3.56.44 \(\int \frac {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} (11 x-x^2-2 x^3+(4-19 x-9 x^2-2 x^3) \log (x)+(8+8 x+2 x^2) \log ^2(x))}{-2 x^3+(-14 x^2-4 x^3) \log (x)+(-16 x-12 x^2-2 x^3) \log ^2(x)+(32+16 x+2 x^2) \log ^3(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^x x (x-\log (x))^{3/2}}{x+(4+x) \log (x)} \]

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Rubi [F]  time = 9.93, 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 {e^{\frac {1}{2} (2 x+3 \log (x-\log (x)))} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*x + 3*Log[x - Log[x]])/2)*(11*x - x^2 - 2*x^3 + (4 - 19*x - 9*x^2 - 2*x^3)*Log[x] + (8 + 8*x + 2*x^
2)*Log[x]^2))/(-2*x^3 + (-14*x^2 - 4*x^3)*Log[x] + (-16*x - 12*x^2 - 2*x^3)*Log[x]^2 + (32 + 16*x + 2*x^2)*Log
[x]^3),x]

[Out]

-Defer[Int][E^x*Sqrt[x - Log[x]], x] - 4*Defer[Int][(E^x*Sqrt[x - Log[x]])/(4 + x)^2, x] + 4*Defer[Int][(E^x*S
qrt[x - Log[x]])/(4 + x), x] + 12*Defer[Int][(E^x*Sqrt[x - Log[x]])/(x + 4*Log[x] + x*Log[x])^2, x] - 9*Defer[
Int][(E^x*x*Sqrt[x - Log[x]])/(x + 4*Log[x] + x*Log[x])^2, x] - Defer[Int][(E^x*x^2*Sqrt[x - Log[x]])/(x + 4*L
og[x] + x*Log[x])^2, x] - 64*Defer[Int][(E^x*Sqrt[x - Log[x]])/((4 + x)^2*(x + 4*Log[x] + x*Log[x])^2), x] - 3
2*Defer[Int][(E^x*Sqrt[x - Log[x]])/((4 + x)*(x + 4*Log[x] + x*Log[x])^2), x] - Defer[Int][(E^x*Sqrt[x - Log[x
]])/(x + 4*Log[x] + x*Log[x]), x]/2 + (5*Defer[Int][(E^x*x*Sqrt[x - Log[x]])/(x + 4*Log[x] + x*Log[x]), x])/2
+ Defer[Int][(E^x*x^2*Sqrt[x - Log[x]])/(x + 4*Log[x] + x*Log[x]), x] - 32*Defer[Int][(E^x*Sqrt[x - Log[x]])/(
(4 + x)^2*(x + 4*Log[x] + x*Log[x])), x] + 8*Defer[Int][(E^x*Sqrt[x - Log[x]])/((4 + x)*(x + 4*Log[x] + x*Log[
x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x (x-\log (x))^{3/2} \left (11 x-x^2-2 x^3+\left (4-19 x-9 x^2-2 x^3\right ) \log (x)+\left (8+8 x+2 x^2\right ) \log ^2(x)\right )}{-2 x^3+\left (-14 x^2-4 x^3\right ) \log (x)+\left (-16 x-12 x^2-2 x^3\right ) \log ^2(x)+\left (32+16 x+2 x^2\right ) \log ^3(x)} \, dx\\ &=\int \frac {e^x \sqrt {x-\log (x)} \left (x \left (-11+x+2 x^2\right )+\left (-4+19 x+9 x^2+2 x^3\right ) \log (x)-2 (2+x)^2 \log ^2(x)\right )}{2 (x+(4+x) \log (x))^2} \, dx\\ &=\frac {1}{2} \int \frac {e^x \sqrt {x-\log (x)} \left (x \left (-11+x+2 x^2\right )+\left (-4+19 x+9 x^2+2 x^3\right ) \log (x)-2 (2+x)^2 \log ^2(x)\right )}{(x+(4+x) \log (x))^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {2 e^x (2+x)^2 \sqrt {x-\log (x)}}{(4+x)^2}-\frac {2 e^x x \left (80+76 x+17 x^2+x^3\right ) \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))^2}+\frac {e^x \left (-16+88 x+71 x^2+21 x^3+2 x^4\right ) \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^x \left (-16+88 x+71 x^2+21 x^3+2 x^4\right ) \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))} \, dx-\int \frac {e^x (2+x)^2 \sqrt {x-\log (x)}}{(4+x)^2} \, dx-\int \frac {e^x x \left (80+76 x+17 x^2+x^3\right ) \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {e^x \sqrt {x-\log (x)}}{x+4 \log (x)+x \log (x)}+\frac {5 e^x x \sqrt {x-\log (x)}}{x+4 \log (x)+x \log (x)}+\frac {2 e^x x^2 \sqrt {x-\log (x)}}{x+4 \log (x)+x \log (x)}-\frac {64 e^x \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))}+\frac {16 e^x \sqrt {x-\log (x)}}{(4+x) (x+4 \log (x)+x \log (x))}\right ) \, dx-\int \left (e^x \sqrt {x-\log (x)}+\frac {4 e^x \sqrt {x-\log (x)}}{(4+x)^2}-\frac {4 e^x \sqrt {x-\log (x)}}{4+x}\right ) \, dx-\int \left (-\frac {12 e^x \sqrt {x-\log (x)}}{(x+4 \log (x)+x \log (x))^2}+\frac {9 e^x x \sqrt {x-\log (x)}}{(x+4 \log (x)+x \log (x))^2}+\frac {e^x x^2 \sqrt {x-\log (x)}}{(x+4 \log (x)+x \log (x))^2}+\frac {64 e^x \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))^2}+\frac {32 e^x \sqrt {x-\log (x)}}{(4+x) (x+4 \log (x)+x \log (x))^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^x \sqrt {x-\log (x)}}{x+4 \log (x)+x \log (x)} \, dx\right )+\frac {5}{2} \int \frac {e^x x \sqrt {x-\log (x)}}{x+4 \log (x)+x \log (x)} \, dx-4 \int \frac {e^x \sqrt {x-\log (x)}}{(4+x)^2} \, dx+4 \int \frac {e^x \sqrt {x-\log (x)}}{4+x} \, dx+8 \int \frac {e^x \sqrt {x-\log (x)}}{(4+x) (x+4 \log (x)+x \log (x))} \, dx-9 \int \frac {e^x x \sqrt {x-\log (x)}}{(x+4 \log (x)+x \log (x))^2} \, dx+12 \int \frac {e^x \sqrt {x-\log (x)}}{(x+4 \log (x)+x \log (x))^2} \, dx-32 \int \frac {e^x \sqrt {x-\log (x)}}{(4+x) (x+4 \log (x)+x \log (x))^2} \, dx-32 \int \frac {e^x \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))} \, dx-64 \int \frac {e^x \sqrt {x-\log (x)}}{(4+x)^2 (x+4 \log (x)+x \log (x))^2} \, dx-\int e^x \sqrt {x-\log (x)} \, dx-\int \frac {e^x x^2 \sqrt {x-\log (x)}}{(x+4 \log (x)+x \log (x))^2} \, dx+\int \frac {e^x x^2 \sqrt {x-\log (x)}}{x+4 \log (x)+x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.63, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^x x (x-\log (x))^{3/2}}{x+(4+x) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*x + 3*Log[x - Log[x]])/2)*(11*x - x^2 - 2*x^3 + (4 - 19*x - 9*x^2 - 2*x^3)*Log[x] + (8 + 8*x
+ 2*x^2)*Log[x]^2))/(-2*x^3 + (-14*x^2 - 4*x^3)*Log[x] + (-16*x - 12*x^2 - 2*x^3)*Log[x]^2 + (32 + 16*x + 2*x^
2)*Log[x]^3),x]

[Out]

(E^x*x*(x - Log[x])^(3/2))/(x + (4 + x)*Log[x])

