3.1329 \(\int \frac{(1-2 x)^2}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=33 \[ \frac{44}{125 (5 x+3)}-\frac{121}{250 (5 x+3)^2}+\frac{4}{125} \log (5 x+3) \]

[Out]

-121/(250*(3 + 5*x)^2) + 44/(125*(3 + 5*x)) + (4*Log[3 + 5*x])/125

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Rubi [A]  time = 0.011919, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ \frac{44}{125 (5 x+3)}-\frac{121}{250 (5 x+3)^2}+\frac{4}{125} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^2/(3 + 5*x)^3,x]

[Out]

-121/(250*(3 + 5*x)^2) + 44/(125*(3 + 5*x)) + (4*Log[3 + 5*x])/125

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(1-2 x)^2}{(3+5 x)^3} \, dx &=\int \left (\frac{121}{25 (3+5 x)^3}-\frac{44}{25 (3+5 x)^2}+\frac{4}{25 (3+5 x)}\right ) \, dx\\ &=-\frac{121}{250 (3+5 x)^2}+\frac{44}{125 (3+5 x)}+\frac{4}{125} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0112523, size = 31, normalized size = 0.94 \[ \frac{440 x+8 (5 x+3)^2 \log (10 x+6)+143}{250 (5 x+3)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x)^2/(3 + 5*x)^3,x]

[Out]

(143 + 440*x + 8*(3 + 5*x)^2*Log[6 + 10*x])/(250*(3 + 5*x)^2)

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Maple [A]  time = 0.005, size = 28, normalized size = 0.9 \begin{align*} -{\frac{121}{250\, \left ( 3+5\,x \right ) ^{2}}}+{\frac{44}{375+625\,x}}+{\frac{4\,\ln \left ( 3+5\,x \right ) }{125}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/250/(3+5*x)^2+44/125/(3+5*x)+4/125*ln(3+5*x)

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Maxima [A]  time = 1.1466, size = 38, normalized size = 1.15 \begin{align*} \frac{11 \,{\left (40 \, x + 13\right )}}{250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac{4}{125} \, \log \left (5 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/250*(40*x + 13)/(25*x^2 + 30*x + 9) + 4/125*log(5*x + 3)

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Fricas [A]  time = 1.2269, size = 105, normalized size = 3.18 \begin{align*} \frac{8 \,{\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) + 440 \, x + 143}{250 \,{\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/(3+5*x)^3,x, algorithm="fricas")

[Out]

1/250*(8*(25*x^2 + 30*x + 9)*log(5*x + 3) + 440*x + 143)/(25*x^2 + 30*x + 9)

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Sympy [A]  time = 0.107233, size = 24, normalized size = 0.73 \begin{align*} \frac{440 x + 143}{6250 x^{2} + 7500 x + 2250} + \frac{4 \log{\left (5 x + 3 \right )}}{125} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(440*x + 143)/(6250*x**2 + 7500*x + 2250) + 4*log(5*x + 3)/125

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Giac [A]  time = 1.45829, size = 32, normalized size = 0.97 \begin{align*} \frac{11 \,{\left (40 \, x + 13\right )}}{250 \,{\left (5 \, x + 3\right )}^{2}} + \frac{4}{125} \, \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/(3+5*x)^3,x, algorithm="giac")

[Out]

11/250*(40*x + 13)/(5*x + 3)^2 + 4/125*log(abs(5*x + 3))