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

Optimal. Leaf size=45 \[ \frac {50 x^2}{27}-\frac {20 x}{3}+\frac {518}{243 (3 x+2)}-\frac {49}{486 (3 x+2)^2}+\frac {503}{81} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 45, 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 {50 x^2}{27}-\frac {20 x}{3}+\frac {518}{243 (3 x+2)}-\frac {49}{486 (3 x+2)^2}+\frac {503}{81} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-20*x)/3 + (50*x^2)/27 - 49/(486*(2 + 3*x)^2) + 518/(243*(2 + 3*x)) + (503*Log[2 + 3*x])/81

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)^2 (3+5 x)^2}{(2+3 x)^3} \, dx &=\int \left (-\frac {20}{3}+\frac {100 x}{27}+\frac {49}{81 (2+3 x)^3}-\frac {518}{81 (2+3 x)^2}+\frac {503}{27 (2+3 x)}\right ) \, dx\\ &=-\frac {20 x}{3}+\frac {50 x^2}{27}-\frac {49}{486 (2+3 x)^2}+\frac {518}{243 (2+3 x)}+\frac {503}{81} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.93 \begin {gather*} \frac {503}{81} \log (3 x+2)-\frac {-900 x^4+2040 x^3+6480 x^2+4508 x+913}{54 (3 x+2)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/54*(913 + 4508*x + 6480*x^2 + 2040*x^3 - 900*x^4)/(2 + 3*x)^2 + (503*Log[2 + 3*x])/81

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.72, size = 52, normalized size = 1.16 \begin {gather*} \frac {8100 \, x^{4} - 18360 \, x^{3} - 35280 \, x^{2} + 3018 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 9852 \, x + 2023}{486 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/486*(8100*x^4 - 18360*x^3 - 35280*x^2 + 3018*(9*x^2 + 12*x + 4)*log(3*x + 2) - 9852*x + 2023)/(9*x^2 + 12*x
+ 4)

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giac [A]  time = 0.81, size = 32, normalized size = 0.71 \begin {gather*} \frac {50}{27} \, x^{2} - \frac {20}{3} \, x + \frac {7 \, {\left (444 \, x + 289\right )}}{486 \, {\left (3 \, x + 2\right )}^{2}} + \frac {503}{81} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/27*x^2 - 20/3*x + 7/486*(444*x + 289)/(3*x + 2)^2 + 503/81*log(abs(3*x + 2))

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maple [A]  time = 0.01, size = 36, normalized size = 0.80 \begin {gather*} \frac {50 x^{2}}{27}-\frac {20 x}{3}+\frac {503 \ln \left (3 x +2\right )}{81}-\frac {49}{486 \left (3 x +2\right )^{2}}+\frac {518}{243 \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-20/3*x+50/27*x^2-49/486/(3*x+2)^2+518/243/(3*x+2)+503/81*ln(3*x+2)

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maxima [A]  time = 0.53, size = 36, normalized size = 0.80 \begin {gather*} \frac {50}{27} \, x^{2} - \frac {20}{3} \, x + \frac {7 \, {\left (444 \, x + 289\right )}}{486 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {503}{81} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/27*x^2 - 20/3*x + 7/486*(444*x + 289)/(9*x^2 + 12*x + 4) + 503/81*log(3*x + 2)

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mupad [B]  time = 0.03, size = 31, normalized size = 0.69 \begin {gather*} \frac {503\,\ln \left (x+\frac {2}{3}\right )}{81}-\frac {20\,x}{3}+\frac {\frac {518\,x}{729}+\frac {2023}{4374}}{x^2+\frac {4\,x}{3}+\frac {4}{9}}+\frac {50\,x^2}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(503*log(x + 2/3))/81 - (20*x)/3 + ((518*x)/729 + 2023/4374)/((4*x)/3 + x^2 + 4/9) + (50*x^2)/27

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sympy [A]  time = 0.13, size = 36, normalized size = 0.80 \begin {gather*} \frac {50 x^{2}}{27} - \frac {20 x}{3} + \frac {3108 x + 2023}{4374 x^{2} + 5832 x + 1944} + \frac {503 \log {\left (3 x + 2 \right )}}{81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50*x**2/27 - 20*x/3 + (3108*x + 2023)/(4374*x**2 + 5832*x + 1944) + 503*log(3*x + 2)/81

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