3.61.52 \(\int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 (2 x-8 x^2)}{-1+4 x} \, dx\)

Optimal. Leaf size=34 \[ -3+x \left (x \left (-e^2+x\right )-\frac {-5+e^{e^{2 x}}+\log \left (-\frac {1}{4}+x\right )}{x}\right ) \]

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Rubi [A]  time = 0.18, antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 4, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6742, 2282, 2194, 1850} \begin {gather*} x^3-e^2 x^2-e^{e^{2 x}}-\log (1-4 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + E^(E^(2*x) + 2*x)*(2 - 8*x) - 3*x^2 + 12*x^3 + E^2*(2*x - 8*x^2))/(-1 + 4*x),x]

[Out]

-E^E^(2*x) - E^2*x^2 + x^3 - Log[1 - 4*x]

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])

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 e^{e^{2 x}+2 x}+\frac {4-2 e^2 x+\left (3+8 e^2\right ) x^2-12 x^3}{1-4 x}\right ) \, dx\\ &=-\left (2 \int e^{e^{2 x}+2 x} \, dx\right )+\int \frac {4-2 e^2 x+\left (3+8 e^2\right ) x^2-12 x^3}{1-4 x} \, dx\\ &=\int \left (-2 e^2 x+3 x^2-\frac {4}{-1+4 x}\right ) \, dx-\operatorname {Subst}\left (\int e^x \, dx,x,e^{2 x}\right )\\ &=-e^{e^{2 x}}-e^2 x^2+x^3-\log (1-4 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 29, normalized size = 0.85 \begin {gather*} -e^{e^{2 x}}-e^2 x^2+x^3-\log (1-4 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + E^(E^(2*x) + 2*x)*(2 - 8*x) - 3*x^2 + 12*x^3 + E^2*(2*x - 8*x^2))/(-1 + 4*x),x]

[Out]

-E^E^(2*x) - E^2*x^2 + x^3 - Log[1 - 4*x]

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fricas [A]  time = 0.77, size = 45, normalized size = 1.32 \begin {gather*} {\left ({\left (x^{3} - x^{2} e^{2}\right )} e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x - 1\right ) - e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(4*x-1),x, algorithm="fricas")

[Out]

((x^3 - x^2*e^2)*e^(2*x) - e^(2*x)*log(4*x - 1) - e^(2*x + e^(2*x)))*e^(-2*x)

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giac [A]  time = 0.15, size = 48, normalized size = 1.41 \begin {gather*} {\left (x^{3} e^{\left (2 \, x\right )} - x^{2} e^{\left (2 \, x + 2\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x - 1\right ) - e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(4*x-1),x, algorithm="giac")

[Out]

(x^3*e^(2*x) - x^2*e^(2*x + 2) - e^(2*x)*log(4*x - 1) - e^(2*x + e^(2*x)))*e^(-2*x)

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maple [A]  time = 0.32, size = 27, normalized size = 0.79

method	result	size
norman	\(x^{3}-x^{2} {\mathrm e}^{2}-{\mathrm e}^{{\mathrm e}^{2 x}}-\ln \left (4 x -1\right )\)	\(27\)
risch	\(x^{3}-x^{2} {\mathrm e}^{2}-{\mathrm e}^{{\mathrm e}^{2 x}}-\ln \left (4 x -1\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(4*x-1),x,method=_RETURNVERBOSE)

[Out]

x^3-x^2*exp(2)-exp(exp(x)^2)-ln(4*x-1)

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maxima [A]  time = 0.42, size = 52, normalized size = 1.53 \begin {gather*} x^{3} - \frac {1}{8} \, {\left (8 \, x^{2} + 4 \, x + \log \left (4 \, x - 1\right )\right )} e^{2} + \frac {1}{8} \, {\left (4 \, x + \log \left (4 \, x - 1\right )\right )} e^{2} - e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (4 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(4*x-1),x, algorithm="maxima")

[Out]

x^3 - 1/8*(8*x^2 + 4*x + log(4*x - 1))*e^2 + 1/8*(4*x + log(4*x - 1))*e^2 - e^(e^(2*x)) - log(4*x - 1)

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mupad [B]  time = 4.41, size = 24, normalized size = 0.71 \begin {gather*} x^3-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x^2\,{\mathrm {e}}^2-\ln \left (x-\frac {1}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^2 - exp(2)*(2*x - 8*x^2) - 12*x^3 + exp(2*x)*exp(exp(2*x))*(8*x - 2) + 4)/(4*x - 1),x)

[Out]

x^3 - exp(exp(2*x)) - x^2*exp(2) - log(x - 1/4)

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sympy [A]  time = 0.19, size = 22, normalized size = 0.65 \begin {gather*} x^{3} - x^{2} e^{2} - e^{e^{2 x}} - \log {\left (4 x - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x+2)*exp(x)**2*exp(exp(x)**2)+(-8*x**2+2*x)*exp(2)+12*x**3-3*x**2-4)/(4*x-1),x)

[Out]

x**3 - x**2*exp(2) - exp(exp(2*x)) - log(4*x - 1)

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