3.12.70 \(\int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^4} \, dx\) [1170]

Optimal. Leaf size=44 \[ -\frac {7}{243 (2+3 x)^3}+\frac {4}{9 (2+3 x)^2}-\frac {65}{27 (2+3 x)}-\frac {50}{81} \log (2+3 x) \]

[Out]

-7/243/(2+3*x)^3+4/9/(2+3*x)^2-65/27/(2+3*x)-50/81*ln(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 44, 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 = {78} \begin {gather*} -\frac {65}{27 (3 x+2)}+\frac {4}{9 (3 x+2)^2}-\frac {7}{243 (3 x+2)^3}-\frac {50}{81} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)*(3 + 5*x)^2)/(2 + 3*x)^4,x]

[Out]

-7/(243*(2 + 3*x)^3) + 4/(9*(2 + 3*x)^2) - 65/(27*(2 + 3*x)) - (50*Log[2 + 3*x])/81

Rule 78

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+5 x)^2}{(2+3 x)^4} \, dx &=\int \left (\frac {7}{27 (2+3 x)^4}-\frac {8}{3 (2+3 x)^3}+\frac {65}{9 (2+3 x)^2}-\frac {50}{27 (2+3 x)}\right ) \, dx\\ &=-\frac {7}{243 (2+3 x)^3}+\frac {4}{9 (2+3 x)^2}-\frac {65}{27 (2+3 x)}-\frac {50}{81} \log (2+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 36, normalized size = 0.82 \begin {gather*} -\frac {2131+6696 x+5265 x^2+150 (2+3 x)^3 \log (2+3 x)}{243 (2+3 x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)*(3 + 5*x)^2)/(2 + 3*x)^4,x]

[Out]

-1/243*(2131 + 6696*x + 5265*x^2 + 150*(2 + 3*x)^3*Log[2 + 3*x])/(2 + 3*x)^3

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Maple [A]
time = 0.08, size = 37, normalized size = 0.84

method	result	size
risch	\(\frac {-\frac {65}{3} x^{2}-\frac {248}{9} x -\frac {2131}{243}}{\left (2+3 x \right )^{3}}-\frac {50 \ln \left (2+3 x \right )}{81}\)	\(29\)
norman	\(\frac {\frac {643}{54} x +\frac {1351}{36} x^{2}+\frac {2131}{72} x^{3}}{\left (2+3 x \right )^{3}}-\frac {50 \ln \left (2+3 x \right )}{81}\)	\(32\)
default	\(-\frac {7}{243 \left (2+3 x \right )^{3}}+\frac {4}{9 \left (2+3 x \right )^{2}}-\frac {65}{27 \left (2+3 x \right )}-\frac {50 \ln \left (2+3 x \right )}{81}\)	\(37\)
meijerg	\(\frac {3 x \left (\frac {9}{4} x^{2}+\frac {9}{2} x +3\right )}{16 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {x^{2} \left (3+\frac {3 x}{2}\right )}{8 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {35 x^{3}}{48 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {25 x \left (\frac {99}{2} x^{2}+45 x +12\right )}{324 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {50 \ln \left (1+\frac {3 x}{2}\right )}{81}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-2*x)*(3+5*x)^2/(2+3*x)^4,x,method=_RETURNVERBOSE)

[Out]

-7/243/(2+3*x)^3+4/9/(2+3*x)^2-65/27/(2+3*x)-50/81*ln(2+3*x)

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Maxima [A]
time = 0.28, size = 38, normalized size = 0.86 \begin {gather*} -\frac {5265 \, x^{2} + 6696 \, x + 2131}{243 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} - \frac {50}{81} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^2/(2+3*x)^4,x, algorithm="maxima")

[Out]

-1/243*(5265*x^2 + 6696*x + 2131)/(27*x^3 + 54*x^2 + 36*x + 8) - 50/81*log(3*x + 2)

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Fricas [A]
time = 0.66, size = 52, normalized size = 1.18 \begin {gather*} -\frac {5265 \, x^{2} + 150 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) + 6696 \, x + 2131}{243 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^2/(2+3*x)^4,x, algorithm="fricas")

[Out]

-1/243*(5265*x^2 + 150*(27*x^3 + 54*x^2 + 36*x + 8)*log(3*x + 2) + 6696*x + 2131)/(27*x^3 + 54*x^2 + 36*x + 8)

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Sympy [A]
time = 0.05, size = 36, normalized size = 0.82 \begin {gather*} - \frac {5265 x^{2} + 6696 x + 2131}{6561 x^{3} + 13122 x^{2} + 8748 x + 1944} - \frac {50 \log {\left (3 x + 2 \right )}}{81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)**2/(2+3*x)**4,x)

[Out]

-(5265*x**2 + 6696*x + 2131)/(6561*x**3 + 13122*x**2 + 8748*x + 1944) - 50*log(3*x + 2)/81

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Giac [A]
time = 3.58, size = 29, normalized size = 0.66 \begin {gather*} -\frac {5265 \, x^{2} + 6696 \, x + 2131}{243 \, {\left (3 \, x + 2\right )}^{3}} - \frac {50}{81} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^2/(2+3*x)^4,x, algorithm="giac")

[Out]

-1/243*(5265*x^2 + 6696*x + 2131)/(3*x + 2)^3 - 50/81*log(abs(3*x + 2))

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Mupad [B]
time = 0.04, size = 34, normalized size = 0.77 \begin {gather*} -\frac {50\,\ln \left (x+\frac {2}{3}\right )}{81}-\frac {\frac {65\,x^2}{81}+\frac {248\,x}{243}+\frac {2131}{6561}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x - 1)*(5*x + 3)^2)/(3*x + 2)^4,x)

[Out]

- (50*log(x + 2/3))/81 - ((248*x)/243 + (65*x^2)/81 + 2131/6561)/((4*x)/3 + 2*x^2 + x^3 + 8/27)

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