3.78.40 $\int \frac{-5e^7+e^6(-5e^4+20e^{2}x)+e^6(20e^{2}x+10e^{4}x)\log (x)+(-e^{4}x+e^3(-2e^{4}x+8e^2{x}^2)+e^6(4e^{4}x+8e^2{x}^2-16x^3)){\log }^2(x)+(2e^7{x}^2+e^6(2e^4{x}^2-8e^2{x}^3)){\log }^3(x)-e^{10}{x}^3{\log }^4(x)}{(e^{4}x+e^3(2e^{4}x-8e^2{x}^2)+e^6(e^{4}x-8e^2{x}^2+16x^3)){\log }^2(x)+(-2e^7{x}^2+e^6(-2e^4{x}^2+8e^2{x}^3)){\log }^3(x)+e^{10}{x}^3{\log }^4(x)}dx$

Optimal. Leaf size=28 $$-x+\frac{5}{\log (x)(1+\frac{1}{e^3}-x(\frac{4}{e^2}+\log (x)))}$$

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Rubi [F]  time = 2.13, 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 = {} $$\begin{array}{c}\int \frac{-5e^7+e^6(-5e^4+20e^{2}x)+e^6(20e^{2}x+10e^{4}x)\log (x)+(-e^{4}x+e^3(-2e^{4}x+8e^2{x}^2)+e^6(4e^{4}x+8e^2{x}^2-16x^3)){\log }^2(x)+(2e^7{x}^2+e^6(2e^4{x}^2-8e^2{x}^3)){\log }^3(x)-e^{10}{x}^3{\log }^4(x)}{(e^{4}x+e^3(2e^{4}x-8e^2{x}^2)+e^6(e^{4}x-8e^2{x}^2+16x^3)){\log }^2(x)+(-2e^7{x}^2+e^6(-2e^4{x}^2+8e^2{x}^3)){\log }^3(x)+e^{10}{x}^3{\log }^4(x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-5*E^7 + E^6*(-5*E^4 + 20*E^2*x) + E^6*(20*E^2*x + 10*E^4*x)*Log[x] + (-(E^4*x) + E^3*(-2*E^4*x + 8*E^2*x
^2) + E^6*(4*E^4*x + 8*E^2*x^2 - 16*x^3))*Log[x]^2 + (2*E^7*x^2 + E^6*(2*E^4*x^2 - 8*E^2*x^3))*Log[x]^3 - E^10
*x^3*Log[x]^4)/((E^4*x + E^3*(2*E^4*x - 8*E^2*x^2) + E^6*(E^4*x - 8*E^2*x^2 + 16*x^3))*Log[x]^2 + (-2*E^7*x^2
+ E^6*(-2*E^4*x^2 + 8*E^2*x^3))*Log[x]^3 + E^10*x^3*Log[x]^4),x]

[Out]

-x + 5*E^3*Defer[Int][1/(x*(-1 - E^3 + 4*E*x)*Log[x]^2), x] + 20*E^4*Defer[Int][1/((1 + E^3 - 4*E*x)^2*Log[x])
, x] - (5*E^8*Defer[Int][(-1 - E^3 + 4*E*x + E^3*x*Log[x])^(-2), x])/4 + (5*E^6*(4 + E^2 + 4*E^3 + E^5)*Defer[
Int][1/((1 + E^3 - 4*E*x)*(-1 - E^3 + 4*E*x + E^3*x*Log[x])^2), x])/4 - 5*E^6*(1 + E^3)*Defer[Int][1/((1 + E^3
- 4*E*x)^2*(-1 - E^3 + 4*E*x + E^3*x*Log[x])), x] + 5*E^6*Defer[Int][1/((1 + E^3 - 4*E*x)*(-1 - E^3 + 4*E*x +
E^3*x*Log[x])), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-5e^3(1+e^3-4ex)+10e^4(2+e^2)x\log (x)-x(1+2e^3-4e^6-8ex-8e^{4}x+16e^2{x}^2){\log }^2(x)+2e^3{x}^2(1+e^3-4ex){\log }^3(x)-e^6{x}^3{\log }^4(x)}{x{\log }^2(x){(1+e^3-4ex-e^{3}x\log (x))}^2}dx\\ & =\int (-1+\frac{5e^3}{x(-1-e^3+4ex){\log }^2(x)}+\frac{20e^4}{{(1+e^3-4ex)}^2\log (x)}+\frac{5(e^6+e^9+e^{9}x)}{(1+e^3-4ex){(-1-e^3+4ex+e^{3}x\log (x))}^2}-\frac{20e^{7}x}{{(1+e^3-4ex)}^2(-1-e^3+4ex+e^{3}x\log (x))})dx\\ & =-x+5\int \frac{e^6+e^9+e^{9}x}{(1+e^3-4ex){(-1-e^3+4ex+e^{3}x\log (x))}^2}dx+\left(5e^3\right)\int \frac{1}{x(-1-e^3+4ex){\log }^2(x)}dx+\left(20e^4\right)\int \frac{1}{{(1+e^3-4ex)}^2\log (x)}dx-\left(20e^7\right)\int \frac{x}{{(1+e^3-4ex)}^2(-1-e^3+4ex+e^{3}x\log (x))}dx\\ & =-x+5\int (-\frac{e^8}{4{(-1-e^3+4ex+e^{3}x\log (x))}^2}+\frac{e^6(4+e^2+4e^3+e^5)}{4(1+e^3-4ex){(-1-e^3+4ex+e^{3}x\log (x))}^2})dx+\left(5e^3\right)\int \frac{1}{x(-1-e^3+4ex){\log }^2(x)}dx+\left(20e^4\right)\int \frac{1}{{(1+e^3-4ex)}^2\log (x)}dx-\left(20e^7\right)\int (\frac{1+e^3}{4e{(1+e^3-4ex)}^2(-1-e^3+4ex+e^{3}x\log (x))}-\frac{1}{4e(1+e^3-4ex)(-1-e^3+4ex+e^{3}x\log (x))})dx\\ & =-x+\left(5e^3\right)\int \frac{1}{x(-1-e^3+4ex){\log }^2(x)}dx+\left(20e^4\right)\int \frac{1}{{(1+e^3-4ex)}^2\log (x)}dx+\left(5e^6\right)\int \frac{1}{(1+e^3-4ex)(-1-e^3+4ex+e^{3}x\log (x))}dx-\frac{1}{4}\left(5e^8\right)\int \frac{1}{{(-1-e^3+4ex+e^{3}x\log (x))}^2}dx-\left(5e^6(1+e^3)\right)\int \frac{1}{{(1+e^3-4ex)}^2(-1-e^3+4ex+e^{3}x\log (x))}dx+\frac{1}{4}\left(5e^6(4+e^2+4e^3+e^5)\right)\int \frac{1}{(1+e^3-4ex){(-1-e^3+4ex+e^{3}x\log (x))}^2}dx\end{array}\end{array}$$

