3.100.84 \(\int (18+2 x+e^3 (-18-42 x-6 x^2)+e^6 (4 x+10 x^2+4 x^3)+(e^3 (18+4 x)+e^6 (-6 x-6 x^2)) \log (x)+2 e^6 x \log ^2(x)) \, dx\)

Optimal. Leaf size=18 \[ \left (9+x-e^3 x (2+x-\log (x))\right )^2 \]

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Rubi [B]  time = 0.09, antiderivative size = 126, normalized size of antiderivative = 7.00, number of steps used = 10, number of rules used = 4, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {2356, 2295, 2304, 2305} \begin {gather*} e^6 x^4+4 e^6 x^3-2 e^3 x^3-2 e^6 x^3 \log (x)-\frac {1}{2} e^3 \left (2-3 e^3\right ) x^2+\frac {5 e^6 x^2}{2}-21 e^3 x^2+x^2+e^6 x^2 \log ^2(x)+e^3 \left (2-3 e^3\right ) x^2 \log (x)-e^6 x^2 \log (x)-36 e^3 x+18 x+18 e^3 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[18 + 2*x + E^3*(-18 - 42*x - 6*x^2) + E^6*(4*x + 10*x^2 + 4*x^3) + (E^3*(18 + 4*x) + E^6*(-6*x - 6*x^2))*L
og[x] + 2*E^6*x*Log[x]^2,x]

[Out]

18*x - 36*E^3*x + x^2 - 21*E^3*x^2 + (5*E^6*x^2)/2 - (E^3*(2 - 3*E^3)*x^2)/2 - 2*E^3*x^3 + 4*E^6*x^3 + E^6*x^4
 + 18*E^3*x*Log[x] - E^6*x^2*Log[x] + E^3*(2 - 3*E^3)*x^2*Log[x] - 2*E^6*x^3*Log[x] + E^6*x^2*Log[x]^2

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=18 x+x^2+e^3 \int \left (-18-42 x-6 x^2\right ) \, dx+e^6 \int \left (4 x+10 x^2+4 x^3\right ) \, dx+\left (2 e^6\right ) \int x \log ^2(x) \, dx+\int \left (e^3 (18+4 x)+e^6 \left (-6 x-6 x^2\right )\right ) \log (x) \, dx\\ &=18 x-18 e^3 x+x^2-21 e^3 x^2+2 e^6 x^2-2 e^3 x^3+\frac {10 e^6 x^3}{3}+e^6 x^4+e^6 x^2 \log ^2(x)-\left (2 e^6\right ) \int x \log (x) \, dx+\int \left (18 e^3 \log (x)-2 e^3 \left (-2+3 e^3\right ) x \log (x)-6 e^6 x^2 \log (x)\right ) \, dx\\ &=18 x-18 e^3 x+x^2-21 e^3 x^2+\frac {5 e^6 x^2}{2}-2 e^3 x^3+\frac {10 e^6 x^3}{3}+e^6 x^4-e^6 x^2 \log (x)+e^6 x^2 \log ^2(x)+\left (18 e^3\right ) \int \log (x) \, dx-\left (6 e^6\right ) \int x^2 \log (x) \, dx+\left (2 e^3 \left (2-3 e^3\right )\right ) \int x \log (x) \, dx\\ &=18 x-36 e^3 x+x^2-21 e^3 x^2+\frac {5 e^6 x^2}{2}-\frac {1}{2} e^3 \left (2-3 e^3\right ) x^2-2 e^3 x^3+4 e^6 x^3+e^6 x^4+18 e^3 x \log (x)-e^6 x^2 \log (x)+e^3 \left (2-3 e^3\right ) x^2 \log (x)-2 e^6 x^3 \log (x)+e^6 x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 101, normalized size = 5.61 \begin {gather*} 18 x-36 e^3 x+x^2-22 e^3 x^2+4 e^6 x^2-2 e^3 x^3+4 e^6 x^3+e^6 x^4+18 e^3 x \log (x)+2 e^3 x^2 \log (x)-4 e^6 x^2 \log (x)-2 e^6 x^3 \log (x)+e^6 x^2 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[18 + 2*x + E^3*(-18 - 42*x - 6*x^2) + E^6*(4*x + 10*x^2 + 4*x^3) + (E^3*(18 + 4*x) + E^6*(-6*x - 6*x
^2))*Log[x] + 2*E^6*x*Log[x]^2,x]

[Out]

18*x - 36*E^3*x + x^2 - 22*E^3*x^2 + 4*E^6*x^2 - 2*E^3*x^3 + 4*E^6*x^3 + E^6*x^4 + 18*E^3*x*Log[x] + 2*E^3*x^2
*Log[x] - 4*E^6*x^2*Log[x] - 2*E^6*x^3*Log[x] + E^6*x^2*Log[x]^2

