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

Optimal. Leaf size=54 \[ \frac {14}{1331 (1-2 x)}-\frac {1}{1331 (5 x+3)}+\frac {49}{484 (1-2 x)^2}-\frac {72 \log (1-2 x)}{14641}+\frac {72 \log (5 x+3)}{14641} \]

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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 {14}{1331 (1-2 x)}-\frac {1}{1331 (5 x+3)}+\frac {49}{484 (1-2 x)^2}-\frac {72 \log (1-2 x)}{14641}+\frac {72 \log (5 x+3)}{14641} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(484*(1 - 2*x)^2) + 14/(1331*(1 - 2*x)) - 1/(1331*(3 + 5*x)) - (72*Log[1 - 2*x])/14641 + (72*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)^3 (3+5 x)^2} \, dx &=\int \left (-\frac {49}{121 (-1+2 x)^3}+\frac {28}{1331 (-1+2 x)^2}-\frac {144}{14641 (-1+2 x)}+\frac {5}{1331 (3+5 x)^2}+\frac {360}{14641 (3+5 x)}\right ) \, dx\\ &=\frac {49}{484 (1-2 x)^2}+\frac {14}{1331 (1-2 x)}-\frac {1}{1331 (3+5 x)}-\frac {72 \log (1-2 x)}{14641}+\frac {72 \log (3+5 x)}{14641}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 48, normalized size = 0.89 \begin {gather*} \frac {\frac {616}{1-2 x}-\frac {44}{5 x+3}+\frac {5929}{(1-2 x)^2}-288 \log (1-2 x)+288 \log (10 x+6)}{58564} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5929/(1 - 2*x)^2 + 616/(1 - 2*x) - 44/(3 + 5*x) - 288*Log[1 - 2*x] + 288*Log[6 + 10*x])/58564

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.11, size = 75, normalized size = 1.39 \begin {gather*} -\frac {6336 \, x^{2} - 288 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 288 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 29205 \, x - 19591}{58564 \, {\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)^2/(1-2*x)^3/(3+5*x)^2,x, algorithm="fricas")

[Out]

-1/58564*(6336*x^2 - 288*(20*x^3 - 8*x^2 - 7*x + 3)*log(5*x + 3) + 288*(20*x^3 - 8*x^2 - 7*x + 3)*log(2*x - 1)
 - 29205*x - 19591)/(20*x^3 - 8*x^2 - 7*x + 3)

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giac [A]  time = 1.19, size = 51, normalized size = 0.94 \begin {gather*} -\frac {1}{1331 \, {\left (5 \, x + 3\right )}} + \frac {35 \, {\left (\frac {429}{5 \, x + 3} - 43\right )}}{14641 \, {\left (\frac {11}{5 \, x + 3} - 2\right )}^{2}} - \frac {72}{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)^2/(1-2*x)^3/(3+5*x)^2,x, algorithm="giac")

[Out]

-1/1331/(5*x + 3) + 35/14641*(429/(5*x + 3) - 43)/(11/(5*x + 3) - 2)^2 - 72/14641*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 {72 \ln \left (2 x -1\right )}{14641}+\frac {72 \ln \left (5 x +3\right )}{14641}-\frac {1}{1331 \left (5 x +3\right )}+\frac {49}{484 \left (2 x -1\right )^{2}}-\frac {14}{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)^3/(5*x+3)^2,x)

[Out]

-1/1331/(5*x+3)+72/14641*ln(5*x+3)+49/484/(2*x-1)^2-14/1331/(2*x-1)-72/14641*ln(2*x-1)

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maxima [A]  time = 0.45, size = 46, normalized size = 0.85 \begin {gather*} -\frac {576 \, x^{2} - 2655 \, x - 1781}{5324 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} + \frac {72}{14641} \, \log \left (5 \, x + 3\right ) - \frac {72}{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)^3/(3+5*x)^2,x, algorithm="maxima")

[Out]

-1/5324*(576*x^2 - 2655*x - 1781)/(20*x^3 - 8*x^2 - 7*x + 3) + 72/14641*log(5*x + 3) - 72/14641*log(2*x - 1)

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mupad [B]  time = 1.08, size = 38, normalized size = 0.70 \begin {gather*} \frac {144\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}-\frac {-\frac {36\,x^2}{6655}+\frac {531\,x}{21296}+\frac {1781}{106480}}{-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)^2/((2*x - 1)^3*(5*x + 3)^2),x)

[Out]

(144*atanh((20*x)/11 + 1/11))/14641 - ((531*x)/21296 - (36*x^2)/6655 + 1781/106480)/((7*x)/20 + (2*x^2)/5 - x^
3 - 3/20)

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sympy [A]  time = 0.17, size = 44, normalized size = 0.81 \begin {gather*} - \frac {576 x^{2} - 2655 x - 1781}{106480 x^{3} - 42592 x^{2} - 37268 x + 15972} - \frac {72 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {72 \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)**3/(3+5*x)**2,x)

[Out]

-(576*x**2 - 2655*x - 1781)/(106480*x**3 - 42592*x**2 - 37268*x + 15972) - 72*log(x - 1/2)/14641 + 72*log(x +
3/5)/14641

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