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

Optimal. Leaf size=54 \[ \frac {20}{1331 (1-2 x)}-\frac {25}{1331 (5 x+3)}+\frac {1}{121 (1-2 x)^2}-\frac {150 \log (1-2 x)}{14641}+\frac {150 \log (5 x+3)}{14641} \]

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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 = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {44} \begin {gather*} \frac {20}{1331 (1-2 x)}-\frac {25}{1331 (5 x+3)}+\frac {1}{121 (1-2 x)^2}-\frac {150 \log (1-2 x)}{14641}+\frac {150 \log (5 x+3)}{14641} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - 2*x)^3*(3 + 5*x)^2),x]

[Out]

1/(121*(1 - 2*x)^2) + 20/(1331*(1 - 2*x)) - 25/(1331*(3 + 5*x)) - (150*Log[1 - 2*x])/14641 + (150*Log[3 + 5*x]
)/14641

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {1}{(1-2 x)^3 (3+5 x)^2} \, dx &=\int \left (-\frac {4}{121 (-1+2 x)^3}+\frac {40}{1331 (-1+2 x)^2}-\frac {300}{14641 (-1+2 x)}+\frac {125}{1331 (3+5 x)^2}+\frac {750}{14641 (3+5 x)}\right ) \, dx\\ &=\frac {1}{121 (1-2 x)^2}+\frac {20}{1331 (1-2 x)}-\frac {25}{1331 (3+5 x)}-\frac {150 \log (1-2 x)}{14641}+\frac {150 \log (3+5 x)}{14641}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 62, normalized size = 1.15 \begin {gather*} \frac {50}{1331 (5 (1-2 x)-11)}+\frac {20}{1331 (1-2 x)}+\frac {1}{121 (1-2 x)^2}+\frac {150 \log (11-5 (1-2 x))}{14641}-\frac {150 \log (1-2 x)}{14641} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

50/(1331*(-11 + 5*(1 - 2*x))) + 1/(121*(1 - 2*x)^2) + 20/(1331*(1 - 2*x)) + (150*Log[11 - 5*(1 - 2*x)])/14641
- (150*Log[1 - 2*x])/14641

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 75, normalized size = 1.39 \begin {gather*} -\frac {3300 \, x^{2} - 150 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 150 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 1485 \, x - 748}{14641 \, {\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(1/(1-2*x)^3/(3+5*x)^2,x, algorithm="fricas")

[Out]

-1/14641*(3300*x^2 - 150*(20*x^3 - 8*x^2 - 7*x + 3)*log(5*x + 3) + 150*(20*x^3 - 8*x^2 - 7*x + 3)*log(2*x - 1)
- 1485*x - 748)/(20*x^3 - 8*x^2 - 7*x + 3)

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giac [A]  time = 1.18, size = 51, normalized size = 0.94 \begin {gather*} -\frac {25}{1331 \, {\left (5 \, x + 3\right )}} + \frac {100 \, {\left (\frac {33}{5 \, x + 3} - 5\right )}}{14641 \, {\left (\frac {11}{5 \, x + 3} - 2\right )}^{2}} - \frac {150}{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(1/(1-2*x)^3/(3+5*x)^2,x, algorithm="giac")

[Out]

-25/1331/(5*x + 3) + 100/14641*(33/(5*x + 3) - 5)/(11/(5*x + 3) - 2)^2 - 150/14641*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 {150 \ln \left (2 x -1\right )}{14641}+\frac {150 \ln \left (5 x +3\right )}{14641}-\frac {25}{1331 \left (5 x +3\right )}+\frac {1}{121 \left (2 x -1\right )^{2}}-\frac {20}{1331 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/1331/(5*x+3)+150/14641*ln(5*x+3)+1/121/(2*x-1)^2-20/1331/(2*x-1)-150/14641*ln(2*x-1)

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maxima [A]  time = 0.59, size = 46, normalized size = 0.85 \begin {gather*} -\frac {300 \, x^{2} - 135 \, x - 68}{1331 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} + \frac {150}{14641} \, \log \left (5 \, x + 3\right ) - \frac {150}{14641} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1331*(300*x^2 - 135*x - 68)/(20*x^3 - 8*x^2 - 7*x + 3) + 150/14641*log(5*x + 3) - 150/14641*log(2*x - 1)

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mupad [B]  time = 1.10, size = 38, normalized size = 0.70 \begin {gather*} \frac {300\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}-\frac {-\frac {15\,x^2}{1331}+\frac {27\,x}{5324}+\frac {17}{6655}}{-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(-1/((2*x - 1)^3*(5*x + 3)^2),x)

[Out]

(300*atanh((20*x)/11 + 1/11))/14641 - ((27*x)/5324 - (15*x^2)/1331 + 17/6655)/((7*x)/20 + (2*x^2)/5 - x^3 - 3/
20)

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sympy [A]  time = 0.18, size = 44, normalized size = 0.81 \begin {gather*} - \frac {300 x^{2} - 135 x - 68}{26620 x^{3} - 10648 x^{2} - 9317 x + 3993} - \frac {150 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {150 \log {\left (x + \frac {3}{5} \right )}}{14641} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(300*x**2 - 135*x - 68)/(26620*x**3 - 10648*x**2 - 9317*x + 3993) - 150*log(x - 1/2)/14641 + 150*log(x + 3/5)
/14641

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