3.84.44 \(\int \frac {32-128 x+128 x^2+e^4 (8+7 x-28 x^2+28 x^3)}{e^4 (8-32 x+32 x^2)} \, dx\)

Optimal. Leaf size=26 \[ -\frac {9 x^2}{16}+x \left (\frac {4}{e^4}+x-\frac {1}{-1+2 x}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 27, 1850} \begin {gather*} \frac {7 x^2}{16}+\frac {4 x}{e^4}+\frac {1}{2 (1-2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 - 128*x + 128*x^2 + E^4*(8 + 7*x - 28*x^2 + 28*x^3))/(E^4*(8 - 32*x + 32*x^2)),x]

[Out]

1/(2*(1 - 2*x)) + (4*x)/E^4 + (7*x^2)/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {32-128 x+128 x^2+e^4 \left (8+7 x-28 x^2+28 x^3\right )}{8-32 x+32 x^2} \, dx}{e^4}\\ &=\frac {\int \frac {32-128 x+128 x^2+e^4 \left (8+7 x-28 x^2+28 x^3\right )}{8 (-1+2 x)^2} \, dx}{e^4}\\ &=\frac {\int \frac {32-128 x+128 x^2+e^4 \left (8+7 x-28 x^2+28 x^3\right )}{(-1+2 x)^2} \, dx}{8 e^4}\\ &=\frac {\int \left (32+7 e^4 x+\frac {8 e^4}{(-1+2 x)^2}\right ) \, dx}{8 e^4}\\ &=\frac {1}{2 (1-2 x)}+\frac {4 x}{e^4}+\frac {7 x^2}{16}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.92 \begin {gather*} -\frac {7}{64}+\frac {1}{2-4 x}+\frac {4 x}{e^4}+\frac {7 x^2}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 128*x + 128*x^2 + E^4*(8 + 7*x - 28*x^2 + 28*x^3))/(E^4*(8 - 32*x + 32*x^2)),x]

[Out]

-7/64 + (2 - 4*x)^(-1) + (4*x)/E^4 + (7*x^2)/16

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fricas [A]  time = 0.52, size = 35, normalized size = 1.35 \begin {gather*} \frac {{\left (128 \, x^{2} + {\left (14 \, x^{3} - 7 \, x^{2} - 8\right )} e^{4} - 64 \, x\right )} e^{\left (-4\right )}}{16 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((28*x^3-28*x^2+7*x+8)*exp(4)+128*x^2-128*x+32)/(32*x^2-32*x+8)/exp(4),x, algorithm="fricas")

[Out]

1/16*(128*x^2 + (14*x^3 - 7*x^2 - 8)*e^4 - 64*x)*e^(-4)/(2*x - 1)

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giac [A]  time = 0.17, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{16} \, {\left (7 \, x^{2} e^{4} + 64 \, x - \frac {8 \, e^{4}}{2 \, x - 1}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((28*x^3-28*x^2+7*x+8)*exp(4)+128*x^2-128*x+32)/(32*x^2-32*x+8)/exp(4),x, algorithm="giac")

[Out]

1/16*(7*x^2*e^4 + 64*x - 8*e^4/(2*x - 1))*e^(-4)

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maple [A]  time = 0.40, size = 19, normalized size = 0.73

method	result	size
risch	\(\frac {7 x^{2}}{16}+4 x \,{\mathrm e}^{-4}-\frac {1}{4 \left (x -\frac {1}{2}\right )}\)	\(19\)
default	\(\frac {{\mathrm e}^{-4} \left (32 x +\frac {7 x^{2} {\mathrm e}^{4}}{2}-\frac {4 \,{\mathrm e}^{4}}{2 x -1}\right )}{8}\)	\(29\)
gosper	\(\frac {\left (14 x^{3} {\mathrm e}^{4}-7 x^{2} {\mathrm e}^{4}+128 x^{2}-8 \,{\mathrm e}^{4}-32\right ) {\mathrm e}^{-4}}{32 x -16}\)	\(39\)
norman	\(\frac {-\left (4+{\mathrm e}^{4}\right ) {\mathrm e}^{-4} x +\frac {7 x^{3}}{8}-\frac {\left (7 \,{\mathrm e}^{4}-128\right ) {\mathrm e}^{-4} x^{2}}{16}}{2 x -1}\)	\(41\)
meijerg	\(\frac {4 \,{\mathrm e}^{-4} x}{1-2 x}+\frac {7 x \left (-8 x^{2}-12 x +12\right )}{64 \left (1-2 x \right )}+\frac {21 \ln \left (1-2 x \right )}{32}-\frac {\left (-28 \,{\mathrm e}^{4}+128\right ) {\mathrm e}^{-4} \left (-\frac {2 x \left (6-6 x \right )}{3 \left (1-2 x \right )}-2 \ln \left (1-2 x \right )\right )}{64}+\frac {\left (7 \,{\mathrm e}^{4}-128\right ) {\mathrm e}^{-4} \left (\frac {2 x}{1-2 x}+\ln \left (1-2 x \right )\right )}{32}+\frac {x}{1-2 x}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((28*x^3-28*x^2+7*x+8)*exp(4)+128*x^2-128*x+32)/(32*x^2-32*x+8)/exp(4),x,method=_RETURNVERBOSE)

[Out]

7/16*x^2+4*x*exp(-4)-1/4/(x-1/2)

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maxima [A]  time = 0.36, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{16} \, {\left (7 \, x^{2} e^{4} + 64 \, x - \frac {8 \, e^{4}}{2 \, x - 1}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((28*x^3-28*x^2+7*x+8)*exp(4)+128*x^2-128*x+32)/(32*x^2-32*x+8)/exp(4),x, algorithm="maxima")

[Out]

1/16*(7*x^2*e^4 + 64*x - 8*e^4/(2*x - 1))*e^(-4)

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mupad [B]  time = 0.09, size = 20, normalized size = 0.77 \begin {gather*} 4\,x\,{\mathrm {e}}^{-4}-\frac {1}{4\,\left (x-\frac {1}{2}\right )}+\frac {7\,x^2}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4)*(exp(4)*(7*x - 28*x^2 + 28*x^3 + 8) - 128*x + 128*x^2 + 32))/(32*x^2 - 32*x + 8),x)

[Out]

4*x*exp(-4) - 1/(4*(x - 1/2)) + (7*x^2)/16

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sympy [A]  time = 0.17, size = 19, normalized size = 0.73 \begin {gather*} \frac {7 x^{2}}{16} + \frac {4 x}{e^{4}} - \frac {1}{4 x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((28*x**3-28*x**2+7*x+8)*exp(4)+128*x**2-128*x+32)/(32*x**2-32*x+8)/exp(4),x)

[Out]

7*x**2/16 + 4*x*exp(-4) - 1/(4*x - 2)

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