3.2.84 \(\int \frac {-60+90 x-x^3+2 e^{2 x} x^3+2 x^4+e^x (-2 x^3-2 x^4)+(-30 x+2 x^2-4 e^{2 x} x^2-4 x^3+e^x (4 x^2+4 x^3)) \log (x)+(-x+2 e^{2 x} x+2 x^2+e^x (-2 x-2 x^2)) \log ^2(x)}{x^3-2 x^2 \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ -2+\left (e^x-x\right )^2-x-\frac {15 (4-2 x)}{x-\log (x)} \]

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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 {gather*} \int \frac {-60+90 x-x^3+2 e^{2 x} x^3+2 x^4+e^x \left (-2 x^3-2 x^4\right )+\left (-30 x+2 x^2-4 e^{2 x} x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )\right ) \log (x)+\left (-x+2 e^{2 x} x+2 x^2+e^x \left (-2 x-2 x^2\right )\right ) \log ^2(x)}{x^3-2 x^2 \log (x)+x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-60 + 90*x - x^3 + 2*E^(2*x)*x^3 + 2*x^4 + E^x*(-2*x^3 - 2*x^4) + (-30*x + 2*x^2 - 4*E^(2*x)*x^2 - 4*x^3
+ E^x*(4*x^2 + 4*x^3))*Log[x] + (-x + 2*E^(2*x)*x + 2*x^2 + E^x*(-2*x - 2*x^2))*Log[x]^2)/(x^3 - 2*x^2*Log[x]
+ x*Log[x]^2),x]

[Out]

2*E^x + E^(2*x) - x + x^2 - 2*E^x*(1 + x) + 90*Defer[Int][(x - Log[x])^(-2), x] - 60*Defer[Int][1/(x*(x - Log[
x])^2), x] - 30*Defer[Int][x/(x - Log[x])^2, x] - 30*Defer[Int][(-x + Log[x])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-60+90 x-x^3+2 e^{2 x} x^3+2 x^4+e^x \left (-2 x^3-2 x^4\right )+\left (-30 x+2 x^2-4 e^{2 x} x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )\right ) \log (x)+\left (-x+2 e^{2 x} x+2 x^2+e^x \left (-2 x-2 x^2\right )\right ) \log ^2(x)}{x (x-\log (x))^2} \, dx\\ &=\int \left (2 e^{2 x}-2 e^x (1+x)+\frac {90}{(x-\log (x))^2}-\frac {60}{x (x-\log (x))^2}-\frac {x^2}{(x-\log (x))^2}+\frac {2 x^3}{(x-\log (x))^2}-\frac {30 \log (x)}{(x-\log (x))^2}+\frac {2 x \log (x)}{(x-\log (x))^2}-\frac {4 x^2 \log (x)}{(x-\log (x))^2}-\frac {\log ^2(x)}{(x-\log (x))^2}+\frac {2 x \log ^2(x)}{(x-\log (x))^2}\right ) \, dx\\ &=2 \int e^{2 x} \, dx-2 \int e^x (1+x) \, dx+2 \int \frac {x^3}{(x-\log (x))^2} \, dx+2 \int \frac {x \log (x)}{(x-\log (x))^2} \, dx+2 \int \frac {x \log ^2(x)}{(x-\log (x))^2} \, dx-4 \int \frac {x^2 \log (x)}{(x-\log (x))^2} \, dx-30 \int \frac {\log (x)}{(x-\log (x))^2} \, dx-60 \int \frac {1}{x (x-\log (x))^2} \, dx+90 \int \frac {1}{(x-\log (x))^2} \, dx-\int \frac {x^2}{(x-\log (x))^2} \, dx-\int \frac {\log ^2(x)}{(x-\log (x))^2} \, dx\\ &=e^{2 x}-2 e^x (1+x)+2 \int e^x \, dx+2 \int \left (\frac {x^2}{(x-\log (x))^2}-\frac {x}{x-\log (x)}\right ) \, dx+2 \int \left (x+\frac {x^3}{(x-\log (x))^2}-\frac {2 x^2}{x-\log (x)}\right ) \, dx+2 \int \frac {x^3}{(x-\log (x))^2} \, dx-4 \int \left (\frac {x^3}{(x-\log (x))^2}-\frac {x^2}{x-\log (x)}\right ) \, dx-30 \int \left (\frac {x}{(x-\log (x))^2}+\frac {1}{-x+\log (x)}\right ) \, dx-60 \int \frac {1}{x (x-\log (x))^2} \, dx+90 \int \frac {1}{(x-\log (x))^2} \, dx-\int \left (1+\frac {x^2}{(x-\log (x))^2}-\frac {2 x}{x-\log (x)}\right ) \, dx-\int \frac {x^2}{(x-\log (x))^2} \, dx\\ &=2 e^x+e^{2 x}-x+x^2-2 e^x (1+x)+2 \int \frac {x^2}{(x-\log (x))^2} \, dx+2 \left (2 \int \frac {x^3}{(x-\log (x))^2} \, dx\right )-4 \int \frac {x^3}{(x-\log (x))^2} \, dx-30 \int \frac {x}{(x-\log (x))^2} \, dx-30 \int \frac {1}{-x+\log (x)} \, dx-60 \int \frac {1}{x (x-\log (x))^2} \, dx+90 \int \frac {1}{(x-\log (x))^2} \, dx-2 \int \frac {x^2}{(x-\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 31, normalized size = 1.07 \begin {gather*} e^{2 x}-x-2 e^x x+x^2-\frac {30 (-2+x)}{-x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60 + 90*x - x^3 + 2*E^(2*x)*x^3 + 2*x^4 + E^x*(-2*x^3 - 2*x^4) + (-30*x + 2*x^2 - 4*E^(2*x)*x^2 -
4*x^3 + E^x*(4*x^2 + 4*x^3))*Log[x] + (-x + 2*E^(2*x)*x + 2*x^2 + E^x*(-2*x - 2*x^2))*Log[x]^2)/(x^3 - 2*x^2*L
og[x] + x*Log[x]^2),x]

