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

Optimal. Leaf size=45 \[ \frac {18 x^2}{125}-\frac {264 x}{625}-\frac {682}{3125 (5 x+3)}-\frac {121}{6250 (5 x+3)^2}+\frac {829 \log (5 x+3)}{3125} \]

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Rubi [A]  time = 0.02, antiderivative size = 45, 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 {18 x^2}{125}-\frac {264 x}{625}-\frac {682}{3125 (5 x+3)}-\frac {121}{6250 (5 x+3)^2}+\frac {829 \log (5 x+3)}{3125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-264*x)/625 + (18*x^2)/125 - 121/(6250*(3 + 5*x)^2) - 682/(3125*(3 + 5*x)) + (829*Log[3 + 5*x])/3125

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 {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^3} \, dx &=\int \left (-\frac {264}{625}+\frac {36 x}{125}+\frac {121}{625 (3+5 x)^3}+\frac {682}{625 (3+5 x)^2}+\frac {829}{625 (3+5 x)}\right ) \, dx\\ &=-\frac {264 x}{625}+\frac {18 x^2}{125}-\frac {121}{6250 (3+5 x)^2}-\frac {682}{3125 (3+5 x)}+\frac {829 \log (3+5 x)}{3125}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.93 \begin {gather*} \frac {\frac {5 \left (4500 x^4-7800 x^3-23760 x^2-17564 x-4277\right )}{(5 x+3)^2}+1658 \log (5 x+3)}{6250} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*(-4277 - 17564*x - 23760*x^2 - 7800*x^3 + 4500*x^4))/(3 + 5*x)^2 + 1658*Log[3 + 5*x])/6250

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.21, size = 52, normalized size = 1.16 \begin {gather*} \frac {22500 \, x^{4} - 39000 \, x^{3} - 71100 \, x^{2} + 1658 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 30580 \, x - 4213}{6250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6250*(22500*x^4 - 39000*x^3 - 71100*x^2 + 1658*(25*x^2 + 30*x + 9)*log(5*x + 3) - 30580*x - 4213)/(25*x^2 +
30*x + 9)

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giac [A]  time = 0.96, size = 32, normalized size = 0.71 \begin {gather*} \frac {18}{125} \, x^{2} - \frac {264}{625} \, x - \frac {11 \, {\left (620 \, x + 383\right )}}{6250 \, {\left (5 \, x + 3\right )}^{2}} + \frac {829}{3125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18/125*x^2 - 264/625*x - 11/6250*(620*x + 383)/(5*x + 3)^2 + 829/3125*log(abs(5*x + 3))

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maple [A]  time = 0.01, size = 36, normalized size = 0.80 \begin {gather*} \frac {18 x^{2}}{125}-\frac {264 x}{625}+\frac {829 \ln \left (5 x +3\right )}{3125}-\frac {121}{6250 \left (5 x +3\right )^{2}}-\frac {682}{3125 \left (5 x +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-264/625*x+18/125*x^2-121/6250/(5*x+3)^2-682/3125/(5*x+3)+829/3125*ln(5*x+3)

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maxima [A]  time = 0.56, size = 36, normalized size = 0.80 \begin {gather*} \frac {18}{125} \, x^{2} - \frac {264}{625} \, x - \frac {11 \, {\left (620 \, x + 383\right )}}{6250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {829}{3125} \, \log \left (5 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18/125*x^2 - 264/625*x - 11/6250*(620*x + 383)/(25*x^2 + 30*x + 9) + 829/3125*log(5*x + 3)

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mupad [B]  time = 0.04, size = 32, normalized size = 0.71 \begin {gather*} \frac {829\,\ln \left (x+\frac {3}{5}\right )}{3125}-\frac {264\,x}{625}-\frac {\frac {682\,x}{15625}+\frac {4213}{156250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}}+\frac {18\,x^2}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(829*log(x + 3/5))/3125 - (264*x)/625 - ((682*x)/15625 + 4213/156250)/((6*x)/5 + x^2 + 9/25) + (18*x^2)/125

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sympy [A]  time = 0.13, size = 37, normalized size = 0.82 \begin {gather*} \frac {18 x^{2}}{125} - \frac {264 x}{625} + \frac {- 6820 x - 4213}{156250 x^{2} + 187500 x + 56250} + \frac {829 \log {\left (5 x + 3 \right )}}{3125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18*x**2/125 - 264*x/625 + (-6820*x - 4213)/(156250*x**2 + 187500*x + 56250) + 829*log(5*x + 3)/3125

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