3.69.5 $\int (-36x+12e^3{x}^2-e^6{x}^3+(-12x+2e^3{x}^2)\log (4)-x{\log }^2(4)+(-72x+36e^3{x}^2-4e^6{x}^3+(-24x+6e^3{x}^2)\log (4)-2x{\log }^2(4))\log (x))dx$

Optimal. Leaf size=21 $$5-x^2{(6-e^{3}x+\log (4))}^2\log (x)$$

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Rubi [B]  time = 0.09, antiderivative size = 105, normalized size of antiderivative = 5.00, number of steps used = 9, number of rules used = 3, integrand size = 87, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.034, Rules used = {6, 2356, 2304} $$\begin{array}{c}-e^6{x}^4\log (x)+4e^3{x}^3+2e^3{x}^3(6+\log (4))\log (x)-\frac{2}{3}{e}^3{x}^3(6+\log (4))+\frac{2}{3}{e}^3{x}^3\log (4)-\frac{1}{2}{x}^2(36+{\log }^2(4))-x^2(6+\log (4){)}^2\log (x)+\frac{1}{2}{x}^2(6+\log (4){)}^2-6x^2\log (4)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[-36*x + 12*E^3*x^2 - E^6*x^3 + (-12*x + 2*E^3*x^2)*Log[4] - x*Log[4]^2 + (-72*x + 36*E^3*x^2 - 4*E^6*x^3 +
(-24*x + 6*E^3*x^2)*Log[4] - 2*x*Log[4]^2)*Log[x],x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (12e^3{x}^2-e^6{x}^3+(-12x+2e^3{x}^2)\log (4)+x(-36-{\log }^2(4))+(-72x+36e^3{x}^2-4e^6{x}^3+(-24x+6e^3{x}^2)\log (4)-2x{\log }^2(4))\log (x))dx\\ & =4e^3{x}^3-\frac{e^6{x}^4}{4}-\frac{1}{2}{x}^2(36+{\log }^2(4))+\log (4)\int (-12x+2e^3{x}^2)dx+\int (-72x+36e^3{x}^2-4e^6{x}^3+(-24x+6e^3{x}^2)\log (4)-2x{\log }^2(4))\log (x)dx\\ & =4e^3{x}^3-\frac{e^6{x}^4}{4}-6x^2\log (4)+\frac{2}{3}{e}^3{x}^3\log (4)-\frac{1}{2}{x}^2(36+{\log }^2(4))+\int (36e^3{x}^2-4e^6{x}^3+(-24x+6e^3{x}^2)\log (4)+x(-72-2{\log }^2(4)))\log (x)dx\\ & =4e^3{x}^3-\frac{e^6{x}^4}{4}-6x^2\log (4)+\frac{2}{3}{e}^3{x}^3\log (4)-\frac{1}{2}{x}^2(36+{\log }^2(4))+\int (-4e^6{x}^3\log (x)+6e^3{x}^2(6+\log (4))\log (x)-2x(6+\log (4){)}^2\log (x))dx\\ & =4e^3{x}^3-\frac{e^6{x}^4}{4}-6x^2\log (4)+\frac{2}{3}{e}^3{x}^3\log (4)-\frac{1}{2}{x}^2(36+{\log }^2(4))-\left(4e^6\right)\int x^3\log (x)dx+(6e^3(6+\log (4)))\int x^2\log (x)dx-(2(6+\log (4){)}^2)\int x\log (x)dx\\ & =4e^3{x}^3-6x^2\log (4)+\frac{2}{3}{e}^3{x}^3\log (4)-\frac{2}{3}{e}^3{x}^3(6+\log (4))+\frac{1}{2}{x}^2(6+\log (4){)}^2-\frac{1}{2}{x}^2(36+{\log }^2(4))-e^6{x}^4\log (x)+2e^3{x}^3(6+\log (4))\log (x)-x^2(6+\log (4){)}^2\log (x)\end{array}\end{array}$$

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Mathematica [B]  time = 0.09, size = 69, normalized size = 3.29 $$\begin{array}{c}-36x^2\log (x)+12e^3{x}^3\log (x)-e^6{x}^4\log (x)-12x^2\log (2)\log (x)-6x^2\log (4)\log (x)+2e^3{x}^3\log (4)\log (x)-2x^2\log (2)\log (4)\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[-36*x + 12*E^3*x^2 - E^6*x^3 + (-12*x + 2*E^3*x^2)*Log[4] - x*Log[4]^2 + (-72*x + 36*E^3*x^2 - 4*E^6
*x^3 + (-24*x + 6*E^3*x^2)*Log[4] - 2*x*Log[4]^2)*Log[x],x]

[Out]

-36*x^2*Log[x] + 12*E^3*x^3*Log[x] - E^6*x^4*Log[x] - 12*x^2*Log[2]*Log[x] - 6*x^2*Log[4]*Log[x] + 2*E^3*x^3*L
og[4]*Log[x] - 2*x^2*Log[2]*Log[4]*Log[x]

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fricas [B]  time = 0.63, size = 48, normalized size = 2.29 $$\begin{array}{c}-(x^4{e}^6-12x^3{e}^3+4x^2\log (2{)}^2+36x^2-4(x^3{e}^3-6x^2)\log (2))\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(2)^2+2*(6*x^2*exp(3)-24*x)*log(2)-4*x^3*exp(3)^2+36*x^2*exp(3)-72*x)*log(x)-4*x*log(2)^2+2
*(2*x^2*exp(3)-12*x)*log(2)-x^3*exp(3)^2+12*x^2*exp(3)-36*x,x, algorithm="fricas")

[Out]

-(x^4*e^6 - 12*x^3*e^3 + 4*x^2*log(2)^2 + 36*x^2 - 4*(x^3*e^3 - 6*x^2)*log(2))*log(x)

