3.74.88 \(\int \frac {1}{9} (18 x-27 x^2+2 x \log ^2(x)-4 x \log (x) \log (3 x+e^4 x)+2 x \log ^2(3 x+e^4 x)) \, dx\)

Optimal. Leaf size=28 \[ x^2 \left (1-x+\frac {1}{9} \left (\log (x)-\log \left (\left (3+e^4\right ) x\right )\right )^2\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 80, normalized size of antiderivative = 2.86, number of steps used = 11, number of rules used = 6, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 2305, 2304, 2380, 2366, 2421} \begin {gather*} -x^3+\frac {19 x^2}{18}+\frac {1}{9} x^2 \log ^2(x)+\frac {1}{9} x^2 \log ^2\left (\left (3+e^4\right ) x\right )-\frac {1}{18} x^2 (1-2 \log (x))-\frac {1}{9} x^2 \log (x)-\frac {2}{9} x^2 \log (x) \log \left (\left (3+e^4\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18*x - 27*x^2 + 2*x*Log[x]^2 - 4*x*Log[x]*Log[3*x + E^4*x] + 2*x*Log[3*x + E^4*x]^2)/9,x]

[Out]

(19*x^2)/18 - x^3 - (x^2*(1 - 2*Log[x]))/18 - (x^2*Log[x])/9 + (x^2*Log[x]^2)/9 - (2*x^2*Log[x]*Log[(3 + E^4)*
x])/9 + (x^2*Log[(3 + E^4)*x]^2)/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 2380

Int[Log[(d_.)*(u_)^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> Int[(g*
x)^q*Log[d*ExpandToSum[u, x]^r]*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, g, r, n, p, q}, x] && BinomialQ
[u, x] &&  !BinomialMatchQ[u, x]

Rule 2421

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \left (18 x-27 x^2+2 x \log ^2(x)-4 x \log (x) \log \left (3 x+e^4 x\right )+2 x \log ^2\left (3 x+e^4 x\right )\right ) \, dx\\ &=x^2-x^3+\frac {2}{9} \int x \log ^2(x) \, dx+\frac {2}{9} \int x \log ^2\left (3 x+e^4 x\right ) \, dx-\frac {4}{9} \int x \log (x) \log \left (3 x+e^4 x\right ) \, dx\\ &=x^2-x^3+\frac {1}{9} x^2 \log ^2(x)-\frac {2}{9} \int x \log (x) \, dx+\frac {2}{9} \int x \log ^2\left (\left (3+e^4\right ) x\right ) \, dx-\frac {4}{9} \int x \log (x) \log \left (\left (3+e^4\right ) x\right ) \, dx\\ &=\frac {19 x^2}{18}-x^3-\frac {1}{9} x^2 \log (x)+\frac {1}{9} x^2 \log ^2(x)+\frac {1}{9} x^2 \log \left (\left (3+e^4\right ) x\right )-\frac {2}{9} x^2 \log (x) \log \left (\left (3+e^4\right ) x\right )+\frac {1}{9} x^2 \log ^2\left (\left (3+e^4\right ) x\right )-\frac {2}{9} \int x \log \left (\left (3+e^4\right ) x\right ) \, dx+\frac {4}{9} \int \frac {1}{4} x (-1+2 \log (x)) \, dx\\ &=\frac {10 x^2}{9}-x^3-\frac {1}{9} x^2 \log (x)+\frac {1}{9} x^2 \log ^2(x)-\frac {2}{9} x^2 \log (x) \log \left (\left (3+e^4\right ) x\right )+\frac {1}{9} x^2 \log ^2\left (\left (3+e^4\right ) x\right )+\frac {1}{9} \int x (-1+2 \log (x)) \, dx\\ &=\frac {19 x^2}{18}-x^3-\frac {1}{18} x^2 (1-2 \log (x))-\frac {1}{9} x^2 \log (x)+\frac {1}{9} x^2 \log ^2(x)-\frac {2}{9} x^2 \log (x) \log \left (\left (3+e^4\right ) x\right )+\frac {1}{9} x^2 \log ^2\left (\left (3+e^4\right ) x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.82 \begin {gather*} -x^3+\frac {1}{9} x^2 \left (9+\log ^2\left (3+e^4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18*x - 27*x^2 + 2*x*Log[x]^2 - 4*x*Log[x]*Log[3*x + E^4*x] + 2*x*Log[3*x + E^4*x]^2)/9,x]

