3.17.10 $\int \frac{-e^{e^2+\frac{e^{e^2}}{10x^2}}-240x^3+160x^4+e^{3x}(-80x^3+220x^4-120x^5)+e^{6x}(20x^3-50x^4+30x^5)+e^{\frac{e^{e^2}}{20x^2}}(e^{e^2}(6+e^{3x}(-1+x)-4x)+40x^3+e^{3x}(20x^3-30x^4))}{5x^3}dx$

Optimal. Leaf size=31 $${(-2+e^{\frac{e^{e^2}}{20x^2}}+(4-e^{3x})(-1+x))}^2$$

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Rubi [F]  time = 3.24, 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 = {} $$\begin{array}{c}\int \frac{-e^{e^2+\frac{e^{e^2}}{10x^2}}-240x^3+160x^4+e^{3x}(-80x^3+220x^4-120x^5)+e^{6x}(20x^3-50x^4+30x^5)+e^{\frac{e^{e^2}}{20x^2}}(e^{e^2}(6+e^{3x}(-1+x)-4x)+40x^3+e^{3x}(20x^3-30x^4))}{5x^3}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-E^(E^2 + E^E^2/(10*x^2)) - 240*x^3 + 160*x^4 + E^(3*x)*(-80*x^3 + 220*x^4 - 120*x^5) + E^(6*x)*(20*x^3 -
50*x^4 + 30*x^5) + E^(E^E^2/(20*x^2))*(E^E^2*(6 + E^(3*x)*(-1 + x) - 4*x) + 40*x^3 + E^(3*x)*(20*x^3 - 30*x^4
)))/(5*x^3),x]

[Out]

