3.41.77 $\int \frac{1}{5}(9+6x-3x^2+e^x(-15+5x)+(9+15x-11x^2+e^x(-15-5x+5x^2))\log (x)+(-3x+x^2+(-6x+3x^2)\log (x))\log (x\log (x)))dx$

Optimal. Leaf size=33 $$(3-x)x\log (x)(-e^x+x+\frac{1}{5}(3-x-x\log (x\log (x))))$$

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Rubi [F]  time = 0.58, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.000, Rules used = {} $$\begin{array}{c}\int \frac{1}{5}(9+6x-3x^2+e^x(-15+5x)+(9+15x-11x^2+e^x(-15-5x+5x^2))\log (x)+(-3x+x^2+(-6x+3x^2)\log (x))\log (x\log (x)))dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(9 + 6*x - 3*x^2 + E^x*(-15 + 5*x) + (9 + 15*x - 11*x^2 + E^x*(-15 - 5*x + 5*x^2))*Log[x] + (-3*x + x^2 +
(-6*x + 3*x^2)*Log[x])*Log[x*Log[x]])/5,x]

[Out]

x^3/45 + (9*x*Log[x])/5 - 3*E^x*x*Log[x] + (3*x^2*Log[x])/2 + E^x*x^2*Log[x] - (11*x^3*Log[x])/15 - (3*ExpInte
gralEi[2*Log[x]]*Log[x])/10 + (ExpIntegralEi[3*Log[x]]*Log[x])/15 + (3*ExpIntegralEi[2*Log[x]]*(1 + Log[x]))/1
0 - (ExpIntegralEi[3*Log[x]]*(1 + Log[x]))/15 - (3*x^2*Log[x*Log[x]])/10 + (x^3*Log[x*Log[x]])/15 - (6*Defer[I
nt][x*Log[x]*Log[x*Log[x]], x])/5 + (3*Defer[Int][x^2*Log[x]*Log[x*Log[x]], x])/5

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{5}\int (9+6x-3x^2+e^x(-15+5x)+(9+15x-11x^2+e^x(-15-5x+5x^2))\log (x)+(-3x+x^2+(-6x+3x^2)\log (x))\log (x\log (x)))dx\\ & =\frac{9x}{5}+\frac{3x^2}{5}-\frac{x^3}{5}+\frac{1}{5}\int e^x(-15+5x)dx+\frac{1}{5}\int (9+15x-11x^2+e^x(-15-5x+5x^2))\log (x)dx+\frac{1}{5}\int (-3x+x^2+(-6x+3x^2)\log (x))\log (x\log (x))dx\\ & =-e^x(3-x)+\frac{9x}{5}+\frac{3x^2}{5}-\frac{x^3}{5}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)-\frac{1}{5}\int (9+5e^x(-3+x)+\frac{15x}{2}-\frac{11x^2}{3})dx+\frac{1}{5}\int x(-3+x+3(-2+x)\log (x))\log (x\log (x))dx-\int e^{x}dx\\ & =-e^x-e^x(3-x)-\frac{3x^2}{20}+\frac{2x^3}{45}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)+\frac{1}{5}\int (-3x\log (x\log (x))+x^2\log (x\log (x))-6x\log (x)\log (x\log (x))+3x^2\log (x)\log (x\log (x)))dx-\int e^x(-3+x)dx\\ & =-e^x-\frac{3x^2}{20}+\frac{2x^3}{45}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)+\frac{1}{5}\int x^2\log (x\log (x))dx-\frac{3}{5}\int x\log (x\log (x))dx+\frac{3}{5}\int x^2\log (x)\log (x\log (x))dx-\frac{6}{5}\int x\log (x)\log (x\log (x))dx+\int e^{x}dx\\ & =-\frac{3x^2}{20}+\frac{2x^3}{45}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)-\frac{3}{10}{x}^2\log (x\log (x))+\frac{1}{15}{x}^3\log (x\log (x))-\frac{1}{5}\int \frac{x^2(1+\log (x))}{3\log (x)}dx+\frac{3}{5}\int \frac{x(1+\log (x))}{2\log (x)}dx+\frac{3}{5}\int x^2\log (x)\log (x\log (x))dx-\frac{6}{5}\int x\log (x)\log (x\log (x))dx\\ & =-\frac{3x^2}{20}+\frac{2x^3}{45}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)-\frac{3}{10}{x}^2\log (x\log (x))+\frac{1}{15}{x}^3\log (x\log (x))-\frac{1}{15}\int \frac{x^2(1+\log (x))}{\log (x)}dx+\frac{3}{10}\int \frac{x(1+\log (x))}{\log (x)}dx+\frac{3}{5}\int x^2\log (x)\log (x\log (x))dx-\frac{6}{5}\int x\log (x)\log (x\log (x))dx\\ & =-\frac{3x^2}{20}+\frac{2x^3}{45}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)+\frac{3}{10}\text{Ei}(2\log (x))(1+\log (x))-\frac{1}{15}\text{Ei}(3\log (x))(1+\log (x))-\frac{3}{10}{x}^2\log (x\log (x))+\frac{1}{15}{x}^3\log (x\log (x))+\frac{1}{15}\int \frac{\text{Ei}(3\log (x))}{x}dx-\frac{3}{10}\int \frac{\text{Ei}(2\log (x))}{x}dx+\frac{3}{5}\int x^2\log (x)\log (x\log (x))dx-\frac{6}{5}\int x\log (x)\log (x\log (x))dx\\ & =-\frac{3x^2}{20}+\frac{2x^3}{45}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)+\frac{3}{10}\text{Ei}(2\log (x))(1+\log (x))-\frac{1}{15}\text{Ei}(3\log (x))(1+\log (x))-\frac{3}{10}{x}^2\log (x\log (x))+\frac{1}{15}{x}^3\log (x\log (x))+\frac{1}{15}\mathrm{Subst}(\int \text{Ei}(3x)dx,x,\log (x))-\frac{3}{10}\mathrm{Subst}(\int \text{Ei}(2x)dx,x,\log (x))+\frac{3}{5}\int x^2\log (x)\log (x\log (x))dx-\frac{6}{5}\int x\log (x)\log (x\log (x))dx\\ & =\frac{x^3}{45}+\frac{9}{5}x\log (x)-3e^{x}x\log (x)+\frac{3}{2}{x}^2\log (x)+e^x{x}^2\log (x)-\frac{11}{15}{x}^3\log (x)-\frac{3}{10}\text{Ei}(2\log (x))\log (x)+\frac{1}{15}\text{Ei}(3\log (x))\log (x)+\frac{3}{10}\text{Ei}(2\log (x))(1+\log (x))-\frac{1}{15}\text{Ei}(3\log (x))(1+\log (x))-\frac{3}{10}{x}^2\log (x\log (x))+\frac{1}{15}{x}^3\log (x\log (x))+\frac{3}{5}\int x^2\log (x)\log (x\log (x))dx-\frac{6}{5}\int x\log (x)\log (x\log (x))dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.30, size = 27, normalized size = 0.82 $$\begin{array}{c}\frac{1}{5}(-3+x)x\log (x)(-3+5e^x-4x+x\log (x\log (x)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(9 + 6*x - 3*x^2 + E^x*(-15 + 5*x) + (9 + 15*x - 11*x^2 + E^x*(-15 - 5*x + 5*x^2))*Log[x] + (-3*x +
x^2 + (-6*x + 3*x^2)*Log[x])*Log[x*Log[x]])/5,x]

