3.73.28 $\int (-9x^2+2x\log (x)+2x{\log }^2(x)-e^x\log (1+\log (4)))dx$

Optimal. Leaf size=24 $$-3x^3+x^2{\log }^2(x)-e^x\log (1+\log (4))$$

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 28, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.107, Rules used = {2304, 2305, 2194} $$\begin{array}{c}-3x^3+x^2{\log }^2(x)-e^x\log (1+\log (4))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[-9*x^2 + 2*x*Log[x] + 2*x*Log[x]^2 - E^x*Log[1 + Log[4]],x]

[Out]

-3*x^3 + x^2*Log[x]^2 - E^x*Log[1 + Log[4]]

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 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]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-3x^3+2\int x\log (x)dx+2\int x{\log }^2(x)dx-\log (1+\log (4))\int e^{x}dx\\ & =-\frac{x^2}{2}-3x^3+x^2\log (x)+x^2{\log }^2(x)-e^x\log (1+\log (4))-2\int x\log (x)dx\\ & =-3x^3+x^2{\log }^2(x)-e^x\log (1+\log (4))\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 $$\begin{array}{c}-3x^3+x^2{\log }^2(x)-e^x\log (1+\log (4))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[-9*x^2 + 2*x*Log[x] + 2*x*Log[x]^2 - E^x*Log[1 + Log[4]],x]

[Out]

-3*x^3 + x^2*Log[x]^2 - E^x*Log[1 + Log[4]]

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fricas [A]  time = 1.14, size = 25, normalized size = 1.04 $$\begin{array}{c}{x}^2\log (x{)}^2-3x^3-e^x\log (2\log (2)+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)*log(1+2*log(2))+2*x*log(x)^2+2*x*log(x)-9*x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 25, normalized size = 1.04 $$\begin{array}{c}{x}^2\log (x{)}^2-3x^3-e^x\log (2\log (2)+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)*log(1+2*log(2))+2*x*log(x)^2+2*x*log(x)-9*x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 26, normalized size = 1.08

method	result	size
default	$x^2\ln (x{)}^2-{\mathrm{e}}^x\ln (1+2\ln (2))-3x^3$	$26$
norman	$x^2\ln (x{)}^2-{\mathrm{e}}^x\ln (1+2\ln (2))-3x^3$	$26$
risch	$x^2\ln (x{)}^2-{\mathrm{e}}^x\ln (1+2\ln (2))-3x^3$	$26$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x)*ln(1+2*ln(2))+2*x*ln(x)^2+2*x*ln(x)-9*x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 45, normalized size = 1.88 $$\begin{array}{c}\frac{1}{2}(2\log (x{)}^2-2\log (x)+1)x^2-3x^3+x^2\log (x)-\frac{1}{2}{x}^2-e^x\log (2\log (2)+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)*log(1+2*log(2))+2*x*log(x)^2+2*x*log(x)-9*x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.32, size = 25, normalized size = 1.04 $$\begin{array}{c}{x}^2{\ln (x)}^2-{\mathrm{e}}^x\ln (2\ln (2)+1)-3x^3\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*log(x)^2 + 2*x*log(x) - exp(x)*log(2*log(2) + 1) - 9*x^2,x)

[Out]

x^2*log(x)^2 - exp(x)*log(2*log(2) + 1) - 3*x^3

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sympy [A]  time = 0.30, size = 24, normalized size = 1.00 $$\begin{array}{c}-3x^3+x^2\log {(x)}^2-e^x\log (1+2\log (2))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)*ln(1+2*ln(2))+2*x*ln(x)**2+2*x*ln(x)-9*x**2,x)

[Out]

-3*x**3 + x**2*log(x)**2 - exp(x)*log(1 + 2*log(2))

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