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

Optimal. Leaf size=54 \[ \frac {14}{1331 (1-2 x)}-\frac {37}{1331 (5 x+3)}-\frac {1}{242 (5 x+3)^2}-\frac {144 \log (1-2 x)}{14641}+\frac {144 \log (5 x+3)}{14641} \]

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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 = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \begin {gather*} \frac {14}{1331 (1-2 x)}-\frac {37}{1331 (5 x+3)}-\frac {1}{242 (5 x+3)^2}-\frac {144 \log (1-2 x)}{14641}+\frac {144 \log (5 x+3)}{14641} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

14/(1331*(1 - 2*x)) - 1/(242*(3 + 5*x)^2) - 37/(1331*(3 + 5*x)) - (144*Log[1 - 2*x])/14641 + (144*Log[3 + 5*x]
)/14641

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {2+3 x}{(1-2 x)^2 (3+5 x)^3} \, dx &=\int \left (\frac {28}{1331 (-1+2 x)^2}-\frac {288}{14641 (-1+2 x)}+\frac {5}{121 (3+5 x)^3}+\frac {185}{1331 (3+5 x)^2}+\frac {720}{14641 (3+5 x)}\right ) \, dx\\ &=\frac {14}{1331 (1-2 x)}-\frac {1}{242 (3+5 x)^2}-\frac {37}{1331 (3+5 x)}-\frac {144 \log (1-2 x)}{14641}+\frac {144 \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 (1440 x^2+936 x+19\right )}{(2 x-1) (5 x+3)^2}-288 \log (1-2 x)+288 \log (10 x+6)}{29282} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-11*(19 + 936*x + 1440*x^2))/((-1 + 2*x)*(3 + 5*x)^2) - 288*Log[1 - 2*x] + 288*Log[6 + 10*x])/29282

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.30, size = 75, normalized size = 1.39 \begin {gather*} -\frac {15840 \, x^{2} - 288 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (5 \, x + 3\right ) + 288 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (2 \, x - 1\right ) + 10296 \, x + 209}{29282 \, {\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)/(1-2*x)^2/(3+5*x)^3,x, algorithm="fricas")

[Out]

-1/29282*(15840*x^2 - 288*(50*x^3 + 35*x^2 - 12*x - 9)*log(5*x + 3) + 288*(50*x^3 + 35*x^2 - 12*x - 9)*log(2*x
- 1) + 10296*x + 209)/(50*x^3 + 35*x^2 - 12*x - 9)

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giac [A]  time = 1.15, size = 51, normalized size = 0.94 \begin {gather*} -\frac {14}{1331 \, {\left (2 \, x - 1\right )}} + \frac {10 \, {\left (\frac {429}{2 \, x - 1} + 190\right )}}{14641 \, {\left (\frac {11}{2 \, x - 1} + 5\right )}^{2}} + \frac {144}{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)/(1-2*x)^2/(3+5*x)^3,x, algorithm="giac")

[Out]

-14/1331/(2*x - 1) + 10/14641*(429/(2*x - 1) + 190)/(11/(2*x - 1) + 5)^2 + 144/14641*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 {144 \ln \left (2 x -1\right )}{14641}+\frac {144 \ln \left (5 x +3\right )}{14641}-\frac {1}{242 \left (5 x +3\right )^{2}}-\frac {37}{1331 \left (5 x +3\right )}-\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)/(1-2*x)^2/(5*x+3)^3,x)

[Out]

-1/242/(5*x+3)^2-37/1331/(5*x+3)+144/14641*ln(5*x+3)-14/1331/(2*x-1)-144/14641*ln(2*x-1)

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maxima [A]  time = 0.48, size = 46, normalized size = 0.85 \begin {gather*} -\frac {1440 \, x^{2} + 936 \, x + 19}{2662 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} + \frac {144}{14641} \, \log \left (5 \, x + 3\right ) - \frac {144}{14641} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2662*(1440*x^2 + 936*x + 19)/(50*x^3 + 35*x^2 - 12*x - 9) + 144/14641*log(5*x + 3) - 144/14641*log(2*x - 1)

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mupad [B]  time = 0.04, size = 37, normalized size = 0.69 \begin {gather*} \frac {288\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}+\frac {\frac {72\,x^2}{6655}+\frac {234\,x}{33275}+\frac {19}{133100}}{-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*x - 1)^2*(5*x + 3)^3),x)

[Out]

(288*atanh((20*x)/11 + 1/11))/14641 + ((234*x)/33275 + (72*x^2)/6655 + 19/133100)/((6*x)/25 - (7*x^2)/10 - x^3
+ 9/50)

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sympy [A]  time = 0.15, size = 46, normalized size = 0.85 \begin {gather*} \frac {- 1440 x^{2} - 936 x - 19}{133100 x^{3} + 93170 x^{2} - 31944 x - 23958} - \frac {144 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {144 \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)/(1-2*x)**2/(3+5*x)**3,x)

[Out]

(-1440*x**2 - 936*x - 19)/(133100*x**3 + 93170*x**2 - 31944*x - 23958) - 144*log(x - 1/2)/14641 + 144*log(x +
3/5)/14641

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