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

Optimal. Leaf size=41 \[ \frac {125 x^3}{4}+\frac {325 x^2}{2}+\frac {8245 x}{16}+\frac {9317}{32 (1-2 x)}+\frac {8349}{16} \log (1-2 x) \]

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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 = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \begin {gather*} \frac {125 x^3}{4}+\frac {325 x^2}{2}+\frac {8245 x}{16}+\frac {9317}{32 (1-2 x)}+\frac {8349}{16} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

9317/(32*(1 - 2*x)) + (8245*x)/16 + (325*x^2)/2 + (125*x^3)/4 + (8349*Log[1 - 2*x])/16

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 {(2+3 x) (3+5 x)^3}{(1-2 x)^2} \, dx &=\int \left (\frac {8245}{16}+325 x+\frac {375 x^2}{4}+\frac {9317}{16 (-1+2 x)^2}+\frac {8349}{8 (-1+2 x)}\right ) \, dx\\ &=\frac {9317}{32 (1-2 x)}+\frac {8245 x}{16}+\frac {325 x^2}{2}+\frac {125 x^3}{4}+\frac {8349}{16} \log (1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 1.00 \begin {gather*} \frac {2000 x^4+9400 x^3+27780 x^2-35830 x+16698 (2 x-1) \log (1-2 x)+353}{64 x-32} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(353 - 35830*x + 27780*x^2 + 9400*x^3 + 2000*x^4 + 16698*(-1 + 2*x)*Log[1 - 2*x])/(-32 + 64*x)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.78, size = 42, normalized size = 1.02 \begin {gather*} \frac {2000 \, x^{4} + 9400 \, x^{3} + 27780 \, x^{2} + 16698 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 16490 \, x - 9317}{32 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(2000*x^4 + 9400*x^3 + 27780*x^2 + 16698*(2*x - 1)*log(2*x - 1) - 16490*x - 9317)/(2*x - 1)

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giac [A]  time = 1.04, size = 57, normalized size = 1.39 \begin {gather*} \frac {5}{32} \, {\left (2 \, x - 1\right )}^{3} {\left (\frac {335}{2 \, x - 1} + \frac {2244}{{\left (2 \, x - 1\right )}^{2}} + 25\right )} - \frac {9317}{32 \, {\left (2 \, x - 1\right )}} - \frac {8349}{16} \, \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)*(3+5*x)^3/(1-2*x)^2,x, algorithm="giac")

[Out]

5/32*(2*x - 1)^3*(335/(2*x - 1) + 2244/(2*x - 1)^2 + 25) - 9317/32/(2*x - 1) - 8349/16*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 {125 x^{3}}{4}+\frac {325 x^{2}}{2}+\frac {8245 x}{16}+\frac {8349 \ln \left (2 x -1\right )}{16}-\frac {9317}{32 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/4*x^3+325/2*x^2+8245/16*x-9317/32/(2*x-1)+8349/16*ln(2*x-1)

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maxima [A]  time = 0.52, size = 31, normalized size = 0.76 \begin {gather*} \frac {125}{4} \, x^{3} + \frac {325}{2} \, x^{2} + \frac {8245}{16} \, x - \frac {9317}{32 \, {\left (2 \, x - 1\right )}} + \frac {8349}{16} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/4*x^3 + 325/2*x^2 + 8245/16*x - 9317/32/(2*x - 1) + 8349/16*log(2*x - 1)

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mupad [B]  time = 0.03, size = 29, normalized size = 0.71 \begin {gather*} \frac {8245\,x}{16}+\frac {8349\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {9317}{64\,\left (x-\frac {1}{2}\right )}+\frac {325\,x^2}{2}+\frac {125\,x^3}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8245*x)/16 + (8349*log(x - 1/2))/16 - 9317/(64*(x - 1/2)) + (325*x^2)/2 + (125*x^3)/4

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sympy [A]  time = 0.11, size = 34, normalized size = 0.83 \begin {gather*} \frac {125 x^{3}}{4} + \frac {325 x^{2}}{2} + \frac {8245 x}{16} + \frac {8349 \log {\left (2 x - 1 \right )}}{16} - \frac {9317}{64 x - 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125*x**3/4 + 325*x**2/2 + 8245*x/16 + 8349*log(2*x - 1)/16 - 9317/(64*x - 32)

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