-12*E^(3*x) + E^(6*x) + (6 - E^(E^E^2/(20*x^2)) - 4*x)^2 + 20*E^(3*x)*x - 2*E^(6*x)*x - 8*E^(3*x)*x^2 + E^(6*x
)*x^2 + 4*Defer[Int][E^((E^E^2 + 60*x^3)/(20*x^2)), x] - Defer[Int][E^(E^2 + (E^E^2 + 60*x^3)/(20*x^2))/x^3, x
]/5 + Defer[Int][E^(E^2 + (E^E^2 + 60*x^3)/(20*x^2))/x^2, x]/5 - 6*Defer[Int][E^((E^E^2 + 60*x^3)/(20*x^2))*x,
x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{5}\int \frac{-e^{e^2+\frac{e^{e^2}}{10x^2}}-240x^3+160x^4+e^{3x}(-80x^3+220x^4-120x^5)+e^{6x}(20x^3-50x^4+30x^5)+e^{\frac{e^{e^2}}{20x^2}}(e^{e^2}(6+e^{3x}(-1+x)-4x)+40x^3+e^{3x}(20x^3-30x^4))}{x^3}dx\\ & =\frac{1}{5}\int (10e^{6x}(-1+x)(-2+3x)+\frac{(-6+e^{\frac{e^{e^2}}{20x^2}}+4x)(-e^{e^2+\frac{e^{e^2}}{20x^2}}+40x^3)}{x^3}-\frac{e^{3x}(e^{e^2+\frac{e^{e^2}}{20x^2}}-e^{e^2+\frac{e^{e^2}}{20x^2}}x+80x^3-20e^{\frac{e^{e^2}}{20x^2}}{x}^3-220x^4+30e^{\frac{e^{e^2}}{20x^2}}{x}^4+120x^5)}{x^3})dx\\ & =\frac{1}{5}\int \frac{(-6+e^{\frac{e^{e^2}}{20x^2}}+4x)(-e^{e^2+\frac{e^{e^2}}{20x^2}}+40x^3)}{x^3}dx-\frac{1}{5}\int \frac{e^{3x}(e^{e^2+\frac{e^{e^2}}{20x^2}}-e^{e^2+\frac{e^{e^2}}{20x^2}}x+80x^3-20e^{\frac{e^{e^2}}{20x^2}}{x}^3-220x^4+30e^{\frac{e^{e^2}}{20x^2}}{x}^4+120x^5)}{x^3}dx+2\int e^{6x}(-1+x)(-2+3x)dx\\ & ={(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2-\frac{1}{5}\int \frac{e^{3x}(-e^{e^2+\frac{e^{e^2}}{20x^2}}(-1+x)+10e^{\frac{e^{e^2}}{20x^2}}{x}^3(-2+3x)+20x^3(4-11x+6x^2))}{x^3}dx+2\int (2e^{6x}-5e^{6x}x+3e^{6x}{x}^2)dx\\ & ={(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2-\frac{1}{5}\int (20e^{3x}(-1+2x)(-4+3x)+\frac{e^{\frac{e^{e^2}}{20x^2}+3x}(e^{e^2}-e^{e^2}x-20x^3+30x^4)}{x^3})dx+4\int e^{6x}dx+6\int e^{6x}{x}^{2}dx-10\int e^{6x}xdx\\ & =\frac{2e^{6x}}{3}+{(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2-\frac{5}{3}{e}^{6x}x+e^{6x}{x}^2-\frac{1}{5}\int \frac{e^{\frac{e^{e^2}}{20x^2}+3x}(e^{e^2}-e^{e^2}x-20x^3+30x^4)}{x^3}dx+\frac{5}{3}\int e^{6x}dx-2\int e^{6x}xdx-4\int e^{3x}(-1+2x)(-4+3x)dx\\ & =\frac{17e^{6x}}{18}+{(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2-2e^{6x}x+e^{6x}{x}^2-\frac{1}{5}\int \frac{e^{\frac{e^{e^2}+60x^3}{20x^2}}(e^{e^2}-e^{e^2}x-20x^3+30x^4)}{x^3}dx+\frac{1}{3}\int e^{6x}dx-4\int (4e^{3x}-11e^{3x}x+6e^{3x}{x}^2)dx\\ & =e^{6x}+{(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2-2e^{6x}x+e^{6x}{x}^2-\frac{1}{5}\int (-20e^{\frac{e^{e^2}+60x^3}{20x^2}}+\frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^3}-\frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^2}+30e^{\frac{e^{e^2}+60x^3}{20x^2}}x)dx-16\int e^{3x}dx-24\int e^{3x}{x}^{2}dx+44\int e^{3x}xdx\\ & =-\frac{16e^{3x}}{3}+e^{6x}+{(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2+\frac{44}{3}{e}^{3x}x-2e^{6x}x-8e^{3x}{x}^2+e^{6x}{x}^2-\frac{1}{5}\int \frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^3}dx+\frac{1}{5}\int \frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^2}dx+4\int e^{\frac{e^{e^2}+60x^3}{20x^2}}dx-6\int e^{\frac{e^{e^2}+60x^3}{20x^2}}xdx-\frac{44}{3}\int e^{3x}dx+16\int e^{3x}xdx\\ & =-\frac{92e^{3x}}{9}+e^{6x}+{(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2+20e^{3x}x-2e^{6x}x-8e^{3x}{x}^2+e^{6x}{x}^2-\frac{1}{5}\int \frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^3}dx+\frac{1}{5}\int \frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^2}dx+4\int e^{\frac{e^{e^2}+60x^3}{20x^2}}dx-\frac{16}{3}\int e^{3x}dx-6\int e^{\frac{e^{e^2}+60x^3}{20x^2}}xdx\\ & =-12e^{3x}+e^{6x}+{(6-e^{\frac{e^{e^2}}{20x^2}}-4x)}^2+20e^{3x}x-2e^{6x}x-8e^{3x}{x}^2+e^{6x}{x}^2-\frac{1}{5}\int \frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^3}dx+\frac{1}{5}\int \frac{e^{e^2+\frac{e^{e^2}+60x^3}{20x^2}}}{x^2}dx+4\int e^{\frac{e^{e^2}+60x^3}{20x^2}}dx-6\int e^{\frac{e^{e^2}+60x^3}{20x^2}}xdx\end{array}\end{array}$$