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fricas [B]  time = 2.02, size = 78, normalized size = 4.33 \begin {gather*} x^{2} e^{6} \log \relax (x)^{2} + x^{2} + {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} e^{6} - 2 \, {\left (x^{3} + 11 \, x^{2} + 18 \, x\right )} e^{3} - 2 \, {\left ({\left (x^{3} + 2 \, x^{2}\right )} e^{6} - {\left (x^{2} + 9 \, x\right )} e^{3}\right )} \log \relax (x) + 18 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)^2*log(x)^2+((-6*x^2-6*x)*exp(3)^2+(4*x+18)*exp(3))*log(x)+(4*x^3+10*x^2+4*x)*exp(3)^2+(-6
*x^2-42*x-18)*exp(3)+2*x+18,x, algorithm="fricas")

[Out]

x^2*e^6*log(x)^2 + x^2 + (x^4 + 4*x^3 + 4*x^2)*e^6 - 2*(x^3 + 11*x^2 + 18*x)*e^3 - 2*((x^3 + 2*x^2)*e^6 - (x^2
 + 9*x)*e^3)*log(x) + 18*x

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giac [B]  time = 0.31, size = 129, normalized size = 7.17 \begin {gather*} -2 \, x^{3} e^{6} \log \relax (x) + \frac {2}{3} \, x^{3} e^{6} - 3 \, x^{2} e^{6} \log \relax (x) + 2 \, x^{2} e^{3} \log \relax (x) + \frac {3}{2} \, x^{2} e^{6} - x^{2} e^{3} + 18 \, x e^{3} \log \relax (x) + x^{2} + \frac {1}{3} \, {\left (3 \, x^{4} + 10 \, x^{3} + 6 \, x^{2}\right )} e^{6} + \frac {1}{2} \, {\left (2 \, x^{2} \log \relax (x)^{2} - 2 \, x^{2} \log \relax (x) + x^{2}\right )} e^{6} - {\left (2 \, x^{3} + 21 \, x^{2} + 18 \, x\right )} e^{3} - 18 \, x e^{3} + 18 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)^2*log(x)^2+((-6*x^2-6*x)*exp(3)^2+(4*x+18)*exp(3))*log(x)+(4*x^3+10*x^2+4*x)*exp(3)^2+(-6
*x^2-42*x-18)*exp(3)+2*x+18,x, algorithm="giac")

[Out]

-2*x^3*e^6*log(x) + 2/3*x^3*e^6 - 3*x^2*e^6*log(x) + 2*x^2*e^3*log(x) + 3/2*x^2*e^6 - x^2*e^3 + 18*x*e^3*log(x
) + x^2 + 1/3*(3*x^4 + 10*x^3 + 6*x^2)*e^6 + 1/2*(2*x^2*log(x)^2 - 2*x^2*log(x) + x^2)*e^6 - (2*x^3 + 21*x^2 +
 18*x)*e^3 - 18*x*e^3 + 18*x

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maple [B]  time = 0.04, size = 91, normalized size = 5.06




method result size



risch \(x^{2} {\mathrm e}^{6} \ln \relax (x )^{2}-2 \,{\mathrm e}^{6} x^{3} \ln \relax (x )+x^{4} {\mathrm e}^{6}-4 \ln \relax (x ) {\mathrm e}^{6} x^{2}+4 x^{3} {\mathrm e}^{6}+2 \,{\mathrm e}^{3} \ln \relax (x ) x^{2}+4 x^{2} {\mathrm e}^{6}-2 x^{3} {\mathrm e}^{3}+18 x \,{\mathrm e}^{3} \ln \relax (x )-22 x^{2} {\mathrm e}^{3}-36 x \,{\mathrm e}^{3}+x^{2}+18 x\) \(91\)
norman \(x^{4} {\mathrm e}^{6}+\left (4 \,{\mathrm e}^{6}-2 \,{\mathrm e}^{3}\right ) x^{3}+\left (-36 \,{\mathrm e}^{3}+18\right ) x +\left (4 \,{\mathrm e}^{6}-22 \,{\mathrm e}^{3}+1\right ) x^{2}+x^{2} {\mathrm e}^{6} \ln \relax (x )^{2}+\left (-4 \,{\mathrm e}^{6}+2 \,{\mathrm e}^{3}\right ) x^{2} \ln \relax (x )+18 x \,{\mathrm e}^{3} \ln \relax (x )-2 \,{\mathrm e}^{6} x^{3} \ln \relax (x )\) \(96\)
default \(18 x -2 \,{\mathrm e}^{6} x^{3} \ln \relax (x )+\frac {2 x^{3} {\mathrm e}^{6}}{3}-4 \ln \relax (x ) {\mathrm e}^{6} x^{2}+2 x^{2} {\mathrm e}^{6}+2 \,{\mathrm e}^{3} \ln \relax (x ) x^{2}-x^{2} {\mathrm e}^{3}+18 x \,{\mathrm e}^{3} \ln \relax (x )-18 x \,{\mathrm e}^{3}+{\mathrm e}^{3} \left (-2 x^{3}-21 x^{2}-18 x \right )+{\mathrm e}^{6} \left (x^{4}+\frac {10}{3} x^{3}+2 x^{2}\right )+x^{2}+x^{2} {\mathrm e}^{6} \ln \relax (x )^{2}\) \(124\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*exp(3)^2*ln(x)^2+((-6*x^2-6*x)*exp(3)^2+(4*x+18)*exp(3))*ln(x)+(4*x^3+10*x^2+4*x)*exp(3)^2+(-6*x^2-42*
x-18)*exp(3)+2*x+18,x,method=_RETURNVERBOSE)

