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

Optimal. Leaf size=44 \[ \frac {100 x^5}{3}+\frac {50 x^4}{9}-\frac {2515 x^3}{81}-\frac {559 x^2}{162}+\frac {3305 x}{243}-\frac {49}{729} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 44, 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^5}{3}+\frac {50 x^4}{9}-\frac {2515 x^3}{81}-\frac {559 x^2}{162}+\frac {3305 x}{243}-\frac {49}{729} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3305*x)/243 - (559*x^2)/162 - (2515*x^3)/81 + (50*x^4)/9 + (100*x^5)/3 - (49*Log[2 + 3*x])/729

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)^3}{2+3 x} \, dx &=\int \left (\frac {3305}{243}-\frac {559 x}{81}-\frac {2515 x^2}{27}+\frac {200 x^3}{9}+\frac {500 x^4}{3}-\frac {49}{243 (2+3 x)}\right ) \, dx\\ &=\frac {3305 x}{243}-\frac {559 x^2}{162}-\frac {2515 x^3}{81}+\frac {50 x^4}{9}+\frac {100 x^5}{3}-\frac {49}{729} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.84 \begin {gather*} \frac {145800 x^5+24300 x^4-135810 x^3-15093 x^2+59490 x-294 \log (3 x+2)+20528}{4374} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(20528 + 59490*x - 15093*x^2 - 135810*x^3 + 24300*x^4 + 145800*x^5 - 294*Log[2 + 3*x])/4374

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 32, normalized size = 0.73 \begin {gather*} \frac {100}{3} \, x^{5} + \frac {50}{9} \, x^{4} - \frac {2515}{81} \, x^{3} - \frac {559}{162} \, x^{2} + \frac {3305}{243} \, x - \frac {49}{729} \, \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)^3/(2+3*x),x, algorithm="fricas")

[Out]

100/3*x^5 + 50/9*x^4 - 2515/81*x^3 - 559/162*x^2 + 3305/243*x - 49/729*log(3*x + 2)

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giac [A]  time = 0.99, size = 33, normalized size = 0.75 \begin {gather*} \frac {100}{3} \, x^{5} + \frac {50}{9} \, x^{4} - \frac {2515}{81} \, x^{3} - \frac {559}{162} \, x^{2} + \frac {3305}{243} \, x - \frac {49}{729} \, \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)^3/(2+3*x),x, algorithm="giac")

[Out]

100/3*x^5 + 50/9*x^4 - 2515/81*x^3 - 559/162*x^2 + 3305/243*x - 49/729*log(abs(3*x + 2))

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maple [A]  time = 0.00, size = 33, normalized size = 0.75 \begin {gather*} \frac {100 x^{5}}{3}+\frac {50 x^{4}}{9}-\frac {2515 x^{3}}{81}-\frac {559 x^{2}}{162}+\frac {3305 x}{243}-\frac {49 \ln \left (3 x +2\right )}{729} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3305/243*x-559/162*x^2-2515/81*x^3+50/9*x^4+100/3*x^5-49/729*ln(3*x+2)

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maxima [A]  time = 0.57, size = 32, normalized size = 0.73 \begin {gather*} \frac {100}{3} \, x^{5} + \frac {50}{9} \, x^{4} - \frac {2515}{81} \, x^{3} - \frac {559}{162} \, x^{2} + \frac {3305}{243} \, x - \frac {49}{729} \, \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)^3/(2+3*x),x, algorithm="maxima")

[Out]

100/3*x^5 + 50/9*x^4 - 2515/81*x^3 - 559/162*x^2 + 3305/243*x - 49/729*log(3*x + 2)

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mupad [B]  time = 0.03, size = 30, normalized size = 0.68 \begin {gather*} \frac {3305\,x}{243}-\frac {49\,\ln \left (x+\frac {2}{3}\right )}{729}-\frac {559\,x^2}{162}-\frac {2515\,x^3}{81}+\frac {50\,x^4}{9}+\frac {100\,x^5}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3305*x)/243 - (49*log(x + 2/3))/729 - (559*x^2)/162 - (2515*x^3)/81 + (50*x^4)/9 + (100*x^5)/3

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sympy [A]  time = 0.10, size = 41, normalized size = 0.93 \begin {gather*} \frac {100 x^{5}}{3} + \frac {50 x^{4}}{9} - \frac {2515 x^{3}}{81} - \frac {559 x^{2}}{162} + \frac {3305 x}{243} - \frac {49 \log {\left (3 x + 2 \right )}}{729} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

100*x**5/3 + 50*x**4/9 - 2515*x**3/81 - 559*x**2/162 + 3305*x/243 - 49*log(3*x + 2)/729

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