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Mathematica [B]  time = 1.76, size = 91, normalized size = 2.94 $$\begin{array}{c}\frac{1}{5}(5e^{\frac{e^{e^2}}{10x^2}}-e^{\frac{e^{e^2}}{20x^2}}(60-40x)+5e^{6x}(-1+x{)}^2-240x+80x^2-10e^{3x}(-1+x)(-6+e^{\frac{e^{e^2}}{20x^2}}+4x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-E^(E^2 + E^E^2/(10*x^2)) - 240*x^3 + 160*x^4 + E^(3*x)*(-80*x^3 + 220*x^4 - 120*x^5) + E^(6*x)*(20
*x^3 - 50*x^4 + 30*x^5) + E^(E^E^2/(20*x^2))*(E^E^2*(6 + E^(3*x)*(-1 + x) - 4*x) + 40*x^3 + E^(3*x)*(20*x^3 -
30*x^4)))/(5*x^3),x]

[Out]

(5*E^(E^E^2/(10*x^2)) - E^(E^E^2/(20*x^2))*(60 - 40*x) + 5*E^(6*x)*(-1 + x)^2 - 240*x + 80*x^2 - 10*E^(3*x)*(-
1 + x)*(-6 + E^(E^E^2/(20*x^2)) + 4*x))/5

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fricas [B]  time = 0.74, size = 71, normalized size = 2.29 $$\begin{array}{c}16x^2+(x^2-2x+1)e^{\left(6x\right)}-4(2x^2-5x+3)e^{\left(3x\right)}-2((x-1)e^{\left(3x\right)}-4x+6)e^{\left(\frac{e^{\left(e^2\right)}}{20x^2}\right)}-48x+e^{\left(\frac{e^{\left(e^2\right)}}{10x^2}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(2))*exp(1/20*exp(exp(2))/x^2)^2+(((x-1)*exp(3*x)+6-4*x)*exp(exp(2))+(-30*x^4+20*x^3)*e
xp(3*x)+40*x^3)*exp(1/20*exp(exp(2))/x^2)+(30*x^5-50*x^4+20*x^3)*exp(3*x)^2+(-120*x^5+220*x^4-80*x^3)*exp(3*x)
+160*x^4-240*x^3)/x^3,x, algorithm="fricas")

[Out]

16*x^2 + (x^2 - 2*x + 1)*e^(6*x) - 4*(2*x^2 - 5*x + 3)*e^(3*x) - 2*((x - 1)*e^(3*x) - 4*x + 6)*e^(1/20*e^(e^2)
/x^2) - 48*x + e^(1/10*e^(e^2)/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{160x^4-240x^3+10(3x^5-5x^4+2x^3)e^{\left(6x\right)}-20(6x^5-11x^4+4x^3)e^{\left(3x\right)}+(40x^3-10(3x^4-2x^3)e^{\left(3x\right)}+((x-1)e^{\left(3x\right)}-4x+6)e^{\left(e^2\right)})e^{\left(\frac{e^{\left(e^2\right)}}{20x^2}\right)}-e^{(\frac{e^{\left(e^2\right)}}{10x^2}+e^2)}}{5x^3}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(2))*exp(1/20*exp(exp(2))/x^2)^2+(((x-1)*exp(3*x)+6-4*x)*exp(exp(2))+(-30*x^4+20*x^3)*e
xp(3*x)+40*x^3)*exp(1/20*exp(exp(2))/x^2)+(30*x^5-50*x^4+20*x^3)*exp(3*x)^2+(-120*x^5+220*x^4-80*x^3)*exp(3*x)
+160*x^4-240*x^3)/x^3,x, algorithm="giac")

