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

Optimal. Leaf size=54 \[ -\frac {22}{343 (1-2 x)}-\frac {1}{343 (3 x+2)}+\frac {121}{196 (1-2 x)^2}+\frac {64 \log (1-2 x)}{2401}-\frac {64 \log (3 x+2)}{2401} \]

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Rubi [A]  time = 0.02, antiderivative size = 54, 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 {22}{343 (1-2 x)}-\frac {1}{343 (3 x+2)}+\frac {121}{196 (1-2 x)^2}+\frac {64 \log (1-2 x)}{2401}-\frac {64 \log (3 x+2)}{2401} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

121/(196*(1 - 2*x)^2) - 22/(343*(1 - 2*x)) - 1/(343*(2 + 3*x)) + (64*Log[1 - 2*x])/2401 - (64*Log[2 + 3*x])/24
01

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)^2}{(1-2 x)^3 (2+3 x)^2} \, dx &=\int \left (-\frac {121}{49 (-1+2 x)^3}-\frac {44}{343 (-1+2 x)^2}+\frac {128}{2401 (-1+2 x)}+\frac {3}{343 (2+3 x)^2}-\frac {192}{2401 (2+3 x)}\right ) \, dx\\ &=\frac {121}{196 (1-2 x)^2}-\frac {22}{343 (1-2 x)}-\frac {1}{343 (2+3 x)}+\frac {64 \log (1-2 x)}{2401}-\frac {64 \log (2+3 x)}{2401}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 47, normalized size = 0.87 \begin {gather*} \frac {\frac {7 \left (512 x^2+2645 x+1514\right )}{(1-2 x)^2 (3 x+2)}+256 \log (1-2 x)-256 \log (6 x+4)}{9604} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*(1514 + 2645*x + 512*x^2))/((1 - 2*x)^2*(2 + 3*x)) + 256*Log[1 - 2*x] - 256*Log[4 + 6*x])/9604

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.58, size = 75, normalized size = 1.39 \begin {gather*} \frac {3584 \, x^{2} - 256 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 256 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 18515 \, x + 10598}{9604 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9604*(3584*x^2 - 256*(12*x^3 - 4*x^2 - 5*x + 2)*log(3*x + 2) + 256*(12*x^3 - 4*x^2 - 5*x + 2)*log(2*x - 1) +
18515*x + 10598)/(12*x^3 - 4*x^2 - 5*x + 2)

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giac [A]  time = 1.10, size = 51, normalized size = 0.94 \begin {gather*} -\frac {1}{343 \, {\left (3 \, x + 2\right )}} + \frac {33 \, {\left (\frac {203}{3 \, x + 2} - 25\right )}}{2401 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}^{2}} + \frac {64}{2401} \, \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)^2/(1-2*x)^3/(2+3*x)^2,x, algorithm="giac")

[Out]

-1/343/(3*x + 2) + 33/2401*(203/(3*x + 2) - 25)/(7/(3*x + 2) - 2)^2 + 64/2401*log(abs(-7/(3*x + 2) + 2))

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maple [A]  time = 0.01, size = 45, normalized size = 0.83 \begin {gather*} \frac {64 \ln \left (2 x -1\right )}{2401}-\frac {64 \ln \left (3 x +2\right )}{2401}-\frac {1}{343 \left (3 x +2\right )}+\frac {121}{196 \left (2 x -1\right )^{2}}+\frac {22}{343 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/343/(3*x+2)-64/2401*ln(3*x+2)+121/196/(2*x-1)^2+22/343/(2*x-1)+64/2401*ln(2*x-1)

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maxima [A]  time = 0.71, size = 46, normalized size = 0.85 \begin {gather*} \frac {512 \, x^{2} + 2645 \, x + 1514}{1372 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} - \frac {64}{2401} \, \log \left (3 \, x + 2\right ) + \frac {64}{2401} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1372*(512*x^2 + 2645*x + 1514)/(12*x^3 - 4*x^2 - 5*x + 2) - 64/2401*log(3*x + 2) + 64/2401*log(2*x - 1)

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mupad [B]  time = 0.04, size = 38, normalized size = 0.70 \begin {gather*} -\frac {128\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{2401}-\frac {\frac {32\,x^2}{1029}+\frac {2645\,x}{16464}+\frac {757}{8232}}{-x^3+\frac {x^2}{3}+\frac {5\,x}{12}-\frac {1}{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (128*atanh((12*x)/7 + 1/7))/2401 - ((2645*x)/16464 + (32*x^2)/1029 + 757/8232)/((5*x)/12 + x^2/3 - x^3 - 1/6
)

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sympy [A]  time = 0.17, size = 46, normalized size = 0.85 \begin {gather*} - \frac {- 512 x^{2} - 2645 x - 1514}{16464 x^{3} - 5488 x^{2} - 6860 x + 2744} + \frac {64 \log {\left (x - \frac {1}{2} \right )}}{2401} - \frac {64 \log {\left (x + \frac {2}{3} \right )}}{2401} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-512*x**2 - 2645*x - 1514)/(16464*x**3 - 5488*x**2 - 6860*x + 2744) + 64*log(x - 1/2)/2401 - 64*log(x + 2/3)
/2401

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