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

Optimal. Leaf size=23 \[ -\frac {3 x^2}{5}+\frac {13 x}{25}+\frac {11}{125} \log (5 x+3) \]

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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 = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \begin {gather*} -\frac {3 x^2}{5}+\frac {13 x}{25}+\frac {11}{125} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(13*x)/25 - (3*x^2)/5 + (11*Log[3 + 5*x])/125

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+3 x)}{3+5 x} \, dx &=\int \left (\frac {13}{25}-\frac {6 x}{5}+\frac {11}{25 (3+5 x)}\right ) \, dx\\ &=\frac {13 x}{25}-\frac {3 x^2}{5}+\frac {11}{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 (-75 x^2+65 x+11 \log (5 x+3)+66\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(66 + 65*x - 75*x^2 + 11*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 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 = 0.94, size = 17, normalized size = 0.74 \begin {gather*} -\frac {3}{5} \, x^{2} + \frac {13}{25} \, x + \frac {11}{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*x)/(3+5*x),x, algorithm="fricas")

[Out]

-3/5*x^2 + 13/25*x + 11/125*log(5*x + 3)

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giac [A]  time = 1.22, size = 18, normalized size = 0.78 \begin {gather*} -\frac {3}{5} \, x^{2} + \frac {13}{25} \, x + \frac {11}{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*x)/(3+5*x),x, algorithm="giac")

[Out]

-3/5*x^2 + 13/25*x + 11/125*log(abs(5*x + 3))

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maple [A]  time = 0.00, size = 18, normalized size = 0.78 \begin {gather*} -\frac {3 x^{2}}{5}+\frac {13 x}{25}+\frac {11 \ln \left (5 x +3\right )}{125} \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]

13/25*x-3/5*x^2+11/125*ln(5*x+3)

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maxima [A]  time = 0.47, size = 17, normalized size = 0.74 \begin {gather*} -\frac {3}{5} \, x^{2} + \frac {13}{25} \, x + \frac {11}{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*x)/(3+5*x),x, algorithm="maxima")

[Out]

-3/5*x^2 + 13/25*x + 11/125*log(5*x + 3)

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mupad [B]  time = 0.03, size = 15, normalized size = 0.65 \begin {gather*} \frac {13\,x}{25}+\frac {11\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {3\,x^2}{5} \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]

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

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sympy [A]  time = 0.08, size = 20, normalized size = 0.87 \begin {gather*} - \frac {3 x^{2}}{5} + \frac {13 x}{25} + \frac {11 \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*x)/(3+5*x),x)

[Out]

-3*x**2/5 + 13*x/25 + 11*log(5*x + 3)/125

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