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

Optimal. Leaf size=32 \[ \frac {1}{21 (3 x+2)}-\frac {11}{49} \log (1-2 x)+\frac {11}{49} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 32, 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 {1}{21 (3 x+2)}-\frac {11}{49} \log (1-2 x)+\frac {11}{49} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(21*(2 + 3*x)) - (11*Log[1 - 2*x])/49 + (11*Log[2 + 3*x])/49

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 {3+5 x}{(1-2 x) (2+3 x)^2} \, dx &=\int \left (-\frac {22}{49 (-1+2 x)}-\frac {1}{7 (2+3 x)^2}+\frac {33}{49 (2+3 x)}\right ) \, dx\\ &=\frac {1}{21 (2+3 x)}-\frac {11}{49} \log (1-2 x)+\frac {11}{49} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 0.94 \begin {gather*} \frac {1}{147} \left (\frac {7}{3 x+2}-33 \log (3-6 x)+33 \log (3 x+2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7/(2 + 3*x) - 33*Log[3 - 6*x] + 33*Log[2 + 3*x])/147

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.36, size = 37, normalized size = 1.16 \begin {gather*} \frac {33 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 33 \, {\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 7}{147 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/147*(33*(3*x + 2)*log(3*x + 2) - 33*(3*x + 2)*log(2*x - 1) + 7)/(3*x + 2)

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giac [A]  time = 0.78, size = 25, normalized size = 0.78 \begin {gather*} \frac {1}{21 \, {\left (3 \, x + 2\right )}} - \frac {11}{49} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21/(3*x + 2) - 11/49*log(abs(-7/(3*x + 2) + 2))

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maple [A]  time = 0.01, size = 27, normalized size = 0.84 \begin {gather*} -\frac {11 \ln \left (2 x -1\right )}{49}+\frac {11 \ln \left (3 x +2\right )}{49}+\frac {1}{63 x +42} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21/(3*x+2)+11/49*ln(3*x+2)-11/49*ln(2*x-1)

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maxima [A]  time = 0.49, size = 26, normalized size = 0.81 \begin {gather*} \frac {1}{21 \, {\left (3 \, x + 2\right )}} + \frac {11}{49} \, \log \left (3 \, x + 2\right ) - \frac {11}{49} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21/(3*x + 2) + 11/49*log(3*x + 2) - 11/49*log(2*x - 1)

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mupad [B]  time = 0.04, size = 16, normalized size = 0.50 \begin {gather*} \frac {22\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{49}+\frac {1}{63\,\left (x+\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(22*atanh((12*x)/7 + 1/7))/49 + 1/(63*(x + 2/3))

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sympy [A]  time = 0.13, size = 26, normalized size = 0.81 \begin {gather*} - \frac {11 \log {\left (x - \frac {1}{2} \right )}}{49} + \frac {11 \log {\left (x + \frac {2}{3} \right )}}{49} + \frac {1}{63 x + 42} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11*log(x - 1/2)/49 + 11*log(x + 2/3)/49 + 1/(63*x + 42)

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