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Mathematica [B]  time = 0.09, size = 61, normalized size = 2.18 $$\begin{array}{c}-x+\frac{5e^3}{(1+e^3-4ex)\log (x)}-\frac{5e^{6}x}{(1+e^3-4ex)(-1-e^3+4ex+e^{3}x\log (x))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^7 + E^6*(-5*E^4 + 20*E^2*x) + E^6*(20*E^2*x + 10*E^4*x)*Log[x] + (-(E^4*x) + E^3*(-2*E^4*x + 8
*E^2*x^2) + E^6*(4*E^4*x + 8*E^2*x^2 - 16*x^3))*Log[x]^2 + (2*E^7*x^2 + E^6*(2*E^4*x^2 - 8*E^2*x^3))*Log[x]^3
- E^10*x^3*Log[x]^4)/((E^4*x + E^3*(2*E^4*x - 8*E^2*x^2) + E^6*(E^4*x - 8*E^2*x^2 + 16*x^3))*Log[x]^2 + (-2*E^
7*x^2 + E^6*(-2*E^4*x^2 + 8*E^2*x^3))*Log[x]^3 + E^10*x^3*Log[x]^4),x]

[Out]

-x + (5*E^3)/((1 + E^3 - 4*E*x)*Log[x]) - (5*E^6*x)/((1 + E^3 - 4*E*x)*(-1 - E^3 + 4*E*x + E^3*x*Log[x]))

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fricas [B]  time = 0.86, size = 61, normalized size = 2.18 $$\begin{array}{c}-\frac{x^2{e}^3\log (x{)}^2+(4x^{2}e-x{e}^3-x)\log (x)+5e^3}{x{e}^3\log (x{)}^2+(4xe-e^3-1)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(2)^2*exp(3)^2*log(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*log(
x)^3+((4*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*log(x)^2+(10
*x*exp(2)^2+20*exp(2)*x)*exp(3)^2*log(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*e
xp(3)^2*log(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*log(x)^3+((x*exp(2)^2-8*x^2*e
xp(2)+16*x^3)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*log(x)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.94, size = 66, normalized size = 2.36 $$\begin{array}{c}-\frac{x^2{e}^3\log (x{)}^2+4x^{2}e\log (x)-x{e}^3\log (x)-x\log (x)+5e^3}{x{e}^3\log (x{)}^2+4xe\log (x)-e^3\log (x)-\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(2)^2*exp(3)^2*log(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*log(
x)^3+((4*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*log(x)^2+(10
*x*exp(2)^2+20*exp(2)*x)*exp(3)^2*log(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*e
xp(3)^2*log(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*log(x)^3+((x*exp(2)^2-8*x^2*e
xp(2)+16*x^3)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*log(x)^2),x, algorithm="giac")

[Out]

-(x^2*e^3*log(x)^2 + 4*x^2*e*log(x) - x*e^3*log(x) - x*log(x) + 5*e^3)/(x*e^3*log(x)^2 + 4*x*e*log(x) - e^3*lo
g(x) - log(x))