[Out]

E^(2*x) - x - 2*E^x*x + x^2 - (30*(-2 + x))/(-x + Log[x])

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fricas [B]  time = 0.49, size = 55, normalized size = 1.90 \begin {gather*} \frac {x^{3} - 2 \, x^{2} e^{x} - x^{2} + x e^{\left (2 \, x\right )} - {\left (x^{2} - 2 \, x e^{x} - x + e^{\left (2 \, x\right )}\right )} \log \relax (x) + 30 \, x - 60}{x - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^2+(-2*x^2-2*x)*exp(x)+2*x^2-x)*log(x)^2+(-4*exp(x)^2*x^2+(4*x^3+4*x^2)*exp(x)-4*x^3+2*x
^2-30*x)*log(x)+2*exp(x)^2*x^3+(-2*x^4-2*x^3)*exp(x)+2*x^4-x^3+90*x-60)/(x*log(x)^2-2*x^2*log(x)+x^3),x, algor
ithm="fricas")

[Out]

(x^3 - 2*x^2*e^x - x^2 + x*e^(2*x) - (x^2 - 2*x*e^x - x + e^(2*x))*log(x) + 30*x - 60)/(x - log(x))

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giac [B]  time = 0.58, size = 61, normalized size = 2.10 \begin {gather*} \frac {x^{3} - 2 \, x^{2} e^{x} - x^{2} \log \relax (x) + 2 \, x e^{x} \log \relax (x) - x^{2} + x e^{\left (2 \, x\right )} + x \log \relax (x) - e^{\left (2 \, x\right )} \log \relax (x) + 30 \, x - 60}{x - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^2+(-2*x^2-2*x)*exp(x)+2*x^2-x)*log(x)^2+(-4*exp(x)^2*x^2+(4*x^3+4*x^2)*exp(x)-4*x^3+2*x
^2-30*x)*log(x)+2*exp(x)^2*x^3+(-2*x^4-2*x^3)*exp(x)+2*x^4-x^3+90*x-60)/(x*log(x)^2-2*x^2*log(x)+x^3),x, algor
ithm="giac")

