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

Optimal. Leaf size=55 \[ \frac {740}{243 (3 x+2)}-\frac {503}{162 (3 x+2)^2}+\frac {518}{729 (3 x+2)^3}-\frac {49}{972 (3 x+2)^4}+\frac {100}{243} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 55, 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 {740}{243 (3 x+2)}-\frac {503}{162 (3 x+2)^2}+\frac {518}{729 (3 x+2)^3}-\frac {49}{972 (3 x+2)^4}+\frac {100}{243} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-49/(972*(2 + 3*x)^4) + 518/(729*(2 + 3*x)^3) - 503/(162*(2 + 3*x)^2) + 740/(243*(2 + 3*x)) + (100*Log[2 + 3*x
])/243

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 (3+5 x)^2}{(2+3 x)^5} \, dx &=\int \left (\frac {49}{81 (2+3 x)^5}-\frac {518}{81 (2+3 x)^4}+\frac {503}{27 (2+3 x)^3}-\frac {740}{81 (2+3 x)^2}+\frac {100}{81 (2+3 x)}\right ) \, dx\\ &=-\frac {49}{972 (2+3 x)^4}+\frac {518}{729 (2+3 x)^3}-\frac {503}{162 (2+3 x)^2}+\frac {740}{243 (2+3 x)}+\frac {100}{243} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.75 \begin {gather*} \frac {239760 x^3+398034 x^2+217248 x+1200 (3 x+2)^4 \log (3 x+2)+38821}{2916 (3 x+2)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(38821 + 217248*x + 398034*x^2 + 239760*x^3 + 1200*(2 + 3*x)^4*Log[2 + 3*x])/(2916*(2 + 3*x)^4)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.51, size = 67, normalized size = 1.22 \begin {gather*} \frac {239760 \, x^{3} + 398034 \, x^{2} + 1200 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 217248 \, x + 38821}{2916 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2916*(239760*x^3 + 398034*x^2 + 1200*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 217248*x + 3882
1)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

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giac [A]  time = 0.97, size = 55, normalized size = 1.00 \begin {gather*} \frac {740}{243 \, {\left (3 \, x + 2\right )}} - \frac {503}{162 \, {\left (3 \, x + 2\right )}^{2}} + \frac {518}{729 \, {\left (3 \, x + 2\right )}^{3}} - \frac {49}{972 \, {\left (3 \, x + 2\right )}^{4}} - \frac {100}{243} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

740/243/(3*x + 2) - 503/162/(3*x + 2)^2 + 518/729/(3*x + 2)^3 - 49/972/(3*x + 2)^4 - 100/243*log(1/3*abs(3*x +
2)/(3*x + 2)^2)

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maple [A]  time = 0.01, size = 46, normalized size = 0.84 \begin {gather*} \frac {100 \ln \left (3 x +2\right )}{243}-\frac {49}{972 \left (3 x +2\right )^{4}}+\frac {518}{729 \left (3 x +2\right )^{3}}-\frac {503}{162 \left (3 x +2\right )^{2}}+\frac {740}{243 \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/972/(3*x+2)^4+518/729/(3*x+2)^3-503/162/(3*x+2)^2+740/243/(3*x+2)+100/243*ln(3*x+2)

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maxima [A]  time = 0.52, size = 48, normalized size = 0.87 \begin {gather*} \frac {239760 \, x^{3} + 398034 \, x^{2} + 217248 \, x + 38821}{2916 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} + \frac {100}{243} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2916*(239760*x^3 + 398034*x^2 + 217248*x + 38821)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) + 100/243*log(3*x
+ 2)

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mupad [B]  time = 0.04, size = 43, normalized size = 0.78 \begin {gather*} \frac {100\,\ln \left (x+\frac {2}{3}\right )}{243}+\frac {\frac {740\,x^3}{729}+\frac {91\,x^2}{54}+\frac {18104\,x}{19683}+\frac {38821}{236196}}{x^4+\frac {8\,x^3}{3}+\frac {8\,x^2}{3}+\frac {32\,x}{27}+\frac {16}{81}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(100*log(x + 2/3))/243 + ((18104*x)/19683 + (91*x^2)/54 + (740*x^3)/729 + 38821/236196)/((32*x)/27 + (8*x^2)/3
+ (8*x^3)/3 + x^4 + 16/81)

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sympy [A]  time = 0.16, size = 44, normalized size = 0.80 \begin {gather*} \frac {239760 x^{3} + 398034 x^{2} + 217248 x + 38821}{236196 x^{4} + 629856 x^{3} + 629856 x^{2} + 279936 x + 46656} + \frac {100 \log {\left (3 x + 2 \right )}}{243} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(239760*x**3 + 398034*x**2 + 217248*x + 38821)/(236196*x**4 + 629856*x**3 + 629856*x**2 + 279936*x + 46656) +
100*log(3*x + 2)/243

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