3.33.22 $\int \frac{-25-10x-x^2+25x^3+10x^4+x^5+e^2(-5-15x^2+5x^3)+(10+2x-10x^3-2x^4)\log (-2+2x^3)+(-1+x^3){\log }^2(-2+2x^3)}{-25-10x-x^2+25x^3+10x^4+x^5+(10+2x-10x^3-2x^4)\log (-2+2x^3)+(-1+x^3){\log }^2(-2+2x^3)}dx$

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

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Rubi [F]  time = 36.91, 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 = {} $$\begin{array}{c}\int \frac{-25-10x-x^2+25x^3+10x^4+x^5+e^2(-5-15x^2+5x^3)+(10+2x-10x^3-2x^4)\log (-2+2x^3)+(-1+x^3){\log }^2(-2+2x^3)}{-25-10x-x^2+25x^3+10x^4+x^5+(10+2x-10x^3-2x^4)\log (-2+2x^3)+(-1+x^3){\log }^2(-2+2x^3)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + E^2*(-5 - 15*x^2 + 5*x^3) + (10 + 2*x - 10*x^3 - 2*x^4)*Log[-2
+ 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3]^2)/(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + (10 + 2*x - 10*x^3 - 2*x
^4)*Log[-2 + 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3]^2),x]

[Out]

x - (10*Defer[Int][1/((-1 + x)*(5 + x - Log[2*(-1 + x^3)])), x])/3 - (20*Defer[Int][(-5 - x + Log[2*(-1 + x^3)
])^(-1), x])/3 - (85*Defer[Int][(5 + x - Log[-2 + 2*x^3])^(-2), x])/3 + ((49 - 5*E^2)*Defer[Int][(5 + x - Log[
-2 + 2*x^3])^(-2), x])/3 + (5*(5 + E^2)*Defer[Int][(5 + x - Log[-2 + 2*x^3])^(-2), x])/3 + 2*(2 + 5*E^2)*Defer
[Int][(5 + x - Log[-2 + 2*x^3])^(-2), x] - ((1 + 15*E^2)*Defer[Int][(5 + x - Log[-2 + 2*x^3])^(-2), x])/3 + ((
(22*I)/3)*Defer[Int][1/((-1 + I*Sqrt[3] - 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] - (((2*I)/3)*(49 - 5*
E^2)*Defer[Int][1/((-1 + I*Sqrt[3] - 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] + (((20*I)/3)*(5 + E^2)*De
fer[Int][1/((-1 + I*Sqrt[3] - 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] - ((4*I)*(2 + 5*E^2)*Defer[Int][1
/((-1 + I*Sqrt[3] - 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] + Defer[Int][1/((-1 + x)*(5 + x - Log[-2 +
2*x^3])^2), x]/3 - ((1 + 15*E^2)*Defer[Int][1/((-1 + x)*(5 + x - Log[-2 + 2*x^3])^2), x])/3 - (20*Defer[Int][x
/(5 + x - Log[-2 + 2*x^3])^2, x])/3 - ((49 - 5*E^2)*Defer[Int][x/(5 + x - Log[-2 + 2*x^3])^2, x])/3 + (5*(5 +
E^2)*Defer[Int][x/(5 + x - Log[-2 + 2*x^3])^2, x])/3 + (5*(9 + E^2)*Defer[Int][x/(5 + x - Log[-2 + 2*x^3])^2,
x])/3 - ((1 + 15*E^2)*Defer[Int][x/(5 + x - Log[-2 + 2*x^3])^2, x])/3 + (20*Defer[Int][x^2/(5 + x - Log[-2 + 2
*x^3])^2, x])/3 + (5*(5 + E^2)*Defer[Int][x^2/(5 + x - Log[-2 + 2*x^3])^2, x])/3 - (5*(9 + E^2)*Defer[Int][x^2
/(5 + x - Log[-2 + 2*x^3])^2, x])/3 + (4*(3 + I*Sqrt[3])*Defer[Int][1/((1 - I*Sqrt[3] + 2*x)*(5 + x - Log[-2 +
2*x^3])^2), x])/3 - (2*(3 + I*Sqrt[3])*(2 + 5*E^2)*Defer[Int][1/((1 - I*Sqrt[3] + 2*x)*(5 + x - Log[-2 + 2*x^
3])^2), x])/3 + (((22*I)/3)*Defer[Int][1/((1 + I*Sqrt[3] + 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] + (4
*(3 - I*Sqrt[3])*Defer[Int][1/((1 + I*Sqrt[3] + 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/3 - (((2*I)/3)*(49 - 5*
E^2)*Defer[Int][1/((1 + I*Sqrt[3] + 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] + (((20*I)/3)*(5 + E^2)*Def
er[Int][1/((1 + I*Sqrt[3] + 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] - ((4*I)*(2 + 5*E^2)*Defer[Int][1/(
(1 + I*Sqrt[3] + 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/Sqrt[3] - (2*(3 - I*Sqrt[3])*(2 + 5*E^2)*Defer[Int][1/
((1 + I*Sqrt[3] + 2*x)*(5 + x - Log[-2 + 2*x^3])^2), x])/3 - (20*Defer[Int][(5 + x - Log[-2 + 2*x^3])^(-1), x]
)/3 + (10*Defer[Int][1/((-1 + x)*(5 + x - Log[-2 + 2*x^3])), x])/3

