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

Optimal. Leaf size=46 \[ -\frac {154}{3 x+2}-\frac {121}{5 x+3}-\frac {49}{6 (3 x+2)^2}+1133 \log (3 x+2)-1133 \log (5 x+3) \]

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Rubi [A]  time = 0.02, antiderivative size = 46, 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} \begin {gather*} -\frac {154}{3 x+2}-\frac {121}{5 x+3}-\frac {49}{6 (3 x+2)^2}+1133 \log (3 x+2)-1133 \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-49/(6*(2 + 3*x)^2) - 154/(2 + 3*x) - 121/(3 + 5*x) + 1133*Log[2 + 3*x] - 1133*Log[3 + 5*x]

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 (3+5 x)^2} \, dx &=\int \left (\frac {49}{(2+3 x)^3}+\frac {462}{(2+3 x)^2}+\frac {3399}{2+3 x}+\frac {605}{(3+5 x)^2}-\frac {5665}{3+5 x}\right ) \, dx\\ &=-\frac {49}{6 (2+3 x)^2}-\frac {154}{2+3 x}-\frac {121}{3+5 x}+1133 \log (2+3 x)-1133 \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 48, normalized size = 1.04 \begin {gather*} -\frac {154}{3 x+2}-\frac {121}{5 x+3}-\frac {49}{6 (3 x+2)^2}+1133 \log (5 (3 x+2))-1133 \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-49/(6*(2 + 3*x)^2) - 154/(2 + 3*x) - 121/(3 + 5*x) + 1133*Log[5*(2 + 3*x)] - 1133*Log[3 + 5*x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.95, size = 75, normalized size = 1.63 \begin {gather*} -\frac {20394 \, x^{2} + 6798 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (5 \, x + 3\right ) - 6798 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (3 \, x + 2\right ) + 26513 \, x + 8595}{6 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(20394*x^2 + 6798*(45*x^3 + 87*x^2 + 56*x + 12)*log(5*x + 3) - 6798*(45*x^3 + 87*x^2 + 56*x + 12)*log(3*x
+ 2) + 26513*x + 8595)/(45*x^3 + 87*x^2 + 56*x + 12)

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giac [A]  time = 0.95, size = 49, normalized size = 1.07 \begin {gather*} -\frac {121}{5 \, x + 3} + \frac {35 \, {\left (\frac {202}{5 \, x + 3} + 501\right )}}{2 \, {\left (\frac {1}{5 \, x + 3} + 3\right )}^{2}} + 1133 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/(5*x + 3) + 35/2*(202/(5*x + 3) + 501)/(1/(5*x + 3) + 3)^2 + 1133*log(abs(-1/(5*x + 3) - 3))

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maple [A]  time = 0.01, size = 45, normalized size = 0.98 \begin {gather*} 1133 \ln \left (3 x +2\right )-1133 \ln \left (5 x +3\right )-\frac {49}{6 \left (3 x +2\right )^{2}}-\frac {154}{3 x +2}-\frac {121}{5 x +3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/6/(3*x+2)^2-154/(3*x+2)-121/(5*x+3)+1133*ln(3*x+2)-1133*ln(5*x+3)

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maxima [A]  time = 0.72, size = 46, normalized size = 1.00 \begin {gather*} -\frac {20394 \, x^{2} + 26513 \, x + 8595}{6 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} - 1133 \, \log \left (5 \, x + 3\right ) + 1133 \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(20394*x^2 + 26513*x + 8595)/(45*x^3 + 87*x^2 + 56*x + 12) - 1133*log(5*x + 3) + 1133*log(3*x + 2)

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mupad [B]  time = 1.10, size = 36, normalized size = 0.78 \begin {gather*} 2266\,\mathrm {atanh}\left (30\,x+19\right )-\frac {\frac {1133\,x^2}{15}+\frac {26513\,x}{270}+\frac {191}{6}}{x^3+\frac {29\,x^2}{15}+\frac {56\,x}{45}+\frac {4}{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2266*atanh(30*x + 19) - ((26513*x)/270 + (1133*x^2)/15 + 191/6)/((56*x)/45 + (29*x^2)/15 + x^3 + 4/15)

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sympy [A]  time = 0.17, size = 42, normalized size = 0.91 \begin {gather*} \frac {- 20394 x^{2} - 26513 x - 8595}{270 x^{3} + 522 x^{2} + 336 x + 72} - 1133 \log {\left (x + \frac {3}{5} \right )} + 1133 \log {\left (x + \frac {2}{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-20394*x**2 - 26513*x - 8595)/(270*x**3 + 522*x**2 + 336*x + 72) - 1133*log(x + 3/5) + 1133*log(x + 2/3)

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