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

Optimal. Leaf size=43 \[ -\frac {1}{49 (1-2 x)}+\frac {11}{28 (1-2 x)^2}+\frac {3}{343} \log (1-2 x)-\frac {3}{343} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 43, 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}{49 (1-2 x)}+\frac {11}{28 (1-2 x)^2}+\frac {3}{343} \log (1-2 x)-\frac {3}{343} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

11/(28*(1 - 2*x)^2) - 1/(49*(1 - 2*x)) + (3*Log[1 - 2*x])/343 - (3*Log[2 + 3*x])/343

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)^3 (2+3 x)} \, dx &=\int \left (-\frac {11}{7 (-1+2 x)^3}-\frac {2}{49 (-1+2 x)^2}+\frac {6}{343 (-1+2 x)}-\frac {9}{343 (2+3 x)}\right ) \, dx\\ &=\frac {11}{28 (1-2 x)^2}-\frac {1}{49 (1-2 x)}+\frac {3}{343} \log (1-2 x)-\frac {3}{343} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.81 \begin {gather*} \frac {\frac {7 (8 x+73)}{(1-2 x)^2}+12 \log (1-2 x)-12 \log (6 x+4)}{1372} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*(73 + 8*x))/(1 - 2*x)^2 + 12*Log[1 - 2*x] - 12*Log[4 + 6*x])/1372

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.93, size = 55, normalized size = 1.28 \begin {gather*} -\frac {12 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (3 \, x + 2\right ) - 12 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 56 \, x - 511}{1372 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1372*(12*(4*x^2 - 4*x + 1)*log(3*x + 2) - 12*(4*x^2 - 4*x + 1)*log(2*x - 1) - 56*x - 511)/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.20, size = 33, normalized size = 0.77 \begin {gather*} \frac {8 \, x + 73}{196 \, {\left (2 \, x - 1\right )}^{2}} - \frac {3}{343} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {3}{343} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/196*(8*x + 73)/(2*x - 1)^2 - 3/343*log(abs(3*x + 2)) + 3/343*log(abs(2*x - 1))

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maple [A]  time = 0.01, size = 36, normalized size = 0.84 \begin {gather*} \frac {3 \ln \left (2 x -1\right )}{343}-\frac {3 \ln \left (3 x +2\right )}{343}+\frac {11}{28 \left (2 x -1\right )^{2}}+\frac {1}{98 x -49} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/343*ln(3*x+2)+11/28/(2*x-1)^2+1/49/(2*x-1)+3/343*ln(2*x-1)

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maxima [A]  time = 0.52, size = 36, normalized size = 0.84 \begin {gather*} \frac {8 \, x + 73}{196 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {3}{343} \, \log \left (3 \, x + 2\right ) + \frac {3}{343} \, \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)^3/(2+3*x),x, algorithm="maxima")

[Out]

1/196*(8*x + 73)/(4*x^2 - 4*x + 1) - 3/343*log(3*x + 2) + 3/343*log(2*x - 1)

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mupad [B]  time = 0.04, size = 25, normalized size = 0.58 \begin {gather*} \frac {\frac {x}{98}+\frac {73}{784}}{x^2-x+\frac {1}{4}}-\frac {6\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{343} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x/98 + 73/784)/(x^2 - x + 1/4) - (6*atanh((12*x)/7 + 1/7))/343

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sympy [A]  time = 0.14, size = 36, normalized size = 0.84 \begin {gather*} - \frac {- 8 x - 73}{784 x^{2} - 784 x + 196} + \frac {3 \log {\left (x - \frac {1}{2} \right )}}{343} - \frac {3 \log {\left (x + \frac {2}{3} \right )}}{343} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-8*x - 73)/(784*x**2 - 784*x + 196) + 3*log(x - 1/2)/343 - 3*log(x + 2/3)/343

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