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maple [A]  time = 0.17, size = 32, normalized size = 1.14

method	result	size
risch	$-x-\frac{5{\mathrm{e}}^3}{(x{\mathrm{e}}^3\ln (x)-{\mathrm{e}}^3+4x\mathrm{e}-1)\ln (x)}$	$32$
norman	$\frac{\frac{{\mathrm{e}}^4{\mathrm{e}}^{-3}({\mathrm{e}}^6+2{\mathrm{e}}^3+1)\ln (x)}{4}+(-\frac{{\mathrm{e}}^3{\mathrm{e}}^4}{4}-\frac{{\mathrm{e}}^4}{4})x\ln (x{)}^2-5{\mathrm{e}}^2{\mathrm{e}}^3-4x^2{\mathrm{e}}^3\ln (x)-{\mathrm{e}}^2\ln (x{)}^2{\mathrm{e}}^3{x}^2}{\ln (x)({\mathrm{e}}^3{\mathrm{e}}^{2}x\ln (x)-{\mathrm{e}}^2{\mathrm{e}}^3+4x{\mathrm{e}}^3-{\mathrm{e}}^2)}$	$104$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3*exp(2)^2*exp(3)^2*ln(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*ln(x)^3+((4
*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*ln(x)^2+(10*x*exp(2)
^2+20*exp(2)*x)*exp(3)^2*ln(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*exp(3)^2*ln
(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*ln(x)^3+((x*exp(2)^2-8*x^2*exp(2)+16*x^3
)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.44, size = 60, normalized size = 2.14 $$\begin{array}{c}-\frac{x^2{e}^3\log (x{)}^2+(4x^{2}e-x(e^3+1))\log (x)+5e^3}{x{e}^3\log (x{)}^2+(4xe-e^3-1)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(2)^2*exp(3)^2*log(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*log(
x)^3+((4*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*log(x)^2+(10
*x*exp(2)^2+20*exp(2)*x)*exp(3)^2*log(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*e
xp(3)^2*log(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*log(x)^3+((x*exp(2)^2-8*x^2*e
xp(2)+16*x^3)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*log(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.76, size = 34, normalized size = 1.21 $$\begin{array}{c}-x-\frac{5}{x({\ln (x)}^2-\frac{{\mathrm{e}}^{-3}\ln (x)({\mathrm{e}}^3-4x\mathrm{e}+1)}{x})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(7) + exp(6)*(5*exp(4) - 20*x*exp(2)) + log(x)^2*(x*exp(4) - exp(6)*(4*x*exp(4) + 8*x^2*exp(2) - 16
*x^3) + exp(3)*(2*x*exp(4) - 8*x^2*exp(2))) + log(x)^3*(exp(6)*(8*x^3*exp(2) - 2*x^2*exp(4)) - 2*x^2*exp(7)) -
exp(6)*log(x)*(20*x*exp(2) + 10*x*exp(4)) + x^3*exp(10)*log(x)^4)/(log(x)^2*(exp(6)*(x*exp(4) - 8*x^2*exp(2)
+ 16*x^3) + x*exp(4) + exp(3)*(2*x*exp(4) - 8*x^2*exp(2))) + log(x)^3*(exp(6)*(8*x^3*exp(2) - 2*x^2*exp(4)) -
2*x^2*exp(7)) + x^3*exp(10)*log(x)^4),x)

[Out]

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

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sympy [A]  time = 0.33, size = 32, normalized size = 1.14 $$\begin{array}{c}-x-\frac{5e^3}{x{e}^3\log {(x)}^2+(4ex-e^3-1)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3*exp(2)**2*exp(3)**2*ln(x)**4+((2*x**2*exp(2)**2-8*x**3*exp(2))*exp(3)**2+2*x**2*exp(2)**2*exp
(3))*ln(x)**3+((4*x*exp(2)**2+8*x**2*exp(2)-16*x**3)*exp(3)**2+(-2*x*exp(2)**2+8*x**2*exp(2))*exp(3)-x*exp(2)*
*2)*ln(x)**2+(10*x*exp(2)**2+20*exp(2)*x)*exp(3)**2*ln(x)+(-5*exp(2)**2+20*exp(2)*x)*exp(3)**2-5*exp(2)**2*exp
(3))/(x**3*exp(2)**2*exp(3)**2*ln(x)**4+((-2*x**2*exp(2)**2+8*x**3*exp(2))*exp(3)**2-2*x**2*exp(2)**2*exp(3))*
ln(x)**3+((x*exp(2)**2-8*x**2*exp(2)+16*x**3)*exp(3)**2+(2*x*exp(2)**2-8*x**2*exp(2))*exp(3)+x*exp(2)**2)*ln(x
)**2),x)

[Out]

-x - 5*exp(3)/(x*exp(3)*log(x)**2 + (4*E*x - exp(3) - 1)*log(x))

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