[Out]

integrate(1/5*(160*x^4 - 240*x^3 + 10*(3*x^5 - 5*x^4 + 2*x^3)*e^(6*x) - 20*(6*x^5 - 11*x^4 + 4*x^3)*e^(3*x) +
(40*x^3 - 10*(3*x^4 - 2*x^3)*e^(3*x) + ((x - 1)*e^(3*x) - 4*x + 6)*e^(e^2))*e^(1/20*e^(e^2)/x^2) - e^(1/10*e^(
e^2)/x^2 + e^2))/x^3, x)

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maple [B]  time = 0.14, size = 80, normalized size = 2.58

method	result	size
risch	$16x^2-48x+\frac{(5x^2-10x+5){\mathrm{e}}^{6x}}{5}+\frac{(-40x^2+100x-60){\mathrm{e}}^{3x}}{5}+{\mathrm{e}}^{\frac{{\mathrm{e}}^{{\mathrm{e}}^2}}{10x^2}}+\frac{(-10x{\mathrm{e}}^{3x}+40x+10{\mathrm{e}}^{3x}-60){\mathrm{e}}^{\frac{{\mathrm{e}}^{{\mathrm{e}}^2}}{20x^2}}}{5}$	$80$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-exp(exp(2))*exp(1/20*exp(exp(2))/x^2)^2+(((x-1)*exp(3*x)+6-4*x)*exp(exp(2))+(-30*x^4+20*x^3)*exp(3*x
)+40*x^3)*exp(1/20*exp(exp(2))/x^2)+(30*x^5-50*x^4+20*x^3)*exp(3*x)^2+(-120*x^5+220*x^4-80*x^3)*exp(3*x)+160*x
^4-240*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

16*x^2-48*x+1/5*(5*x^2-10*x+5)*exp(6*x)+1/5*(-40*x^2+100*x-60)*exp(3*x)+exp(1/10*exp(exp(2))/x^2)+1/5*(-10*x*e
xp(3*x)+40*x+10*exp(3*x)-60)*exp(1/20*exp(exp(2))/x^2)

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maxima [C]  time = 0.83, size = 181, normalized size = 5.84 $$\begin{array}{c}2\sqrt{\frac{1}{5}}x\sqrt{-\frac{e^{\left(e^2\right)}}{x^2}}\mathrm{\Gamma }(-\frac{1}{2},-\frac{e^{\left(e^2\right)}}{20x^2})+16x^2+\frac{1}{18}(18x^2-6x+1)e^{\left(6x\right)}-\frac{5}{18}(6x-1)e^{\left(6x\right)}-\frac{8}{9}(9x^2-6x+2)e^{\left(3x\right)}+\frac{44}{9}(3x-1)e^{\left(3x\right)}-2(x-1)e^{(3x+\frac{e^{\left(e^2\right)}}{20x^2})}+\frac{4\sqrt{\frac{1}{5}}\sqrt{\pi }(\mathrm{erf}\left(\frac{1}{2}\sqrt{\frac{1}{5}}\sqrt{-\frac{e^{\left(e^2\right)}}{x^2}}\right)-1)e^{\left(e^2\right)}}{x\sqrt{-\frac{e^{\left(e^2\right)}}{x^2}}}-48x+\frac{2}{3}{e}^{\left(6x\right)}-\frac{16}{3}{e}^{\left(3x\right)}+e^{\left(\frac{e^{\left(e^2\right)}}{10x^2}\right)}-12e^{\left(\frac{e^{\left(e^2\right)}}{20x^2}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(2))*exp(1/20*exp(exp(2))/x^2)^2+(((x-1)*exp(3*x)+6-4*x)*exp(exp(2))+(-30*x^4+20*x^3)*e
xp(3*x)+40*x^3)*exp(1/20*exp(exp(2))/x^2)+(30*x^5-50*x^4+20*x^3)*exp(3*x)^2+(-120*x^5+220*x^4-80*x^3)*exp(3*x)
+160*x^4-240*x^3)/x^3,x, algorithm="maxima")

[Out]

