*x+3)

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maxima [A]  time = 0.69, size = 29, normalized size = 0.78 \begin {gather*} -\frac {8}{75} \, x - \frac {1331}{125 \, {\left (5 \, x + 3\right )}} - \frac {4719}{125} \, \log \left (5 \, x + 3\right ) + \frac {343}{9} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/75*x - 1331/125/(5*x + 3) - 4719/125*log(5*x + 3) + 343/9*log(3*x + 2)

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mupad [B]  time = 0.04, size = 25, normalized size = 0.68 \begin {gather*} \frac {343\,\ln \left (x+\frac {2}{3}\right )}{9}-\frac {8\,x}{75}-\frac {4719\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {1331}{625\,\left (x+\frac {3}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(343*log(x + 2/3))/9 - (8*x)/75 - (4719*log(x + 3/5))/125 - 1331/(625*(x + 3/5))

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sympy [A]  time = 0.15, size = 31, normalized size = 0.84 \begin {gather*} - \frac {8 x}{75} - \frac {4719 \log {\left (x + \frac {3}{5} \right )}}{125} + \frac {343 \log {\left (x + \frac {2}{3} \right )}}{9} - \frac {1331}{625 x + 375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*x/75 - 4719*log(x + 3/5)/125 + 343*log(x + 2/3)/9 - 1331/(625*x + 375)

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