3.28.3 \(\int \frac {1}{9} (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x ((-12-48 x) \log (3)+36 \log ^2(3))+e^{2 x} ((-6-48 x) \log (3)+108 \log ^2(3))+(-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)) \log (x)+18 x \log ^2(x)) \, dx\)

Optimal. Leaf size=22 \[ \left (-\frac {4 x}{3}+\left (1+e^x\right )^2 \log (3)+x \log (x)\right )^2 \]

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Rubi [B]  time = 0.20, antiderivative size = 165, normalized size of antiderivative = 7.50, number of steps used = 18, number of rules used = 9, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {12, 2194, 2187, 2176, 2554, 2282, 43, 2305, 2304} \begin {gather*} \frac {16 x^2}{9}+x^2 \log ^2(x)-\frac {8}{3} x^2 \log (x)+4 e^{3 x} \log ^2(3)+e^{4 x} \log ^2(3)+2 x \log (3) \log (x)-\frac {8}{3} x \log (3)-\frac {4}{3} e^x \log (3) (4 x+1-\log (27))-4 e^x \log (3) \log (x)-e^{2 x} \log (3) \log (x)+4 e^x (x+1) \log (3) \log (x)+e^{2 x} (2 x+1) \log (3) \log (x)+\frac {4}{3} e^x \log (3)+\frac {1}{3} e^{2 x} \log (3)-\frac {1}{3} e^{2 x} \log (3) (8 x+1-18 \log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*x - 6*Log[3] + 108*E^(3*x)*Log[3]^2 + 36*E^(4*x)*Log[3]^2 + E^x*((-12 - 48*x)*Log[3] + 36*Log[3]^2) + E
^(2*x)*((-6 - 48*x)*Log[3] + 108*Log[3]^2) + (-30*x + 18*Log[3] + E^(2*x)*(18 + 36*x)*Log[3] + E^x*(36 + 36*x)
*Log[3])*Log[x] + 18*x*Log[x]^2)/9,x]

[Out]

(16*x^2)/9 + (4*E^x*Log[3])/3 + (E^(2*x)*Log[3])/3 - (8*x*Log[3])/3 - (E^(2*x)*(1 + 8*x - 18*Log[3])*Log[3])/3
+ 4*E^(3*x)*Log[3]^2 + E^(4*x)*Log[3]^2 - (4*E^x*Log[3]*(1 + 4*x - Log[27]))/3 - (8*x^2*Log[x])/3 - 4*E^x*Log
[3]*Log[x] - E^(2*x)*Log[3]*Log[x] + 2*x*Log[3]*Log[x] + 4*E^x*(1 + x)*Log[3]*Log[x] + E^(2*x)*(1 + 2*x)*Log[3
]*Log[x] + x^2*Log[x]^2