[Out]

-x^3 + (x^2*(9 + Log[3 + E^4]^2))/9

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fricas [A]  time = 0.53, size = 21, normalized size = 0.75 \begin {gather*} \frac {1}{9} \, x^{2} \log \left (e^{4} + 3\right )^{2} - x^{3} + x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/9*x*log(x*exp(4)+3*x)^2-4/9*x*log(x)*log(x*exp(4)+3*x)+2/9*x*log(x)^2-3*x^2+2*x,x, algorithm="fric
as")

[Out]

1/9*x^2*log(e^4 + 3)^2 - x^3 + x^2

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giac [B]  time = 0.23, size = 135, normalized size = 4.82 \begin {gather*} -\frac {1}{9} \, x^{2} \log \relax (x)^{2} - \frac {2}{9} \, x^{2} \log \relax (x) \log \left (e^{4} + 3\right ) - x^{3} + \frac {{\left (x e^{4} + 3 \, x\right )}^{2} \log \left (x e^{4} + 3 \, x\right )^{2}}{9 \, {\left (e^{8} + 6 \, e^{4} + 9\right )}} + \frac {1}{9} \, x^{2} \log \relax (x) + \frac {1}{9} \, x^{2} \log \left (e^{4} + 3\right ) + \frac {17}{18} \, x^{2} - \frac {{\left (x e^{4} + 3 \, x\right )}^{2} \log \left (x e^{4} + 3 \, x\right )}{9 \, {\left (e^{8} + 6 \, e^{4} + 9\right )}} + \frac {{\left (x e^{4} + 3 \, x\right )}^{2}}{18 \, {\left (e^{8} + 6 \, e^{4} + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/9*x*log(x*exp(4)+3*x)^2-4/9*x*log(x)*log(x*exp(4)+3*x)+2/9*x*log(x)^2-3*x^2+2*x,x, algorithm="giac
")

[Out]

-1/9*x^2*log(x)^2 - 2/9*x^2*log(x)*log(e^4 + 3) - x^3 + 1/9*(x*e^4 + 3*x)^2*log(x*e^4 + 3*x)^2/(e^8 + 6*e^4 +
9) + 1/9*x^2*log(x) + 1/9*x^2*log(e^4 + 3) + 17/18*x^2 - 1/9*(x*e^4 + 3*x)^2*log(x*e^4 + 3*x)/(e^8 + 6*e^4 + 9
) + 1/18*(x*e^4 + 3*x)^2/(e^8 + 6*e^4 + 9)

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maple [A]  time = 0.05, size = 56, normalized size = 2.00