[Out]

((-3 + x)*x*Log[x]*(-3 + 5*E^x - 4*x + x*Log[x*Log[x]]))/5

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fricas [A]  time = 0.69, size = 48, normalized size = 1.45 $$\begin{array}{c}\frac{1}{5}(x^3-3x^2)\log (x\log (x))\log (x)-\frac{1}{5}(4x^3-9x^2-5(x^2-3x)e^x-9x)\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^2-6*x)*log(x)+x^2-3*x)*log(x*log(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*log(x)+1/5*
(5*x-15)*exp(x)-3/5*x^2+6/5*x+9/5,x, algorithm="fricas")

[Out]

1/5*(x^3 - 3*x^2)*log(x*log(x))*log(x) - 1/5*(4*x^3 - 9*x^2 - 5*(x^2 - 3*x)*e^x - 9*x)*log(x)

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giac [B]  time = 0.20, size = 93, normalized size = 2.82 $$\begin{array}{c}\frac{1}{5}(x^3-3x^2)\log (x{)}^2+\frac{1}{5}(x^3-3x^2)\log (x)\log (\log (x))-(x-1)e^x+(x-4)e^x-\frac{1}{30}(22x^3-45x^2-30(x^2-3x)e^x-54x)\log (x)-\frac{1}{30}(2x^3-9x^2)\log (x)+3e^x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^2-6*x)*log(x)+x^2-3*x)*log(x*log(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*log(x)+1/5*
(5*x-15)*exp(x)-3/5*x^2+6/5*x+9/5,x, algorithm="giac")

[Out]

1/5*(x^3 - 3*x^2)*log(x)^2 + 1/5*(x^3 - 3*x^2)*log(x)*log(log(x)) - (x - 1)*e^x + (x - 4)*e^x - 1/30*(22*x^3 -
45*x^2 - 30*(x^2 - 3*x)*e^x - 54*x)*log(x) - 1/30*(2*x^3 - 9*x^2)*log(x) + 3*e^x

