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

Optimal. Leaf size=54 \[ -\frac {1421}{5324 (1-2 x)}-\frac {1}{6655 (5 x+3)}+\frac {343}{968 (1-2 x)^2}-\frac {21 \log (1-2 x)}{14641}+\frac {21 \log (5 x+3)}{14641} \]

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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 {1421}{5324 (1-2 x)}-\frac {1}{6655 (5 x+3)}+\frac {343}{968 (1-2 x)^2}-\frac {21 \log (1-2 x)}{14641}+\frac {21 \log (5 x+3)}{14641} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

343/(968*(1 - 2*x)^2) - 1421/(5324*(1 - 2*x)) - 1/(6655*(3 + 5*x)) - (21*Log[1 - 2*x])/14641 + (21*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)^3}{(1-2 x)^3 (3+5 x)^2} \, dx &=\int \left (-\frac {343}{242 (-1+2 x)^3}-\frac {1421}{2662 (-1+2 x)^2}-\frac {42}{14641 (-1+2 x)}+\frac {1}{1331 (3+5 x)^2}+\frac {105}{14641 (3+5 x)}\right ) \, dx\\ &=\frac {343}{968 (1-2 x)^2}-\frac {1421}{5324 (1-2 x)}-\frac {1}{6655 (3+5 x)}-\frac {21 \log (1-2 x)}{14641}+\frac {21 \log (3+5 x)}{14641}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 0.87 \begin {gather*} \frac {\frac {11 \left (142068 x^2+108567 x+13957\right )}{(1-2 x)^2 (5 x+3)}-840 \log (1-2 x)+840 \log (10 x+6)}{585640} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((11*(13957 + 108567*x + 142068*x^2))/((1 - 2*x)^2*(3 + 5*x)) - 840*Log[1 - 2*x] + 840*Log[6 + 10*x])/585640

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.01, size = 75, normalized size = 1.39 \begin {gather*} \frac {1562748 \, x^{2} + 840 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 840 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (2 \, x - 1\right ) + 1194237 \, x + 153527}{585640 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/585640*(1562748*x^2 + 840*(20*x^3 - 8*x^2 - 7*x + 3)*log(5*x + 3) - 840*(20*x^3 - 8*x^2 - 7*x + 3)*log(2*x -
1) + 1194237*x + 153527)/(20*x^3 - 8*x^2 - 7*x + 3)

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giac [A]  time = 1.21, size = 51, normalized size = 0.94 \begin {gather*} -\frac {1}{6655 \, {\left (5 \, x + 3\right )}} + \frac {245 \, {\left (\frac {66}{5 \, x + 3} + 23\right )}}{29282 \, {\left (\frac {11}{5 \, x + 3} - 2\right )}^{2}} - \frac {21}{14641} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6655/(5*x + 3) + 245/29282*(66/(5*x + 3) + 23)/(11/(5*x + 3) - 2)^2 - 21/14641*log(abs(-11/(5*x + 3) + 2))

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maple [A]  time = 0.01, size = 45, normalized size = 0.83 \begin {gather*} -\frac {21 \ln \left (2 x -1\right )}{14641}+\frac {21 \ln \left (5 x +3\right )}{14641}-\frac {1}{6655 \left (5 x +3\right )}+\frac {343}{968 \left (2 x -1\right )^{2}}+\frac {1421}{5324 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6655/(5*x+3)+21/14641*ln(5*x+3)+343/968/(2*x-1)^2+1421/5324/(2*x-1)-21/14641*ln(2*x-1)

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maxima [A]  time = 0.61, size = 46, normalized size = 0.85 \begin {gather*} \frac {142068 \, x^{2} + 108567 \, x + 13957}{53240 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} + \frac {21}{14641} \, \log \left (5 \, x + 3\right ) - \frac {21}{14641} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/53240*(142068*x^2 + 108567*x + 13957)/(20*x^3 - 8*x^2 - 7*x + 3) + 21/14641*log(5*x + 3) - 21/14641*log(2*x
- 1)

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mupad [B]  time = 0.04, size = 38, normalized size = 0.70 \begin {gather*} \frac {42\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}-\frac {\frac {35517\,x^2}{266200}+\frac {108567\,x}{1064800}+\frac {13957}{1064800}}{-x^3+\frac {2\,x^2}{5}+\frac {7\,x}{20}-\frac {3}{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(42*atanh((20*x)/11 + 1/11))/14641 - ((108567*x)/1064800 + (35517*x^2)/266200 + 13957/1064800)/((7*x)/20 + (2*
x^2)/5 - x^3 - 3/20)

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sympy [A]  time = 0.18, size = 46, normalized size = 0.85 \begin {gather*} - \frac {- 142068 x^{2} - 108567 x - 13957}{1064800 x^{3} - 425920 x^{2} - 372680 x + 159720} - \frac {21 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {21 \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)**3/(1-2*x)**3/(3+5*x)**2,x)

[Out]

-(-142068*x**2 - 108567*x - 13957)/(1064800*x**3 - 425920*x**2 - 372680*x + 159720) - 21*log(x - 1/2)/14641 +
21*log(x + 3/5)/14641

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