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

Optimal. Leaf size=43 \[ -\frac {407}{196 (1-2 x)}+\frac {121}{56 (1-2 x)^2}-\frac {1}{343} \log (1-2 x)+\frac {1}{343} \log (3 x+2) \]

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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 = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {88} \begin {gather*} -\frac {407}{196 (1-2 x)}+\frac {121}{56 (1-2 x)^2}-\frac {1}{343} \log (1-2 x)+\frac {1}{343} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

121/(56*(1 - 2*x)^2) - 407/(196*(1 - 2*x)) - Log[1 - 2*x]/343 + Log[2 + 3*x]/343

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 {(3+5 x)^2}{(1-2 x)^3 (2+3 x)} \, dx &=\int \left (-\frac {121}{14 (-1+2 x)^3}-\frac {407}{98 (-1+2 x)^2}-\frac {2}{343 (-1+2 x)}+\frac {3}{343 (2+3 x)}\right ) \, dx\\ &=\frac {121}{56 (1-2 x)^2}-\frac {407}{196 (1-2 x)}-\frac {1}{343} \log (1-2 x)+\frac {1}{343} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 35, normalized size = 0.81 \begin {gather*} \frac {\frac {77 (148 x+3)}{(1-2 x)^2}-8 \log (3-6 x)+8 \log (3 x+2)}{2744} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((77*(3 + 148*x))/(1 - 2*x)^2 - 8*Log[3 - 6*x] + 8*Log[2 + 3*x])/2744

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.55, size = 55, normalized size = 1.28 \begin {gather*} \frac {8 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (3 \, x + 2\right ) - 8 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) + 11396 \, x + 231}{2744 \, {\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)^2/(1-2*x)^3/(2+3*x),x, algorithm="fricas")

[Out]

1/2744*(8*(4*x^2 - 4*x + 1)*log(3*x + 2) - 8*(4*x^2 - 4*x + 1)*log(2*x - 1) + 11396*x + 231)/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.19, size = 33, normalized size = 0.77 \begin {gather*} \frac {11 \, {\left (148 \, x + 3\right )}}{392 \, {\left (2 \, x - 1\right )}^{2}} + \frac {1}{343} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {1}{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)^2/(1-2*x)^3/(2+3*x),x, algorithm="giac")

[Out]

11/392*(148*x + 3)/(2*x - 1)^2 + 1/343*log(abs(3*x + 2)) - 1/343*log(abs(2*x - 1))

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maple [A]  time = 0.01, size = 36, normalized size = 0.84 \begin {gather*} -\frac {\ln \left (2 x -1\right )}{343}+\frac {\ln \left (3 x +2\right )}{343}+\frac {121}{56 \left (2 x -1\right )^{2}}+\frac {407}{196 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/343*ln(3*x+2)+121/56/(2*x-1)^2+407/196/(2*x-1)-1/343*ln(2*x-1)

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maxima [A]  time = 0.59, size = 36, normalized size = 0.84 \begin {gather*} \frac {11 \, {\left (148 \, x + 3\right )}}{392 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {1}{343} \, \log \left (3 \, x + 2\right ) - \frac {1}{343} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/392*(148*x + 3)/(4*x^2 - 4*x + 1) + 1/343*log(3*x + 2) - 1/343*log(2*x - 1)

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mupad [B]  time = 0.04, size = 25, normalized size = 0.58 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{343}+\frac {\frac {407\,x}{392}+\frac {33}{1568}}{x^2-x+\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh((12*x)/7 + 1/7))/343 + ((407*x)/392 + 33/1568)/(x^2 - x + 1/4)

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sympy [A]  time = 0.15, size = 32, normalized size = 0.74 \begin {gather*} - \frac {- 1628 x - 33}{1568 x^{2} - 1568 x + 392} - \frac {\log {\left (x - \frac {1}{2} \right )}}{343} + \frac {\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)**2/(1-2*x)**3/(2+3*x),x)

[Out]

-(-1628*x - 33)/(1568*x**2 - 1568*x + 392) - log(x - 1/2)/343 + log(x + 2/3)/343

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