3.1212 \(\int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=34 \[ -\frac {9 x^2}{25}+\frac {33 x}{125}-\frac {11}{625 (5 x+3)}+\frac {64}{625} \log (5 x+3) \]

[Out]

33/125*x-9/25*x^2-11/625/(3+5*x)+64/625*ln(3+5*x)

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Rubi [A]  time = 0.01, antiderivative size = 34, 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} \[ -\frac {9 x^2}{25}+\frac {33 x}{125}-\frac {11}{625 (5 x+3)}+\frac {64}{625} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(33*x)/125 - (9*x^2)/25 - 11/(625*(3 + 5*x)) + (64*Log[3 + 5*x])/625

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 {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx &=\int \left (\frac {33}{125}-\frac {18 x}{25}+\frac {11}{125 (3+5 x)^2}+\frac {64}{125 (3+5 x)}\right ) \, dx\\ &=\frac {33 x}{125}-\frac {9 x^2}{25}-\frac {11}{625 (3+5 x)}+\frac {64}{625} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 1.21 \[ \frac {-1125 x^3+150 x^2+1545 x+64 (5 x+3) \log (-3 (5 x+3))+619}{625 (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(619 + 1545*x + 150*x^2 - 1125*x^3 + 64*(3 + 5*x)*Log[-3*(3 + 5*x)])/(625*(3 + 5*x))

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fricas [A]  time = 0.80, size = 37, normalized size = 1.09 \[ -\frac {1125 \, x^{3} - 150 \, x^{2} - 64 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 495 \, x + 11}{625 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/625*(1125*x^3 - 150*x^2 - 64*(5*x + 3)*log(5*x + 3) - 495*x + 11)/(5*x + 3)

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giac [A]  time = 1.17, size = 48, normalized size = 1.41 \[ \frac {3}{625} \, {\left (5 \, x + 3\right )}^{2} {\left (\frac {29}{5 \, x + 3} - 3\right )} - \frac {11}{625 \, {\left (5 \, x + 3\right )}} - \frac {64}{625} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/625*(5*x + 3)^2*(29/(5*x + 3) - 3) - 11/625/(5*x + 3) - 64/625*log(1/5*abs(5*x + 3)/(5*x + 3)^2)

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maple [A]  time = 0.00, size = 27, normalized size = 0.79 \[ -\frac {9 x^{2}}{25}+\frac {33 x}{125}+\frac {64 \ln \left (5 x +3\right )}{625}-\frac {11}{625 \left (5 x +3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

33/125*x-9/25*x^2-11/625/(5*x+3)+64/625*ln(5*x+3)

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maxima [A]  time = 0.53, size = 26, normalized size = 0.76 \[ -\frac {9}{25} \, x^{2} + \frac {33}{125} \, x - \frac {11}{625 \, {\left (5 \, x + 3\right )}} + \frac {64}{625} \, \log \left (5 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/25*x^2 + 33/125*x - 11/625/(5*x + 3) + 64/625*log(5*x + 3)

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mupad [B]  time = 0.03, size = 24, normalized size = 0.71 \[ \frac {33\,x}{125}+\frac {64\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {11}{3125\,\left (x+\frac {3}{5}\right )}-\frac {9\,x^2}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(33*x)/125 + (64*log(x + 3/5))/625 - 11/(3125*(x + 3/5)) - (9*x^2)/25

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sympy [A]  time = 0.10, size = 27, normalized size = 0.79 \[ - \frac {9 x^{2}}{25} + \frac {33 x}{125} + \frac {64 \log {\left (5 x + 3 \right )}}{625} - \frac {11}{3125 x + 1875} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*x**2/25 + 33*x/125 + 64*log(5*x + 3)/625 - 11/(3125*x + 1875)

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