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

Optimal. Leaf size=23 \[ -\frac {15 x^2}{4}-\frac {53 x}{4}-\frac {77}{8} \log (1-2 x) \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \begin {gather*} -\frac {15 x^2}{4}-\frac {53 x}{4}-\frac {77}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-53*x)/4 - (15*x^2)/4 - (77*Log[1 - 2*x])/8

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(2+3 x) (3+5 x)}{1-2 x} \, dx &=\int \left (-\frac {53}{4}-\frac {15 x}{2}-\frac {77}{4 (-1+2 x)}\right ) \, dx\\ &=-\frac {53 x}{4}-\frac {15 x^2}{4}-\frac {77}{8} \log (1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{16} \left (-60 x^2-212 x-154 \log (1-2 x)+121\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(121 - 212*x - 60*x^2 - 154*Log[1 - 2*x])/16

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.18, size = 17, normalized size = 0.74 \begin {gather*} -\frac {15}{4} \, x^{2} - \frac {53}{4} \, x - \frac {77}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*x^2 - 53/4*x - 77/8*log(2*x - 1)

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giac [A]  time = 0.67, size = 18, normalized size = 0.78 \begin {gather*} -\frac {15}{4} \, x^{2} - \frac {53}{4} \, x - \frac {77}{8} \, \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)*(3+5*x)/(1-2*x),x, algorithm="giac")

[Out]

-15/4*x^2 - 53/4*x - 77/8*log(abs(2*x - 1))

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maple [A]  time = 0.00, size = 18, normalized size = 0.78 \begin {gather*} -\frac {15 x^{2}}{4}-\frac {53 x}{4}-\frac {77 \ln \left (2 x -1\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*x^2-53/4*x-77/8*ln(2*x-1)

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maxima [A]  time = 0.49, size = 17, normalized size = 0.74 \begin {gather*} -\frac {15}{4} \, x^{2} - \frac {53}{4} \, x - \frac {77}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*x^2 - 53/4*x - 77/8*log(2*x - 1)

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mupad [B]  time = 0.03, size = 15, normalized size = 0.65 \begin {gather*} -\frac {53\,x}{4}-\frac {77\,\ln \left (x-\frac {1}{2}\right )}{8}-\frac {15\,x^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (53*x)/4 - (77*log(x - 1/2))/8 - (15*x^2)/4

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sympy [A]  time = 0.09, size = 22, normalized size = 0.96 \begin {gather*} - \frac {15 x^{2}}{4} - \frac {53 x}{4} - \frac {77 \log {\left (2 x - 1 \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*x**2/4 - 53*x/4 - 77*log(2*x - 1)/8

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