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

Optimal. Leaf size=46 \[ \frac {21}{3 x+2}+\frac {68}{5 x+3}-\frac {11}{2 (5 x+3)^2}-309 \log (3 x+2)+309 \log (5 x+3) \]

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Rubi [A]  time = 0.02, antiderivative size = 46, 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 = {77} \begin {gather*} \frac {21}{3 x+2}+\frac {68}{5 x+3}-\frac {11}{2 (5 x+3)^2}-309 \log (3 x+2)+309 \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

21/(2 + 3*x) - 11/(2*(3 + 5*x)^2) + 68/(3 + 5*x) - 309*Log[2 + 3*x] + 309*Log[3 + 5*x]

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)^2 (3+5 x)^3} \, dx &=\int \left (-\frac {63}{(2+3 x)^2}-\frac {927}{2+3 x}+\frac {55}{(3+5 x)^3}-\frac {340}{(3+5 x)^2}+\frac {1545}{3+5 x}\right ) \, dx\\ &=\frac {21}{2+3 x}-\frac {11}{2 (3+5 x)^2}+\frac {68}{3+5 x}-309 \log (2+3 x)+309 \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 1.04 \begin {gather*} \frac {21}{3 x+2}+\frac {68}{5 x+3}-\frac {11}{2 (5 x+3)^2}-309 \log (3 x+2)+309 \log (-3 (5 x+3)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

21/(2 + 3*x) - 11/(2*(3 + 5*x)^2) + 68/(3 + 5*x) - 309*Log[2 + 3*x] + 309*Log[-3*(3 + 5*x)]

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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)^2 (3+5 x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.13, size = 75, normalized size = 1.63 \begin {gather*} \frac {3090 \, x^{2} + 618 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (5 \, x + 3\right ) - 618 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (3 \, x + 2\right ) + 3811 \, x + 1172}{2 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3090*x^2 + 618*(75*x^3 + 140*x^2 + 87*x + 18)*log(5*x + 3) - 618*(75*x^3 + 140*x^2 + 87*x + 18)*log(3*x +
2) + 3811*x + 1172)/(75*x^3 + 140*x^2 + 87*x + 18)

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giac [A]  time = 1.20, size = 49, normalized size = 1.07 \begin {gather*} \frac {21}{3 \, x + 2} - \frac {15 \, {\left (\frac {202}{3 \, x + 2} - 845\right )}}{2 \, {\left (\frac {1}{3 \, x + 2} - 5\right )}^{2}} + 309 \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

21/(3*x + 2) - 15/2*(202/(3*x + 2) - 845)/(1/(3*x + 2) - 5)^2 + 309*log(abs(-1/(3*x + 2) + 5))

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maple [A]  time = 0.01, size = 45, normalized size = 0.98 \begin {gather*} -309 \ln \left (3 x +2\right )+309 \ln \left (5 x +3\right )+\frac {21}{3 x +2}-\frac {11}{2 \left (5 x +3\right )^{2}}+\frac {68}{5 x +3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

21/(3*x+2)-11/2/(5*x+3)^2+68/(5*x+3)-309*ln(3*x+2)+309*ln(5*x+3)

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maxima [A]  time = 0.50, size = 46, normalized size = 1.00 \begin {gather*} \frac {3090 \, x^{2} + 3811 \, x + 1172}{2 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} + 309 \, \log \left (5 \, x + 3\right ) - 309 \, \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)^2/(3+5*x)^3,x, algorithm="maxima")

[Out]

1/2*(3090*x^2 + 3811*x + 1172)/(75*x^3 + 140*x^2 + 87*x + 18) + 309*log(5*x + 3) - 309*log(3*x + 2)

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mupad [B]  time = 0.04, size = 35, normalized size = 0.76 \begin {gather*} \frac {\frac {103\,x^2}{5}+\frac {3811\,x}{150}+\frac {586}{75}}{x^3+\frac {28\,x^2}{15}+\frac {29\,x}{25}+\frac {6}{25}}-618\,\mathrm {atanh}\left (30\,x+19\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3811*x)/150 + (103*x^2)/5 + 586/75)/((29*x)/25 + (28*x^2)/15 + x^3 + 6/25) - 618*atanh(30*x + 19)

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sympy [A]  time = 0.16, size = 42, normalized size = 0.91 \begin {gather*} - \frac {- 3090 x^{2} - 3811 x - 1172}{150 x^{3} + 280 x^{2} + 174 x + 36} + 309 \log {\left (x + \frac {3}{5} \right )} - 309 \log {\left (x + \frac {2}{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-3090*x**2 - 3811*x - 1172)/(150*x**3 + 280*x**2 + 174*x + 36) + 309*log(x + 3/5) - 309*log(x + 2/3)

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