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giac [B]  time = 0.19, size = 89, normalized size = 4.24 $$\begin{array}{c}-x^4{e}^6\log (x)+4x^3{e}^3\log (2)\log (x)-\frac{4}{3}{x}^3{e}^3\log (2)+12x^3{e}^3\log (x)-4x^2\log (2{)}^2\log (x)-24x^2\log (2)\log (x)+12x^2\log (2)-36x^2\log (x)+\frac{4}{3}(x^3{e}^3-9x^2)\log (2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(2)^2+2*(6*x^2*exp(3)-24*x)*log(2)-4*x^3*exp(3)^2+36*x^2*exp(3)-72*x)*log(x)-4*x*log(2)^2+2
*(2*x^2*exp(3)-12*x)*log(2)-x^3*exp(3)^2+12*x^2*exp(3)-36*x,x, algorithm="giac")

[Out]

-x^4*e^6*log(x) + 4*x^3*e^3*log(2)*log(x) - 4/3*x^3*e^3*log(2) + 12*x^3*e^3*log(x) - 4*x^2*log(2)^2*log(x) - 2
4*x^2*log(2)*log(x) + 12*x^2*log(2) - 36*x^2*log(x) + 4/3*(x^3*e^3 - 9*x^2)*log(2)

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maple [A]  time = 0.04, size = 38, normalized size = 1.81

method	result	size
risch	$-x^2(x^2{\mathrm{e}}^6-4{\mathrm{e}}^3\ln (2)x-12x{\mathrm{e}}^3+4\ln (2{)}^2+24\ln (2)+36)\ln (x)$	$38$
norman	$(4{\mathrm{e}}^3\ln (2)+12{\mathrm{e}}^3)x^3\ln (x)+(-4\ln (2{)}^2-24\ln (2)-36)x^2\ln (x)-{\mathrm{e}}^6{x}^4\ln (x)$	$48$
default	$-{\mathrm{e}}^6{x}^4\ln (x)+4{\mathrm{e}}^3\ln (2)x^3\ln (x)+12{\mathrm{e}}^3{x}^3\ln (x)-4\ln (2{)}^2\ln (x)x^2-24x^2\ln (2)\ln (x)-36x^2\ln (x)$	$60$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x*ln(2)^2+2*(6*x^2*exp(3)-24*x)*ln(2)-4*x^3*exp(3)^2+36*x^2*exp(3)-72*x)*ln(x)-4*x*ln(2)^2+2*(2*x^2*ex
p(3)-12*x)*ln(2)-x^3*exp(3)^2+12*x^2*exp(3)-36*x,x,method=_RETURNVERBOSE)

[Out]

-x^2*(x^2*exp(6)-4*exp(3)*ln(2)*x-12*x*exp(3)+4*ln(2)^2+24*ln(2)+36)*ln(x)

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maxima [B]  time = 0.37, size = 116, normalized size = 5.52 $$\begin{array}{c}-\frac{4}{3}(e^3\log (2)+3e^3)x^3+4x^3{e}^3-2x^2\log (2{)}^2+2(\log (2{)}^2+6\log (2)+9)x^2-18x^2+\frac{4}{3}(x^3{e}^3-9x^2)\log (2)-(x^4{e}^6-12x^3{e}^3+4x^2\log (2{)}^2+36x^2-4(x^3{e}^3-6x^2)\log (2))\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(2)^2+2*(6*x^2*exp(3)-24*x)*log(2)-4*x^3*exp(3)^2+36*x^2*exp(3)-72*x)*log(x)-4*x*log(2)^2+2
*(2*x^2*exp(3)-12*x)*log(2)-x^3*exp(3)^2+12*x^2*exp(3)-36*x,x, algorithm="maxima")

[Out]

-4/3*(e^3*log(2) + 3*e^3)*x^3 + 4*x^3*e^3 - 2*x^2*log(2)^2 + 2*(log(2)^2 + 6*log(2) + 9)*x^2 - 18*x^2 + 4/3*(x
^3*e^3 - 9*x^2)*log(2) - (x^4*e^6 - 12*x^3*e^3 + 4*x^2*log(2)^2 + 36*x^2 - 4*(x^3*e^3 - 6*x^2)*log(2))*log(x)

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mupad [B]  time = 4.56, size = 18, normalized size = 0.86 $$\begin{array}{c}-x^2\ln (x){(\ln (4)-x{\mathrm{e}}^3+6)}^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(12*x^2*exp(3) - 36*x - x^3*exp(6) - 2*log(2)*(12*x - 2*x^2*exp(3)) - 4*x*log(2)^2 - log(x)*(72*x - 36*x^2*
exp(3) + 4*x^3*exp(6) + 2*log(2)*(24*x - 6*x^2*exp(3)) + 8*x*log(2)^2),x)

[Out]

-x^2*log(x)*(log(4) - x*exp(3) + 6)^2

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sympy [B]  time = 0.17, size = 53, normalized size = 2.52 $$\begin{array}{c}(-x^4{e}^6+4x^3{e}^3\log (2)+12x^3{e}^3-36x^2-24x^2\log (2)-4x^2\log {(2)}^2)\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*ln(2)**2+2*(6*x**2*exp(3)-24*x)*ln(2)-4*x**3*exp(3)**2+36*x**2*exp(3)-72*x)*ln(x)-4*x*ln(2)**2
+2*(2*x**2*exp(3)-12*x)*ln(2)-x**3*exp(3)**2+12*x**2*exp(3)-36*x,x)

[Out]

(-x**4*exp(6) + 4*x**3*exp(3)*log(2) + 12*x**3*exp(3) - 36*x**2 - 24*x**2*log(2) - 4*x**2*log(2)**2)*log(x)

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