2*sqrt(1/5)*x*sqrt(-e^(e^2)/x^2)*gamma(-1/2, -1/20*e^(e^2)/x^2) + 16*x^2 + 1/18*(18*x^2 - 6*x + 1)*e^(6*x) - 5
/18*(6*x - 1)*e^(6*x) - 8/9*(9*x^2 - 6*x + 2)*e^(3*x) + 44/9*(3*x - 1)*e^(3*x) - 2*(x - 1)*e^(3*x + 1/20*e^(e^
2)/x^2) + 4*sqrt(1/5)*sqrt(pi)*(erf(1/2*sqrt(1/5)*sqrt(-e^(e^2)/x^2)) - 1)*e^(e^2)/(x*sqrt(-e^(e^2)/x^2)) - 48
*x + 2/3*e^(6*x) - 16/3*e^(3*x) + e^(1/10*e^(e^2)/x^2) - 12*e^(1/20*e^(e^2)/x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}\int -\frac{\frac{{\mathrm{e}}^{\frac{{\mathrm{e}}^{{\mathrm{e}}^2}}{10x^2}}{\mathrm{e}}^{{\mathrm{e}}^2}}{5}-\frac{{\mathrm{e}}^{\frac{{\mathrm{e}}^{{\mathrm{e}}^2}}{20x^2}}({\mathrm{e}}^{3x}(20x^3-30x^4)+{\mathrm{e}}^{{\mathrm{e}}^2}({\mathrm{e}}^{3x}(x-1)-4x+6)+40x^3)}{5}-\frac{{\mathrm{e}}^{6x}(30x^5-50x^4+20x^3)}{5}+\frac{{\mathrm{e}}^{3x}(120x^5-220x^4+80x^3)}{5}+48x^3-32x^4}{x^3}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(exp(exp(2))/(10*x^2))*exp(exp(2)))/5 - (exp(exp(exp(2))/(20*x^2))*(exp(3*x)*(20*x^3 - 30*x^4) + exp
(exp(2))*(exp(3*x)*(x - 1) - 4*x + 6) + 40*x^3))/5 - (exp(6*x)*(20*x^3 - 50*x^4 + 30*x^5))/5 + (exp(3*x)*(80*x
^3 - 220*x^4 + 120*x^5))/5 + 48*x^3 - 32*x^4)/x^3,x)

[Out]

int(-((exp(exp(exp(2))/(10*x^2))*exp(exp(2)))/5 - (exp(exp(exp(2))/(20*x^2))*(exp(3*x)*(20*x^3 - 30*x^4) + exp
(exp(2))*(exp(3*x)*(x - 1) - 4*x + 6) + 40*x^3))/5 - (exp(6*x)*(20*x^3 - 50*x^4 + 30*x^5))/5 + (exp(3*x)*(80*x
^3 - 220*x^4 + 120*x^5))/5 + 48*x^3 - 32*x^4)/x^3, x)

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sympy [B]  time = 7.67, size = 80, normalized size = 2.58 $$\begin{array}{c}16x^2-48x+(-8x^2+20x-12)e^{3x}+(x^2-2x+1)e^{6x}+(-2x{e}^{3x}+8x+2e^{3x}-12)e^{\frac{e^{e^2}}{20x^2}}+e^{\frac{e^{e^2}}{10x^2}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(exp(2))*exp(1/20*exp(exp(2))/x**2)**2+(((x-1)*exp(3*x)+6-4*x)*exp(exp(2))+(-30*x**4+20*x**
3)*exp(3*x)+40*x**3)*exp(1/20*exp(exp(2))/x**2)+(30*x**5-50*x**4+20*x**3)*exp(3*x)**2+(-120*x**5+220*x**4-80*x
**3)*exp(3*x)+160*x**4-240*x**3)/x**3,x)

[Out]

16*x**2 - 48*x + (-8*x**2 + 20*x - 12)*exp(3*x) + (x**2 - 2*x + 1)*exp(6*x) + (-2*x*exp(3*x) + 8*x + 2*exp(3*x
) - 12)*exp(exp(exp(2))/(20*x**2)) + exp(exp(exp(2))/(10*x**2))

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