3.16.52 \(\int \frac {e^x (-4 x+x^2+4 x \log (\frac {3+3 x}{x})) (-8-6 x-x^2+x^3+(4+8 x+4 x^2) \log (\frac {3+3 x}{x}))}{-4 x-3 x^2+x^3+(4 x+4 x^2) \log (\frac {3+3 x}{x})} \, dx\)

Optimal. Leaf size=24 \[ e^x x \left (1+x-4 \left (\frac {5}{4}-\log \left (3+\frac {3}{x}\right )\right )\right ) \]

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Rubi [A]  time = 0.64, antiderivative size = 43, normalized size of antiderivative = 1.79, number of steps used = 14, number of rules used = 7, integrand size = 92, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {6688, 6742, 2199, 2194, 2176, 2178, 2554} \begin {gather*} e^x x^2-4 e^x x-4 e^x \log \left (\frac {3}{x}+3\right )+4 e^x (x+1) \log \left (\frac {3}{x}+3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-4*x + x^2 + 4*x*Log[(3 + 3*x)/x])*(-8 - 6*x - x^2 + x^3 + (4 + 8*x + 4*x^2)*Log[(3 + 3*x)/x]))/(-4*
x - 3*x^2 + x^3 + (4*x + 4*x^2)*Log[(3 + 3*x)/x]),x]

[Out]

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 \frac {e^x \left (-8-6 x-x^2+x^3+4 (1+x)^2 \log \left (3+\frac {3}{x}\right )\right )}{1+x} \, dx\\ &=\int \left (\frac {e^x \left (-8-6 x-x^2+x^3\right )}{1+x}+4 e^x (1+x) \log \left (3+\frac {3}{x}\right )\right ) \, dx\\ &=4 \int e^x (1+x) \log \left (3+\frac {3}{x}\right ) \, dx+\int \frac {e^x \left (-8-6 x-x^2+x^3\right )}{1+x} \, dx\\ &=-4 e^x \log \left (3+\frac {3}{x}\right )+4 e^x (1+x) \log \left (3+\frac {3}{x}\right )-4 \int \frac {e^x}{-1-x} \, dx+\int \left (-4 e^x-2 e^x x+e^x x^2-\frac {4 e^x}{1+x}\right ) \, dx\\ &=\frac {4 \text {Ei}(1+x)}{e}-4 e^x \log \left (3+\frac {3}{x}\right )+4 e^x (1+x) \log \left (3+\frac {3}{x}\right )-2 \int e^x x \, dx-4 \int e^x \, dx-4 \int \frac {e^x}{1+x} \, dx+\int e^x x^2 \, dx\\ &=-4 e^x-2 e^x x+e^x x^2-4 e^x \log \left (3+\frac {3}{x}\right )+4 e^x (1+x) \log \left (3+\frac {3}{x}\right )+2 \int e^x \, dx-2 \int e^x x \, dx\\ &=-2 e^x-4 e^x x+e^x x^2-4 e^x \log \left (3+\frac {3}{x}\right )+4 e^x (1+x) \log \left (3+\frac {3}{x}\right )+2 \int e^x \, dx\\ &=-4 e^x x+e^x x^2-4 e^x \log \left (3+\frac {3}{x}\right )+4 e^x (1+x) \log \left (3+\frac {3}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.03, size = 18, normalized size = 0.75 \begin {gather*} e^x x \left (-4+x+4 \log \left (3+\frac {3}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-4*x + x^2 + 4*x*Log[(3 + 3*x)/x])*(-8 - 6*x - x^2 + x^3 + (4 + 8*x + 4*x^2)*Log[(3 + 3*x)/x])
)/(-4*x - 3*x^2 + x^3 + (4*x + 4*x^2)*Log[(3 + 3*x)/x]),x]

[Out]

E^x*x*(-4 + x + 4*Log[3 + 3/x])

