3.52.49 \(\int \frac {-1-64 e^{-1250+2 x}+(3 x+e^{-1250+2 x} (-32+32 x)) \log (x)-3 x^2 \log ^2(x)+x^3 \log ^3(x)}{-1+3 x \log (x)-3 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx\)

Optimal. Leaf size=25 \[ x+\frac {16 e^{-1250+2 x} x^2}{\left (x-x^2 \log (x)\right )^2} \]

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Rubi [F]  time = 2.98, 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 {-1-64 e^{-1250+2 x}+\left (3 x+e^{-1250+2 x} (-32+32 x)\right ) \log (x)-3 x^2 \log ^2(x)+x^3 \log ^3(x)}{-1+3 x \log (x)-3 x^2 \log ^2(x)+x^3 \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 - 64*E^(-1250 + 2*x) + (3*x + E^(-1250 + 2*x)*(-32 + 32*x))*Log[x] - 3*x^2*Log[x]^2 + x^3*Log[x]^3)/(-
1 + 3*x*Log[x] - 3*x^2*Log[x]^2 + x^3*Log[x]^3),x]

[Out]

x - 32*Defer[Int][E^(-1250 + 2*x)/(-1 + x*Log[x])^3, x] - 32*Defer[Int][E^(-1250 + 2*x)/(x*(-1 + x*Log[x])^3),
x] + 32*Defer[Int][E^(-1250 + 2*x)/(-1 + x*Log[x])^2, x] - 32*Defer[Int][E^(-1250 + 2*x)/(x*(-1 + x*Log[x])^2
), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+64 e^{-1250+2 x}-\left (3 x+e^{-1250+2 x} (-32+32 x)\right ) \log (x)+3 x^2 \log ^2(x)-x^3 \log ^3(x)}{(1-x \log (x))^3} \, dx\\ &=\int \left (-\frac {1}{(-1+x \log (x))^3}+\frac {3 x \log (x)}{(-1+x \log (x))^3}-\frac {3 x^2 \log ^2(x)}{(-1+x \log (x))^3}+\frac {x^3 \log ^3(x)}{(-1+x \log (x))^3}+\frac {32 e^{-1250+2 x} (-2-\log (x)+x \log (x))}{(-1+x \log (x))^3}\right ) \, dx\\ &=3 \int \frac {x \log (x)}{(-1+x \log (x))^3} \, dx-3 \int \frac {x^2 \log ^2(x)}{(-1+x \log (x))^3} \, dx+32 \int \frac {e^{-1250+2 x} (-2-\log (x)+x \log (x))}{(-1+x \log (x))^3} \, dx-\int \frac {1}{(-1+x \log (x))^3} \, dx+\int \frac {x^3 \log ^3(x)}{(-1+x \log (x))^3} \, dx\\ &=3 \int \left (\frac {1}{(-1+x \log (x))^3}+\frac {1}{(-1+x \log (x))^2}\right ) \, dx-3 \int \left (\frac {1}{(-1+x \log (x))^3}+\frac {2}{(-1+x \log (x))^2}+\frac {1}{-1+x \log (x)}\right ) \, dx+32 \int \left (\frac {e^{-1250+2 x} (-1-x)}{x (-1+x \log (x))^3}+\frac {e^{-1250+2 x} (-1+x)}{x (-1+x \log (x))^2}\right ) \, dx-\int \frac {1}{(-1+x \log (x))^3} \, dx+\int \left (1+\frac {1}{(-1+x \log (x))^3}+\frac {3}{(-1+x \log (x))^2}+\frac {3}{-1+x \log (x)}\right ) \, dx\\ &=x+2 \left (3 \int \frac {1}{(-1+x \log (x))^2} \, dx\right )-6 \int \frac {1}{(-1+x \log (x))^2} \, dx+32 \int \frac {e^{-1250+2 x} (-1-x)}{x (-1+x \log (x))^3} \, dx+32 \int \frac {e^{-1250+2 x} (-1+x)}{x (-1+x \log (x))^2} \, dx\\ &=x+2 \left (3 \int \frac {1}{(-1+x \log (x))^2} \, dx\right )-6 \int \frac {1}{(-1+x \log (x))^2} \, dx+32 \int \left (-\frac {e^{-1250+2 x}}{(-1+x \log (x))^3}-\frac {e^{-1250+2 x}}{x (-1+x \log (x))^3}\right ) \, dx+32 \int \left (\frac {e^{-1250+2 x}}{(-1+x \log (x))^2}-\frac {e^{-1250+2 x}}{x (-1+x \log (x))^2}\right ) \, dx\\ &=x+2 \left (3 \int \frac {1}{(-1+x \log (x))^2} \, dx\right )-6 \int \frac {1}{(-1+x \log (x))^2} \, dx-32 \int \frac {e^{-1250+2 x}}{(-1+x \log (x))^3} \, dx-32 \int \frac {e^{-1250+2 x}}{x (-1+x \log (x))^3} \, dx+32 \int \frac {e^{-1250+2 x}}{(-1+x \log (x))^2} \, dx-32 \int \frac {e^{-1250+2 x}}{x (-1+x \log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^{1250} x+\frac {16 e^{2 x}}{(-1+x \log (x))^2}}{e^{1250}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 64*E^(-1250 + 2*x) + (3*x + E^(-1250 + 2*x)*(-32 + 32*x))*Log[x] - 3*x^2*Log[x]^2 + x^3*Log[x]
^3)/(-1 + 3*x*Log[x] - 3*x^2*Log[x]^2 + x^3*Log[x]^3),x]

