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

Optimal. Leaf size=37 \[ -\frac {225 x^4}{8}-\frac {455 x^3}{4}-\frac {3529 x^2}{16}-\frac {5353 x}{16}-\frac {5929}{32} \log (1-2 x) \]

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Rubi [A]  time = 0.02, antiderivative size = 37, 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 {225 x^4}{8}-\frac {455 x^3}{4}-\frac {3529 x^2}{16}-\frac {5353 x}{16}-\frac {5929}{32} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5353*x)/16 - (3529*x^2)/16 - (455*x^3)/4 - (225*x^4)/8 - (5929*Log[1 - 2*x])/32

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 {(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx &=\int \left (-\frac {5353}{16}-\frac {3529 x}{8}-\frac {1365 x^2}{4}-\frac {225 x^3}{2}-\frac {5929}{16 (-1+2 x)}\right ) \, dx\\ &=-\frac {5353 x}{16}-\frac {3529 x^2}{16}-\frac {455 x^3}{4}-\frac {225 x^4}{8}-\frac {5929}{32} \log (1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.86 \begin {gather*} \frac {1}{128} \left (-3600 x^4-14560 x^3-28232 x^2-42824 x-23716 \log (1-2 x)+30515\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(30515 - 42824*x - 28232*x^2 - 14560*x^3 - 3600*x^4 - 23716*Log[1 - 2*x])/128

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.49, size = 27, normalized size = 0.73 \begin {gather*} -\frac {225}{8} \, x^{4} - \frac {455}{4} \, x^{3} - \frac {3529}{16} \, x^{2} - \frac {5353}{16} \, x - \frac {5929}{32} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/8*x^4 - 455/4*x^3 - 3529/16*x^2 - 5353/16*x - 5929/32*log(2*x - 1)

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giac [A]  time = 0.94, size = 28, normalized size = 0.76 \begin {gather*} -\frac {225}{8} \, x^{4} - \frac {455}{4} \, x^{3} - \frac {3529}{16} \, x^{2} - \frac {5353}{16} \, x - \frac {5929}{32} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/8*x^4 - 455/4*x^3 - 3529/16*x^2 - 5353/16*x - 5929/32*log(abs(2*x - 1))

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maple [A]  time = 0.00, size = 28, normalized size = 0.76 \begin {gather*} -\frac {225 x^{4}}{8}-\frac {455 x^{3}}{4}-\frac {3529 x^{2}}{16}-\frac {5353 x}{16}-\frac {5929 \ln \left (2 x -1\right )}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/8*x^4-455/4*x^3-3529/16*x^2-5353/16*x-5929/32*ln(2*x-1)

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maxima [A]  time = 0.63, size = 27, normalized size = 0.73 \begin {gather*} -\frac {225}{8} \, x^{4} - \frac {455}{4} \, x^{3} - \frac {3529}{16} \, x^{2} - \frac {5353}{16} \, x - \frac {5929}{32} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/8*x^4 - 455/4*x^3 - 3529/16*x^2 - 5353/16*x - 5929/32*log(2*x - 1)

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mupad [B]  time = 0.02, size = 25, normalized size = 0.68 \begin {gather*} -\frac {5353\,x}{16}-\frac {5929\,\ln \left (x-\frac {1}{2}\right )}{32}-\frac {3529\,x^2}{16}-\frac {455\,x^3}{4}-\frac {225\,x^4}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (5353*x)/16 - (5929*log(x - 1/2))/32 - (3529*x^2)/16 - (455*x^3)/4 - (225*x^4)/8

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sympy [A]  time = 0.10, size = 36, normalized size = 0.97 \begin {gather*} - \frac {225 x^{4}}{8} - \frac {455 x^{3}}{4} - \frac {3529 x^{2}}{16} - \frac {5353 x}{16} - \frac {5929 \log {\left (2 x - 1 \right )}}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225*x**4/8 - 455*x**3/4 - 3529*x**2/16 - 5353*x/16 - 5929*log(2*x - 1)/32

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