3.1328 \(\int \frac{(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=38 \[ \frac{12 x}{125}-\frac{319}{625 (5 x+3)}-\frac{121}{1250 (5 x+3)^2}-\frac{128}{625} \log (5 x+3) \]

[Out]

(12*x)/125 - 121/(1250*(3 + 5*x)^2) - 319/(625*(3 + 5*x)) - (128*Log[3 + 5*x])/625

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Rubi [A]  time = 0.0157388, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {77} \[ \frac{12 x}{125}-\frac{319}{625 (5 x+3)}-\frac{121}{1250 (5 x+3)^2}-\frac{128}{625} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*x)/125 - 121/(1250*(3 + 5*x)^2) - 319/(625*(3 + 5*x)) - (128*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 (2+3 x)}{(3+5 x)^3} \, dx &=\int \left (\frac{12}{125}+\frac{121}{125 (3+5 x)^3}+\frac{319}{125 (3+5 x)^2}-\frac{128}{125 (3+5 x)}\right ) \, dx\\ &=\frac{12 x}{125}-\frac{121}{1250 (3+5 x)^2}-\frac{319}{625 (3+5 x)}-\frac{128}{625} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0168775, size = 37, normalized size = 0.97 \[ \frac{\frac{5 \left (600 x^3+420 x^2-782 x-515\right )}{(5 x+3)^2}-256 \log (10 x+6)}{1250} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*(-515 - 782*x + 420*x^2 + 600*x^3))/(3 + 5*x)^2 - 256*Log[6 + 10*x])/1250

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Maple [A]  time = 0.006, size = 31, normalized size = 0.8 \begin{align*}{\frac{12\,x}{125}}-{\frac{121}{1250\, \left ( 3+5\,x \right ) ^{2}}}-{\frac{319}{1875+3125\,x}}-{\frac{128\,\ln \left ( 3+5\,x \right ) }{625}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12/125*x-121/1250/(3+5*x)^2-319/625/(3+5*x)-128/625*ln(3+5*x)

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Maxima [A]  time = 1.10953, size = 42, normalized size = 1.11 \begin{align*} \frac{12}{125} \, x - \frac{11 \,{\left (58 \, x + 37\right )}}{250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac{128}{625} \, \log \left (5 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12/125*x - 11/250*(58*x + 37)/(25*x^2 + 30*x + 9) - 128/625*log(5*x + 3)

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Fricas [A]  time = 1.30331, size = 142, normalized size = 3.74 \begin{align*} \frac{3000 \, x^{3} + 3600 \, x^{2} - 256 \,{\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 2110 \, x - 2035}{1250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1250*(3000*x^3 + 3600*x^2 - 256*(25*x^2 + 30*x + 9)*log(5*x + 3) - 2110*x - 2035)/(25*x^2 + 30*x + 9)

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Sympy [A]  time = 0.114629, size = 29, normalized size = 0.76 \begin{align*} \frac{12 x}{125} - \frac{638 x + 407}{6250 x^{2} + 7500 x + 2250} - \frac{128 \log{\left (5 x + 3 \right )}}{625} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*x/125 - (638*x + 407)/(6250*x**2 + 7500*x + 2250) - 128*log(5*x + 3)/625

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Giac [A]  time = 2.34214, size = 36, normalized size = 0.95 \begin{align*} \frac{12}{125} \, x - \frac{11 \,{\left (58 \, x + 37\right )}}{250 \,{\left (5 \, x + 3\right )}^{2}} - \frac{128}{625} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12/125*x - 11/250*(58*x + 37)/(5*x + 3)^2 - 128/625*log(abs(5*x + 3))