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fricas [A]  time = 0.68, size = 23, normalized size = 0.96 \begin {gather*} e^{\left (x + \log \left (x^{2} + 4 \, x \log \left (\frac {3 \, {\left (x + 1\right )}}{x}\right ) - 4 \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+8*x+4)*log((3*x+3)/x)+x^3-x^2-6*x-8)*exp(log(4*x*log((3*x+3)/x)+x^2-4*x)+x)/((4*x^2+4*x)*log
((3*x+3)/x)+x^3-3*x^2-4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.36, size = 22, normalized size = 0.92 \begin {gather*} e^{\left (x + \log \left (x^{2} + 4 \, x \log \left (\frac {3}{x} + 3\right ) - 4 \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+8*x+4)*log((3*x+3)/x)+x^3-x^2-6*x-8)*exp(log(4*x*log((3*x+3)/x)+x^2-4*x)+x)/((4*x^2+4*x)*log
((3*x+3)/x)+x^3-3*x^2-4*x),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (4 x^{2}+8 x +4\right ) \ln \left (\frac {3 x +3}{x}\right )+x^{3}-x^{2}-6 x -8\right ) {\mathrm e}^{\ln \left (4 x \ln \left (\frac {3 x +3}{x}\right )+x^{2}-4 x \right )+x}}{\left (4 x^{2}+4 x \right ) \ln \left (\frac {3 x +3}{x}\right )+x^{3}-3 x^{2}-4 x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+8*x+4)*ln((3*x+3)/x)+x^3-x^2-6*x-8)*exp(ln(4*x*ln((3*x+3)/x)+x^2-4*x)+x)/((4*x^2+4*x)*ln((3*x+3)/x
)+x^3-3*x^2-4*x),x)

[Out]

int(((4*x^2+8*x+4)*ln((3*x+3)/x)+x^3-x^2-6*x-8)*exp(ln(4*x*ln((3*x+3)/x)+x^2-4*x)+x)/((4*x^2+4*x)*ln((3*x+3)/x
)+x^3-3*x^2-4*x),x)

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maxima [A]  time = 0.83, size = 29, normalized size = 1.21 \begin {gather*} 4 \, x e^{x} \log \left (x + 1\right ) + {\left (x^{2} + 4 \, x {\left (\log \relax (3) - 1\right )} - 4 \, x \log \relax (x)\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+8*x+4)*log((3*x+3)/x)+x^3-x^2-6*x-8)*exp(log(4*x*log((3*x+3)/x)+x^2-4*x)+x)/((4*x^2+4*x)*log
((3*x+3)/x)+x^3-3*x^2-4*x),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{x+\ln \left (4\,x\,\ln \left (\frac {3\,x+3}{x}\right )-4\,x+x^2\right )}\,\left (6\,x-\ln \left (\frac {3\,x+3}{x}\right )\,\left (4\,x^2+8\,x+4\right )+x^2-x^3+8\right )}{4\,x-\ln \left (\frac {3\,x+3}{x}\right )\,\left (4\,x^2+4\,x\right )+3\,x^2-x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(4*x*log((3*x + 3)/x) - 4*x + x^2))*(6*x - log((3*x + 3)/x)*(8*x + 4*x^2 + 4) + x^2 - x^3 + 8)
)/(4*x - log((3*x + 3)/x)*(4*x + 4*x^2) + 3*x^2 - x^3),x)

[Out]

int((exp(x + log(4*x*log((3*x + 3)/x) - 4*x + x^2))*(6*x - log((3*x + 3)/x)*(8*x + 4*x^2 + 4) + x^2 - x^3 + 8)
)/(4*x - log((3*x + 3)/x)*(4*x + 4*x^2) + 3*x^2 - x^3), x)

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sympy [A]  time = 0.42, size = 20, normalized size = 0.83 \begin {gather*} \left (x^{2} + 4 x \log {\left (\frac {3 x + 3}{x} \right )} - 4 x\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+8*x+4)*ln((3*x+3)/x)+x**3-x**2-6*x-8)*exp(ln(4*x*ln((3*x+3)/x)+x**2-4*x)+x)/((4*x**2+4*x)*l
n((3*x+3)/x)+x**3-3*x**2-4*x),x)

[Out]

(x**2 + 4*x*log((3*x + 3)/x) - 4*x)*exp(x)

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