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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 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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx\\ &=\frac {4 x^2}{9}-\frac {2}{3} x \log (3)+\frac {1}{9} \int e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right ) \, dx+\frac {1}{9} \int e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right ) \, dx+\frac {1}{9} \int \left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x) \, dx+2 \int x \log ^2(x) \, dx+\left (4 \log ^2(3)\right ) \int e^{4 x} \, dx+\left (12 \log ^2(3)\right ) \int e^{3 x} \, dx\\ &=\frac {4 x^2}{9}-\frac {2}{3} x \log (3)+4 e^{3 x} \log ^2(3)+e^{4 x} \log ^2(3)-\frac {5}{3} x^2 \log (x)-4 e^x \log (3) \log (x)-e^{2 x} \log (3) \log (x)+2 x \log (3) \log (x)+4 e^x (1+x) \log (3) \log (x)+e^{2 x} (1+2 x) \log (3) \log (x)+x^2 \log ^2(x)-\frac {1}{9} \int 3 \left (-5 x+6 \left (1+e^x\right )^2 \log (3)\right ) \, dx+\frac {1}{9} \int e^{2 x} (-48 x \log (3)-6 (1-18 \log (3)) \log (3)) \, dx+\frac {1}{9} \int e^x (-48 x \log (3)-12 \log (3) (1-\log (27))) \, dx-2 \int x \log (x) \, dx\\ &=\frac {17 x^2}{18}-\frac {2}{3} x \log (3)-\frac {1}{3} e^{2 x} (1+8 x-18 \log (3)) \log (3)+4 e^{3 x} \log ^2(3)+e^{4 x} \log ^2(3)-\frac {4}{3} e^x \log (3) (1+4 x-\log (27))-\frac {8}{3} x^2 \log (x)-4 e^x \log (3) \log (x)-e^{2 x} \log (3) \log (x)+2 x \log (3) \log (x)+4 e^x (1+x) \log (3) \log (x)+e^{2 x} (1+2 x) \log (3) \log (x)+x^2 \log ^2(x)-\frac {1}{3} \int \left (-5 x+6 \left (1+e^x\right )^2 \log (3)\right ) \, dx+\frac {1}{3} (8 \log (3)) \int e^{2 x} \, dx+\frac {1}{3} (16 \log (3)) \int e^x \, dx\\ &=\frac {16 x^2}{9}+\frac {16}{3} e^x \log (3)+\frac {4}{3} e^{2 x} \log (3)-\frac {2}{3} x \log (3)-\frac {1}{3} e^{2 x} (1+8 x-18 \log (3)) \log (3)+4 e^{3 x} \log ^2(3)+e^{4 x} \log ^2(3)-\frac {4}{3} e^x \log (3) (1+4 x-\log (27))-\frac {8}{3} x^2 \log (x)-4 e^x \log (3) \log (x)-e^{2 x} \log (3) \log (x)+2 x \log (3) \log (x)+4 e^x (1+x) \log (3) \log (x)+e^{2 x} (1+2 x) \log (3) \log (x)+x^2 \log ^2(x)-(2 \log (3)) \int \left (1+e^x\right )^2 \, dx\\ &=\frac {16 x^2}{9}+\frac {16}{3} e^x \log (3)+\frac {4}{3} e^{2 x} \log (3)-\frac {2}{3} x \log (3)-\frac {1}{3} e^{2 x} (1+8 x-18 \log (3)) \log (3)+4 e^{3 x} \log ^2(3)+e^{4 x} \log ^2(3)-\frac {4}{3} e^x \log (3) (1+4 x-\log (27))-\frac {8}{3} x^2 \log (x)-4 e^x \log (3) \log (x)-e^{2 x} \log (3) \log (x)+2 x \log (3) \log (x)+4 e^x (1+x) \log (3) \log (x)+e^{2 x} (1+2 x) \log (3) \log (x)+x^2 \log ^2(x)-(2 \log (3)) \operatorname {Subst}\left (\int \frac {(1+x)^2}{x} \, dx,x,e^x\right )\\ &=\frac {16 x^2}{9}+\frac {16}{3} e^x \log (3)+\frac {4}{3} e^{2 x} \log (3)-\frac {2}{3} x \log (3)-\frac {1}{3} e^{2 x} (1+8 x-18 \log (3)) \log (3)+4 e^{3 x} \log ^2(3)+e^{4 x} \log ^2(3)-\frac {4}{3} e^x \log (3) (1+4 x-\log (27))-\frac {8}{3} x^2 \log (x)-4 e^x \log (3) \log (x)-e^{2 x} \log (3) \log (x)+2 x \log (3) \log (x)+4 e^x (1+x) \log (3) \log (x)+e^{2 x} (1+2 x) \log (3) \log (x)+x^2 \log ^2(x)-(2 \log (3)) \operatorname {Subst}\left (\int \left (2+\frac {1}{x}+x\right ) \, dx,x,e^x\right )\\ &=\frac {16 x^2}{9}+\frac {4}{3} e^x \log (3)+\frac {1}{3} e^{2 x} \log (3)-\frac {8}{3} x \log (3)-\frac {1}{3} e^{2 x} (1+8 x-18 \log (3)) \log (3)+4 e^{3 x} \log ^2(3)+e^{4 x} \log ^2(3)-\frac {4}{3} e^x \log (3) (1+4 x-\log (27))-\frac {8}{3} x^2 \log (x)-4 e^x \log (3) \log (x)-e^{2 x} \log (3) \log (x)+2 x \log (3) \log (x)+4 e^x (1+x) \log (3) \log (x)+e^{2 x} (1+2 x) \log (3) \log (x)+x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 32, normalized size = 1.45 \begin {gather*} \frac {1}{9} \left (-4 x+6 e^x \log (3)+\log (27)+e^{2 x} \log (27)+3 x \log (x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x - 6*Log[3] + 108*E^(3*x)*Log[3]^2 + 36*E^(4*x)*Log[3]^2 + E^x*((-12 - 48*x)*Log[3] + 36*Log[3]^
2) + E^(2*x)*((-6 - 48*x)*Log[3] + 108*Log[3]^2) + (-30*x + 18*Log[3] + E^(2*x)*(18 + 36*x)*Log[3] + E^x*(36 +
36*x)*Log[3])*Log[x] + 18*x*Log[x]^2)/9,x]

[Out]

