3.15.95 \(\int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^3} \, dx\)

Optimal. Leaf size=50 \[ -\frac {135 x^3}{8}-\frac {3861 x^2}{32}-540 x-\frac {57281}{64 (1-2 x)}+\frac {26411}{128 (1-2 x)^2}-\frac {24843}{32} \log (1-2 x) \]

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Rubi [A]  time = 0.02, antiderivative size = 50, 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 {135 x^3}{8}-\frac {3861 x^2}{32}-540 x-\frac {57281}{64 (1-2 x)}+\frac {26411}{128 (1-2 x)^2}-\frac {24843}{32} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x)^3,x]

[Out]

26411/(128*(1 - 2*x)^2) - 57281/(64*(1 - 2*x)) - 540*x - (3861*x^2)/32 - (135*x^3)/8 - (24843*Log[1 - 2*x])/32

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)^4 (3+5 x)}{(1-2 x)^3} \, dx &=\int \left (-540-\frac {3861 x}{16}-\frac {405 x^2}{8}-\frac {26411}{32 (-1+2 x)^3}-\frac {57281}{32 (-1+2 x)^2}-\frac {24843}{16 (-1+2 x)}\right ) \, dx\\ &=\frac {26411}{128 (1-2 x)^2}-\frac {57281}{64 (1-2 x)}-540 x-\frac {3861 x^2}{32}-\frac {135 x^3}{8}-\frac {24843}{32} \log (1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 1.02 \begin {gather*} -\frac {2160 x^5+13284 x^4+54216 x^3-103950 x^2-1310 x+24843 (1-2 x)^2 \log (1-2 x)+12365}{32 (1-2 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x)^3,x]

[Out]

-1/32*(12365 - 1310*x - 103950*x^2 + 54216*x^3 + 13284*x^4 + 2160*x^5 + 24843*(1 - 2*x)^2*Log[1 - 2*x])/(1 - 2
*x)^2

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x)^3,x]

[Out]

IntegrateAlgebraic[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x)^3, x]

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fricas [A]  time = 1.58, size = 57, normalized size = 1.14 \begin {gather*} -\frac {8640 \, x^{5} + 53136 \, x^{4} + 216864 \, x^{3} - 261036 \, x^{2} + 99372 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 160004 \, x + 88151}{128 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(3+5*x)/(1-2*x)^3,x, algorithm="fricas")

[Out]

-1/128*(8640*x^5 + 53136*x^4 + 216864*x^3 - 261036*x^2 + 99372*(4*x^2 - 4*x + 1)*log(2*x - 1) - 160004*x + 881
51)/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.18, size = 37, normalized size = 0.74 \begin {gather*} -\frac {135}{8} \, x^{3} - \frac {3861}{32} \, x^{2} - 540 \, x + \frac {343 \, {\left (668 \, x - 257\right )}}{128 \, {\left (2 \, x - 1\right )}^{2}} - \frac {24843}{32} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(3+5*x)/(1-2*x)^3,x, algorithm="giac")

[Out]

-135/8*x^3 - 3861/32*x^2 - 540*x + 343/128*(668*x - 257)/(2*x - 1)^2 - 24843/32*log(abs(2*x - 1))

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maple [A]  time = 0.01, size = 41, normalized size = 0.82 \begin {gather*} -\frac {135 x^{3}}{8}-\frac {3861 x^{2}}{32}-540 x -\frac {24843 \ln \left (2 x -1\right )}{32}+\frac {26411}{128 \left (2 x -1\right )^{2}}+\frac {57281}{64 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x+2)^4*(5*x+3)/(1-2*x)^3,x)

[Out]

-135/8*x^3-3861/32*x^2-540*x+26411/128/(2*x-1)^2+57281/64/(2*x-1)-24843/32*ln(2*x-1)

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maxima [A]  time = 0.53, size = 41, normalized size = 0.82 \begin {gather*} -\frac {135}{8} \, x^{3} - \frac {3861}{32} \, x^{2} - 540 \, x + \frac {343 \, {\left (668 \, x - 257\right )}}{128 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {24843}{32} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(3+5*x)/(1-2*x)^3,x, algorithm="maxima")

[Out]

-135/8*x^3 - 3861/32*x^2 - 540*x + 343/128*(668*x - 257)/(4*x^2 - 4*x + 1) - 24843/32*log(2*x - 1)

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mupad [B]  time = 0.03, size = 36, normalized size = 0.72 \begin {gather*} \frac {\frac {57281\,x}{128}-\frac {88151}{512}}{x^2-x+\frac {1}{4}}-\frac {24843\,\ln \left (x-\frac {1}{2}\right )}{32}-540\,x-\frac {3861\,x^2}{32}-\frac {135\,x^3}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((3*x + 2)^4*(5*x + 3))/(2*x - 1)^3,x)

[Out]

((57281*x)/128 - 88151/512)/(x^2 - x + 1/4) - (24843*log(x - 1/2))/32 - 540*x - (3861*x^2)/32 - (135*x^3)/8

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sympy [A]  time = 0.13, size = 42, normalized size = 0.84 \begin {gather*} - \frac {135 x^{3}}{8} - \frac {3861 x^{2}}{32} - 540 x - \frac {88151 - 229124 x}{512 x^{2} - 512 x + 128} - \frac {24843 \log {\left (2 x - 1 \right )}}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)**4*(3+5*x)/(1-2*x)**3,x)

[Out]

-135*x**3/8 - 3861*x**2/32 - 540*x - (88151 - 229124*x)/(512*x**2 - 512*x + 128) - 24843*log(2*x - 1)/32

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