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

Optimal. Leaf size=43 \[ \frac {1331}{196 (1-2 x)}+\frac {1}{441 (3 x+2)}+\frac {4719 \log (1-2 x)}{1372}+\frac {101 \log (3 x+2)}{3087} \]

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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 {1331}{196 (1-2 x)}+\frac {1}{441 (3 x+2)}+\frac {4719 \log (1-2 x)}{1372}+\frac {101 \log (3 x+2)}{3087} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1331/(196*(1 - 2*x)) + 1/(441*(2 + 3*x)) + (4719*Log[1 - 2*x])/1372 + (101*Log[2 + 3*x])/3087

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)^2 (2+3 x)^2} \, dx &=\int \left (\frac {1331}{98 (-1+2 x)^2}+\frac {4719}{686 (-1+2 x)}-\frac {1}{147 (2+3 x)^2}+\frac {101}{1029 (2+3 x)}\right ) \, dx\\ &=\frac {1331}{196 (1-2 x)}+\frac {1}{441 (2+3 x)}+\frac {4719 \log (1-2 x)}{1372}+\frac {101 \log (2+3 x)}{3087}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 40, normalized size = 0.93 \begin {gather*} \frac {-\frac {7 (35929 x+23962)}{6 x^2+x-2}+42471 \log (5-10 x)+404 \log (5 (3 x+2))}{12348} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-7*(23962 + 35929*x))/(-2 + x + 6*x^2) + 42471*Log[5 - 10*x] + 404*Log[5*(2 + 3*x)])/12348

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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)^2 (2+3 x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.34, size = 49, normalized size = 1.14 \begin {gather*} \frac {404 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (3 \, x + 2\right ) + 42471 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (2 \, x - 1\right ) - 251503 \, x - 167734}{12348 \, {\left (6 \, x^{2} + x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12348*(404*(6*x^2 + x - 2)*log(3*x + 2) + 42471*(6*x^2 + x - 2)*log(2*x - 1) - 251503*x - 167734)/(6*x^2 + x
- 2)

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giac [A]  time = 1.05, size = 58, normalized size = 1.35 \begin {gather*} \frac {1}{441 \, {\left (3 \, x + 2\right )}} + \frac {3993}{686 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}} - \frac {125}{36} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) + \frac {4719}{1372} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/441/(3*x + 2) + 3993/686/(7/(3*x + 2) - 2) - 125/36*log(1/3*abs(3*x + 2)/(3*x + 2)^2) + 4719/1372*log(abs(-7
/(3*x + 2) + 2))

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maple [A]  time = 0.01, size = 36, normalized size = 0.84 \begin {gather*} \frac {4719 \ln \left (2 x -1\right )}{1372}+\frac {101 \ln \left (3 x +2\right )}{3087}+\frac {1}{1323 x +882}-\frac {1331}{196 \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)^2/(3*x+2)^2,x)

[Out]

1/441/(3*x+2)+101/3087*ln(3*x+2)-1331/196/(2*x-1)+4719/1372*ln(2*x-1)

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maxima [A]  time = 0.49, size = 34, normalized size = 0.79 \begin {gather*} -\frac {35929 \, x + 23962}{1764 \, {\left (6 \, x^{2} + x - 2\right )}} + \frac {101}{3087} \, \log \left (3 \, x + 2\right ) + \frac {4719}{1372} \, \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)^2/(2+3*x)^2,x, algorithm="maxima")

[Out]

-1/1764*(35929*x + 23962)/(6*x^2 + x - 2) + 101/3087*log(3*x + 2) + 4719/1372*log(2*x - 1)

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mupad [B]  time = 0.05, size = 30, normalized size = 0.70 \begin {gather*} \frac {4719\,\ln \left (x-\frac {1}{2}\right )}{1372}+\frac {101\,\ln \left (x+\frac {2}{3}\right )}{3087}-\frac {\frac {35929\,x}{10584}+\frac {11981}{5292}}{x^2+\frac {x}{6}-\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4719*log(x - 1/2))/1372 + (101*log(x + 2/3))/3087 - ((35929*x)/10584 + 11981/5292)/(x/6 + x^2 - 1/3)

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sympy [A]  time = 0.16, size = 36, normalized size = 0.84 \begin {gather*} \frac {- 35929 x - 23962}{10584 x^{2} + 1764 x - 3528} + \frac {4719 \log {\left (x - \frac {1}{2} \right )}}{1372} + \frac {101 \log {\left (x + \frac {2}{3} \right )}}{3087} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-35929*x - 23962)/(10584*x**2 + 1764*x - 3528) + 4719*log(x - 1/2)/1372 + 101*log(x + 2/3)/3087

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