(-4*x + 6*E^x*Log[3] + Log[27] + E^(2*x)*Log[27] + 3*x*Log[x])^2/9

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fricas [B]  time = 0.55, size = 103, normalized size = 4.68 \begin {gather*} x^{2} \log \relax (x)^{2} + e^{\left (4 \, x\right )} \log \relax (3)^{2} + 4 \, e^{\left (3 \, x\right )} \log \relax (3)^{2} + \frac {16}{9} \, x^{2} - \frac {2}{3} \, {\left (4 \, x \log \relax (3) - 9 \, \log \relax (3)^{2}\right )} e^{\left (2 \, x\right )} - \frac {4}{3} \, {\left (4 \, x \log \relax (3) - 3 \, \log \relax (3)^{2}\right )} e^{x} - \frac {8}{3} \, x \log \relax (3) + \frac {2}{3} \, {\left (3 \, x e^{\left (2 \, x\right )} \log \relax (3) + 6 \, x e^{x} \log \relax (3) - 4 \, x^{2} + 3 \, x \log \relax (3)\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*log(x)^2+1/9*((36*x+18)*log(3)*exp(x)^2+(36*x+36)*log(3)*exp(x)+18*log(3)-30*x)*log(x)+4*log(3)^
2*exp(x)^4+12*log(3)^2*exp(x)^3+1/9*(108*log(3)^2+(-48*x-6)*log(3))*exp(x)^2+1/9*(36*log(3)^2+(-48*x-12)*log(3
))*exp(x)-2/3*log(3)+8/9*x,x, algorithm="fricas")

[Out]

x^2*log(x)^2 + e^(4*x)*log(3)^2 + 4*e^(3*x)*log(3)^2 + 16/9*x^2 - 2/3*(4*x*log(3) - 9*log(3)^2)*e^(2*x) - 4/3*
(4*x*log(3) - 3*log(3)^2)*e^x - 8/3*x*log(3) + 2/3*(3*x*e^(2*x)*log(3) + 6*x*e^x*log(3) - 4*x^2 + 3*x*log(3))*
log(x)

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giac [B]  time = 0.19, size = 128, normalized size = 5.82 \begin {gather*} 2 \, x e^{\left (2 \, x\right )} \log \relax (3) \log \relax (x) + 4 \, x e^{x} \log \relax (3) \log \relax (x) + x^{2} \log \relax (x)^{2} + e^{\left (4 \, x\right )} \log \relax (3)^{2} + 4 \, e^{\left (3 \, x\right )} \log \relax (3)^{2} - \frac {8}{3} \, x^{2} \log \relax (x) + 2 \, x \log \relax (3) \log \relax (x) + \frac {16}{9} \, x^{2} - \frac {1}{3} \, {\left (8 \, x \log \relax (3) - 18 \, \log \relax (3)^{2} - 3 \, \log \relax (3)\right )} e^{\left (2 \, x\right )} - \frac {4}{3} \, {\left (4 \, x \log \relax (3) - 3 \, \log \relax (3)^{2} - 3 \, \log \relax (3)\right )} e^{x} - \frac {8}{3} \, x \log \relax (3) - e^{\left (2 \, x\right )} \log \relax (3) - 4 \, e^{x} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*log(x)^2+1/9*((36*x+18)*log(3)*exp(x)^2+(36*x+36)*log(3)*exp(x)+18*log(3)-30*x)*log(x)+4*log(3)^
2*exp(x)^4+12*log(3)^2*exp(x)^3+1/9*(108*log(3)^2+(-48*x-6)*log(3))*exp(x)^2+1/9*(36*log(3)^2+(-48*x-12)*log(3
))*exp(x)-2/3*log(3)+8/9*x,x, algorithm="giac")

[Out]

2*x*e^(2*x)*log(3)*log(x) + 4*x*e^x*log(3)*log(x) + x^2*log(x)^2 + e^(4*x)*log(3)^2 + 4*e^(3*x)*log(3)^2 - 8/3
*x^2*log(x) + 2*x*log(3)*log(x) + 16/9*x^2 - 1/3*(8*x*log(3) - 18*log(3)^2 - 3*log(3))*e^(2*x) - 4/3*(4*x*log(
3) - 3*log(3)^2 - 3*log(3))*e^x - 8/3*x*log(3) - e^(2*x)*log(3) - 4*e^x*log(3)

