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

Optimal. Leaf size=23 \[ \frac {2 x^2}{5}-\frac {32 x}{25}+\frac {121}{125} \log (5 x+3) \]

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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 = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {43} \begin {gather*} \frac {2 x^2}{5}-\frac {32 x}{25}+\frac {121}{125} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^2/(3 + 5*x),x]

[Out]

(-32*x)/25 + (2*x^2)/5 + (121*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} \, dx &=\int \left (-\frac {32}{25}+\frac {4 x}{5}+\frac {121}{25 (3+5 x)}\right ) \, dx\\ &=-\frac {32 x}{25}+\frac {2 x^2}{5}+\frac {121}{125} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{125} \left (50 x^2-160 x+121 \log (5 x+3)-114\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x)^2/(3 + 5*x),x]

[Out]

(-114 - 160*x + 50*x^2 + 121*Log[3 + 5*x])/125

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.50, size = 17, normalized size = 0.74 \begin {gather*} \frac {2}{5} \, x^{2} - \frac {32}{25} \, x + \frac {121}{125} \, \log \left (5 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^2 - 32/25*x + 121/125*log(5*x + 3)

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giac [A]  time = 0.93, size = 18, normalized size = 0.78 \begin {gather*} \frac {2}{5} \, x^{2} - \frac {32}{25} \, x + \frac {121}{125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^2 - 32/25*x + 121/125*log(abs(5*x + 3))

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maple [A]  time = 0.00, size = 18, normalized size = 0.78 \begin {gather*} \frac {2 x^{2}}{5}-\frac {32 x}{25}+\frac {121 \ln \left (5 x +3\right )}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/25*x+2/5*x^2+121/125*ln(5*x+3)

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maxima [A]  time = 0.52, size = 17, normalized size = 0.74 \begin {gather*} \frac {2}{5} \, x^{2} - \frac {32}{25} \, x + \frac {121}{125} \, \log \left (5 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^2 - 32/25*x + 121/125*log(5*x + 3)

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mupad [B]  time = 0.03, size = 15, normalized size = 0.65 \begin {gather*} \frac {121\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {32\,x}{25}+\frac {2\,x^2}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - 1)^2/(5*x + 3),x)

[Out]

(121*log(x + 3/5))/125 - (32*x)/25 + (2*x^2)/5

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sympy [A]  time = 0.08, size = 20, normalized size = 0.87 \begin {gather*} \frac {2 x^{2}}{5} - \frac {32 x}{25} + \frac {121 \log {\left (5 x + 3 \right )}}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**2/5 - 32*x/25 + 121*log(5*x + 3)/125

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