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

Optimal. Leaf size=43 \[ -\frac {1089}{49 (1-2 x)}+\frac {1331}{112 (1-2 x)^2}-\frac {14289 \log (1-2 x)}{2744}-\frac {\log (3 x+2)}{1029} \]

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Rubi [A]  time = 0.02, antiderivative size = 43, 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 {1089}{49 (1-2 x)}+\frac {1331}{112 (1-2 x)^2}-\frac {14289 \log (1-2 x)}{2744}-\frac {\log (3 x+2)}{1029} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1331/(112*(1 - 2*x)^2) - 1089/(49*(1 - 2*x)) - (14289*Log[1 - 2*x])/2744 - Log[2 + 3*x]/1029

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 {(3+5 x)^3}{(1-2 x)^3 (2+3 x)} \, dx &=\int \left (-\frac {1331}{28 (-1+2 x)^3}-\frac {2178}{49 (-1+2 x)^2}-\frac {14289}{1372 (-1+2 x)}-\frac {1}{343 (2+3 x)}\right ) \, dx\\ &=\frac {1331}{112 (1-2 x)^2}-\frac {1089}{49 (1-2 x)}-\frac {14289 \log (1-2 x)}{2744}-\frac {\log (2+3 x)}{1029}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 35, normalized size = 0.81 \begin {gather*} \frac {\frac {2541 (288 x-67)}{(1-2 x)^2}-85734 \log (3-6 x)-16 \log (3 x+2)}{16464} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2541*(-67 + 288*x))/(1 - 2*x)^2 - 85734*Log[3 - 6*x] - 16*Log[2 + 3*x])/16464

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.68, size = 55, normalized size = 1.28 \begin {gather*} -\frac {16 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (3 \, x + 2\right ) + 85734 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 731808 \, x + 170247}{16464 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16464*(16*(4*x^2 - 4*x + 1)*log(3*x + 2) + 85734*(4*x^2 - 4*x + 1)*log(2*x - 1) - 731808*x + 170247)/(4*x^2
- 4*x + 1)

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giac [A]  time = 1.19, size = 33, normalized size = 0.77 \begin {gather*} \frac {121 \, {\left (288 \, x - 67\right )}}{784 \, {\left (2 \, x - 1\right )}^{2}} - \frac {1}{1029} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {14289}{2744} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/784*(288*x - 67)/(2*x - 1)^2 - 1/1029*log(abs(3*x + 2)) - 14289/2744*log(abs(2*x - 1))

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maple [A]  time = 0.01, size = 36, normalized size = 0.84 \begin {gather*} -\frac {14289 \ln \left (2 x -1\right )}{2744}-\frac {\ln \left (3 x +2\right )}{1029}+\frac {1331}{112 \left (2 x -1\right )^{2}}+\frac {1089}{49 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1029*ln(3*x+2)+1331/112/(2*x-1)^2+1089/49/(2*x-1)-14289/2744*ln(2*x-1)

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maxima [A]  time = 0.50, size = 36, normalized size = 0.84 \begin {gather*} \frac {121 \, {\left (288 \, x - 67\right )}}{784 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {1}{1029} \, \log \left (3 \, x + 2\right ) - \frac {14289}{2744} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/784*(288*x - 67)/(4*x^2 - 4*x + 1) - 1/1029*log(3*x + 2) - 14289/2744*log(2*x - 1)

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mupad [B]  time = 0.05, size = 29, normalized size = 0.67 \begin {gather*} \frac {\frac {1089\,x}{98}-\frac {8107}{3136}}{x^2-x+\frac {1}{4}}-\frac {\ln \left (x+\frac {2}{3}\right )}{1029}-\frac {14289\,\ln \left (x-\frac {1}{2}\right )}{2744} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1089*x)/98 - 8107/3136)/(x^2 - x + 1/4) - log(x + 2/3)/1029 - (14289*log(x - 1/2))/2744

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sympy [A]  time = 0.17, size = 34, normalized size = 0.79 \begin {gather*} - \frac {8107 - 34848 x}{3136 x^{2} - 3136 x + 784} - \frac {14289 \log {\left (x - \frac {1}{2} \right )}}{2744} - \frac {\log {\left (x + \frac {2}{3} \right )}}{1029} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8107 - 34848*x)/(3136*x**2 - 3136*x + 784) - 14289*log(x - 1/2)/2744 - log(x + 2/3)/1029

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