\(\int (-6 x^2+x^{2 x} (16+16 \log (x))) \, dx\) [6689]

Optimal result

Integrand size = 18, antiderivative size = 21 \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=\frac {2 \left (x-x^4+4 x^{1+2 x}\right )}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6873, 12, 6874, 2633} \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=8 x^{2 x}-2 x^3 \]

[In]

Int[-6*x^2 + x^(2*x)*(16 + 16*Log[x]),x]

[Out]

-2*x^3 + 8*x^(2*x)

Rule 12

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

Rule 2633

Int[Log[u_]*(u_)^((a_.)*(x_)), x_Symbol] :> Simp[u^(a*x)/a, x] - Int[SimplifyIntegrand[x*u^(a*x - 1)*D[u, x],
x], x] /; FreeQ[a, x] && InverseFunctionFreeQ[u, x]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = -2 x^3+\int x^{2 x} (16+16 \log (x)) \, dx \\ & = -2 x^3+\int 16 x^{2 x} (1+\log (x)) \, dx \\ & = -2 x^3+16 \int x^{2 x} (1+\log (x)) \, dx \\ & = -2 x^3+16 \int \left (x^{2 x}+x^{2 x} \log (x)\right ) \, dx \\ & = -2 x^3+16 \int x^{2 x} \, dx+16 \int x^{2 x} \log (x) \, dx \\ & = -2 x^3+8 x^{2 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=-2 x^3+8 x^{2 x} \]

[In]

Integrate[-6*x^2 + x^(2*x)*(16 + 16*Log[x]),x]

[Out]

-2*x^3 + 8*x^(2*x)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
risch	\(8 x^{2 x}-2 x^{3}\)	\(14\)
default	\(8 \,{\mathrm e}^{2 x \ln \left (x \right )}-2 x^{3}\)	\(16\)
norman	\(8 \,{\mathrm e}^{2 x \ln \left (x \right )}-2 x^{3}\)	\(16\)
parallelrisch	\(8 \,{\mathrm e}^{2 x \ln \left (x \right )}-2 x^{3}\)	\(16\)
parts	\(8 \,{\mathrm e}^{2 x \ln \left (x \right )}-2 x^{3}\)	\(16\)

[In]

int((16*ln(x)+16)*exp(x*ln(x))^2-6*x^2,x,method=_RETURNVERBOSE)

[Out]

8*(x^x)^2-2*x^3

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=-2 \, x^{3} + 8 \, x^{2 \, x} \]

[In]

integrate((16*log(x)+16)*exp(x*log(x))^2-6*x^2,x, algorithm="fricas")

[Out]

-2*x^3 + 8*x^(2*x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=- 2 x^{3} + 8 e^{2 x \log {\left (x \right )}} \]

[In]

integrate((16*ln(x)+16)*exp(x*ln(x))**2-6*x**2,x)

[Out]

-2*x**3 + 8*exp(2*x*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=-2 \, x^{3} + 8 \, x^{2 \, x} \]

[In]

integrate((16*log(x)+16)*exp(x*log(x))^2-6*x^2,x, algorithm="maxima")

[Out]

-2*x^3 + 8*x^(2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=-2 \, x^{3} + 8 \, x^{2 \, x} \]

[In]

integrate((16*log(x)+16)*exp(x*log(x))^2-6*x^2,x, algorithm="giac")

[Out]

-2*x^3 + 8*x^(2*x)

Mupad [B] (verification not implemented)

Time = 12.44 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \left (-6 x^2+x^{2 x} (16+16 \log (x))\right ) \, dx=8\,x^{2\,x}-2\,x^3 \]

[In]

int(exp(2*x*log(x))*(16*log(x) + 16) - 6*x^2,x)

[Out]

8*x^(2*x) - 2*x^3