\(\int (1+2 x^2 \log (x)+3 x^2 \log (e^{\log ^2(x)} (28+\log (3)))) \, dx\) [6904]

Optimal result

Integrand size = 26, antiderivative size = 18 \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=x+x^3 \log \left (e^{\log ^2(x)} (28+\log (3))\right ) \]

[Out]

x^3*ln((ln(3)+28)*exp(ln(x)^2))+x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2341, 30, 2635, 12} \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=x^3 \log \left ((28+\log (3)) e^{\log ^2(x)}\right )+x \]

[In]

Int[1 + 2*x^2*Log[x] + 3*x^2*Log[E^Log[x]^2*(28 + Log[3])],x]

[Out]

x + x^3*Log[E^Log[x]^2*(28 + Log[3])]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2341

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 2635

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

Rubi steps \begin{align*} \text {integral}& = x+2 \int x^2 \log (x) \, dx+3 \int x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right ) \, dx \\ & = x-\frac {2 x^3}{9}+\frac {2}{3} x^3 \log (x)+x^3 \log \left (e^{\log ^2(x)} (28+\log (3))\right )-3 \int \frac {2}{3} x^2 \log (x) \, dx \\ & = x-\frac {2 x^3}{9}+\frac {2}{3} x^3 \log (x)+x^3 \log \left (e^{\log ^2(x)} (28+\log (3))\right )-2 \int x^2 \log (x) \, dx \\ & = x+x^3 \log \left (e^{\log ^2(x)} (28+\log (3))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=x+x^3 \log \left (e^{\log ^2(x)} (28+\log (3))\right ) \]

[In]

Integrate[1 + 2*x^2*Log[x] + 3*x^2*Log[E^Log[x]^2*(28 + Log[3])],x]

[Out]

x + x^3*Log[E^Log[x]^2*(28 + Log[3])]

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(x^{3} \ln \left (\left (\ln \left (3\right )+28\right ) {\mathrm e}^{\ln \left (x \right )^{2}}\right )+x\)	\(18\)
risch	\(x^{3} \ln \left ({\mathrm e}^{\ln \left (x \right )^{2}}\right )+\ln \left (\ln \left (3\right )+28\right ) x^{3}+x\)	\(22\)
default	\(x +x^{3} \ln \left (x \right )^{2}+\left (\ln \left (\left (\ln \left (3\right )+28\right ) {\mathrm e}^{\ln \left (x \right )^{2}}\right )-\ln \left (x \right )^{2}\right ) x^{3}\)	\(33\)
parts	\(x +x^{3} \ln \left (x \right )^{2}+\left (\ln \left (\left (\ln \left (3\right )+28\right ) {\mathrm e}^{\ln \left (x \right )^{2}}\right )-\ln \left (x \right )^{2}\right ) x^{3}\)	\(33\)

[In]

int(3*x^2*ln((ln(3)+28)*exp(ln(x)^2))+2*x^2*ln(x)+1,x,method=_RETURNVERBOSE)

[Out]

x^3*ln((ln(3)+28)*exp(ln(x)^2))+x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=x^{3} \log \left (x\right )^{2} + x^{3} \log \left (\log \left (3\right ) + 28\right ) + x \]

[In]

integrate(3*x^2*log((log(3)+28)*exp(log(x)^2))+2*x^2*log(x)+1,x, algorithm="fricas")

[Out]

x^3*log(x)^2 + x^3*log(log(3) + 28) + x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=x^{3} \log {\left (x \right )}^{2} + x^{3} \log {\left (\log {\left (3 \right )} + 28 \right )} + x \]

[In]

integrate(3*x**2*ln((ln(3)+28)*exp(ln(x)**2))+2*x**2*ln(x)+1,x)

[Out]

x**3*log(x)**2 + x**3*log(log(3) + 28) + x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (17) = 34\).

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.22 \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=-\frac {2}{9} \, x^{3} {\left (3 \, \log \left (x\right ) - 1\right )} + x^{3} \log \left ({\left (\log \left (3\right ) + 28\right )} e^{\left (\log \left (x\right )^{2}\right )}\right ) + \frac {2}{3} \, x^{3} \log \left (x\right ) - \frac {2}{9} \, x^{3} + x \]

[In]

integrate(3*x^2*log((log(3)+28)*exp(log(x)^2))+2*x^2*log(x)+1,x, algorithm="maxima")

[Out]

-2/9*x^3*(3*log(x) - 1) + x^3*log((log(3) + 28)*e^(log(x)^2)) + 2/3*x^3*log(x) - 2/9*x^3 + x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=x^{3} \log \left (x\right )^{2} + x^{3} \log \left (\log \left (3\right ) + 28\right ) + x \]

[In]

integrate(3*x^2*log((log(3)+28)*exp(log(x)^2))+2*x^2*log(x)+1,x, algorithm="giac")

[Out]

x^3*log(x)^2 + x^3*log(log(3) + 28) + x

Mupad [B] (verification not implemented)

Time = 11.41 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \left (1+2 x^2 \log (x)+3 x^2 \log \left (e^{\log ^2(x)} (28+\log (3))\right )\right ) \, dx=x+x^3\,\ln \left ({\mathrm {e}}^{{\ln \left (x\right )}^2}\right )+x^3\,\ln \left (\ln \left (3\right )+28\right ) \]

[In]

int(2*x^2*log(x) + 3*x^2*log(exp(log(x)^2)*(log(3) + 28)) + 1,x)

[Out]

x + x^3*log(exp(log(x)^2)) + x^3*log(log(3) + 28)