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

Optimal. Leaf size=43 \[ -\frac {68}{3025 (5 x+3)}-\frac {1}{550 (5 x+3)^2}-\frac {49 \log (1-2 x)}{1331}+\frac {49 \log (5 x+3)}{1331} \]

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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 = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {88} \begin {gather*} -\frac {68}{3025 (5 x+3)}-\frac {1}{550 (5 x+3)^2}-\frac {49 \log (1-2 x)}{1331}+\frac {49 \log (5 x+3)}{1331} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(550*(3 + 5*x)^2) - 68/(3025*(3 + 5*x)) - (49*Log[1 - 2*x])/1331 + (49*Log[3 + 5*x])/1331

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)^3} \, dx &=\int \left (-\frac {98}{1331 (-1+2 x)}+\frac {1}{55 (3+5 x)^3}+\frac {68}{605 (3+5 x)^2}+\frac {245}{1331 (3+5 x)}\right ) \, dx\\ &=-\frac {1}{550 (3+5 x)^2}-\frac {68}{3025 (3+5 x)}-\frac {49 \log (1-2 x)}{1331}+\frac {49 \log (3+5 x)}{1331}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 35, normalized size = 0.81 \begin {gather*} \frac {-\frac {11 (680 x+419)}{(5 x+3)^2}-2450 \log (1-2 x)+2450 \log (10 x+6)}{66550} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-11*(419 + 680*x))/(3 + 5*x)^2 - 2450*Log[1 - 2*x] + 2450*Log[6 + 10*x])/66550

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.66, size = 55, normalized size = 1.28 \begin {gather*} \frac {2450 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 2450 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (2 \, x - 1\right ) - 7480 \, x - 4609}{66550 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/66550*(2450*(25*x^2 + 30*x + 9)*log(5*x + 3) - 2450*(25*x^2 + 30*x + 9)*log(2*x - 1) - 7480*x - 4609)/(25*x^
2 + 30*x + 9)

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giac [A]  time = 0.81, size = 33, normalized size = 0.77 \begin {gather*} -\frac {680 \, x + 419}{6050 \, {\left (5 \, x + 3\right )}^{2}} + \frac {49}{1331} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {49}{1331} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6050*(680*x + 419)/(5*x + 3)^2 + 49/1331*log(abs(5*x + 3)) - 49/1331*log(abs(2*x - 1))

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maple [A]  time = 0.01, size = 36, normalized size = 0.84 \begin {gather*} -\frac {49 \ln \left (2 x -1\right )}{1331}+\frac {49 \ln \left (5 x +3\right )}{1331}-\frac {1}{550 \left (5 x +3\right )^{2}}-\frac {68}{3025 \left (5 x +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/550/(5*x+3)^2-68/3025/(5*x+3)+49/1331*ln(5*x+3)-49/1331*ln(2*x-1)

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maxima [A]  time = 0.59, size = 36, normalized size = 0.84 \begin {gather*} -\frac {680 \, x + 419}{6050 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {49}{1331} \, \log \left (5 \, x + 3\right ) - \frac {49}{1331} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6050*(680*x + 419)/(25*x^2 + 30*x + 9) + 49/1331*log(5*x + 3) - 49/1331*log(2*x - 1)

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mupad [B]  time = 0.04, size = 26, normalized size = 0.60 \begin {gather*} \frac {98\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{1331}-\frac {\frac {68\,x}{15125}+\frac {419}{151250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(98*atanh((20*x)/11 + 1/11))/1331 - ((68*x)/15125 + 419/151250)/((6*x)/5 + x^2 + 9/25)

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sympy [A]  time = 0.15, size = 34, normalized size = 0.79 \begin {gather*} - \frac {680 x + 419}{151250 x^{2} + 181500 x + 54450} - \frac {49 \log {\left (x - \frac {1}{2} \right )}}{1331} + \frac {49 \log {\left (x + \frac {3}{5} \right )}}{1331} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(680*x + 419)/(151250*x**2 + 181500*x + 54450) - 49*log(x - 1/2)/1331 + 49*log(x + 3/5)/1331

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