[Out]

x^2*exp(6)*ln(x)^2-2*exp(6)*x^3*ln(x)+x^4*exp(6)-4*ln(x)*exp(6)*x^2+4*x^3*exp(6)+2*exp(3)*ln(x)*x^2+4*x^2*exp(
6)-2*x^3*exp(3)+18*x*exp(3)*ln(x)-22*x^2*exp(3)-36*x*exp(3)+x^2+18*x

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maxima [B]  time = 0.36, size = 120, normalized size = 6.67 \begin {gather*} \frac {1}{2} \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} e^{6} + \frac {2}{3} \, x^{3} e^{6} + \frac {1}{2} \, x^{2} {\left (3 \, e^{6} - 2 \, e^{3}\right )} + x^{2} + \frac {1}{3} \, {\left (3 \, x^{4} + 10 \, x^{3} + 6 \, x^{2}\right )} e^{6} - {\left (2 \, x^{3} + 21 \, x^{2} + 18 \, x\right )} e^{3} - 18 \, x e^{3} - {\left ({\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{6} - 2 \, {\left (x^{2} + 9 \, x\right )} e^{3}\right )} \log \relax (x) + 18 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)^2*log(x)^2+((-6*x^2-6*x)*exp(3)^2+(4*x+18)*exp(3))*log(x)+(4*x^3+10*x^2+4*x)*exp(3)^2+(-6
*x^2-42*x-18)*exp(3)+2*x+18,x, algorithm="maxima")

[Out]

1/2*(2*log(x)^2 - 2*log(x) + 1)*x^2*e^6 + 2/3*x^3*e^6 + 1/2*x^2*(3*e^6 - 2*e^3) + x^2 + 1/3*(3*x^4 + 10*x^3 +
6*x^2)*e^6 - (2*x^3 + 21*x^2 + 18*x)*e^3 - 18*x*e^3 - ((2*x^3 + 3*x^2)*e^6 - 2*(x^2 + 9*x)*e^3)*log(x) + 18*x

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mupad [B]  time = 7.70, size = 74, normalized size = 4.11 \begin {gather*} x^2\,\left ({\mathrm {e}}^6\,{\ln \relax (x)}^2+\left (2\,{\mathrm {e}}^3-4\,{\mathrm {e}}^6\right )\,\ln \relax (x)-22\,{\mathrm {e}}^3+4\,{\mathrm {e}}^6+1\right )+x^4\,{\mathrm {e}}^6+x\,\left (18\,{\mathrm {e}}^3\,\ln \relax (x)-36\,{\mathrm {e}}^3+18\right )-x^3\,\left (2\,{\mathrm {e}}^3-4\,{\mathrm {e}}^6+2\,{\mathrm {e}}^6\,\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - exp(3)*(42*x + 6*x^2 + 18) + exp(6)*(4*x + 10*x^2 + 4*x^3) - log(x)*(exp(6)*(6*x + 6*x^2) - exp(3)*(
4*x + 18)) + 2*x*exp(6)*log(x)^2 + 18,x)

[Out]

x^2*(4*exp(6) - 22*exp(3) + exp(6)*log(x)^2 + log(x)*(2*exp(3) - 4*exp(6)) + 1) + x^4*exp(6) + x*(18*exp(3)*lo
g(x) - 36*exp(3) + 18) - x^3*(2*exp(3) - 4*exp(6) + 2*exp(6)*log(x))

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sympy [B]  time = 0.19, size = 90, normalized size = 5.00 \begin {gather*} x^{4} e^{6} + x^{3} \left (- 2 e^{3} + 4 e^{6}\right ) + x^{2} e^{6} \log {\relax (x )}^{2} + x^{2} \left (- 22 e^{3} + 1 + 4 e^{6}\right ) + x \left (18 - 36 e^{3}\right ) + \left (- 2 x^{3} e^{6} - 4 x^{2} e^{6} + 2 x^{2} e^{3} + 18 x e^{3}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)**2*ln(x)**2+((-6*x**2-6*x)*exp(3)**2+(4*x+18)*exp(3))*ln(x)+(4*x**3+10*x**2+4*x)*exp(3)**
2+(-6*x**2-42*x-18)*exp(3)+2*x+18,x)

[Out]

x**4*exp(6) + x**3*(-2*exp(3) + 4*exp(6)) + x**2*exp(6)*log(x)**2 + x**2*(-22*exp(3) + 1 + 4*exp(6)) + x*(18 -
 36*exp(3)) + (-2*x**3*exp(6) - 4*x**2*exp(6) + 2*x**2*exp(3) + 18*x*exp(3))*log(x)

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