3.1313 \(\int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac {121}{25 (5 x+3)}+\frac {49}{3} \log (3 x+2)-\frac {407}{25} \log (5 x+3) \]

[Out]

-121/25/(3+5*x)+49/3*ln(2+3*x)-407/25*ln(3+5*x)

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Rubi [A]  time = 0.01, antiderivative size = 32, 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} \[ -\frac {121}{25 (5 x+3)}+\frac {49}{3} \log (3 x+2)-\frac {407}{25} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-121/(25*(3 + 5*x)) + (49*Log[2 + 3*x])/3 - (407*Log[3 + 5*x])/25

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}{(2+3 x) (3+5 x)^2} \, dx &=\int \left (\frac {49}{2+3 x}+\frac {121}{5 (3+5 x)^2}-\frac {407}{5 (3+5 x)}\right ) \, dx\\ &=-\frac {121}{25 (3+5 x)}+\frac {49}{3} \log (2+3 x)-\frac {407}{25} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 1.00 \[ -\frac {121}{125 x+75}+\frac {49}{3} \log (3 x+2)-\frac {407}{25} \log (-3 (5 x+3)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-121/(75 + 125*x) + (49*Log[2 + 3*x])/3 - (407*Log[-3*(3 + 5*x)])/25

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fricas [A]  time = 0.51, size = 37, normalized size = 1.16 \[ -\frac {1221 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1225 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 363}{75 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/75*(1221*(5*x + 3)*log(5*x + 3) - 1225*(5*x + 3)*log(3*x + 2) + 363)/(5*x + 3)

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giac [A]  time = 1.06, size = 43, normalized size = 1.34 \[ -\frac {121}{25 \, {\left (5 \, x + 3\right )}} - \frac {4}{75} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) + \frac {49}{3} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/25/(5*x + 3) - 4/75*log(1/5*abs(5*x + 3)/(5*x + 3)^2) + 49/3*log(abs(-1/(5*x + 3) - 3))

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maple [A]  time = 0.01, size = 27, normalized size = 0.84 \[ \frac {49 \ln \left (3 x +2\right )}{3}-\frac {407 \ln \left (5 x +3\right )}{25}-\frac {121}{25 \left (5 x +3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/25/(5*x+3)+49/3*ln(3*x+2)-407/25*ln(5*x+3)

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maxima [A]  time = 0.55, size = 26, normalized size = 0.81 \[ -\frac {121}{25 \, {\left (5 \, x + 3\right )}} - \frac {407}{25} \, \log \left (5 \, x + 3\right ) + \frac {49}{3} \, \log \left (3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/25/(5*x + 3) - 407/25*log(5*x + 3) + 49/3*log(3*x + 2)

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mupad [B]  time = 1.10, size = 22, normalized size = 0.69 \[ \frac {49\,\ln \left (x+\frac {2}{3}\right )}{3}-\frac {407\,\ln \left (x+\frac {3}{5}\right )}{25}-\frac {121}{125\,\left (x+\frac {3}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(49*log(x + 2/3))/3 - (407*log(x + 3/5))/25 - 121/(125*(x + 3/5))

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sympy [A]  time = 0.14, size = 26, normalized size = 0.81 \[ - \frac {407 \log {\left (x + \frac {3}{5} \right )}}{25} + \frac {49 \log {\left (x + \frac {2}{3} \right )}}{3} - \frac {121}{125 x + 75} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-407*log(x + 3/5)/25 + 49*log(x + 2/3)/3 - 121/(125*x + 75)

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