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

Optimal. Leaf size=21 \[ \frac {11}{5} \log (5 x+3)-\frac {7}{3} \log (3 x+2) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, 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 = {72} \begin {gather*} \frac {11}{5} \log (5 x+3)-\frac {7}{3} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*Log[2 + 3*x])/3 + (11*Log[3 + 5*x])/5

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx &=\int \left (-\frac {7}{2+3 x}+\frac {11}{3+5 x}\right ) \, dx\\ &=-\frac {7}{3} \log (2+3 x)+\frac {11}{5} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \begin {gather*} \frac {11}{5} \log (5 x+3)-\frac {7}{3} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*Log[2 + 3*x])/3 + (11*Log[3 + 5*x])/5

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.16, size = 17, normalized size = 0.81 \begin {gather*} \frac {11}{5} \, \log \left (5 \, x + 3\right ) - \frac {7}{3} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/5*log(5*x + 3) - 7/3*log(3*x + 2)

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giac [A]  time = 1.20, size = 19, normalized size = 0.90 \begin {gather*} \frac {11}{5} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {7}{3} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/5*log(abs(5*x + 3)) - 7/3*log(abs(3*x + 2))

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maple [A]  time = 0.00, size = 18, normalized size = 0.86 \begin {gather*} -\frac {7 \ln \left (3 x +2\right )}{3}+\frac {11 \ln \left (5 x +3\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/3*ln(3*x+2)+11/5*ln(5*x+3)

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maxima [A]  time = 0.53, size = 17, normalized size = 0.81 \begin {gather*} \frac {11}{5} \, \log \left (5 \, x + 3\right ) - \frac {7}{3} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/5*log(5*x + 3) - 7/3*log(3*x + 2)

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mupad [B]  time = 0.06, size = 13, normalized size = 0.62 \begin {gather*} \frac {11\,\ln \left (x+\frac {3}{5}\right )}{5}-\frac {7\,\ln \left (x+\frac {2}{3}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*log(x + 3/5))/5 - (7*log(x + 2/3))/3

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sympy [A]  time = 0.12, size = 19, normalized size = 0.90 \begin {gather*} \frac {11 \log {\left (x + \frac {3}{5} \right )}}{5} - \frac {7 \log {\left (x + \frac {2}{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11*log(x + 3/5)/5 - 7*log(x + 2/3)/3

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