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

Optimal. Leaf size=54 \[ \frac {31}{343 (1-2 x)}+\frac {3}{343 (3 x+2)}+\frac {11}{98 (1-2 x)^2}-\frac {87 \log (1-2 x)}{2401}+\frac {87 \log (3 x+2)}{2401} \]

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Rubi [A]  time = 0.03, 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 {31}{343 (1-2 x)}+\frac {3}{343 (3 x+2)}+\frac {11}{98 (1-2 x)^2}-\frac {87 \log (1-2 x)}{2401}+\frac {87 \log (3 x+2)}{2401} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

11/(98*(1 - 2*x)^2) + 31/(343*(1 - 2*x)) + 3/(343*(2 + 3*x)) - (87*Log[1 - 2*x])/2401 + (87*Log[2 + 3*x])/2401

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)^2} \, dx &=\int \left (-\frac {22}{49 (-1+2 x)^3}+\frac {62}{343 (-1+2 x)^2}-\frac {174}{2401 (-1+2 x)}-\frac {9}{343 (2+3 x)^2}+\frac {261}{2401 (2+3 x)}\right ) \, dx\\ &=\frac {11}{98 (1-2 x)^2}+\frac {31}{343 (1-2 x)}+\frac {3}{343 (2+3 x)}-\frac {87 \log (1-2 x)}{2401}+\frac {87 \log (2+3 x)}{2401}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 47, normalized size = 0.87 \begin {gather*} \frac {\frac {7 \left (-348 x^2+145 x+284\right )}{(1-2 x)^2 (3 x+2)}-174 \log (1-2 x)+174 \log (6 x+4)}{4802} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*(284 + 145*x - 348*x^2))/((1 - 2*x)^2*(2 + 3*x)) - 174*Log[1 - 2*x] + 174*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)^3 (2+3 x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.53, size = 75, normalized size = 1.39 \begin {gather*} -\frac {2436 \, x^{2} - 174 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 174 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (2 \, x - 1\right ) - 1015 \, x - 1988}{4802 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\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)^2,x, algorithm="fricas")

[Out]

-1/4802*(2436*x^2 - 174*(12*x^3 - 4*x^2 - 5*x + 2)*log(3*x + 2) + 174*(12*x^3 - 4*x^2 - 5*x + 2)*log(2*x - 1)
- 1015*x - 1988)/(12*x^3 - 4*x^2 - 5*x + 2)

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giac [A]  time = 1.18, size = 51, normalized size = 0.94 \begin {gather*} \frac {3}{343 \, {\left (3 \, x + 2\right )}} + \frac {6 \, {\left (\frac {448}{3 \, x + 2} - 95\right )}}{2401 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}^{2}} - \frac {87}{2401} \, \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)^3/(2+3*x)^2,x, algorithm="giac")

[Out]

3/343/(3*x + 2) + 6/2401*(448/(3*x + 2) - 95)/(7/(3*x + 2) - 2)^2 - 87/2401*log(abs(-7/(3*x + 2) + 2))

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maple [A]  time = 0.01, size = 45, normalized size = 0.83 \begin {gather*} -\frac {87 \ln \left (2 x -1\right )}{2401}+\frac {87 \ln \left (3 x +2\right )}{2401}+\frac {3}{343 \left (3 x +2\right )}+\frac {11}{98 \left (2 x -1\right )^{2}}-\frac {31}{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)^3/(3*x+2)^2,x)

[Out]

3/343/(3*x+2)+87/2401*ln(3*x+2)+11/98/(2*x-1)^2-31/343/(2*x-1)-87/2401*ln(2*x-1)

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maxima [A]  time = 0.55, size = 46, normalized size = 0.85 \begin {gather*} -\frac {348 \, x^{2} - 145 \, x - 284}{686 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} + \frac {87}{2401} \, \log \left (3 \, x + 2\right ) - \frac {87}{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)^3/(2+3*x)^2,x, algorithm="maxima")

[Out]

-1/686*(348*x^2 - 145*x - 284)/(12*x^3 - 4*x^2 - 5*x + 2) + 87/2401*log(3*x + 2) - 87/2401*log(2*x - 1)

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mupad [B]  time = 1.07, size = 38, normalized size = 0.70 \begin {gather*} \frac {174\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{2401}-\frac {-\frac {29\,x^2}{686}+\frac {145\,x}{8232}+\frac {71}{2058}}{-x^3+\frac {x^2}{3}+\frac {5\,x}{12}-\frac {1}{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(174*atanh((12*x)/7 + 1/7))/2401 - ((145*x)/8232 - (29*x^2)/686 + 71/2058)/((5*x)/12 + x^2/3 - x^3 - 1/6)

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sympy [A]  time = 0.16, size = 44, normalized size = 0.81 \begin {gather*} - \frac {348 x^{2} - 145 x - 284}{8232 x^{3} - 2744 x^{2} - 3430 x + 1372} - \frac {87 \log {\left (x - \frac {1}{2} \right )}}{2401} + \frac {87 \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)**3/(2+3*x)**2,x)

[Out]

-(348*x**2 - 145*x - 284)/(8232*x**3 - 2744*x**2 - 3430*x + 1372) - 87*log(x - 1/2)/2401 + 87*log(x + 2/3)/240
1

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