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

Optimal. Leaf size=21 \[ \frac {2 \left (x-x^4+4 x^{1+2 x}\right )}{x} \]

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Rubi [A]  time = 0.06, antiderivative size = 13, normalized size of antiderivative = 0.62, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6741, 12, 6742, 2553} \begin {gather*} 8 x^{2 x}-2 x^3 \end {gather*}

Antiderivative was successfully verified.

[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 2553

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 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 13, normalized size = 0.62 \begin {gather*} -2 x^3+8 x^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*x^3 + 8*x^(2*x)

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fricas [A]  time = 0.54, size = 13, normalized size = 0.62 \begin {gather*} -2 \, x^{3} + 8 \, x^{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 13, normalized size = 0.62 \begin {gather*} -2 \, x^{3} + 8 \, x^{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 14, normalized size = 0.67

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.40, size = 13, normalized size = 0.62 \begin {gather*} -2 \, x^{3} + 8 \, x^{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.96, size = 13, normalized size = 0.62 \begin {gather*} 8\,x^{2\,x}-2\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*x^(2*x) - 2*x^3

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sympy [A]  time = 0.27, size = 14, normalized size = 0.67 \begin {gather*} - 2 x^{3} + 8 e^{2 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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