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

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

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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 {gather*} \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx \end {gather*}

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

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

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 {gather*} \frac {x^{2} - x \log \left (2 \, x^{3} - 2\right ) + 5 \, x - 5 \, e^{2}}{x - \log \left (2 \, x^{3} - 2\right ) + 5} \end {gather*}

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 {gather*} \frac {x^{2} - x \log \left (2 \, x^{3} - 2\right ) + 5 \, x - 5 \, e^{2}}{x - \log \left (2 \, x^{3} - 2\right ) + 5} \end {gather*}

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	\(x -\frac {5 \,{\mathrm e}^{2}}{5+x -\ln \left (2 x^{3}-2\right )}\)	\(22\)
norman	\(\frac {x^{2}-25+5 \ln \left (2 x^{3}-2\right )-\ln \left (2 x^{3}-2\right ) x -5 \,{\mathrm e}^{2}}{5+x -\ln \left (2 x^{3}-2\right )}\)	\(47\)

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 {gather*} \frac {x^{2} - x {\left (\log \relax (2) - 5\right )} - x \log \left (x^{2} + x + 1\right ) - x \log \left (x - 1\right ) - 5 \, e^{2}}{x - \log \relax (2) - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) + 5} \end {gather*}

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 {gather*} x-\frac {5\,{\mathrm {e}}^2}{x-\ln \left (2\,x^3-2\right )+5} \end {gather*}

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 {gather*} x + \frac {5 e^{2}}{- x + \log {\left (2 x^{3} - 2 \right )} - 5} \end {gather*}

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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