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

Optimal. Leaf size=55 \[ -\frac {428}{243 (3 x+2)}+\frac {259}{81 (3 x+2)^2}-\frac {2009}{729 (3 x+2)^3}+\frac {343}{972 (3 x+2)^4}-\frac {40}{243} \log (3 x+2) \]

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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 = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \begin {gather*} -\frac {428}{243 (3 x+2)}+\frac {259}{81 (3 x+2)^2}-\frac {2009}{729 (3 x+2)^3}+\frac {343}{972 (3 x+2)^4}-\frac {40}{243} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

343/(972*(2 + 3*x)^4) - 2009/(729*(2 + 3*x)^3) + 259/(81*(2 + 3*x)^2) - 428/(243*(2 + 3*x)) - (40*Log[2 + 3*x]
)/243

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 {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx &=\int \left (-\frac {343}{81 (2+3 x)^5}+\frac {2009}{81 (2+3 x)^4}-\frac {518}{27 (2+3 x)^3}+\frac {428}{81 (2+3 x)^2}-\frac {40}{81 (2+3 x)}\right ) \, dx\\ &=\frac {343}{972 (2+3 x)^4}-\frac {2009}{729 (2+3 x)^3}+\frac {259}{81 (2+3 x)^2}-\frac {428}{243 (2+3 x)}-\frac {40}{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 {138672 x^3+193428 x^2+97116 x+480 (3 x+2)^4 \log (6 x+4)+18835}{2916 (3 x+2)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2916*(18835 + 97116*x + 193428*x^2 + 138672*x^3 + 480*(2 + 3*x)^4*Log[4 + 6*x])/(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)^3 (3+5 x)}{(2+3 x)^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.15, size = 67, normalized size = 1.22 \begin {gather*} -\frac {138672 \, x^{3} + 193428 \, x^{2} + 480 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 97116 \, x + 18835}{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)^3*(3+5*x)/(2+3*x)^5,x, algorithm="fricas")

[Out]

-1/2916*(138672*x^3 + 193428*x^2 + 480*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 97116*x + 18835
)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

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giac [A]  time = 1.11, size = 55, normalized size = 1.00 \begin {gather*} -\frac {428}{243 \, {\left (3 \, x + 2\right )}} + \frac {259}{81 \, {\left (3 \, x + 2\right )}^{2}} - \frac {2009}{729 \, {\left (3 \, x + 2\right )}^{3}} + \frac {343}{972 \, {\left (3 \, x + 2\right )}^{4}} + \frac {40}{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)^3*(3+5*x)/(2+3*x)^5,x, algorithm="giac")

[Out]

-428/243/(3*x + 2) + 259/81/(3*x + 2)^2 - 2009/729/(3*x + 2)^3 + 343/972/(3*x + 2)^4 + 40/243*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 {40 \ln \left (3 x +2\right )}{243}+\frac {343}{972 \left (3 x +2\right )^{4}}-\frac {2009}{729 \left (3 x +2\right )^{3}}+\frac {259}{81 \left (3 x +2\right )^{2}}-\frac {428}{243 \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

343/972/(3*x+2)^4-2009/729/(3*x+2)^3+259/81/(3*x+2)^2-428/243/(3*x+2)-40/243*ln(3*x+2)

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maxima [A]  time = 0.55, size = 48, normalized size = 0.87 \begin {gather*} -\frac {138672 \, x^{3} + 193428 \, x^{2} + 97116 \, x + 18835}{2916 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} - \frac {40}{243} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2916*(138672*x^3 + 193428*x^2 + 97116*x + 18835)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) - 40/243*log(3*x
+ 2)

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mupad [B]  time = 1.09, size = 44, normalized size = 0.80 \begin {gather*} -\frac {40\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {\frac {428\,x^3}{729}+\frac {199\,x^2}{243}+\frac {8093\,x}{19683}+\frac {18835}{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)^3*(5*x + 3))/(3*x + 2)^5,x)

[Out]

- (40*log(x + 2/3))/243 - ((8093*x)/19683 + (199*x^2)/243 + (428*x^3)/729 + 18835/236196)/((32*x)/27 + (8*x^2)
/3 + (8*x^3)/3 + x^4 + 16/81)

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sympy [A]  time = 0.15, size = 46, normalized size = 0.84 \begin {gather*} - \frac {138672 x^{3} + 193428 x^{2} + 97116 x + 18835}{236196 x^{4} + 629856 x^{3} + 629856 x^{2} + 279936 x + 46656} - \frac {40 \log {\left (3 x + 2 \right )}}{243} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(138672*x**3 + 193428*x**2 + 97116*x + 18835)/(236196*x**4 + 629856*x**3 + 629856*x**2 + 279936*x + 46656) -
40*log(3*x + 2)/243

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