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

Optimal. Leaf size=46 \[ \frac {405 x^4}{16}+144 x^3+\frac {13419 x^2}{32}+\frac {16203 x}{16}+\frac {26411}{64 (1-2 x)}+\frac {57281}{64} \log (1-2 x) \]

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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 = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \begin {gather*} \frac {405 x^4}{16}+144 x^3+\frac {13419 x^2}{32}+\frac {16203 x}{16}+\frac {26411}{64 (1-2 x)}+\frac {57281}{64} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

26411/(64*(1 - 2*x)) + (16203*x)/16 + (13419*x^2)/32 + 144*x^3 + (405*x^4)/16 + (57281*Log[1 - 2*x])/64

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)^2} \, dx &=\int \left (\frac {16203}{16}+\frac {13419 x}{16}+432 x^2+\frac {405 x^3}{4}+\frac {26411}{32 (-1+2 x)^2}+\frac {57281}{32 (-1+2 x)}\right ) \, dx\\ &=\frac {26411}{64 (1-2 x)}+\frac {16203 x}{16}+\frac {13419 x^2}{32}+144 x^3+\frac {405 x^4}{16}+\frac {57281}{64} \log (1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.07 \begin {gather*} \frac {12960 x^5+67248 x^4+177840 x^3+411144 x^2-582198 x+229124 (2 x-1) \log (1-2 x)+55831}{256 (2 x-1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(55831 - 582198*x + 411144*x^2 + 177840*x^3 + 67248*x^4 + 12960*x^5 + 229124*(-1 + 2*x)*Log[1 - 2*x])/(256*(-1
+ 2*x))

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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)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.08, size = 47, normalized size = 1.02 \begin {gather*} \frac {3240 \, x^{5} + 16812 \, x^{4} + 44460 \, x^{3} + 102786 \, x^{2} + 57281 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 64812 \, x - 26411}{64 \, {\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)^2,x, algorithm="fricas")

[Out]

1/64*(3240*x^5 + 16812*x^4 + 44460*x^3 + 102786*x^2 + 57281*(2*x - 1)*log(2*x - 1) - 64812*x - 26411)/(2*x - 1
)

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giac [A]  time = 0.94, size = 66, normalized size = 1.43 \begin {gather*} \frac {3}{256} \, {\left (2 \, x - 1\right )}^{4} {\left (\frac {2076}{2 \, x - 1} + \frac {14364}{{\left (2 \, x - 1\right )}^{2}} + \frac {66248}{{\left (2 \, x - 1\right )}^{3}} + 135\right )} - \frac {26411}{64 \, {\left (2 \, x - 1\right )}} - \frac {57281}{64} \, \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)^4*(3+5*x)/(1-2*x)^2,x, algorithm="giac")

[Out]

3/256*(2*x - 1)^4*(2076/(2*x - 1) + 14364/(2*x - 1)^2 + 66248/(2*x - 1)^3 + 135) - 26411/64/(2*x - 1) - 57281/
64*log(1/2*abs(2*x - 1)/(2*x - 1)^2)

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maple [A]  time = 0.01, size = 37, normalized size = 0.80 \begin {gather*} \frac {405 x^{4}}{16}+144 x^{3}+\frac {13419 x^{2}}{32}+\frac {16203 x}{16}+\frac {57281 \ln \left (2 x -1\right )}{64}-\frac {26411}{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)^2,x)

[Out]

405/16*x^4+144*x^3+13419/32*x^2+16203/16*x-26411/64/(2*x-1)+57281/64*ln(2*x-1)

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maxima [A]  time = 0.55, size = 36, normalized size = 0.78 \begin {gather*} \frac {405}{16} \, x^{4} + 144 \, x^{3} + \frac {13419}{32} \, x^{2} + \frac {16203}{16} \, x - \frac {26411}{64 \, {\left (2 \, x - 1\right )}} + \frac {57281}{64} \, \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)^2,x, algorithm="maxima")

[Out]

405/16*x^4 + 144*x^3 + 13419/32*x^2 + 16203/16*x - 26411/64/(2*x - 1) + 57281/64*log(2*x - 1)

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mupad [B]  time = 0.03, size = 34, normalized size = 0.74 \begin {gather*} \frac {16203\,x}{16}+\frac {57281\,\ln \left (x-\frac {1}{2}\right )}{64}-\frac {26411}{128\,\left (x-\frac {1}{2}\right )}+\frac {13419\,x^2}{32}+144\,x^3+\frac {405\,x^4}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16203*x)/16 + (57281*log(x - 1/2))/64 - 26411/(128*(x - 1/2)) + (13419*x^2)/32 + 144*x^3 + (405*x^4)/16

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sympy [A]  time = 0.11, size = 39, normalized size = 0.85 \begin {gather*} \frac {405 x^{4}}{16} + 144 x^{3} + \frac {13419 x^{2}}{32} + \frac {16203 x}{16} + \frac {57281 \log {\left (2 x - 1 \right )}}{64} - \frac {26411}{128 x - 64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

405*x**4/16 + 144*x**3 + 13419*x**2/32 + 16203*x/16 + 57281*log(2*x - 1)/64 - 26411/(128*x - 64)

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