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maple [B]  time = 0.07, size = 107, normalized size = 4.86

method	result	size
default	\(\frac {16 x^{2}}{9}+x^{2} \ln \relax (x )^{2}-\frac {8 x^{2} \ln \relax (x )}{3}+4 \ln \relax (3)^{2} {\mathrm e}^{x}-\frac {16 x \ln \relax (3) {\mathrm e}^{x}}{3}+6 \ln \relax (3)^{2} {\mathrm e}^{2 x}-\frac {8 \ln \relax (3) {\mathrm e}^{2 x} x}{3}+2 x \ln \relax (3) \ln \relax (x )-\frac {8 x \ln \relax (3)}{3}+4 x \ln \relax (3) {\mathrm e}^{x} \ln \relax (x )+2 \ln \relax (x ) {\mathrm e}^{2 x} \ln \relax (3) x +4 \ln \relax (3)^{2} {\mathrm e}^{3 x}+\ln \relax (3)^{2} {\mathrm e}^{4 x}\)	\(107\)
risch	\(\frac {16 x^{2}}{9}+x^{2} \ln \relax (x )^{2}-\frac {8 x^{2} \ln \relax (x )}{3}+4 \ln \relax (3)^{2} {\mathrm e}^{x}-\frac {16 x \ln \relax (3) {\mathrm e}^{x}}{3}+6 \ln \relax (3)^{2} {\mathrm e}^{2 x}-\frac {8 \ln \relax (3) {\mathrm e}^{2 x} x}{3}+2 x \ln \relax (3) \ln \relax (x )-\frac {8 x \ln \relax (3)}{3}+4 x \ln \relax (3) {\mathrm e}^{x} \ln \relax (x )+2 \ln \relax (x ) {\mathrm e}^{2 x} \ln \relax (3) x +4 \ln \relax (3)^{2} {\mathrm e}^{3 x}+\ln \relax (3)^{2} {\mathrm e}^{4 x}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*ln(x)^2+1/9*((36*x+18)*ln(3)*exp(x)^2+(36*x+36)*ln(3)*exp(x)+18*ln(3)-30*x)*ln(x)+4*ln(3)^2*exp(x)^4+1
2*ln(3)^2*exp(x)^3+1/9*(108*ln(3)^2+(-48*x-6)*ln(3))*exp(x)^2+1/9*(36*ln(3)^2+(-48*x-12)*ln(3))*exp(x)-2/3*ln(
3)+8/9*x,x,method=_RETURNVERBOSE)

[Out]

16/9*x^2+x^2*ln(x)^2-8/3*x^2*ln(x)+4*ln(3)^2*exp(x)-16/3*x*ln(3)*exp(x)+6*ln(3)^2*exp(x)^2-8/3*ln(3)*exp(x)^2*
x+2*x*ln(3)*ln(x)-8/3*x*ln(3)+4*x*ln(3)*exp(x)*ln(x)+2*ln(x)*exp(x)^2*ln(3)*x+4*ln(3)^2*exp(x)^3+ln(3)^2*exp(x
)^4

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maxima [B]  time = 0.63, size = 134, normalized size = 6.09 \begin {gather*} \frac {1}{2} \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + e^{\left (4 \, x\right )} \log \relax (3)^{2} + 4 \, e^{\left (3 \, x\right )} \log \relax (3)^{2} + \frac {23}{18} \, x^{2} - \frac {1}{3} \, {\left (8 \, x \log \relax (3) - 18 \, \log \relax (3)^{2} - 3 \, \log \relax (3)\right )} e^{\left (2 \, x\right )} - \frac {4}{3} \, {\left (4 \, x \log \relax (3) - 3 \, \log \relax (3)^{2} - 3 \, \log \relax (3)\right )} e^{x} - \frac {8}{3} \, x \log \relax (3) - e^{\left (2 \, x\right )} \log \relax (3) - 4 \, e^{x} \log \relax (3) + \frac {1}{3} \, {\left (6 \, x e^{\left (2 \, x\right )} \log \relax (3) + 12 \, x e^{x} \log \relax (3) - 5 \, x^{2} + 6 \, x \log \relax (3)\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*log(x)^2+1/9*((36*x+18)*log(3)*exp(x)^2+(36*x+36)*log(3)*exp(x)+18*log(3)-30*x)*log(x)+4*log(3)^
2*exp(x)^4+12*log(3)^2*exp(x)^3+1/9*(108*log(3)^2+(-48*x-6)*log(3))*exp(x)^2+1/9*(36*log(3)^2+(-48*x-12)*log(3
))*exp(x)-2/3*log(3)+8/9*x,x, algorithm="maxima")

[Out]

