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

Optimal. Leaf size=54 \[ \frac {49}{1331 (1-2 x)}-\frac {14}{1331 (5 x+3)}-\frac {1}{1210 (5 x+3)^2}-\frac {273 \log (1-2 x)}{14641}+\frac {273 \log (5 x+3)}{14641} \]

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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 = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {88} \begin {gather*} \frac {49}{1331 (1-2 x)}-\frac {14}{1331 (5 x+3)}-\frac {1}{1210 (5 x+3)^2}-\frac {273 \log (1-2 x)}{14641}+\frac {273 \log (5 x+3)}{14641} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(1331*(1 - 2*x)) - 1/(1210*(3 + 5*x)^2) - 14/(1331*(3 + 5*x)) - (273*Log[1 - 2*x])/14641 + (273*Log[3 + 5*x
])/14641

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)^2 (3+5 x)^3} \, dx &=\int \left (\frac {98}{1331 (-1+2 x)^2}-\frac {546}{14641 (-1+2 x)}+\frac {1}{121 (3+5 x)^3}+\frac {70}{1331 (3+5 x)^2}+\frac {1365}{14641 (3+5 x)}\right ) \, dx\\ &=\frac {49}{1331 (1-2 x)}-\frac {1}{1210 (3+5 x)^2}-\frac {14}{1331 (3+5 x)}-\frac {273 \log (1-2 x)}{14641}+\frac {273 \log (3+5 x)}{14641}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 47, normalized size = 0.87 \begin {gather*} \frac {-\frac {11 \left (13650 x^2+14862 x+3979\right )}{(2 x-1) (5 x+3)^2}-2730 \log (1-2 x)+2730 \log (10 x+6)}{146410} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-11*(3979 + 14862*x + 13650*x^2))/((-1 + 2*x)*(3 + 5*x)^2) - 2730*Log[1 - 2*x] + 2730*Log[6 + 10*x])/146410

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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)^2 (3+5 x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.50, size = 75, normalized size = 1.39 \begin {gather*} -\frac {150150 \, x^{2} - 2730 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (5 \, x + 3\right ) + 2730 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (2 \, x - 1\right ) + 163482 \, x + 43769}{146410 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/146410*(150150*x^2 - 2730*(50*x^3 + 35*x^2 - 12*x - 9)*log(5*x + 3) + 2730*(50*x^3 + 35*x^2 - 12*x - 9)*log
(2*x - 1) + 163482*x + 43769)/(50*x^3 + 35*x^2 - 12*x - 9)

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giac [A]  time = 1.25, size = 51, normalized size = 0.94 \begin {gather*} -\frac {49}{1331 \, {\left (2 \, x - 1\right )}} + \frac {2 \, {\left (\frac {792}{2 \, x - 1} + 355\right )}}{14641 \, {\left (\frac {11}{2 \, x - 1} + 5\right )}^{2}} + \frac {273}{14641} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/1331/(2*x - 1) + 2/14641*(792/(2*x - 1) + 355)/(11/(2*x - 1) + 5)^2 + 273/14641*log(abs(-11/(2*x - 1) - 5)
)

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maple [A]  time = 0.01, size = 45, normalized size = 0.83 \begin {gather*} -\frac {273 \ln \left (2 x -1\right )}{14641}+\frac {273 \ln \left (5 x +3\right )}{14641}-\frac {1}{1210 \left (5 x +3\right )^{2}}-\frac {14}{1331 \left (5 x +3\right )}-\frac {49}{1331 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1210/(5*x+3)^2-14/1331/(5*x+3)+273/14641*ln(5*x+3)-49/1331/(2*x-1)-273/14641*ln(2*x-1)

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maxima [A]  time = 0.59, size = 46, normalized size = 0.85 \begin {gather*} -\frac {13650 \, x^{2} + 14862 \, x + 3979}{13310 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} + \frac {273}{14641} \, \log \left (5 \, x + 3\right ) - \frac {273}{14641} \, \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)^2/(3+5*x)^3,x, algorithm="maxima")

[Out]

-1/13310*(13650*x^2 + 14862*x + 3979)/(50*x^3 + 35*x^2 - 12*x - 9) + 273/14641*log(5*x + 3) - 273/14641*log(2*
x - 1)

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mupad [B]  time = 0.04, size = 37, normalized size = 0.69 \begin {gather*} \frac {546\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}+\frac {\frac {273\,x^2}{13310}+\frac {7431\,x}{332750}+\frac {3979}{665500}}{-x^3-\frac {7\,x^2}{10}+\frac {6\,x}{25}+\frac {9}{50}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(546*atanh((20*x)/11 + 1/11))/14641 + ((7431*x)/332750 + (273*x^2)/13310 + 3979/665500)/((6*x)/25 - (7*x^2)/10
- x^3 + 9/50)

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sympy [A]  time = 0.16, size = 46, normalized size = 0.85 \begin {gather*} \frac {- 13650 x^{2} - 14862 x - 3979}{665500 x^{3} + 465850 x^{2} - 159720 x - 119790} - \frac {273 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {273 \log {\left (x + \frac {3}{5} \right )}}{14641} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-13650*x**2 - 14862*x - 3979)/(665500*x**3 + 465850*x**2 - 159720*x - 119790) - 273*log(x - 1/2)/14641 + 273*
log(x + 3/5)/14641

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