[Out]

(x^3 - 2*x^2*e^x - x^2*log(x) + 2*x*e^x*log(x) - x^2 + x*e^(2*x) + x*log(x) - e^(2*x)*log(x) + 30*x - 60)/(x -
log(x))

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maple [A]  time = 0.06, size = 30, normalized size = 1.03

method	result	size
risch	\(x^{2}-2 \,{\mathrm e}^{x} x +{\mathrm e}^{2 x}-x +\frac {30 x -60}{x -\ln \relax (x )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(x)^2+(-2*x^2-2*x)*exp(x)+2*x^2-x)*ln(x)^2+(-4*exp(x)^2*x^2+(4*x^3+4*x^2)*exp(x)-4*x^3+2*x^2-30*x
)*ln(x)+2*exp(x)^2*x^3+(-2*x^4-2*x^3)*exp(x)+2*x^4-x^3+90*x-60)/(x*ln(x)^2-2*x^2*ln(x)+x^3),x,method=_RETURNVE
RBOSE)

[Out]

x^2-2*exp(x)*x+exp(2*x)-x+30*(x-2)/(x-ln(x))

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maxima [B]  time = 0.50, size = 57, normalized size = 1.97 \begin {gather*} \frac {x^{3} - x^{2} + {\left (x - \log \relax (x)\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{2} - x \log \relax (x)\right )} e^{x} - {\left (x^{2} - x\right )} \log \relax (x) + 30 \, x - 60}{x - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^2+(-2*x^2-2*x)*exp(x)+2*x^2-x)*log(x)^2+(-4*exp(x)^2*x^2+(4*x^3+4*x^2)*exp(x)-4*x^3+2*x
^2-30*x)*log(x)+2*exp(x)^2*x^3+(-2*x^4-2*x^3)*exp(x)+2*x^4-x^3+90*x-60)/(x*log(x)^2-2*x^2*log(x)+x^3),x, algor
ithm="maxima")

[Out]

(x^3 - x^2 + (x - log(x))*e^(2*x) - 2*(x^2 - x*log(x))*e^x - (x^2 - x)*log(x) + 30*x - 60)/(x - log(x))

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mupad [B]  time = 0.47, size = 56, normalized size = 1.93 \begin {gather*} {\mathrm {e}}^{2\,x}-x+\frac {30}{x-1}-2\,x\,{\mathrm {e}}^x+x^2-\frac {\frac {30\,\left (3\,x-2\right )}{x-1}-\frac {30\,x\,\ln \relax (x)}{x-1}}{x-\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(2*x^3 + 2*x^4) - 90*x + log(x)*(30*x - exp(x)*(4*x^2 + 4*x^3) + 4*x^2*exp(2*x) - 2*x^2 + 4*x^3)
+ log(x)^2*(x - 2*x*exp(2*x) + exp(x)*(2*x + 2*x^2) - 2*x^2) - 2*x^3*exp(2*x) + x^3 - 2*x^4 + 60)/(x*log(x)^2
- 2*x^2*log(x) + x^3),x)

[Out]

exp(2*x) - x + 30/(x - 1) - 2*x*exp(x) + x^2 - ((30*(3*x - 2))/(x - 1) - (30*x*log(x))/(x - 1))/(x - log(x))

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sympy [A]  time = 0.40, size = 26, normalized size = 0.90 \begin {gather*} x^{2} - 2 x e^{x} - x + \frac {60 - 30 x}{- x + \log {\relax (x )}} + e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)**2+(-2*x**2-2*x)*exp(x)+2*x**2-x)*ln(x)**2+(-4*exp(x)**2*x**2+(4*x**3+4*x**2)*exp(x)-4*
x**3+2*x**2-30*x)*ln(x)+2*exp(x)**2*x**3+(-2*x**4-2*x**3)*exp(x)+2*x**4-x**3+90*x-60)/(x*ln(x)**2-2*x**2*ln(x)
+x**3),x)

[Out]

x**2 - 2*x*exp(x) - x + (60 - 30*x)/(-x + log(x)) + exp(2*x)

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