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

Optimal. Leaf size=41 \[ \frac {75 x^3}{4}+\frac {795 x^2}{8}+\frac {5119 x}{16}+\frac {5929}{32 (1-2 x)}+\frac {1309}{4} \log (1-2 x) \]

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Rubi [A]  time = 0.02, antiderivative size = 41, 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 {75 x^3}{4}+\frac {795 x^2}{8}+\frac {5119 x}{16}+\frac {5929}{32 (1-2 x)}+\frac {1309}{4} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5929/(32*(1 - 2*x)) + (5119*x)/16 + (795*x^2)/8 + (75*x^3)/4 + (1309*Log[1 - 2*x])/4

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 {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx &=\int \left (\frac {5119}{16}+\frac {795 x}{4}+\frac {225 x^2}{4}+\frac {5929}{16 (-1+2 x)^2}+\frac {1309}{2 (-1+2 x)}\right ) \, dx\\ &=\frac {5929}{32 (1-2 x)}+\frac {5119 x}{16}+\frac {795 x^2}{8}+\frac {75 x^3}{4}+\frac {1309}{4} \log (1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 1.00 \begin {gather*} \frac {300 x^4+1440 x^3+4324 x^2-5554 x+2618 (2 x-1) \log (1-2 x)+15}{16 x-8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15 - 5554*x + 4324*x^2 + 1440*x^3 + 300*x^4 + 2618*(-1 + 2*x)*Log[1 - 2*x])/(-8 + 16*x)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.36, size = 42, normalized size = 1.02 \begin {gather*} \frac {1200 \, x^{4} + 5760 \, x^{3} + 17296 \, x^{2} + 10472 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 10238 \, x - 5929}{32 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(1200*x^4 + 5760*x^3 + 17296*x^2 + 10472*(2*x - 1)*log(2*x - 1) - 10238*x - 5929)/(2*x - 1)

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giac [A]  time = 0.90, size = 57, normalized size = 1.39 \begin {gather*} \frac {1}{32} \, {\left (2 \, x - 1\right )}^{3} {\left (\frac {1020}{2 \, x - 1} + \frac {6934}{{\left (2 \, x - 1\right )}^{2}} + 75\right )} - \frac {5929}{32 \, {\left (2 \, x - 1\right )}} - \frac {1309}{4} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(2*x - 1)^3*(1020/(2*x - 1) + 6934/(2*x - 1)^2 + 75) - 5929/32/(2*x - 1) - 1309/4*log(1/2*abs(2*x - 1)/(2
*x - 1)^2)

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maple [A]  time = 0.01, size = 32, normalized size = 0.78 \begin {gather*} \frac {75 x^{3}}{4}+\frac {795 x^{2}}{8}+\frac {5119 x}{16}+\frac {1309 \ln \left (2 x -1\right )}{4}-\frac {5929}{32 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

75/4*x^3+795/8*x^2+5119/16*x-5929/32/(2*x-1)+1309/4*ln(2*x-1)

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maxima [A]  time = 0.67, size = 31, normalized size = 0.76 \begin {gather*} \frac {75}{4} \, x^{3} + \frac {795}{8} \, x^{2} + \frac {5119}{16} \, x - \frac {5929}{32 \, {\left (2 \, x - 1\right )}} + \frac {1309}{4} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

75/4*x^3 + 795/8*x^2 + 5119/16*x - 5929/32/(2*x - 1) + 1309/4*log(2*x - 1)

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mupad [B]  time = 0.03, size = 29, normalized size = 0.71 \begin {gather*} \frac {5119\,x}{16}+\frac {1309\,\ln \left (x-\frac {1}{2}\right )}{4}-\frac {5929}{64\,\left (x-\frac {1}{2}\right )}+\frac {795\,x^2}{8}+\frac {75\,x^3}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5119*x)/16 + (1309*log(x - 1/2))/4 - 5929/(64*(x - 1/2)) + (795*x^2)/8 + (75*x^3)/4

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sympy [A]  time = 0.11, size = 34, normalized size = 0.83 \begin {gather*} \frac {75 x^{3}}{4} + \frac {795 x^{2}}{8} + \frac {5119 x}{16} + \frac {1309 \log {\left (2 x - 1 \right )}}{4} - \frac {5929}{64 x - 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

75*x**3/4 + 795*x**2/8 + 5119*x/16 + 1309*log(2*x - 1)/4 - 5929/(64*x - 32)

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