3.13.27 \(\int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx\) [1227]

Optimal. Leaf size=38 \[ -\frac {18 x}{125}-\frac {11}{1250 (3+5 x)^2}-\frac {64}{625 (3+5 x)}+\frac {87}{625} \log (3+5 x) \]

[Out]

-18/125*x-11/1250/(3+5*x)^2-64/625/(3+5*x)+87/625*ln(3+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 38, 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 {18 x}{125}-\frac {64}{625 (5 x+3)}-\frac {11}{1250 (5 x+3)^2}+\frac {87}{625} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-18*x)/125 - 11/(1250*(3 + 5*x)^2) - 64/(625*(3 + 5*x)) + (87*Log[3 + 5*x])/625

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) (2+3 x)^2}{(3+5 x)^3} \, dx &=\int \left (-\frac {18}{125}+\frac {11}{125 (3+5 x)^3}+\frac {64}{125 (3+5 x)^2}+\frac {87}{125 (3+5 x)}\right ) \, dx\\ &=-\frac {18 x}{125}-\frac {11}{1250 (3+5 x)^2}-\frac {64}{625 (3+5 x)}+\frac {87}{625} \log (3+5 x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 39, normalized size = 1.03 \begin {gather*} -\frac {295+1172 x+1680 x^2+900 x^3}{250 (3+5 x)^2}+\frac {87}{625} \log (-3 (3+5 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/250*(295 + 1172*x + 1680*x^2 + 900*x^3)/(3 + 5*x)^2 + (87*Log[-3*(3 + 5*x)])/625

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Maple [A]
time = 0.11, size = 31, normalized size = 0.82

method	result	size
risch	\(-\frac {18 x}{125}+\frac {-\frac {64 x}{125}-\frac {79}{250}}{\left (3+5 x \right )^{2}}+\frac {87 \ln \left (3+5 x \right )}{625}\)	\(27\)
default	\(-\frac {18 x}{125}-\frac {11}{1250 \left (3+5 x \right )^{2}}-\frac {64}{625 \left (3+5 x \right )}+\frac {87 \ln \left (3+5 x \right )}{625}\)	\(31\)
norman	\(\frac {-\frac {283}{375} x -\frac {1549}{450} x^{2}-\frac {18}{5} x^{3}}{\left (3+5 x \right )^{2}}+\frac {87 \ln \left (3+5 x \right )}{625}\)	\(32\)
meijerg	\(\frac {2 x \left (\frac {5 x}{3}+2\right )}{27 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {2 x^{2}}{27 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {x \left (15 x +6\right )}{30 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {87 \ln \left (1+\frac {5 x}{3}\right )}{625}-\frac {9 x \left (\frac {100}{9} x^{2}+30 x +12\right )}{250 \left (1+\frac {5 x}{3}\right )^{2}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-18/125*x-11/1250/(3+5*x)^2-64/625/(3+5*x)+87/625*ln(3+5*x)

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Maxima [A]
time = 0.33, size = 31, normalized size = 0.82 \begin {gather*} -\frac {18}{125} \, x - \frac {128 \, x + 79}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {87}{625} \, \log \left (5 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-18/125*x - 1/250*(128*x + 79)/(25*x^2 + 30*x + 9) + 87/625*log(5*x + 3)

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Fricas [A]
time = 0.94, size = 47, normalized size = 1.24 \begin {gather*} -\frac {4500 \, x^{3} + 5400 \, x^{2} - 174 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) + 2260 \, x + 395}{1250 \, {\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+3*x)^2/(3+5*x)^3,x, algorithm="fricas")

[Out]

-1/1250*(4500*x^3 + 5400*x^2 - 174*(25*x^2 + 30*x + 9)*log(5*x + 3) + 2260*x + 395)/(25*x^2 + 30*x + 9)

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Sympy [A]
time = 0.04, size = 29, normalized size = 0.76 \begin {gather*} - \frac {18 x}{125} - \frac {128 x + 79}{6250 x^{2} + 7500 x + 2250} + \frac {87 \log {\left (5 x + 3 \right )}}{625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-18*x/125 - (128*x + 79)/(6250*x**2 + 7500*x + 2250) + 87*log(5*x + 3)/625

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Giac [A]
time = 0.68, size = 27, normalized size = 0.71 \begin {gather*} -\frac {18}{125} \, x - \frac {128 \, x + 79}{250 \, {\left (5 \, x + 3\right )}^{2}} + \frac {87}{625} \, \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+3*x)^2/(3+5*x)^3,x, algorithm="giac")

[Out]

-18/125*x - 1/250*(128*x + 79)/(5*x + 3)^2 + 87/625*log(abs(5*x + 3))

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Mupad [B]
time = 0.04, size = 27, normalized size = 0.71 \begin {gather*} \frac {87\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {18\,x}{125}-\frac {\frac {64\,x}{3125}+\frac {79}{6250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(87*log(x + 3/5))/625 - (18*x)/125 - ((64*x)/3125 + 79/6250)/((6*x)/5 + x^2 + 9/25)

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