[Out]

(E^1250*x + (16*E^(2*x))/(-1 + x*Log[x])^2)/E^1250

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fricas [A]  time = 0.56, size = 43, normalized size = 1.72 \begin {gather*} \frac {x^{3} \log \relax (x)^{2} - 2 \, x^{2} \log \relax (x) + x + 16 \, e^{\left (2 \, x - 1250\right )}}{x^{2} \log \relax (x)^{2} - 2 \, x \log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*log(x)^3-3*x^2*log(x)^2+((32*x-32)*exp(x-625)^2+3*x)*log(x)-64*exp(x-625)^2-1)/(x^3*log(x)^3-3*
x^2*log(x)^2+3*x*log(x)-1),x, algorithm="fricas")

[Out]

(x^3*log(x)^2 - 2*x^2*log(x) + x + 16*e^(2*x - 1250))/(x^2*log(x)^2 - 2*x*log(x) + 1)

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giac [B]  time = 1.99, size = 53, normalized size = 2.12 \begin {gather*} \frac {x^{3} e^{1250} \log \relax (x)^{2} - 2 \, x^{2} e^{1250} \log \relax (x) + x e^{1250} + 16 \, e^{\left (2 \, x\right )}}{x^{2} e^{1250} \log \relax (x)^{2} - 2 \, x e^{1250} \log \relax (x) + e^{1250}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*log(x)^3-3*x^2*log(x)^2+((32*x-32)*exp(x-625)^2+3*x)*log(x)-64*exp(x-625)^2-1)/(x^3*log(x)^3-3*
x^2*log(x)^2+3*x*log(x)-1),x, algorithm="giac")

[Out]

(x^3*e^1250*log(x)^2 - 2*x^2*e^1250*log(x) + x*e^1250 + 16*e^(2*x))/(x^2*e^1250*log(x)^2 - 2*x*e^1250*log(x) +
e^1250)

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maple [A]  time = 0.05, size = 19, normalized size = 0.76

method	result	size
risch	\(x +\frac {16 \,{\mathrm e}^{2 x -1250}}{\left (x \ln \relax (x )-1\right )^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*ln(x)^3-3*x^2*ln(x)^2+((32*x-32)*exp(x-625)^2+3*x)*ln(x)-64*exp(x-625)^2-1)/(x^3*ln(x)^3-3*x^2*ln(x)^
2+3*x*ln(x)-1),x,method=_RETURNVERBOSE)

[Out]

x+16*exp(2*x-1250)/(x*ln(x)-1)^2

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maxima [B]  time = 0.38, size = 53, normalized size = 2.12 \begin {gather*} \frac {x^{3} e^{1250} \log \relax (x)^{2} - 2 \, x^{2} e^{1250} \log \relax (x) + x e^{1250} + 16 \, e^{\left (2 \, x\right )}}{x^{2} e^{1250} \log \relax (x)^{2} - 2 \, x e^{1250} \log \relax (x) + e^{1250}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*log(x)^3-3*x^2*log(x)^2+((32*x-32)*exp(x-625)^2+3*x)*log(x)-64*exp(x-625)^2-1)/(x^3*log(x)^3-3*
x^2*log(x)^2+3*x*log(x)-1),x, algorithm="maxima")

[Out]

(x^3*e^1250*log(x)^2 - 2*x^2*e^1250*log(x) + x*e^1250 + 16*e^(2*x))/(x^2*e^1250*log(x)^2 - 2*x*e^1250*log(x) +
e^1250)

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mupad [B]  time = 3.31, size = 34, normalized size = 1.36 \begin {gather*} \frac {x+16\,{\mathrm {e}}^{2\,x-1250}-2\,x^2\,\ln \relax (x)+x^3\,{\ln \relax (x)}^2}{{\left (x\,\ln \relax (x)-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*exp(2*x - 1250) + 3*x^2*log(x)^2 - x^3*log(x)^3 - log(x)*(3*x + exp(2*x - 1250)*(32*x - 32)) + 1)/(3*x
^2*log(x)^2 - x^3*log(x)^3 - 3*x*log(x) + 1),x)

[Out]

(x + 16*exp(2*x - 1250) - 2*x^2*log(x) + x^3*log(x)^2)/(x*log(x) - 1)^2

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sympy [A]  time = 0.30, size = 26, normalized size = 1.04 \begin {gather*} x + \frac {16 e^{2 x - 1250}}{x^{2} \log {\relax (x )}^{2} - 2 x \log {\relax (x )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3*ln(x)**3-3*x**2*ln(x)**2+((32*x-32)*exp(x-625)**2+3*x)*ln(x)-64*exp(x-625)**2-1)/(x**3*ln(x)**
3-3*x**2*ln(x)**2+3*x*ln(x)-1),x)

[Out]

x + 16*exp(2*x - 1250)/(x**2*log(x)**2 - 2*x*log(x) + 1)

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