3.85.54 \(\int \frac {16 x-16 e x}{-1+e^3+12 x-48 x^2+64 x^3+e^2 (-3+12 x)+e (3-24 x+48 x^2)} \, dx\)

Optimal. Leaf size=18 \[ 2-\frac {8}{\left (4-\frac {1-e}{x}\right )^2} \]

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Rubi [B]  time = 0.05, antiderivative size = 38, normalized size of antiderivative = 2.11, number of steps used = 4, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 12, 2074} \begin {gather*} \frac {1-e}{-4 x-e+1}-\frac {(1-e)^2}{2 (-4 x-e+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*x - 16*E*x)/(-1 + E^3 + 12*x - 48*x^2 + 64*x^3 + E^2*(-3 + 12*x) + E*(3 - 24*x + 48*x^2)),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(16-16 e) x}{-1+e^3+12 x-48 x^2+64 x^3+e^2 (-3+12 x)+e \left (3-24 x+48 x^2\right )} \, dx\\ &=(16 (1-e)) \int \frac {x}{-1+e^3+12 x-48 x^2+64 x^3+e^2 (-3+12 x)+e \left (3-24 x+48 x^2\right )} \, dx\\ &=(16 (1-e)) \int \left (\frac {1-e}{4 (-1+e+4 x)^3}+\frac {1}{4 (-1+e+4 x)^2}\right ) \, dx\\ &=-\frac {(1-e)^2}{2 (1-e-4 x)^2}+\frac {1-e}{1-e-4 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.17 \begin {gather*} \frac {(-1+e) (-1+e+8 x)}{2 (-1+e+4 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x - 16*E*x)/(-1 + E^3 + 12*x - 48*x^2 + 64*x^3 + E^2*(-3 + 12*x) + E*(3 - 24*x + 48*x^2)),x]

[Out]

((-1 + E)*(-1 + E + 8*x))/(2*(-1 + E + 4*x)^2)

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fricas [B]  time = 0.60, size = 41, normalized size = 2.28 \begin {gather*} \frac {2 \, {\left (4 \, x - 1\right )} e - 8 \, x + e^{2} + 1}{2 \, {\left (16 \, x^{2} + 2 \, {\left (4 \, x - 1\right )} e - 8 \, x + e^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(1)+16*x)/(exp(1)^3+(12*x-3)*exp(1)^2+(48*x^2-24*x+3)*exp(1)+64*x^3-48*x^2+12*x-1),x, algo
rithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {16 \, {\left (x e - x\right )}}{64 \, x^{3} - 48 \, x^{2} + 3 \, {\left (4 \, x - 1\right )} e^{2} + 3 \, {\left (16 \, x^{2} - 8 \, x + 1\right )} e + 12 \, x + e^{3} - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(1)+16*x)/(exp(1)^3+(12*x-3)*exp(1)^2+(48*x^2-24*x+3)*exp(1)+64*x^3-48*x^2+12*x-1),x, algo
rithm="giac")

[Out]

integrate(-16*(x*e - x)/(64*x^3 - 48*x^2 + 3*(4*x - 1)*e^2 + 3*(16*x^2 - 8*x + 1)*e + 12*x + e^3 - 1), x)

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maple [A]  time = 0.06, size = 31, normalized size = 1.72

method	result	size
norman	\(\frac {\left (4 \,{\mathrm e}-4\right ) x +\frac {{\mathrm e}^{2}}{2}-{\mathrm e}+\frac {1}{2}}{\left ({\mathrm e}+4 x -1\right )^{2}}\)	\(31\)
gosper	\(\frac {\left ({\mathrm e}-1\right ) \left ({\mathrm e}+8 x -1\right )}{2 \,{\mathrm e}^{2}+16 x \,{\mathrm e}+32 x^{2}-4 \,{\mathrm e}-16 x +2}\)	\(39\)
risch	\(\frac {\left (4 \,{\mathrm e}-4\right ) x +\frac {{\mathrm e}^{2}}{2}-{\mathrm e}+\frac {1}{2}}{{\mathrm e}^{2}+8 x \,{\mathrm e}+16 x^{2}-2 \,{\mathrm e}-8 x +1}\)	\(43\)
default	\(-\frac {4 \left ({\mathrm e}-1\right ) \left (\munderset {\textit {\_R} =\RootOf \left (64 \textit {\_Z}^{3}+\left (48 \,{\mathrm e}-48\right ) \textit {\_Z}^{2}+\left (12 \,{\mathrm e}^{2}-24 \,{\mathrm e}+12\right ) \textit {\_Z} +{\mathrm e}^{3}-1-3 \,{\mathrm e}^{2}+3 \,{\mathrm e}\right )}{\sum }\frac {\textit {\_R} \ln \left (x -\textit {\_R} \right )}{1+{\mathrm e}^{2}+8 \textit {\_R} \,{\mathrm e}+16 \textit {\_R}^{2}-2 \,{\mathrm e}-8 \textit {\_R}}\right )}{3}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x*exp(1)+16*x)/(exp(1)^3+(12*x-3)*exp(1)^2+(48*x^2-24*x+3)*exp(1)+64*x^3-48*x^2+12*x-1),x,method=_RET
URNVERBOSE)

[Out]

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

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maxima [B]  time = 0.35, size = 39, normalized size = 2.17 \begin {gather*} \frac {8 \, x {\left (e - 1\right )} + e^{2} - 2 \, e + 1}{2 \, {\left (16 \, x^{2} + 8 \, x {\left (e - 1\right )} + e^{2} - 2 \, e + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(1)+16*x)/(exp(1)^3+(12*x-3)*exp(1)^2+(48*x^2-24*x+3)*exp(1)+64*x^3-48*x^2+12*x-1),x, algo
rithm="maxima")

[Out]

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

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mupad [B]  time = 0.14, size = 38, normalized size = 2.11 \begin {gather*} \frac {\left (\mathrm {e}-1\right )\,\left (8\,x+\mathrm {e}-1\right )}{2\,\left (16\,x^2+\left (8\,\mathrm {e}-8\right )\,x-2\,\mathrm {e}+{\mathrm {e}}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x - 16*x*exp(1))/(12*x + exp(3) + exp(1)*(48*x^2 - 24*x + 3) - 48*x^2 + 64*x^3 + exp(2)*(12*x - 3) - 1
),x)

[Out]

((exp(1) - 1)*(8*x + exp(1) - 1))/(2*(exp(2) - 2*exp(1) + 16*x^2 + x*(8*exp(1) - 8) + 1))

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sympy [B]  time = 0.22, size = 39, normalized size = 2.17 \begin {gather*} \frac {\left (16 - 16 e\right ) \left (- 8 x - e + 1\right )}{512 x^{2} + x \left (-256 + 256 e\right ) - 64 e + 32 + 32 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(1)+16*x)/(exp(1)**3+(12*x-3)*exp(1)**2+(48*x**2-24*x+3)*exp(1)+64*x**3-48*x**2+12*x-1),x)

[Out]

(16 - 16*E)*(-8*x - E + 1)/(512*x**2 + x*(-256 + 256*E) - 64*E + 32 + 32*exp(2))

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