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-(5+x{)}^2(-1+x^3)-5e^2(-1-3x^2+x^3)+2(-5-x+5x^3+x^4)\log \left(2(-1+x^3)\right)-(-1+x^3){\log }^2\left(2(-1+x^3)\right)}{(1-x^3){(5+x-\log (-2+2x^3))}^2}dx\\ & =\int (\frac{25(1+\frac{e^2}{5})+10x+(1+15e^2)x^2-25(1+\frac{e^2}{5})x^3-10x^4-x^5-10\log \left(2(-1+x^3)\right)-2x\log \left(2(-1+x^3)\right)+10x^3\log \left(2(-1+x^3)\right)+2x^4\log \left(2(-1+x^3)\right)+{\log }^2\left(2(-1+x^3)\right)-x^3{\log }^2\left(2(-1+x^3)\right)}{3(1-x){(5+x-\log (-2+2x^3))}^2}+\frac{(2+x)(25(1+\frac{e^2}{5})+10x+(1+15e^2)x^2-25(1+\frac{e^2}{5})x^3-10x^4-x^5-10\log \left(2(-1+x^3)\right)-2x\log \left(2(-1+x^3)\right)+10x^3\log \left(2(-1+x^3)\right)+2x^4\log \left(2(-1+x^3)\right)+{\log }^2\left(2(-1+x^3)\right)-x^3{\log }^2\left(2(-1+x^3)\right))}{3(1+x+x^2){(5+x-\log (-2+2x^3))}^2})dx\\ & =\frac{1}{3}\int \frac{25(1+\frac{e^2}{5})+10x+(1+15e^2)x^2-25(1+\frac{e^2}{5})x^3-10x^4-x^5-10\log \left(2(-1+x^3)\right)-2x\log \left(2(-1+x^3)\right)+10x^3\log \left(2(-1+x^3)\right)+2x^4\log \left(2(-1+x^3)\right)+{\log }^2\left(2(-1+x^3)\right)-x^3{\log }^2\left(2(-1+x^3)\right)}{(1-x){(5+x-\log (-2+2x^3))}^2}dx+\frac{1}{3}\int \frac{(2+x)(25(1+\frac{e^2}{5})+10x+(1+15e^2)x^2-25(1+\frac{e^2}{5})x^3-10x^4-x^5-10\log \left(2(-1+x^3)\right)-2x\log \left(2(-1+x^3)\right)+10x^3\log \left(2(-1+x^3)\right)+2x^4\log \left(2(-1+x^3)\right)+{\log }^2\left(2(-1+x^3)\right)-x^3{\log }^2\left(2(-1+x^3)\right))}{(1+x+x^2){(5+x-\log (-2+2x^3))}^2}dx\\ & =\frac{1}{3}\int \frac{-(5+x{)}^2(-1+x^3)-5e^2(-1-3x^2+x^3)+2(-5-x+5x^3+x^4)\log \left(2(-1+x^3)\right)-(-1+x^3){\log }^2\left(2(-1+x^3)\right)}{(1-x){(5+x-\log (-2+2x^3))}^2}dx+\frac{1}{3}\int \frac{(2+x)(-(5+x{)}^2(-1+x^3)-5e^2(-1-3x^2+x^3)+2(-5-x+5x^3+x^4)\log \left(2(-1+x^3)\right)-(-1+x^3){\log }^2\left(2(-1+x^3)\right))}{(1+x+x^2){(5+x-\log (-2+2x^3))}^2}dx\\ & =\text{Rest of rules removed due to large latex content}\end{array}\end{array}$$

