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

Optimal. Leaf size=54 \[ \frac {22}{343 (1-2 x)}-\frac {31}{343 (3 x+2)}+\frac {1}{98 (3 x+2)^2}-\frac {128 \log (1-2 x)}{2401}+\frac {128 \log (3 x+2)}{2401} \]

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Rubi [A]  time = 0.02, antiderivative size = 54, 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 {22}{343 (1-2 x)}-\frac {31}{343 (3 x+2)}+\frac {1}{98 (3 x+2)^2}-\frac {128 \log (1-2 x)}{2401}+\frac {128 \log (3 x+2)}{2401} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

22/(343*(1 - 2*x)) + 1/(98*(2 + 3*x)^2) - 31/(343*(2 + 3*x)) - (128*Log[1 - 2*x])/2401 + (128*Log[2 + 3*x])/24
01

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 (2+3 x)^3} \, dx &=\int \left (\frac {44}{343 (-1+2 x)^2}-\frac {256}{2401 (-1+2 x)}-\frac {3}{49 (2+3 x)^3}+\frac {93}{343 (2+3 x)^2}+\frac {384}{2401 (2+3 x)}\right ) \, dx\\ &=\frac {22}{343 (1-2 x)}+\frac {1}{98 (2+3 x)^2}-\frac {31}{343 (2+3 x)}-\frac {128 \log (1-2 x)}{2401}+\frac {128 \log (2+3 x)}{2401}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 47, normalized size = 0.87 \begin {gather*} \frac {-\frac {7 \left (768 x^2+576 x+59\right )}{(2 x-1) (3 x+2)^2}-256 \log (1-2 x)+256 \log (6 x+4)}{4802} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-7*(59 + 576*x + 768*x^2))/((-1 + 2*x)*(2 + 3*x)^2) - 256*Log[1 - 2*x] + 256*Log[4 + 6*x])/4802

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.37, size = 75, normalized size = 1.39 \begin {gather*} -\frac {5376 \, x^{2} - 256 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (3 \, x + 2\right ) + 256 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (2 \, x - 1\right ) + 4032 \, x + 413}{4802 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4802*(5376*x^2 - 256*(18*x^3 + 15*x^2 - 4*x - 4)*log(3*x + 2) + 256*(18*x^3 + 15*x^2 - 4*x - 4)*log(2*x - 1
) + 4032*x + 413)/(18*x^3 + 15*x^2 - 4*x - 4)

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giac [A]  time = 0.99, size = 51, normalized size = 0.94 \begin {gather*} -\frac {22}{343 \, {\left (2 \, x - 1\right )}} + \frac {6 \, {\left (\frac {203}{2 \, x - 1} + 90\right )}}{2401 \, {\left (\frac {7}{2 \, x - 1} + 3\right )}^{2}} + \frac {128}{2401} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-22/343/(2*x - 1) + 6/2401*(203/(2*x - 1) + 90)/(7/(2*x - 1) + 3)^2 + 128/2401*log(abs(-7/(2*x - 1) - 3))

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maple [A]  time = 0.01, size = 45, normalized size = 0.83 \begin {gather*} -\frac {128 \ln \left (2 x -1\right )}{2401}+\frac {128 \ln \left (3 x +2\right )}{2401}+\frac {1}{98 \left (3 x +2\right )^{2}}-\frac {31}{343 \left (3 x +2\right )}-\frac {22}{343 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/98/(3*x+2)^2-31/343/(3*x+2)+128/2401*ln(3*x+2)-22/343/(2*x-1)-128/2401*ln(2*x-1)

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maxima [A]  time = 0.57, size = 46, normalized size = 0.85 \begin {gather*} -\frac {768 \, x^{2} + 576 \, x + 59}{686 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} + \frac {128}{2401} \, \log \left (3 \, x + 2\right ) - \frac {128}{2401} \, \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/(2+3*x)^3,x, algorithm="maxima")

[Out]

-1/686*(768*x^2 + 576*x + 59)/(18*x^3 + 15*x^2 - 4*x - 4) + 128/2401*log(3*x + 2) - 128/2401*log(2*x - 1)

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mupad [B]  time = 1.08, size = 37, normalized size = 0.69 \begin {gather*} \frac {256\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{2401}+\frac {\frac {64\,x^2}{1029}+\frac {16\,x}{343}+\frac {59}{12348}}{-x^3-\frac {5\,x^2}{6}+\frac {2\,x}{9}+\frac {2}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(256*atanh((12*x)/7 + 1/7))/2401 + ((16*x)/343 + (64*x^2)/1029 + 59/12348)/((2*x)/9 - (5*x^2)/6 - x^3 + 2/9)

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sympy [A]  time = 0.15, size = 46, normalized size = 0.85 \begin {gather*} \frac {- 768 x^{2} - 576 x - 59}{12348 x^{3} + 10290 x^{2} - 2744 x - 2744} - \frac {128 \log {\left (x - \frac {1}{2} \right )}}{2401} + \frac {128 \log {\left (x + \frac {2}{3} \right )}}{2401} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-768*x**2 - 576*x - 59)/(12348*x**3 + 10290*x**2 - 2744*x - 2744) - 128*log(x - 1/2)/2401 + 128*log(x + 2/3)/
2401

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