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fricas [A]  time = 0.62, size = 24, normalized size = 0.96 \begin {gather*} \frac {x e^{\left (x + \frac {3}{2} \, \log \left (x - \log \relax (x)\right )\right )}}{{\left (x + 4\right )} \log \relax (x) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x+8)*log(x)^2+(-2*x^3-9*x^2-19*x+4)*log(x)-2*x^3-x^2+11*x)*exp(3/2*log(x-log(x))+x)/((2*x^
2+16*x+32)*log(x)^3+(-2*x^3-12*x^2-16*x)*log(x)^2+(-4*x^3-14*x^2)*log(x)-2*x^3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, x^{3} - 2 \, {\left (x^{2} + 4 \, x + 4\right )} \log \relax (x)^{2} + x^{2} + {\left (2 \, x^{3} + 9 \, x^{2} + 19 \, x - 4\right )} \log \relax (x) - 11 \, x\right )} e^{\left (x + \frac {3}{2} \, \log \left (x - \log \relax (x)\right )\right )}}{2 \, {\left ({\left (x^{2} + 8 \, x + 16\right )} \log \relax (x)^{3} - x^{3} - {\left (x^{3} + 6 \, x^{2} + 8 \, x\right )} \log \relax (x)^{2} - {\left (2 \, x^{3} + 7 \, x^{2}\right )} \log \relax (x)\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x+8)*log(x)^2+(-2*x^3-9*x^2-19*x+4)*log(x)-2*x^3-x^2+11*x)*exp(3/2*log(x-log(x))+x)/((2*x^
2+16*x+32)*log(x)^3+(-2*x^3-12*x^2-16*x)*log(x)^2+(-4*x^3-14*x^2)*log(x)-2*x^3),x, algorithm="giac")

[Out]

integrate(-1/2*(2*x^3 - 2*(x^2 + 4*x + 4)*log(x)^2 + x^2 + (2*x^3 + 9*x^2 + 19*x - 4)*log(x) - 11*x)*e^(x + 3/
2*log(x - log(x)))/((x^2 + 8*x + 16)*log(x)^3 - x^3 - (x^3 + 6*x^2 + 8*x)*log(x)^2 - (2*x^3 + 7*x^2)*log(x)),
x)

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maple [A]  time = 0.05, size = 25, normalized size = 1.00

method	result	size
risch	\(\frac {x \left (x -\ln \relax (x )\right )^{\frac {3}{2}} {\mathrm e}^{x}}{x \ln \relax (x )+4 \ln \relax (x )+x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+8*x+8)*ln(x)^2+(-2*x^3-9*x^2-19*x+4)*ln(x)-2*x^3-x^2+11*x)*exp(3/2*ln(x-ln(x))+x)/((2*x^2+16*x+32)
*ln(x)^3+(-2*x^3-12*x^2-16*x)*ln(x)^2+(-4*x^3-14*x^2)*ln(x)-2*x^3),x,method=_RETURNVERBOSE)

[Out]

x/(x*ln(x)+4*ln(x)+x)*(x-ln(x))^(3/2)*exp(x)

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maxima [A]  time = 0.44, size = 30, normalized size = 1.20 \begin {gather*} \frac {{\left (x^{2} - x \log \relax (x)\right )} \sqrt {x - \log \relax (x)} e^{x}}{{\left (x + 4\right )} \log \relax (x) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x+8)*log(x)^2+(-2*x^3-9*x^2-19*x+4)*log(x)-2*x^3-x^2+11*x)*exp(3/2*log(x-log(x))+x)/((2*x^
2+16*x+32)*log(x)^3+(-2*x^3-12*x^2-16*x)*log(x)^2+(-4*x^3-14*x^2)*log(x)-2*x^3),x, algorithm="maxima")

[Out]

(x^2 - x*log(x))*sqrt(x - log(x))*e^x/((x + 4)*log(x) + x)

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mupad [B]  time = 3.78, size = 24, normalized size = 0.96 \begin {gather*} \frac {x\,{\mathrm {e}}^x\,{\left (x-\ln \relax (x)\right )}^{3/2}}{x+4\,\ln \relax (x)+x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + (3*log(x - log(x)))/2)*(x^2 - log(x)^2*(8*x + 2*x^2 + 8) - 11*x + 2*x^3 + log(x)*(19*x + 9*x^2 +
2*x^3 - 4)))/(log(x)*(14*x^2 + 4*x^3) - log(x)^3*(16*x + 2*x^2 + 32) + log(x)^2*(16*x + 12*x^2 + 2*x^3) + 2*x^
3),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+8*x+8)*ln(x)**2+(-2*x**3-9*x**2-19*x+4)*ln(x)-2*x**3-x**2+11*x)*exp(3/2*ln(x-ln(x))+x)/((2*
x**2+16*x+32)*ln(x)**3+(-2*x**3-12*x**2-16*x)*ln(x)**2+(-4*x**3-14*x**2)*ln(x)-2*x**3),x)

[Out]

Timed out

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