method	result	size
risch	\(\frac {x^{2} \ln \left (x \,{\mathrm e}^{4}+3 x \right )^{2}}{9}-\frac {x^{2} \ln \left (x \,{\mathrm e}^{4}+3 x \right )}{9}-\frac {x^{2} \ln \relax (x )^{2}}{9}+\frac {x^{2} \ln \relax (x )}{9}-x^{3}+x^{2}\)	\(56\)
default	\(-x^{3}+\frac {17 x^{2}}{18}+\frac {x^{2} \ln \relax (x )^{2}}{9}+\frac {{\mathrm e}^{8} \ln \left (\left ({\mathrm e}^{4}+3\right ) x \right )^{2} x^{2}}{9 \left ({\mathrm e}^{4}+3\right )^{2}}+\frac {2 \,{\mathrm e}^{4} \ln \left (\left ({\mathrm e}^{4}+3\right ) x \right )^{2} x^{2}}{3 \left ({\mathrm e}^{4}+3\right )^{2}}+\frac {\ln \left (\left ({\mathrm e}^{4}+3\right ) x \right )^{2} x^{2}}{\left ({\mathrm e}^{4}+3\right )^{2}}-\frac {{\mathrm e}^{8} \ln \left (\left ({\mathrm e}^{4}+3\right ) x \right ) x^{2}}{9 \left ({\mathrm e}^{4}+3\right )^{2}}-\frac {2 \,{\mathrm e}^{4} \ln \left (\left ({\mathrm e}^{4}+3\right ) x \right ) x^{2}}{3 \left ({\mathrm e}^{4}+3\right )^{2}}-\frac {\ln \left (\left ({\mathrm e}^{4}+3\right ) x \right ) x^{2}}{\left ({\mathrm e}^{4}+3\right )^{2}}+\frac {{\mathrm e}^{8} x^{2}}{18 \left ({\mathrm e}^{4}+3\right )^{2}}+\frac {{\mathrm e}^{4} x^{2}}{3 \left ({\mathrm e}^{4}+3\right )^{2}}+\frac {x^{2}}{2 \left ({\mathrm e}^{4}+3\right )^{2}}+\frac {x^{2} \ln \left (x \,{\mathrm e}^{4}+3 x \right )}{9}-\frac {2 x^{2} \ln \relax (x ) \ln \left (x \,{\mathrm e}^{4}+3 x \right )}{9}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/9*x*ln(x*exp(4)+3*x)^2-4/9*x*ln(x)*ln(x*exp(4)+3*x)+2/9*x*ln(x)^2-3*x^2+2*x,x,method=_RETURNVERBOSE)

[Out]

1/9*x^2*ln(x*exp(4)+3*x)^2-1/9*x^2*ln(x*exp(4)+3*x)-1/9*x^2*ln(x)^2+1/9*x^2*ln(x)-x^3+x^2

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maxima [B]  time = 0.47, size = 113, normalized size = 4.04 \begin {gather*} \frac {1}{18} \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} - x^{3} + \frac {1}{9} \, x^{2} {\left (\log \left (e^{4} + 3\right ) - 1\right )} + \frac {1}{9} \, x^{2} \log \relax (x) + x^{2} - \frac {1}{9} \, {\left (2 \, x^{2} \log \left (x e^{4} + 3 \, x\right ) - x^{2}\right )} \log \relax (x) + \frac {{\left (x e^{4} + 3 \, x\right )}^{2} {\left (2 \, \log \left (x e^{4} + 3 \, x\right )^{2} - 2 \, \log \left (x e^{4} + 3 \, x\right ) + 1\right )}}{18 \, {\left (e^{4} + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/9*x*log(x*exp(4)+3*x)^2-4/9*x*log(x)*log(x*exp(4)+3*x)+2/9*x*log(x)^2-3*x^2+2*x,x, algorithm="maxi
ma")

[Out]

1/18*(2*log(x)^2 - 2*log(x) + 1)*x^2 - x^3 + 1/9*x^2*(log(e^4 + 3) - 1) + 1/9*x^2*log(x) + x^2 - 1/9*(2*x^2*lo
g(x*e^4 + 3*x) - x^2)*log(x) + 1/18*(x*e^4 + 3*x)^2*(2*log(x*e^4 + 3*x)^2 - 2*log(x*e^4 + 3*x) + 1)/(e^4 + 3)^
2

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mupad [B]  time = 4.74, size = 17, normalized size = 0.61 \begin {gather*} \frac {x^2\,\left ({\ln \left ({\mathrm {e}}^4+3\right )}^2-9\,x+9\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x + (2*x*log(3*x + x*exp(4))^2)/9 + (2*x*log(x)^2)/9 - 3*x^2 - (4*x*log(3*x + x*exp(4))*log(x))/9,x)

[Out]

(x^2*(log(exp(4) + 3)^2 - 9*x + 9))/9

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sympy [A]  time = 0.22, size = 17, normalized size = 0.61 \begin {gather*} - x^{3} + x^{2} \left (1 + \frac {\log {\left (3 + e^{4} \right )}^{2}}{9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/9*x*ln(x*exp(4)+3*x)**2-4/9*x*ln(x)*ln(x*exp(4)+3*x)+2/9*x*ln(x)**2-3*x**2+2*x,x)

[Out]

-x**3 + x**2*(1 + log(3 + exp(4))**2/9)

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