3.1458 \(\int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx\)

Optimal. Leaf size=30 \[ -\frac {25 x^3}{2}-\frac {355 x^2}{8}-\frac {703 x}{8}-\frac {847}{16} \log (1-2 x) \]

[Out]

-703/8*x-355/8*x^2-25/2*x^3-847/16*ln(1-2*x)

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ -\frac {25 x^3}{2}-\frac {355 x^2}{8}-\frac {703 x}{8}-\frac {847}{16} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-703*x)/8 - (355*x^2)/8 - (25*x^3)/2 - (847*Log[1 - 2*x])/16

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)^2}{1-2 x} \, dx &=\int \left (-\frac {703}{8}-\frac {355 x}{4}-\frac {75 x^2}{2}-\frac {847}{8 (-1+2 x)}\right ) \, dx\\ &=-\frac {703 x}{8}-\frac {355 x^2}{8}-\frac {25 x^3}{2}-\frac {847}{16} \log (1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.90 \[ \frac {1}{32} \left (-400 x^3-1420 x^2-2812 x-1694 \log (1-2 x)+1811\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1811 - 2812*x - 1420*x^2 - 400*x^3 - 1694*Log[1 - 2*x])/32

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fricas [A]  time = 0.86, size = 22, normalized size = 0.73 \[ -\frac {25}{2} \, x^{3} - \frac {355}{8} \, x^{2} - \frac {703}{8} \, x - \frac {847}{16} \, \log \left (2 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/2*x^3 - 355/8*x^2 - 703/8*x - 847/16*log(2*x - 1)

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giac [A]  time = 0.89, size = 23, normalized size = 0.77 \[ -\frac {25}{2} \, x^{3} - \frac {355}{8} \, x^{2} - \frac {703}{8} \, x - \frac {847}{16} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/2*x^3 - 355/8*x^2 - 703/8*x - 847/16*log(abs(2*x - 1))

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maple [A]  time = 0.00, size = 23, normalized size = 0.77 \[ -\frac {25 x^{3}}{2}-\frac {355 x^{2}}{8}-\frac {703 x}{8}-\frac {847 \ln \left (2 x -1\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/2*x^3-355/8*x^2-703/8*x-847/16*ln(2*x-1)

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maxima [A]  time = 0.51, size = 22, normalized size = 0.73 \[ -\frac {25}{2} \, x^{3} - \frac {355}{8} \, x^{2} - \frac {703}{8} \, x - \frac {847}{16} \, \log \left (2 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/2*x^3 - 355/8*x^2 - 703/8*x - 847/16*log(2*x - 1)

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mupad [B]  time = 0.03, size = 20, normalized size = 0.67 \[ -\frac {703\,x}{8}-\frac {847\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {355\,x^2}{8}-\frac {25\,x^3}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (703*x)/8 - (847*log(x - 1/2))/16 - (355*x^2)/8 - (25*x^3)/2

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sympy [A]  time = 0.10, size = 29, normalized size = 0.97 \[ - \frac {25 x^{3}}{2} - \frac {355 x^{2}}{8} - \frac {703 x}{8} - \frac {847 \log {\left (2 x - 1 \right )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25*x**3/2 - 355*x**2/8 - 703*x/8 - 847*log(2*x - 1)/16

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