3.1197 \(\int \frac {(1-2 x) (2+3 x)^2}{3+5 x} \, dx\)

Optimal. Leaf size=30 \[ -\frac {6 x^3}{5}-\frac {21 x^2}{50}+\frac {163 x}{125}+\frac {11}{625} \log (5 x+3) \]

[Out]

163/125*x-21/50*x^2-6/5*x^3+11/625*ln(3+5*x)

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Rubi [A]  time = 0.01, antiderivative size = 30, 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} \[ -\frac {6 x^3}{5}-\frac {21 x^2}{50}+\frac {163 x}{125}+\frac {11}{625} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(163*x)/125 - (21*x^2)/50 - (6*x^3)/5 + (11*Log[3 + 5*x])/625

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) (2+3 x)^2}{3+5 x} \, dx &=\int \left (\frac {163}{125}-\frac {21 x}{25}-\frac {18 x^2}{5}+\frac {11}{125 (3+5 x)}\right ) \, dx\\ &=\frac {163 x}{125}-\frac {21 x^2}{50}-\frac {6 x^3}{5}+\frac {11}{625} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 0.90 \[ \frac {-1500 x^3-525 x^2+1630 x+22 \log (5 x+3)+843}{1250} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(843 + 1630*x - 525*x^2 - 1500*x^3 + 22*Log[3 + 5*x])/1250

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fricas [A]  time = 0.72, size = 22, normalized size = 0.73 \[ -\frac {6}{5} \, x^{3} - \frac {21}{50} \, x^{2} + \frac {163}{125} \, x + \frac {11}{625} \, \log \left (5 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/5*x^3 - 21/50*x^2 + 163/125*x + 11/625*log(5*x + 3)

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giac [A]  time = 1.16, size = 23, normalized size = 0.77 \[ -\frac {6}{5} \, x^{3} - \frac {21}{50} \, x^{2} + \frac {163}{125} \, x + \frac {11}{625} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/5*x^3 - 21/50*x^2 + 163/125*x + 11/625*log(abs(5*x + 3))

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maple [A]  time = 0.00, size = 23, normalized size = 0.77 \[ -\frac {6 x^{3}}{5}-\frac {21 x^{2}}{50}+\frac {163 x}{125}+\frac {11 \ln \left (5 x +3\right )}{625} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

163/125*x-21/50*x^2-6/5*x^3+11/625*ln(5*x+3)

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maxima [A]  time = 0.50, size = 22, normalized size = 0.73 \[ -\frac {6}{5} \, x^{3} - \frac {21}{50} \, x^{2} + \frac {163}{125} \, x + \frac {11}{625} \, \log \left (5 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/5*x^3 - 21/50*x^2 + 163/125*x + 11/625*log(5*x + 3)

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mupad [B]  time = 0.03, size = 20, normalized size = 0.67 \[ \frac {163\,x}{125}+\frac {11\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {21\,x^2}{50}-\frac {6\,x^3}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(163*x)/125 + (11*log(x + 3/5))/625 - (21*x^2)/50 - (6*x^3)/5

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sympy [A]  time = 0.09, size = 27, normalized size = 0.90 \[ - \frac {6 x^{3}}{5} - \frac {21 x^{2}}{50} + \frac {163 x}{125} + \frac {11 \log {\left (5 x + 3 \right )}}{625} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*x**3/5 - 21*x**2/50 + 163*x/125 + 11*log(5*x + 3)/625

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