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maple [B]  time = 0.09, size = 60, normalized size = 1.82

method	result	size
default	$\frac{9x^2\ln (x)}{5}-\frac{4x^3\ln (x)}{5}-\frac{3\ln (x)\ln (x\ln (x))x^2}{5}+\frac{\ln (x)x^3\ln (x\ln (x))}{5}+\frac{9x\ln (x)}{5}+x^2{\mathrm{e}}^x\ln (x)-3x{\mathrm{e}}^x\ln (x)$	$60$
risch	$\frac{x^3\ln (x)\ln (\ln (x))}{5}-\frac{3x^2\ln (x)\ln (\ln (x))}{5}+\frac{x^3\ln (x{)}^2}{5}-\frac{3x^2\ln (x{)}^2}{5}-\frac{i\pi x^3\mathrm{csgn}{(ix\ln (x))}^3\ln (x)}{10}+\frac{i\pi x^3\mathrm{csgn}(i\ln (x))\mathrm{csgn}{(ix\ln (x))}^2\ln (x)}{10}+\frac{i\pi x^3\mathrm{csgn}\left(ix\right)\mathrm{csgn}{(ix\ln (x))}^2\ln (x)}{10}+\frac{3i\pi x^2\mathrm{csgn}\left(ix\right)\mathrm{csgn}(i\ln (x))\mathrm{csgn}(ix\ln (x))\ln (x)}{10}-\frac{3i\pi x^2\mathrm{csgn}\left(ix\right)\mathrm{csgn}{(ix\ln (x))}^2\ln (x)}{10}-\frac{i\pi x^3\mathrm{csgn}\left(ix\right)\mathrm{csgn}(i\ln (x))\mathrm{csgn}(ix\ln (x))\ln (x)}{10}-\frac{3i\pi x^2\mathrm{csgn}(i\ln (x))\mathrm{csgn}{(ix\ln (x))}^2\ln (x)}{10}+\frac{3i\pi x^2\mathrm{csgn}{(ix\ln (x))}^3\ln (x)}{10}-\frac{4x^3\ln (x)}{5}+\frac{9x^2\ln (x)}{5}+\frac{9x\ln (x)}{5}+x^2{\mathrm{e}}^x\ln (x)-3x{\mathrm{e}}^x\ln (x)$	$258$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((3*x^2-6*x)*ln(x)+x^2-3*x)*ln(x*ln(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*ln(x)+1/5*(5*x-15)*e
xp(x)-3/5*x^2+6/5*x+9/5,x,method=_RETURNVERBOSE)

[Out]

9/5*x^2*ln(x)-4/5*x^3*ln(x)-3/5*ln(x)*ln(x*ln(x))*x^2+1/5*ln(x)*x^3*ln(x*ln(x))+9/5*x*ln(x)+x^2*exp(x)*ln(x)-3
*x*exp(x)*ln(x)

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maxima [B]  time = 0.39, size = 63, normalized size = 1.91 $$\begin{array}{c}\frac{1}{5}(x^3-3x^2)\log (x\log (x))\log (x)-\frac{1}{30}(22x^3-45x^2-30(x^2-3x)e^x-54x)\log (x)-\frac{1}{30}(2x^3-9x^2)\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^2-6*x)*log(x)+x^2-3*x)*log(x*log(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*log(x)+1/5*
(5*x-15)*exp(x)-3/5*x^2+6/5*x+9/5,x, algorithm="maxima")

[Out]

1/5*(x^3 - 3*x^2)*log(x*log(x))*log(x) - 1/30*(22*x^3 - 45*x^2 - 30*(x^2 - 3*x)*e^x - 54*x)*log(x) - 1/30*(2*x
^3 - 9*x^2)*log(x)

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mupad [B]  time = 3.33, size = 25, normalized size = 0.76 $$\begin{array}{c}-\frac{x\ln (x)(x-3)(4x-5{\mathrm{e}}^x-x\ln (x\ln (x))+3)}{5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x)/5 + (log(x)*(15*x - exp(x)*(5*x - 5*x^2 + 15) - 11*x^2 + 9))/5 + (exp(x)*(5*x - 15))/5 - (log(x*log(
x))*(3*x + log(x)*(6*x - 3*x^2) - x^2))/5 - (3*x^2)/5 + 9/5,x)

[Out]

-(x*log(x)*(x - 3)*(4*x - 5*exp(x) - x*log(x*log(x)) + 3))/5

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sympy [B]  time = 0.76, size = 63, normalized size = 1.91 $$\begin{array}{c}(x^2\log (x)-3x\log (x))e^x+(\frac{x^3\log (x)}{5}-\frac{3x^2\log (x)}{5})\log (x\log (x))+(-\frac{4x^3}{5}+\frac{9x^2}{5}+\frac{9x}{5})\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x**2-6*x)*ln(x)+x**2-3*x)*ln(x*ln(x))+1/5*((5*x**2-5*x-15)*exp(x)-11*x**2+15*x+9)*ln(x)+1/5*
(5*x-15)*exp(x)-3/5*x**2+6/5*x+9/5,x)

[Out]

(x**2*log(x) - 3*x*log(x))*exp(x) + (x**3*log(x)/5 - 3*x**2*log(x)/5)*log(x*log(x)) + (-4*x**3/5 + 9*x**2/5 +
9*x/5)*log(x)

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