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Mathematica [A]  time = 0.07, size = 22, normalized size = 0.79 $$\begin{array}{c}x+\frac{5e^2}{-5-x+\log \left(2(-1+x^3)\right)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + E^2*(-5 - 15*x^2 + 5*x^3) + (10 + 2*x - 10*x^3 - 2*x^4)*
Log[-2 + 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3]^2)/(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + (10 + 2*x - 10*x^3
- 2*x^4)*Log[-2 + 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3]^2),x]

[Out]

x + (5*E^2)/(-5 - x + Log[2*(-1 + x^3)])

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fricas [A]  time = 0.56, size = 38, normalized size = 1.36 $$\begin{array}{c}\frac{x^2-x\log (2x^3-2)+5x-5e^2}{x-\log (2x^3-2)+5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+(5*x^3-15*x^2-5)*exp(2)+x^5+10*x^4+25*x^
3-x^2-10*x-25)/((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x, a
lgorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 38, normalized size = 1.36 $$\begin{array}{c}\frac{x^2-x\log (2x^3-2)+5x-5e^2}{x-\log (2x^3-2)+5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+(5*x^3-15*x^2-5)*exp(2)+x^5+10*x^4+25*x^
3-x^2-10*x-25)/((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x, a
lgorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 22, normalized size = 0.79

method	result	size
risch	$x-\frac{5{\mathrm{e}}^2}{5+x-\ln (2x^3-2)}$	$22$
norman	$\frac{x^2-25+5\ln (2x^3-2)-\ln (2x^3-2)x-5{\mathrm{e}}^2}{5+x-\ln (2x^3-2)}$	$47$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-1)*ln(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*ln(2*x^3-2)+(5*x^3-15*x^2-5)*exp(2)+x^5+10*x^4+25*x^3-x^2-10
*x-25)/((x^3-1)*ln(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*ln(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x,method=_RETU
RNVERBOSE)

[Out]

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

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maxima [B]  time = 0.93, size = 57, normalized size = 2.04 $$\begin{array}{c}\frac{x^2-x(\log (2)-5)-x\log (x^2+x+1)-x\log (x-1)-5e^2}{x-\log (2)-\log (x^2+x+1)-\log (x-1)+5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+(5*x^3-15*x^2-5)*exp(2)+x^5+10*x^4+25*x^
3-x^2-10*x-25)/((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x, a
lgorithm="maxima")

[Out]

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

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mupad [B]  time = 2.20, size = 21, normalized size = 0.75 $$\begin{array}{c}x-\frac{5{\mathrm{e}}^2}{x-\ln (2x^3-2)+5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2*x^3 - 2)*(2*x - 10*x^3 - 2*x^4 + 10) - 10*x - exp(2)*(15*x^2 - 5*x^3 + 5) - x^2 + 25*x^3 + 10*x^4 +
x^5 + log(2*x^3 - 2)^2*(x^3 - 1) - 25)/(log(2*x^3 - 2)*(2*x - 10*x^3 - 2*x^4 + 10) - 10*x - x^2 + 25*x^3 + 10
*x^4 + x^5 + log(2*x^3 - 2)^2*(x^3 - 1) - 25),x)

[Out]

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

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sympy [A]  time = 0.20, size = 17, normalized size = 0.61 $$\begin{array}{c}x+\frac{5e^2}{-x+\log (2x^3-2)-5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-1)*ln(2*x**3-2)**2+(-2*x**4-10*x**3+2*x+10)*ln(2*x**3-2)+(5*x**3-15*x**2-5)*exp(2)+x**5+10*x*
*4+25*x**3-x**2-10*x-25)/((x**3-1)*ln(2*x**3-2)**2+(-2*x**4-10*x**3+2*x+10)*ln(2*x**3-2)+x**5+10*x**4+25*x**3-
x**2-10*x-25),x)

[Out]

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

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