3.12.26 \(\int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^4} \, dx\)

Optimal. Leaf size=49 \[ \frac {100 x}{81}-\frac {503}{81 (3 x+2)}+\frac {259}{243 (3 x+2)^2}-\frac {49}{729 (3 x+2)^3}-\frac {740}{243} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 49, 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 {100 x}{81}-\frac {503}{81 (3 x+2)}+\frac {259}{243 (3 x+2)^2}-\frac {49}{729 (3 x+2)^3}-\frac {740}{243} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)^2*(3 + 5*x)^2)/(2 + 3*x)^4,x]

[Out]

(100*x)/81 - 49/(729*(2 + 3*x)^3) + 259/(243*(2 + 3*x)^2) - 503/(81*(2 + 3*x)) - (740*Log[2 + 3*x])/243

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)^4} \, dx &=\int \left (\frac {100}{81}+\frac {49}{81 (2+3 x)^4}-\frac {518}{81 (2+3 x)^3}+\frac {503}{27 (2+3 x)^2}-\frac {740}{81 (2+3 x)}\right ) \, dx\\ &=\frac {100 x}{81}-\frac {49}{729 (2+3 x)^3}+\frac {259}{243 (2+3 x)^2}-\frac {503}{81 (2+3 x)}-\frac {740}{243} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 0.94 \begin {gather*} \frac {24300 x^4+64800 x^3+24057 x^2-23193 x-2220 (3 x+2)^3 \log (3 x+2)-11803}{729 (3 x+2)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)^2*(3 + 5*x)^2)/(2 + 3*x)^4,x]

[Out]

(-11803 - 23193*x + 24057*x^2 + 64800*x^3 + 24300*x^4 - 2220*(2 + 3*x)^3*Log[2 + 3*x])/(729*(2 + 3*x)^3)

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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)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((1 - 2*x)^2*(3 + 5*x)^2)/(2 + 3*x)^4,x]

[Out]

IntegrateAlgebraic[((1 - 2*x)^2*(3 + 5*x)^2)/(2 + 3*x)^4, x]

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fricas [A]  time = 1.07, size = 62, normalized size = 1.27 \begin {gather*} \frac {24300 \, x^{4} + 48600 \, x^{3} - 8343 \, x^{2} - 2220 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) - 44793 \, x - 16603}{729 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\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)^4,x, algorithm="fricas")

[Out]

1/729*(24300*x^4 + 48600*x^3 - 8343*x^2 - 2220*(27*x^3 + 54*x^2 + 36*x + 8)*log(3*x + 2) - 44793*x - 16603)/(2
7*x^3 + 54*x^2 + 36*x + 8)

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giac [A]  time = 0.92, size = 32, normalized size = 0.65 \begin {gather*} \frac {100}{81} \, x - \frac {40743 \, x^{2} + 51993 \, x + 16603}{729 \, {\left (3 \, x + 2\right )}^{3}} - \frac {740}{243} \, \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)^4,x, algorithm="giac")

[Out]

100/81*x - 1/729*(40743*x^2 + 51993*x + 16603)/(3*x + 2)^3 - 740/243*log(abs(3*x + 2))

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maple [A]  time = 0.01, size = 40, normalized size = 0.82 \begin {gather*} \frac {100 x}{81}-\frac {740 \ln \left (3 x +2\right )}{243}-\frac {49}{729 \left (3 x +2\right )^{3}}+\frac {259}{243 \left (3 x +2\right )^{2}}-\frac {503}{81 \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)^4,x)

[Out]

100/81*x-49/729/(3*x+2)^3+259/243/(3*x+2)^2-503/81/(3*x+2)-740/243*ln(3*x+2)

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maxima [A]  time = 0.56, size = 41, normalized size = 0.84 \begin {gather*} \frac {100}{81} \, x - \frac {40743 \, x^{2} + 51993 \, x + 16603}{729 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} - \frac {740}{243} \, \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)^4,x, algorithm="maxima")

[Out]

100/81*x - 1/729*(40743*x^2 + 51993*x + 16603)/(27*x^3 + 54*x^2 + 36*x + 8) - 740/243*log(3*x + 2)

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mupad [B]  time = 1.12, size = 37, normalized size = 0.76 \begin {gather*} \frac {100\,x}{81}-\frac {740\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {\frac {503\,x^2}{243}+\frac {5777\,x}{2187}+\frac {16603}{19683}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{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)^4,x)

[Out]

(100*x)/81 - (740*log(x + 2/3))/243 - ((5777*x)/2187 + (503*x^2)/243 + 16603/19683)/((4*x)/3 + 2*x^2 + x^3 + 8
/27)

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sympy [A]  time = 0.15, size = 41, normalized size = 0.84 \begin {gather*} \frac {100 x}{81} + \frac {- 40743 x^{2} - 51993 x - 16603}{19683 x^{3} + 39366 x^{2} + 26244 x + 5832} - \frac {740 \log {\left (3 x + 2 \right )}}{243} \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)**4,x)

[Out]

100*x/81 + (-40743*x**2 - 51993*x - 16603)/(19683*x**3 + 39366*x**2 + 26244*x + 5832) - 740*log(3*x + 2)/243

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