1/2*(2*log(x)^2 - 2*log(x) + 1)*x^2 + e^(4*x)*log(3)^2 + 4*e^(3*x)*log(3)^2 + 23/18*x^2 - 1/3*(8*x*log(3) - 18
*log(3)^2 - 3*log(3))*e^(2*x) - 4/3*(4*x*log(3) - 3*log(3)^2 - 3*log(3))*e^x - 8/3*x*log(3) - e^(2*x)*log(3) -
4*e^x*log(3) + 1/3*(6*x*e^(2*x)*log(3) + 12*x*e^x*log(3) - 5*x^2 + 6*x*log(3))*log(x)

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mupad [B]  time = 1.85, size = 106, normalized size = 4.82 \begin {gather*} 4\,{\mathrm {e}}^x\,{\ln \relax (3)}^2-\frac {8\,x^2\,\ln \relax (x)}{3}-\frac {8\,x\,\ln \relax (3)}{3}+6\,{\mathrm {e}}^{2\,x}\,{\ln \relax (3)}^2+4\,{\mathrm {e}}^{3\,x}\,{\ln \relax (3)}^2+{\mathrm {e}}^{4\,x}\,{\ln \relax (3)}^2+x^2\,{\ln \relax (x)}^2+\frac {16\,x^2}{9}-\frac {16\,x\,{\mathrm {e}}^x\,\ln \relax (3)}{3}+2\,x\,\ln \relax (3)\,\ln \relax (x)-\frac {8\,x\,{\mathrm {e}}^{2\,x}\,\ln \relax (3)}{3}+4\,x\,{\mathrm {e}}^x\,\ln \relax (3)\,\ln \relax (x)+2\,x\,{\mathrm {e}}^{2\,x}\,\ln \relax (3)\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x)/9 - (2*log(3))/3 + 2*x*log(x)^2 - (exp(2*x)*(log(3)*(48*x + 6) - 108*log(3)^2))/9 + (log(x)*(18*log(
3) - 30*x + exp(2*x)*log(3)*(36*x + 18) + exp(x)*log(3)*(36*x + 36)))/9 + 12*exp(3*x)*log(3)^2 + 4*exp(4*x)*lo
g(3)^2 - (exp(x)*(log(3)*(48*x + 12) - 36*log(3)^2))/9,x)

[Out]

4*exp(x)*log(3)^2 - (8*x^2*log(x))/3 - (8*x*log(3))/3 + 6*exp(2*x)*log(3)^2 + 4*exp(3*x)*log(3)^2 + exp(4*x)*l
og(3)^2 + x^2*log(x)^2 + (16*x^2)/9 - (16*x*exp(x)*log(3))/3 + 2*x*log(3)*log(x) - (8*x*exp(2*x)*log(3))/3 + 4
*x*exp(x)*log(3)*log(x) + 2*x*exp(2*x)*log(3)*log(x)

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sympy [B]  time = 0.56, size = 121, normalized size = 5.50 \begin {gather*} x^{2} \log {\relax (x )}^{2} + \frac {16 x^{2}}{9} - \frac {8 x \log {\relax (3 )}}{3} + \left (- \frac {8 x^{2}}{3} + 2 x \log {\relax (3 )}\right ) \log {\relax (x )} + \frac {\left (18 x \log {\relax (3 )} \log {\relax (x )} - 24 x \log {\relax (3 )} + 54 \log {\relax (3 )}^{2}\right ) e^{2 x}}{9} + \frac {\left (36 x \log {\relax (3 )} \log {\relax (x )} - 48 x \log {\relax (3 )} + 36 \log {\relax (3 )}^{2}\right ) e^{x}}{9} + e^{4 x} \log {\relax (3 )}^{2} + 4 e^{3 x} \log {\relax (3 )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*ln(x)**2+1/9*((36*x+18)*ln(3)*exp(x)**2+(36*x+36)*ln(3)*exp(x)+18*ln(3)-30*x)*ln(x)+4*ln(3)**2*e
xp(x)**4+12*ln(3)**2*exp(x)**3+1/9*(108*ln(3)**2+(-48*x-6)*ln(3))*exp(x)**2+1/9*(36*ln(3)**2+(-48*x-12)*ln(3))
*exp(x)-2/3*ln(3)+8/9*x,x)

[Out]

x**2*log(x)**2 + 16*x**2/9 - 8*x*log(3)/3 + (-8*x**2/3 + 2*x*log(3))*log(x) + (18*x*log(3)*log(x) - 24*x*log(3
) + 54*log(3)**2)*exp(2*x)/9 + (36*x*log(3)*log(x) - 48*x*log(3) + 36*log(3)**2)*exp(x)/9 + exp(4*x)*log(3)**2
+ 4*exp(3*x)*log(3)**2

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