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

Optimal. Leaf size=22 \[ -\frac {11}{25 (5 x+3)}-\frac {2}{25} \log (5 x+3) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {43} \begin {gather*} -\frac {11}{25 (5 x+3)}-\frac {2}{25} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)/(3 + 5*x)^2,x]

[Out]

-11/(25*(3 + 5*x)) - (2*Log[3 + 5*x])/25

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}{(3+5 x)^2} \, dx &=\int \left (\frac {11}{5 (3+5 x)^2}-\frac {2}{5 (3+5 x)}\right ) \, dx\\ &=-\frac {11}{25 (3+5 x)}-\frac {2}{25} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 22, normalized size = 1.00 \begin {gather*} -\frac {11}{25 (5 x+3)}-\frac {2}{25} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x)/(3 + 5*x)^2,x]

[Out]

-11/(25*(3 + 5*x)) - (2*Log[3 + 5*x])/25

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.24, size = 24, normalized size = 1.09 \begin {gather*} -\frac {2 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 11}{25 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(2*(5*x + 3)*log(5*x + 3) + 11)/(5*x + 3)

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giac [A]  time = 1.17, size = 28, normalized size = 1.27 \begin {gather*} -\frac {11}{25 \, {\left (5 \, x + 3\right )}} + \frac {2}{25} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/25/(5*x + 3) + 2/25*log(1/5*abs(5*x + 3)/(5*x + 3)^2)

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maple [A]  time = 0.01, size = 19, normalized size = 0.86 \begin {gather*} -\frac {2 \ln \left (5 x +3\right )}{25}-\frac {11}{25 \left (5 x +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/25/(5*x+3)-2/25*ln(5*x+3)

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maxima [A]  time = 0.58, size = 18, normalized size = 0.82 \begin {gather*} -\frac {11}{25 \, {\left (5 \, x + 3\right )}} - \frac {2}{25} \, \log \left (5 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/25/(5*x + 3) - 2/25*log(5*x + 3)

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mupad [B]  time = 0.03, size = 16, normalized size = 0.73 \begin {gather*} -\frac {2\,\ln \left (x+\frac {3}{5}\right )}{25}-\frac {11}{125\,\left (x+\frac {3}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - 1)/(5*x + 3)^2,x)

[Out]

- (2*log(x + 3/5))/25 - 11/(125*(x + 3/5))

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sympy [A]  time = 0.09, size = 17, normalized size = 0.77 \begin {gather*} - \frac {2 \log {\left (5 x + 3 \right )}}{25} - \frac {11}{125 x + 75} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(5*x + 3)/25 - 11/(125*x + 75)

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